GTM148 抄书笔记 Part IV. (Chapter VII)
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Contents
Chapter VII. Extensions and Cohomology
The Extension Problem
Definition 7.1.1 If \(K\) and \(Q\) are groups, then an extension of \(K\) by \(Q\) is a group \(G\) having a normal subgroup \(K_1\cong K\) with \(G/K_1\cong Q\).
Example 7.1.2
- Both \(\mathbf Z_6\) and \(\mathrm S_3\) are extensions of \(\mathbf Z_3\) by \(\mathbf Z_2\). However, \(\mathbf Z_6\) is an extension of \(\mathbf Z_2\) by \(\mathbf Z_3\), but \(\mathrm S_3\) is not such an extension.
- For any groups \(K\) and \(Q\), the direct product \(K\times Q\) is an extension of \(K\) by \(Q\) as well as an extension of \(Q\) by \(K\).
Proposition 7.1.3
- If \(K\) and \(Q\) are finite, then every extension \(G\) of \(K\) by \(Q\) has order \(|K||Q|\).
- If \(G\) has a normal series with factor groups \(Q_n,\ldots,Q_1\), then \(|G|=\prod|Q_i|\).
Proposition 7.1.4 If \((a,b)=1\) and \(K\) and \(Q\) are abelian groups of orders \(a\) and \(b\), respectively, then there is only one (to isomorphism) abelian extension of \(K\) by \(Q\).
Proposition 7.1.5 If \(p\) is prime, every nonabelian group of order \(p^3\) is an extension of \(\mathbb Z_p\) by \(\mathbb Z_p\times\mathbb Z_p\).
Automorphism Groups
Definition 7.2.1 The automorphism group of a group \(G\), denoted by \(\mathrm{Aut}(G)\), is the set of all the automorphisms of \(G\) under the operation of composition.
It is easy to check that \(\mathrm{Aut}(G)\) is a group; indeed, it is a subgroup of the symmetric group \(\mathrm S_G\).
Definition 7.2.2 An automorphism \(\varphi\) of \(G\) is inner if it is conjugation by some element of \(G\); otherwise, it is outer. Denote the set of all inner automorphisms of \(G\) by \(\mathrm{Inn}(G)\).
Theorem 7.2.3
- N/C Lemma If \(H\le G\), then \(\mathrm C_G(H)\lhd\mathrm N_G(H)\) and \(\mathrm N_G(H)/\mathrm C_G(H)\) can be imbedded in \(\mathrm {Aut}(H)\).
- \(\mathrm {Inn}(G)\lhd\mathrm {Aut}(G)\) and \(G/\mathrm Z(G)\cong\mathrm{Inn}(G)\).
Proof
- If \(a\in G\), let \(\gamma_a\) denote conjugation by \(a\). Define \(\varphi\!:\mathrm N_G(H)\rightarrow\mathrm{Aut}(H)\) by \(a\mapsto\gamma_a\big|_H\) (note that \(\gamma_a\big|_H\in\mathrm{Aut}(H)\) because \(a\in\mathrm N_G(H)\)); \(\varphi\) is easily seen to be a homomorphism. The following statements are equivalent: \(a\in\mathrm{ker}\varphi\); \(\gamma_a\big|_H\) is the identity on \(H\); \(aha^{-1}=h\) for all \(h\in H\); \(a\in\mathrm C_G(H)\). By the first isomorphism theorem, \(\mathrm C_G(H)\lhd\mathrm N_G(H)\) and \(\mathrm N_G(H)/\mathrm C_G(H)\cong\mathrm{im}\varphi\le\mathrm{Aut}(H)\).
- If \(H=G\), then \(\mathrm N_G(G)=G, \mathrm C_G(G)=\mathrm Z(G)\), and \(\mathrm{im}\varphi=\mathrm{Inn}(G)\). Therefore, \(G/\mathrm Z(G)\cong\mathrm{Inn}(G)\) is a special case of the isomorphism just established. To see that \(\mathrm{Inn}(G)\lhd\mathrm{Aut}(G)\), take \(\gamma_a\in\mathrm{Inn}(G)\) and \(\varphi\in\mathrm{Aut}(G)\). Then \(\varphi\gamma_a\varphi^{-1}=\gamma_{\varphi a}\in\mathrm{Inn}(G)\).
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Definition 7.2.4 The group \(\mathrm{Aut}(G)/\mathrm{Inn}(G)\) is called the outer automorphism group of \(G\).
Example 7.2.6 Nonisomorphic groups can have isomorphic automorphism groups: \(\mathrm{Aut}(\mathbf V)\cong\mathrm S_3\cong\mathrm{Aut}(\mathrm S_3)\).
Example 7.2.7 If \(G\) is an elementary abelian group of order \(p^n\), then \(\mathrm{Aut}(G)\cong\mathrm {GL}(n,p)\). (This follows from Proposition 4.1.17: \(G\) is a vector space over \(\mathbb Z_p\) and every automorphism is a nonsingular linear transformation.)
Theorem 7.2.8 If \(G\) is a cyclic group of order \(n\), then \(\mathrm{Aut}(G)\cong\mathrm U(\mathbb Z_n)\).
Proof
Let \(G=\langle a\rangle\). If \(\varphi\in\mathrm{Aut}(G)\), then \(\varphi(a)=a^k\) for some \(k\); moreover, \(a^k\) must be a generator of \(G\), so that \((k,n)=1\), and \([k]\in\mathrm U(\mathbb Z_n)\). It is routine to show that \(\Theta\!:\mathrm{Aut}(G)\rightarrow\mathrm U(\mathbb Z_n)\), defined by \(\Theta(\varphi)=[k]\), is an isomorphism.
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Proposition 7.2.9
- \(\mathrm{Aut}(\mathbb Z_2)=\mathbf 1\); \(\mathrm{Aut}(\mathbb Z_4)\cong\mathbb Z_2\); if \(m\ge 3\), then \(\mathrm {Aut}(\mathbb Z_{2^m})\cong\mathbb Z_2\times\mathbb Z_{2^{m-2}}\); \(\mathrm{Aut}(\mathbb Z)\cong\mathbb Z_2\).
- If \(p\) is an odd prime, then \(\mathrm{Aut}(\mathbb Z_{p^m})\cong\mathbb Z_i\), where \(l=(p-1)p^{m-1}\).
- If \(n=p_1^{e_1}\ldots p_t^{e_t}\), where the \(p_i\) are distinct primes and the \(e_i>0\), then \(\mathrm{Aut}(\mathbb Z_n)\cong\prod_i\mathrm{Aut}(\mathbb Z_{q_i})\), where \(q_i=p_i^{e_i}\).
Definition 7.2.10 A group \(G\) is complete if it is centerless and every automorphism of \(G\) is inner.
It follows that \(\mathrm {Aut}(G)\cong G\) for every complete group.
Lemma 7.2.11 An automorphism \(\varphi\) of \(\mathrm S_n\) preserves transpositions (\(\varphi(\tau)\) is a transposition whenever \(\tau\) is) if and only if \(\varphi\) is inner.
Proof
If \(\varphi\) is inner, then it preserves the cycle structure of every permutation, by Theorem 3.2.3.
We prove by induction on \(t\ge 2\), that there exist conjugations \(\gamma_2,\ldots,\gamma_t\) such that \(\gamma_t^{-1}\ldots\gamma_2^{-1}\varphi\) fixes \((1\ \ 2),\ldots,(1\ \ t)\). If \(\pi\in\mathrm S_n\), we will denote \(\varphi(\pi)\) by \(\pi^\varphi\) in this proof. By hypothesis, \((1\ \ 2)^\varphi=(i\ \ j)\) for some \(i,j\); define \(\gamma_2\) to be conjugation by \((1\ \ i)(2\ \ j)\). By Lemma 3.2.2, the quick way of computing conjugates in \(\mathrm S_n\), we see that \((1\ \ 2)^\varphi=(1\ \ 2)^{\gamma_2}\), and so \(\gamma_2^{-1}\varphi\) fixes \((1\ \ 2)\).
Let \(\gamma_2,\ldots,\gamma_t\) be given by the inductive hypothesis, so that \(\psi=\gamma_t^{-1}\ldots\gamma_2^{-1}\varphi\) fixes \((1\ \ 2),\ldots,(1\ \ t)\). Since \(\psi\) preserves transpositions, \((1\ \ t+1)^\psi=(l\ \ k)\). Now \((1\ \ 2)\) and \((l\ \ k)\) cannot be disjoint, lest \([(1\ \ 2)(1\ \ t+1)]^{\psi}=(1\ \ 2)^\psi(1\ \ t+1)^\psi=(1\ \ 2)(l\ \ k)\) have order \(2\), while \((1\ \ 2)(1\ \ t+1)\) has order \(3\). Thus, \((1\ \ t+1)^\psi=(1\ \ k)\) or \((1\ \ t+1)^\psi=(2\ \ k)\). If \(k\le t\), then \((1\ \ t+1)^\psi\in\langle(1\ \ 2),\ldots,(1\ \ t)\rangle\), and hence it is fixed by \(\psi\); this contradicts \(\psi\) being injective, for either \((1\ \ t+1)^\psi=(1\ \ k)=(1\ \ k)^\psi=(2\ \ k)=(2\ \ k)^\psi\). Hence, \(k\ge t+1\). Define \(\gamma_{t+1}\) to be conjugation by \((k\ \ t+1)\). Now \(\gamma_{t+1}\) fixes \((1\ \ 2),\ldots,(1\ \ t)\) and \((1\ \ t+1)^{\gamma_{t+1}}=(1\ \ t+1)^\psi\), so that \(\gamma_{t+1}^{-1}\cdots\gamma_2^{-1}\varphi\) fixes \((1\ \ 2),\ldots,(1\ \ t+1)\) and the induction is complete. It follows that \(\gamma_n^{-1}\cdots\gamma_{2}^{-1}\varphi\) fixes \((1\ \ 2),\ldots,(1\ \ n)\). But these transpositions generate \(\mathrm S_n\), by Proposition 2.1.20, and so \(\gamma_n^{-1}\ldots\gamma_2^{-1}\varphi\) is the identity. Therefore, \(\varphi=\gamma_2\cdots\gamma_n\in\mathrm{Inn}(\mathrm S_n)\).
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Theorem 7.2.12 If \(n\neq 2\) or \(n\neq 6\), then \(\mathrm S_n\) is complete.
Proof
Let \(T_k\) denote the conjugacy class in \(\mathrm S_n\) consisting of all products of \(k\) disjoint transpositions. By Proposition 1.3.4, a permutation in \(\mathrm S_n\) is an involultion if and only if it lies in some \(T_k\). It follows that if \(\theta\in\mathrm{Aut}(\mathrm S_n)\), then \(\theta(T_1)=T_k\) for some \(k\). We shall show that if \(n\neq 6\), then \(|T_k|\neq|T_1|\) for \(k\neq 1\). Assuming this, then \(\theta(T_1)=T_1\), and Lemma 7.2.11 completes the proof since \(\mathrm S_{n}\) is centerless for \(n\ge 3\) by Proposition 3.1.19.
Now \(|T_1|=n(n-1)/2\). To count \(T_k\), observe first that there are
\[\frac 12n(n-1)\times \frac 12(n-2)(n-3)\times\cdots\times\frac 12(n-2k+2)(n-2k+1) \]\(k\)-tuples of disjoint transpositions. Since disjoint transpositions commute and there are \(k!\) orderings obtained from any \(k\)-tuple, we have
\[|T_k|=n(n-1)(n-2)\cdots(n-2k+1)/k!2^k. \]The question whether \(|T_1|=|T_k|\) leads to the question whether there is some \(k>1\) such that
\[\begin{align} (n-2)(n-3)\cdots(n-2k+1)=k!2^{k-1}. \end{align} \]Since the right side of \((1)\) is positive, we must have \(n\ge 2k\). Therefore, for fixed \(n\),
\[\text{left side}\ge(2k-2)(2k-3)\cdots(2k-2k+1)=(2k-2)!. \]An easy induction shows that if \(k\ge 4\), then \((2k-2)!>k!2^{k-1}\), and so \((1)\) can hold only if \(k=2\) or \(k=3\). When \(k=2\), the right side is \(4\), and it is easy to see that equality never holds; we may assume, therefore, that \(k=3\). Since \(n\ge 2k\), we must have \(n\ge 6\). If \(n> 6\), then the left side of \((1)\ge 5\times 4\times 3\times 2=120\), while the right side is \(24\). We have shown that if \(n\neq 6\), then \(|T_1|\neq|T_k|\) for all \(k>1\), as desired.
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Corollary 7.2.13 If \(\theta\) is an outer automorphism of \(\mathrm S_6\), and if \(\tau\in\mathrm S_6\) is a transposition, then \(\theta(\tau)\) is a product of three disjoint transpositions.
Corollary 7.2.14 If \(n\neq 2\) or \(n\neq 6\), then \(\mathrm{Aut}(\mathrm S_n)\cong \mathrm S_n\).
Lemma 7.2.15 There exists a transitive subgroup \(K\le\mathrm S_6\) of order \(120\) which contains no transpositions.
Proof
If \(\sigma\) is a \(5\)-cycle, then \(P=\langle\sigma\rangle\) is a Sylow \(5\)-subgroup of \(\mathrm S_5\). The Sylow theorem says that if \(r\) is the number of conjugates of \(P\), then \(r\equiv 1\pmod 5\) and \(r\) is a divisor of \(120\); it follows easily that \(r=6\). The representation of \(\mathrm S_5\) on \(X\), the set of all left cosets of \(N=\mathrm N_{\mathrm S_5}(P)\), is a homomorphism \(\rho\!:\mathrm S_5\rightarrow\mathrm S_X\cong\mathrm S_6\). Now \(X\) is a transitive \(\mathrm S_5\)-set, by Proposition 4.2.13, and so \(|\mathrm{ker}\rho|\le|\mathrm S_5|/r=|\mathrm S_5|/6=20\), by Proposition 3.5.13(iii). Since the only normal subgroups of \(\mathrm S_5\) are \(\mathrm S_5,\mathrm A_5\), and \(\mathbf 1\), it follows that \(\mathrm{ker}\rho=\mathbf 1\) and \(\rho\) is an injection. Therefore, \(\mathrm{im}\rho\cong\mathrm S_5\) is a transtive subgroup of \(\mathrm S_X\) of order \(120\).
For notational convenience, let us write \(K\le\mathrm S_6\) instead of \(\mathrm{im}\rho\le\mathrm S_X\). Now \(K\) contains an element \(\alpha\) of order \(5\) which must be a \(5\)-cycle; say, \(\alpha=(1\ \ 2\ \ 3\ \ 4\ \ 5)\). If \(K\) contains a transposition \((i\ \ j)\), then transitivity of \(K\) provides \(\beta\in K\) with \(\beta(j)=6\), and so \(\beta(i\ \ j)\beta^{-1}=(\beta i\ \ \beta j)=(l\ \ 6)\) for some \(l\neq 6\) (of course, \(l=\beta i\)). Conjugating \((l\ \ 6)\in K\) by the powers of \(\alpha\) shows that \(K\) contains \((1\ \ 6), (2\ \ 6), (3\ \ 6), (4\ \ 6)\), and \((5\ \ 6)\). But these transpositions generate \(\mathrm S_6\), and this contradicts \(K(\cong\mathrm S_5)\) being a proper subgroup of \(\mathrm S_6\).
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Theorem 7.2.16
- There exists an outer automorphism of \(\mathrm S_6\).
- \(\mathrm{Aut}(\mathrm S_6)/\mathrm{Inn}(\mathrm S_6)\cong\mathbb Z_2\), and so \(|\mathrm{Aut}(\mathrm S_6)|=1440\).
Proof
- Let \(K\) be a transitive subgroup of \(\mathrm S_6\) of order \(120\), and let \(Y\) be the family of its left cosets: \(Y=\{\alpha_1K,\ldots,\alpha_6K\}\). If \(\theta\!:\mathrm S_6\rightarrow\mathrm S_Y\) is the representation of \(\mathrm S_6\) on the left cosets of \(K\), then \(\mathrm{ker}\theta\le K\) is a normal subgroup of \(\mathrm S_6\). But \(\mathrm A_6\) is the only proper normal subgroup of \(\mathrm S_6\), so that \(\mathrm{ker}\theta=\mathbf 1\) and \(\theta\) is an injection. Since \(\mathrm S_6\) is finite, \(\theta\) must be a bijection, and so \(\theta\in\mathrm{Aut}(\mathrm S_6)\), for \(\mathrm S_Y\cong\mathrm S_6\). Were \(\theta\) inner, then it would preserve the cycle structure of every permutation in \(\mathrm S_6\). In particular, \(\theta_{(1\ \ 2)}\), defined by \(\theta_{(1\ \ 2)}\!:\alpha_iK\mapsto(1\ \ 2)\alpha_iK\) for all \(i\), is a transposition, and hence \(\theta\) fixes \(\alpha_iK\) for four different \(i\). But if \(\theta_{(1\ \ 2)}\) fixes even one left coset, say \(\alpha_iK=(1\ \ 2)\alpha_iK\), then \(\alpha_i^{-1}(1\ \ 2)\alpha_i\) is a transposition in \(K\). This contradiction shows that \(\theta\) is an outer automorphism.
- Let \(T_1\) be the class of all transpositions in \(\mathrm S_6\), and let \(T_3\) be the class of all products of \(3\) disjoint transpositions. If \(\theta\) and \(\psi\) are outer automorphisms of \(\mathrm S_6\), then both interchange \(T_1\) and \(T_3\), by Corollary 7.2.13, and so \(\theta^{-1}\psi(T_1)=T_1\). Therefore, \(\theta^{-1}\psi\in\mathrm{Inn}(\mathrm S_6)\), by Lemma 7.2.11, and \(\mathrm{Aut}(S_6)/\mathrm{Inn}(\mathrm S_6)\) has order \(2\).
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Definition 7.2.17 A syntheme is a product of \(3\) disjoint transpositions. A pentad is a family of \(5\) synthemes, no two of which have a common transposition.
Proposition 7.2.18 \(\mathrm S_6\) contains exactly \(6\) pentads. They are:
Theorem 7.2.19 If \(\{\sigma_2,\ldots,\sigma_6\}\) is a pentad in some ordering, then there is a unique outer automorphism \(\theta\) of \(\mathrm S_6\) with \(\theta\!:(1\ \ i)\mapsto\sigma_i\) for \(i=2,3,4,5,6\). Moreover, every outer automorphism of \(\mathrm S_6\) has this form.
Proof
Let \(X=\{(1\ \ 2),(1\ \ 3),(1\ \ 4),(1\ \ 5),(1\ \ 6)\}\). If \(\theta\) is an outer automorphism of \(\mathrm S_6\), then Corollary 7.2.13 shows that each \(\theta((1\ \ i))\) is a syntheme. Since \((1\ \ i)\) and \((1\ \ j)\) do not commute for \(i\neq j\), it follows that \(\theta((1\ \ i))\) and \(\theta((1\ \ j))\) do not commute; hence, \(\theta(X)\) is a pentad. Let us count the number of possible functions from \(X\) to pentads arising from outer automorphisms. Given an outer automorphism \(\theta\), there are \(6\) choices of pentad for \(\theta(X)\); given such a pentad \(P\), there are \(5!=120\) bijections \(X\rightarrow P\). Hence, there are at most \(720\) bijections from \(X\) to pentads which can possibly arise as restrictions of outer automorphisms. But there are exactly \(720\) outer automorphisms, by Theorem 7.2.16, and no two of them can restrict to the same bijection because \(X\) generates \(\mathrm S_6\). The statements of the theorem follow.
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Corollary 7.2.20 There is an outer automorphism of \(\mathrm S_6\) which has order \(2\).
Proposition 7.2.21 If \(G\) is a finite nonabelian \(p\)-group, then \(p^2\big||\mathrm{Aut}(G)|\).
Proposition 7.2.22 If \(G\) is a finite abelian group, then \(\mathrm{Aut}(G)\) is abelian if and only if \(G\) is cyclic.
Proposition 7.2.23
- If \(G\) is a finite abelian group with \(|G|>2\), then \(\mathrm{Aut}(G)\) has even order.
- If \(G\) is not abelian, then \(\mathrm{Aut}(G)\) is not cyclic.
- There is no finite group \(G\) with \(\mathrm{Aut}(G)\) cyclic of odd order \(>1\).
Proposition 7.2.24 If \(G\) is a finite group and \(\mathrm{Aut}(G)\) acts transitively on \(G^\#=G-\{\mathbf 1\}\), then \(G\) is an elementary abelian group.
Proposition 7.2.25 If \(H\) and \(K\) are finite groups whose orders are relatively prime, then \(\mathrm{Aut}(H\times K)\cong\mathrm{Aut}(H)\times\mathrm{Aut}(K)\).
Theorem 7.2.26 If \(G\) is a nonabelian simple group, then \(\mathrm {Aut}(G)\) is complete.
Proof
Let \(I=\mathrm{Inn}(G)\lhd\mathrm{Aut}(G)=A\). Now \(\mathrm Z(G)=\mathbf 1\), because \(G\) is simple and nonabelian, and so Theorem 7.2.3(i) gives \(I\cong G\). Now \(\mathrm Z(A)\le\mathrm C_A(I)=\{\varphi\in A:\varphi\gamma_g=\gamma_g\varphi\text{ for all }g\in G\}\). We claim that \(\mathrm C_A(I)=\mathbf 1\); it will then follow that \(A\) is centerless. If \(\varphi\gamma_g=\gamma_g\varphi\) for all \(g\in G\), then \(\gamma_g=\varphi\gamma_g\varphi^{-1}=\gamma_{\varphi(g)}\). Therefore, \(\varphi(g)g^{-1}\in\mathrm Z(G)=\mathbf 1\) for all \(g\in G\), and so \(\varphi=\mathbf 1\).
It remains to show that every \(\sigma\in\mathrm{Aut}(A)\) is inner. Now \(\sigma(I)\lhd A\), bcause \(I\lhd A\), and so \(I\cap\sigma(I)\lhd\sigma(I)\). But \(\sigma(I)\cong I\cong G\) is simple, so that either \(I\cap\sigma(I)=\mathbf 1\) or \(I\cap\sigma(I)=\sigma(I)\). Since both \(I\) and \(\sigma(I)\) are normal, \([I,\sigma(I)]\le I\cap\sigma(I)\). In the first case, we have \([I,\sigma(I)]=\mathbf 1\); that is, \(\sigma(I)\le\mathrm C_A(I)=\mathbf 1\), and this contradicts \(\sigma(I)\cong I\). Hence, \(I\cap\sigma(I)=\sigma(I)\), and so \(\sigma(I)\le I\). This inclusion holds for every \(\sigma\in\mathrm{Aut}(A)\); in particular, \(\sigma^{-1}(I)\le I\), and so \(\sigma(I)=I\) for every \(\sigma\in\mathrm{Aut}(A)\). If \(g\in G\), then \(\gamma_g\in I\); there is thus \(\alpha(g)\in G\) with \(\sigma(\gamma_g)=\gamma_{\alpha(g)}\). It may be easily checked that the function \(\alpha\!:G\rightarrow G\) is a bijection. We now show that \(\alpha\) is an automorphism of \(G\); that is, \(\alpha\in A\). If \(g,h\in G\), then \(\sigma(\gamma_g\gamma_h)=\sigma(\gamma_{gh})=\gamma_{\alpha(gh)}\). On the other hand, \(\sigma(\gamma_g\gamma_h)=\sigma(\gamma_g)\sigma(\gamma_h)=\gamma_{\alpha(g)}\gamma_{\alpha(h)}=\gamma_{\alpha(g)\alpha(h)}\); hence \(\alpha(gh)=\alpha(g)\alpha(h)\).
We claim that \(\sigma=\Gamma_\alpha\), conjugation by \(\alpha\). To this end, define \(\tau=\sigma\circ\Gamma_{\alpha}^{-1}\). Observe, for all \(h\in G\), that
\[\begin{aligned} \tau(\gamma_h)&=\sigma\Gamma_\alpha^{-1}(\gamma_h)=\sigma(\alpha^{-1}\gamma_h\alpha)\\&=\sigma(\gamma_{\alpha^{-1}(h)})\\&=\gamma_{\alpha\alpha^{-1}(h)}=\gamma_h. \end{aligned} \]Thus, \(\tau\) fixes everything in \(I\). If \(\beta\in A\), then for every \(g\in G\),
\[\begin{aligned} \beta\gamma_g\beta^{-1}&=\tau(\beta\gamma_g\beta^{-1})\qquad&&(\text{because }\beta\gamma_g\beta^{-1}\in I\text{ and }\tau\text{ fixes }I)\\ &=\tau(\beta)\gamma_g\tau(\beta)^{-1}&&(\text{because }\tau\text{ fixes }I). \end{aligned} \]Hence \(\tau(\beta)\beta^{-1}\in\mathrm C_A(I)=\mathbf 1\), and \(\tau(\beta)=\beta\). Therefore, \(\tau=\mathbf 1,\sigma=\Gamma_\alpha\), and \(A\) is complete.
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It follows, for every nonabelian simple group \(G\), that \(\mathrm{Aut}(G)\cong\mathrm{Aut}(\mathrm{Aut}(G))\). We know, by Theorem 7.2.3, that every centerless group \(G\) can be imbedded in \(\mathrm{Aut}(G)\). Moreover, \(\mathrm{Aut}(G)\) is also centerless, and so it can also be imbedded in its automorphism group \(\mathrm{Aut}(\mathrm{Aut}(G))\). This process may thus be iterated to give the automorphism tower of \(G\):
Wielandt proved, for every finite centerless group \(G\), that this tower is constant from some point on. Since the last term of an automorphism tower is a complete group, it follows that every finite centerless group can be imbedded in a complete group.
Theorem 7.2.27 If \(K\lhd G\) and \(K\) is complete, then \(K\) is a direct factor of \(G\); that is, there is a normal subgroup \(Q\) of \(G\) with \(G=K\times Q\).
Proof
Define \(Q=\mathrm C_G(K)=\{g\in G:gk=kg\text{ for all }k\in K\}\). Now \(K\cap Q\le\mathrm Z(K)=\mathbf 1\), and so \(K\cap Q=\mathbf 1\). To see that \(G=KQ\), take \(g\in G\). Now \(\gamma_g(K)=K\), because \(K\lhd G\), and so \(\gamma_g\big|_K\in\mathrm{Aut}(K)=\mathrm{Inn}(K)\). There is thus \(k\in K\) with \(gxg^{-1}=kxk^{-1}\) for all \(x\in K\). Hence \(k^{-1}g\in\bigcap_{x\in K}\mathrm C_G(x)=\mathrm C_G(K)=Q\), and so \(g=k(k^{-1}g)\in KQ\). Finally, \(Q\lhd G\), for \(gQg^{-1}=k(k^{-1}g)Q(k^{-1}g)k^{-1}=kQk^{-1}\) (because \(k^{-1}g\in Q\)), and so \(g=k(k^{-1}g)\in KQ\). Finally, \(Q\lhd G\), for \(gQg^{-1}=k(k^{-1}g)Q(g^{-1}g)^{-1}k^{-1}=kQk^{-1}\) (because \(k^{-1}g\in Q\)), and \(kQk^{-1}=Q\) (because every element of \(Q\) commutes with \(k\)). Therefore, \(G=K\times Q\).
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Recall that if \(a\in K\), then \(L_a\!:K\rightarrow K\) denotes left translation by \(a\); that is, \(L_a(x)=ax\) for all \(a\in K\). As in the Cayley theorem, \(K\) is isomorphic to \(K^l=\{L_a:a\in K\}\), which is a subgroup of \(\mathrm S_K\). Similarly, if \(R_a\!:K\rightarrow K\) denotes right translation by \(a\), that is, \(R_a\!:x\mapsto xa^{-1}\), then \(K^r=\{R_a:a\in K\}\) is also a subgroup of \(\mathrm S_K\) isomorphic to \(K\).
Definition 7.2.28 The holomorph of a group \(K\), denoted by \(\mathrm{Hol}(K)\), is the subgroup of \(\mathrm S_K\) generated by \(K^l\) and \(\mathrm{Aut}(K)\).
Notice, for all \(a\in K\), that \(R_a=L_{a^{-1}}\gamma_a\), so that \(K^r\le\mathrm{Hol}(K)\); indeed, it is easy to see that \(\mathrm{Hol}(K)=\langle K^r,\mathrm{Aut}(K)\rangle\).
Lemma 7.2.29 Let \(K\) be a group.
- \(K^l\lhd\mathrm{Hol}(K),K^l\mathrm{Aut}(K)=\mathrm{Hol}(K)\), and \(K^l\cap\mathrm{Aut}(K)=\mathbf 1\).
- \(\mathrm{Hol}(K)/K^l\cong\mathrm{Aut}(K)\).
- \(\mathrm C_{\mathrm{Hol}(K)}(K^l)=K^r\).
- If \(K\) is complete, then \(\mathrm{Hol}(K)=K^l\times K^r\).
- Every automorphism of \(K\) is the restriction of an inner automorphism of \(\mathrm{Hol}(K)\).
- Let \(f\in\mathrm S_K\), then \(f\in\mathrm{Hol}(K)\) if and only if \(f(xy^{-1}z)=f(x)f(y)^{-1}f(z)\) for all \(x,y,z\in K\).
Proof
- It is easy to see that \(\varphi L_a\varphi^{-1}=L_{\varphi(a)}\), and that it lies in \(K^l\) for every \(a\in K\) and \(\varphi\in\mathrm{Aut}(K)\); since \(\mathrm{Hol}(K)=\langle K^l,\mathrm{Aut}(K)\rangle\), it follows that \(K^l\lhd\mathrm{Hol}(K)\) and that \(\mathrm{Hol}(K)=K^l\mathrm{Aut}(K)\). If \(a\in K\), then \(L_a(\mathbf 1)=a\); therefore, \(\mathrm L_a\in\mathrm{Aut}(K)\) if and only if \(a=\mathbf 1\); that is, \(K^l\cap\mathrm{Aut}(K)=\mathbf 1\).
- \(\mathrm{Hol}(K)/K^l=K^l\mathrm{Aut}(K)/K^l\cong\mathrm{Aut}(K)/(K^l\cap\mathrm{Aut}(K))\cong\mathrm{Aut}(K)\).
- If \(a,b,x\in K\), then \(L_aR_b(x)=a(xb^{-1})\) and \(R_bL_a(x)=(ax)b^{-1}\), so that associativity gives \(K^r\le\mathrm C_{\mathrm{Hol}(K)}(K^l)\). For the reverse inclusion, assume that \(\eta\in\mathrm{Hol}(K)\) satisfies \(\eta L_a=L_a\eta\) for all \(a\in K\). Now \(\eta=L_b\varphi\) for some \(b\in K\) and \(\varphi\in\mathrm{Aut}(K)\). If \(x\in K\), then \(\eta L_a(x)=L_b\varphi L_a(x)=b\varphi(a)\varphi(x)\) and \(L_a\eta(x)=L_aL_b\varphi(x)=ab\varphi(x)\). Hence, \(b\varphi(a)=ab\) for all \(a\in K\); that is, \(\varphi=\gamma_{b^{-1}}\). It follows that \(\eta(x)=L_b\varphi(x)=b(b^{-1}xb)=xb\), and so \(\eta=R_{b^{-1}}\in K^r\), as desired.
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Theorem 7.2.30 If a group \(K\) is a direct factor whenever it is (isomorphic to) a normal subgroup of a group, then \(K\) is complete.
Proof
We identify \(K\) with the subgroup \(K^l\le\mathrm{Hol}(K)\). Since \(K^l\) is normal, the hypothesis gives a subgroup \(B\) with \(\mathrm{Hol}(K)=K^l\times B\). Now \(B\le\mathrm C_{\mathrm{Hol}(K)}(K^l)=K^r\), because every element of \(B\) commutes with each element of \(K^l\). It follows that if \(\varphi\in\mathrm{Aut}(K)\le\mathrm{Hol}(K)\), then \(\varphi=L_aR_b\) for some \(a,b\in K\). Hence, \(\varphi(x)=axb^{-1}\) for all \(x\in K\). But now \(axyb^{-1}=\varphi(x)\varphi(y)=axb^{-1}ayb^{-1}\), so that \(\mathbf 1=b^{-1}a\); therefore, \(\varphi=\gamma_a\in\mathrm{Inn}(K)\).
Since \(\mathrm{Hol}(K)=K^l\times B\) and \(B\le K^r\le\mathrm{Hol}(K)\), we have \(K^r=B\times(K^r\cap K^l)\). If \(\varphi\in K^l\cap K^r\), then \(\varphi=L_a=R_b\), for \(a,b\in K\). For all \(c\in K\), \(L_a(c)=R_b(c)\) gives \(ac=cb^{-1}\); if \(c=\mathbf 1\), then \(a=b^{-1}\), from which it follows that \(a\in\mathrm Z(K)\). Therefore, \(K^r\cap K^l=\mathrm Z(K)\) and \(K\cong B\times\mathrm Z(K)\). If \(\mathbf 1\neq\varphi\in\mathrm{Aut}(\mathrm Z(K))\), then it is easy to see that \(\tilde\varphi\!:B\times\mathrm Z(K)\rightarrow B\times\mathrm Z(K)\), defined by \((b,z)\mapsto (b,\varphi z)\), is an automorphism of \(K\); \(\tilde\varphi\) must be outer, for conjugation by \((\beta,\zeta)\in B\times\mathrm Z(K)\cong K\) sends \((b,z)\) into \((\beta,\zeta)(b,z)(\beta^{-1},\zeta^{-1})=(\beta b\beta^{-1},z)\). But \(K\) has no outer automorphisms, so that \(\mathrm{Aut}(\mathrm Z(K))=\mathbf 1\) and \(|\mathrm Z(K)|\le 2\). If \(\mathrm Z(K)\cong\mathbb Z_2\), then it is isomorphic to a normal subgroup \(N\) of \(\mathbb Z_4\) which is not a direct factor. But \(K\) is isomorphic to the normal subgroup \(B\times N\) of \(B\times\mathbb Z_4\) which is not a direct factor, contradicting the hypothesis. Therefore, \(\mathrm Z(K)=\mathbf 1\) and \(K\) is complete.
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The holomorph allows one to extend commutator notation. Recall that the commutator \([a,x]=axa^{-1}x^{-1}=x^ax^{-1}\). Now let \(G\) be a group and let \(A=\mathrm{Aut}(G)\). We may regard \(G\) and \(A\) as subgroups of \(\mathrm{Hol}(G)\) (by identifying \(G\) with \(G^l\)). For \(x\in G\) and \(\alpha\in A\), define
and define
Lemma 7.2.31 Let \(G\) and \(A\) be subgroups of a group \(H\), and let \(G=G_0\ge G_1\ge \cdots\) be a series of normal subgroups of \(G\) such that \([A,G_i]\le G_{i+1}\) for all \(i\). Define \(A_1=A\) and
Then \([A_j,A_l]\le A_{j+l}\) for all \(j\) and \(l\), and \([\mathrm C_{j}(A), G_i]\le G_{i+j+1}\) for all \(i\) and \(j\).
Proof
The definition of \(A_j\) gives \([A_j,G_i]\le G_{i+j}\) for all \(i\). It follows that \([A_j,A_l,G_i]=[A_j,[A_l,G_i]]\le[A_j,G_{l+i}]\le G_{j+l+i}\). Similarly, \([A_l,A_j,G_i]\le G_{j+l+i}\). Now \(G_{j+l+i}\lhd\langle G,A\rangle\), because both \(G\) and \(A\) normalize each \(G_i\). Since \([A_j,A_l,G_i][A_l,A_j,G_i]\le G_{j+l+i}\), the three subgroups lemma gives \([G_i,[A_j,A_l]]=[[A_j,A_l],G_i]\le G_{j+l+i}\). Therefore, \([A_j,A_l]\le A_{j+l}\), by definition of \(A_{j+l}\). It follows, for all \(j\), that \(A_j\lhd A\), because \([A_j,A]=[A_j,A_l]\le[A_{j+l}]\le A_j\), and so \(A=A_1\ge A_2\ge\cdots\) is a central series for \(A\). By Proposition 5.5.23(ii), \(\mathrm C_{j}(A)\le A_{j+1}\) for all \(j\), so that, for all \(i\), \([\mathrm C_{j}(A),G_i]\le[A_{j+1},G_i]\le G_{j+i+1}\), as desired.
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Definition 7.2.32 Let \(G=G_0\ge G_1\ge\cdots\ge G_r=\mathbf 1\) be a series of normal subgroups of a group \(G\). An automorphism \(\alpha\in\mathrm{Aut}(G)\) stabilizes this series if \(\alpha(G_ix)=G_ix\) for all \(i\) and all \(x\in G_{i-1}\). The stabilizer \(A\) of this series is the subgroup
Thus, \(\alpha\) stabilizes a normal series \(G=G_0\ge G_1\ge\cdots\ge G_r=\mathbf 1\) if and only if \(\alpha(G_i)\le G_i\) and the induced map \(G_i/G_{i+1}\rightarrow G_i/G_{i+1}\), defined by \(G_{i+1}x\mapsto G_{i+1}\alpha(x)\), is the identity map for each \(i\).
Theorem 7.2.33 The stabilizer \(A\) of a series of normal subgroups \(G=G_0\ge G_1\ge\cdots\ge G_r=\mathbf 1\) is a nilpotent group of class \(\le r-1\).
Proof
Regard both \(G\) and \(A\) as subgroups of \(\mathrm{Hol}(G)\). For all \(i\), if \(x\in G_i\) and \(\alpha\in A\), then \(\alpha(x)=g_{i+1}x\) for some \(g_{i+1}\in G_{i+1}\), and so \(\alpha(x)x^{-1}\in G_{i+1}\). In commutator notation, \([A, G_i]\le G_{i+1}\). By Lemma 7.2.31, \([\mathrm C_{j-1}(A),G_i]\le G_{i+j}\) for all \(i\) and \(j\). In particular, for \(i=0\) and \(j=r\), we have \([\mathrm C_{r-1}(A),G]\le G_r=\mathbf 1\); that is, for all \(x\in G\) and \(\alpha\in\mathrm C_{r-1}(A)\), we have \(\alpha(x)x^{-1}=\mathbf 1\). Therefore, \(\mathrm C_{r-1}(A)=\mathbf 1\) and \(A\) is nilpotent of class \(\le r-1\).
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Example 7.2.34 Let \(\{v_1,\ldots,v_n\}\) be a basis of a vector space \(V\) over a field \(k\), and define \(V_{i-1}=\langle v_i,v_{i+1},\ldots,v_n\rangle\). Hence,
is a series of normal subgroups of the (additive abelian) group \(V\). If \(A\le\mathrm{GL}(V)\) is the group of automorphisms stabilizing this series, then \(A\) is a nilpotent group of class \(\le n-1\). If each \(\alpha\in A\cap\mathrm{GL}(V)\) is regarded as a matrix (relative to the given basis), then it is easy to see that \(A\cap\mathrm{GL}(V)=\mathrm{UT}(n,k)\), the group of a unitriangular matrices. Therefore, \(\mathrm{UT}(n,k)\) is allso nilpotent of class \(\le n-1\).
Proposition 7.2.35 (P.Hall) If \(G=G_0\ge G_1\ge\cdots\ge G_r=\mathbf 1\) is any (not necessarily normal) series of a group \(G\) (i.e., \(G_i\) need not be a normal subgroup of \(G_{i-1}\)), then the stabilizer of this series is always nilpotent of class \(\le\frac 12r(r-1)\).
Semidirect Products
Definition 7.3.1 Let \(K\) be a (not necessarily normal) subgroup of a group \(G\). Then a subgroup \(Q\le G\) is a complement of \(K\) in \(G\) if \(K\cap Q=\mathbf 1\) and \(KQ=G\).
A normal subgroup \(K\) of a group \(G\) need not have a complement and, even if it does, a complement need not be unique. On the other hand, if they exist, complements are unique to isomorphism, for
A group \(G\) is the direct product of two normal subgroups \(K\) and \(Q\) if \(K\cap Q=\mathbf 1\) and \(KQ=G\).
Definition 7.3.2 A group \(G\) is a semidirect product of \(K\) by \(Q\), denoted by \(G=K\rtimes Q\), if \(K\lhd G\) and \(K\) has a complement \(Q_1\cong Q\). One also says that \(G\) splits over \(K\).
Lemma 7.3.3 If \(K\) is a normal subgroup of a group \(G\), then the following statements are equivalent:
- \(G\) is a semidirect product of \(K\) by \(G/K\) (i.e., \(K\) has a complement in \(G\));
- there is a subgroup \(Q\le G\) so that every element \(g\in G\) has a unique expression \(g=ax\), where \(a\in K\) and \(x\in Q\);
- there exists a homomorphism \(s\!:G/K\rightarrow G\) with \(vs=\mathbf 1_{G/K}\), where \(v\!:G\rightarrow G/K\) is the natural map; and
- there exists a homomorphism \(\pi\!:G\rightarrow G\) with \(\mathrm{ker}\pi=K\) and \(\pi(x)=x\) for all \(x\in\mathrm{im}\pi\) (such a map \(\pi\) is called a retraction of \(G\) and \(\mathrm{im}\pi\) is called a retract of \(G\)).
Proof
(i)\(\Rightarrow\)(ii) Let \(Q\) be a complement of \(K\) in \(G\). Let \(g\in G\). Since \(G=KQ\), there exist \(a\in K\) and \(x\in Q\) with \(g=ax\). If \(g=by\) is a second such factorization, then \(xy^{-1}=a^{-1}b\in K\cap Q=\mathbf 1\). Hence \(b=a\) and \(y=x\).
(ii)\(\Rightarrow\)(iii) Each \(g\in G\) has a unique expression \(g=ax\), where \(a\in K\) and \(x\in Q\). If \(Kg\in G/K\), then \(K=g=Kax=Kx\); define \(s\!:G/K\rightarrow G\) by \(s(Kg)=x\). The routine verification that \(s\) is a well defined homomorphism with \(vs=\mathbf 1_{G/K}\) is easy to check.
(iii)\(\Rightarrow\)(iv) Define \(\pi\!:G\rightarrow G\) by \(\pi=sv\). If \(x=\pi(g)\), then \(\pi(x)=\pi(\pi(g))=svsv(g)=sv(g)=\pi(g)=x\) (because \(vs=\mathbf 1_{G/K}\)). If \(a\in K\), then \(\pi(a)=sv(a)=\mathbf 1\), for \(K=\mathrm{ker}v\). For the reverse inclusion, assume that \(\mathbf 1=\pi(g)=sv(g)=s(Kg)\). Now \(s\) is an injection, so that \(Kg=K\) and so \(g\in K\).
(iv)\(\Rightarrow\)(i) Define \(Q=\mathrm{im}\pi\). If \(g\in G\), then \(\pi(g)=g\); if \(g\in K\), then \(\pi(g)=\mathbf 1\); a fortiori, if \(g\in K\cap Q\), then \(g=\mathbf 1\). If \(g\in G\), then \(g\pi(g^{-1})\in K=\mathrm{ker}\pi\), for \(\pi(g\pi(g^{-1}))=\mathbf 1\). Since \(\pi(g)\in Q\), we have \(g=[g\pi(g^{-1})]\pi(g)\in KQ\). Therefore, \(Q\) is a complement of \(K\) in \(G\) and \(G\) is a semidirect product of \(K\) by \(Q\).
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Example 7.3.4
- \(\mathrm S_n\) is a semidirect product of \(\mathrm A_n\) by \(\mathbb Z_2\).
- \(\mathrm D_{2n}\) is a semidirect product of \(\mathbb Z_n\) by \(\mathbb Z_2\).
- For any group \(K\), \(\mathrm{Hol}(K)\) is a semidirect product of \(K^l\) by \(\mathrm{Aut}(K)\).
- Let \(G\) be a solvable group of order \(mn\), where \((m,n)=1\). If \(G\) contains a normal subgroup of order \(m\), then \(G\) is a semidirect product of \(K\) by a subgroup \(Q\) of order \(n\).
- \(\mathrm{Aut}(\mathrm S_6)\) is a semidirect product of \(\mathrm S_6\) by \(\mathbb Z_2\).
- The group \(\mathrm T\) of order \(12\) (Definition A.1.18) is a semidirect product of \(\mathbb Z_3\) by \(\mathbb Z_4\).
- Both \(\mathrm S_3\) and \(\mathbb Z_6\) are semidirect products of \(\mathbb Z_3\) by \(\mathbb Z_2\).
In contrast to direct product, a semidirect product of \(K\) by \(Q\) is not determined to isomorphism by the two subgroups. However, we see that a semidirect product should depend on "how" \(K\) is normal in \(G\).
Lemma 7.3.5 If \(G\) is a semidirect product of \(K\) by \(Q\), then there is a homomorphism \(\theta\!:Q\rightarrow\mathrm{Aut}(K)\), defined by \(\theta_x=\gamma_x\big|_K\); that is, for all \(x\in Q\) and \(a\in K\),
Moreover, for all \(x,y,\mathbf 1\in Q\) and \(a\in K\),
Definition 7.3.6 Let \(Q\) and \(K\) be groups, and let \(\theta\!:Q\rightarrow\mathrm{Aut}(K)\) be a homomorphism. A semidirect product \(G\) of \(K\) by \(Q\) realizes \(\theta\) if, for all \(x\in Q\) and \(a\in K\),
In this language, Lemma 7.3.5 says that every semidirect product \(G\) of \(K\) by \(Q\) determines some \(\theta\) which it realizes. For example, if \(\theta\) is the trivial map, that is, \(\theta_x=\mathbf 1_K\) for every \(x\in G\), then \(a=\theta_x(a)=xax^{-1}\) for every \(a\in K\), and so \(K\le\mathrm C_G(Q)\).
Definition 7.3.7 Given groups \(Q\) and \(K\) and a homomorphism \(\theta\!:Q\rightarrow\mathrm{Aut}(K)\), define \(G=K\rtimes_\theta Q\) to be the set of all ordered pairs \((a,x)\in K\times Q\) equipped with the operation
Theorem 7.3.8 Given groups \(Q\) and \(K\) and a homomorphism \(\theta\!:Q\rightarrow\mathrm{Aut}(K)\), then \(G=K\rtimes_\theta Q\) is a semidirect product of \(K\) by \(Q\) that realizes \(\theta\).
Proof
It is easy to prove that \(G\) is indeed a group. Define a function \(\pi\!:G\rightarrow Q\) by \((a,x)\mapsto x\). Since the only "twist" occurs in the first coordinate, it is routine to check that \(\pi\) is a normal subgroup of \(G\). We identify \(K\) with \(\mathrm{ker}\pi\) via the isomorphism \(a\mapsto(a,\mathbf 1)\). It is also easy to check that \(\{(\mathbf 1,x):x\in Q\}\) is a subgroup of \(G\) isomorphic to \(Q\) (via \(x\mapsto(\mathbf 1,x)\)), and we identify \(Q\) with this subgroup. Also it is easy to see that \(KQ=G\) and \(K\cap Q=\mathbf 1\), so that \(G\) is a semidirect product of \(K\) by \(Q\).
Finally, \(G\) does realize \(\theta\):
\[(\mathbf 1,x)(a,\mathbf 1)(\mathbf 1,x)^{-1}=(\theta_x(a),x)(\mathbf 1,x^{-1})=(\mathrm\theta_x(a),\mathbf 1). \]■
Since \(K\rtimes_\theta Q\) realizes \(\theta\), that is, \(\theta_x(b)=xbx^{-1}\), there can be no confusion if we write \(b^x=xbx^{-1}\) instead of \(\theta_x(b)\). The operation in \(K\rtimes_\theta Q\) will henceforth be written
Theorem 7.3.9 If \(G\) is a semidirect product of \(K\) by \(Q\), then there exists \(\theta\!:Q\rightarrow\mathrm{Aut}(K)\) with \(G\cong K\rtimes_\theta Q\).
Proof
Define \(\theta_x(a)=xax^{-1}\) (as in Lemma 7.3.5). By Lemma 7.3.3(ii), each \(g\in G\) has a unique expression \(g=ax\) with \(a\in K\) and \(x\in Q\). Since multiplication in \(G\) satisfies
\[(ax)(by)=a(xbx^{-1})xy=ab^xxy, \]it is easy to see that the map \(K\rtimes_\theta Q\rightarrow G\), defined by \((a,x)\mapsto ax\), is an isomorphism.
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Proposition 7.3.10 The group \(\mathbf Q_n\) of generalized quaternions is not a semidirect product.
Proposition 7.3.11 Let \(K\) and \(Q\) be groups.
- If \(K\) and \(Q\) are solvable, then \(K\rtimes_\theta Q\) is also solvable.
- \(K\rtimes_\theta Q\) is the direct product \(K\times Q\) if and only if \(\theta\!:Q\rightarrow\mathrm{Aut}(K)\) is trivial (that is, \(\theta_x=\mathbf 1\) for all \(x\in Q\)).
Proposition 7.3.12 If \(|G|=mn\), where \((m,n)=1\), and if \(K\le G\) has order \(m\), then a subgroup \(Q\le G\) is a complement of \(K\) if and only if \(|Q|=n\).
Wreath Products
Let \(D\) and \(Q\) be groups, let \(\Omega\) be a finite \(Q\)-set, and let \(\{D_\omega:\omega\in\Omega\}\) be a family of isomorphic copies of \(D\) indexed by \(\Omega\).
Definition 7.4.1 Let \(D\) and \(Q\) be groups, let \(\Omega\) be a \(Q\)-set, and let \(K=\prod_{w\in\Omega}D_{\omega}\), where \(D_\omega\cong D\) for all \(\omega\in\Omega\). Then the wreath product of \(D\) by \(Q\), denoted by \(D\wr_\Omega Q\), is the semidirect product of \(K\) by \(Q\), where \(Q\) acts on \(K\) by \(q\cdot(d_\omega)=(d_{q\omega})\) for \(q\in Q\) and \((d_\omega)\in\prod_{w\in\Omega}D_\omega\). The normal subgroup \(K\) of \(D\wr_\Omega Q\) is called the base of the wreath product.
If \(D\) is finite, then \(|K|=|D|^{|\Omega|}\); if \(Q\) is also finite, then \(|D\wr_\Omega Q|=|K\rtimes Q|=|D|^{|\Omega|}|Q|\).
If \(\Lambda\) is a \(D\)-set, then \(\Lambda\times\Omega\) can be made into a \((D\wr_\Omega Q)\)-set. Given \(d\in D\) and \(\omega\in\Omega\), define a permutation \(d_\omega^*\) of \(\Lambda\times\Omega\) as follows: for each \((\lambda,\omega')\in\Lambda\times\Omega\), set
It is easy to see that \(d_{\omega}^*d_\omega'^*=(dd')_\omega^*\), and so \(D_\omega^*\), defined by
is a subgroup of \(\mathrm S_{\Lambda\times\Omega}\); indeed, for each \(\omega\), the map \(D\rightarrow D_\omega^*\), given by \(d\mapsto d_\omega^*\), is an isomorphism.
For each \(q\in Q\), define a permutation \(q^*\) of \(\Lambda\times\Omega\) by
and define \(Q^*=\{q^*:q\in Q\}\). It is easy to see that \(Q^*\) is a subgroup of \(\mathrm S_{\Lambda\times\Omega}\) and that the map \(Q\rightarrow Q^*\), given by \(q\mapsto q^*\), is an isomorphism.
Theorem 7.4.2 Given groups \(D\) and \(Q\), a finite \(Q\)-set \(\Omega\), and a \(D\)-set \(\Lambda\), then the wreath product \(D\wr_\Omega Q\) is isomorphic to the subgroup
and hence \(\Lambda\times\Omega\) is a \((D\wr_\Omega Q)\)-set.
Proof
We show first that \(K^*=\langle\bigcup_{\omega\in\Omega}D_\omega^*\rangle\) is the direct product \(\prod_{\omega\in\Omega}D_\omega^*\). It is easy to see that \(D_\omega^*\) centralizes \(D_\omega^*\) for all \(\omega'\neq\omega\), and so \(D_{\omega}^*\lhd K^*\) for every \(omega\). Each \(d_\omega^*\in D_\omega^*\) fixes all \((\lambda,\omega')\in\Lambda\times\Omega\) with \(\omega'\neq\omega\), while each element of \(\langle\bigcup_{\omega'\neq\omega} D_{\omega'}^*\rangle\) fixes all \((\lambda,\omega)\) for all \(\lambda\in\Lambda\). It follows that if \(d_\omega^*\in D_\omega^*\cap\langle\bigcup_{\omega'\neq\omega} D_{\omega'}^*\rangle\), then \(d_\omega^*=\mathbf 1\).
If \(q\in Q\) and \(\omega\in\Omega\), then a routine computation gives
\[q^*d_\omega^*q^{*-1}=d_{q\omega}^* \]for each \(\omega\in\Omega\). Hence, \(q^*K^*q^{*-1}\le K^*\) for each \(q\in Q\), so that \(K^*\lhd W\) (because \(W=\langle K^*,Q^*\rangle\)); it follows that \(W=K^*Q^*\). To see that \(W\) is a semidirect product of \(K^*\) by \(Q^*\), it suffices to show that \(K^*\cap Q^*=\mathbf 1\). Now \(d_\omega^*(\lambda,\omega')=(d\lambda,\omega')\text{ or }(\lambda,\omega')\); in either case, \(d_\omega^*\) fixes the second coordinate. If \(q^*\in Q^*\), then \(q^*(\lambda,\omega')=(\lambda,q\omega')\) and \(q^*\) fixes the first coordinate. Therefore, any \(g\in K^*\cap Q^*\) fixes every \((\lambda,\omega')\) and hence is the identity.
It is now a simple matter to check that the map \(D\wr_\Omega Q\rightarrow W\), given by \((d_\omega)q\mapsto(d_\omega^*)q^*\), is an isomorphism.
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Call the subgroup \(W\) of \(\mathrm S_{\Lambda\times\Omega}\) the permutation version of \(D\wr_\Omega Q\); when we wish to view \(D\wr_\Omega Q\) acting on \(\Lambda\times \Omega\), then we will think of it as \(W\).
Theorem 7.4.3 Let \(D\) and \(Q\) be groups, let \(\Omega\) be a finite \(Q\)-set, let \(\Lambda\) be a \(D\)-set, and let \(W\le \mathrm S_{\Lambda\times\Omega}\) be the permutation version of \(D\wr_\Omega Q\).
- If \(\Omega\) is a transitive \(Q\)-set and \(\Lambda\) is a transitive \(D\)-set, then \(\Lambda\times\Omega\) is a transitive \((D\wr_\Omega Q)\)-set.
- If \(\omega\in \Omega\), then its stabilizer \(Q_\omega\) acts on \(\Omega-\{\omega\}\). If \((\lambda,\omega)\in\Lambda\times\Omega\) and \(D(\lambda)\le D\) is the stabilizer of \(\lambda\), then the stabilizer \(W_{(\lambda,\omega)}\) of \((\lambda,\omega)\) is isomorphic to \(D(\lambda)\times(D\wr_\Omega Q_\omega)\), and \([W:W_{(\lambda,\omega)}]=[D:D(\lambda)][Q:Q_\omega]\).
Proof
- Let \((\lambda,\omega),(\lambda',\omega')\in\Lambda\times\Omega\). Since \(D\) acts transitively, there is \(d\in D\) with \(d\lambda=\lambda'\); since \(Q\) acts transitively, there is \(q\in Q\) with \(q\omega=\omega'\). It is easy to check that \(q^*d_\omega^*(\lambda,\omega)=(\lambda',\omega')\).
- Each element of \(W\) has the form \((d_\omega^*)q^*\), and \((d_\omega^*)q^*(\lambda,\omega)=(\prod_{\omega'\in\Omega}d_\omega^*)(\lambda,q\omega)=d_{q\omega}^*(\lambda,q\omega)=(d_{q\omega}\lambda,q\omega)\). It follows that \((d_\omega^*)q^*\) fixes \((\lambda,\omega)\) if and only if \(q\) fixes \(\omega\) and \(d_\omega\) fixes \(\lambda\). Let \(D_\omega^*(\lambda)=\{d_\omega^*:d\in D(\lambda)\}\). Now \(D_\omega^*(\lambda)\) is disjoint from \(\langle\prod_{\omega'\neq\omega} D_\omega^*,Q_\omega^*\rangle\) and centralizes it: if \(q^*\in Q_\omega^*\), then \(q^*d_\omega^*q^{*-1}=d_{q\omega}^*=d_\omega^*\); hence
\[\begin{aligned} W_{(\lambda,\omega)}&=\left\langle D_\omega^*(\lambda),\prod_{\omega'\neq\omega}D_\omega^*,Q_\omega^*\right\rangle\\&=D_\omega^*(\lambda)\times\left\langle\prod_{\omega'\neq\omega}D_\omega^*,Q_\omega^*\right\rangle\\ &\cong D(\lambda)\times(D\wr_\Omega Q_\omega). \end{aligned} \]It follows that \(|W_{(\lambda,\omega)}|=|D(\lambda)||D|^{|\Omega|-1}|Q_\omega|\) and
\[[W:W_{(\lambda,\omega)}]=|D|^{|\Omega|}|Q|/|D(\lambda)||D|^{|\Omega|-1}|Q_\omega|=[D:D(\lambda)][Q:Q_\omega]. \]■
Theorem 7.4.4 Wreath product is associative: if both \(\Omega\) and \(\Lambda\) are finite, if \(T\) is a group, and if \(\Delta\) is a \(T\)-set, then \(T\wr_{\Lambda\times\Omega}(D\wr_\Omega Q)\cong(T\wr_\Lambda D)\wr_\Omega Q\).
Proof
The permutation versions of both \(T\wr_{\Lambda\times\Omega}(D\wr_\Omega Q)\) and \((T\wr_\Lambda D)\wr_\Omega Q\) are subgroups of \(\mathrm S_{\Delta\times\Lambda\times\Omega}\); we claim that they coincide. The group \(T\wr_{\Lambda\times\Omega}(D\wr_\Omega Q)\) is generated by all \(t_{(\lambda,\omega)}^*\) and all \(f^*\) (for \(f\in D\wr_{\Omega}Q\)). Note that \(t_{(\lambda,\omega)}^*\!:(\delta',\lambda',\omega')\mapsto(t\delta',\lambda',\omega')\) if \((\lambda',\omega')=(\lambda,\omega)\), and fixes it otherwise; also \(f^*\!:(\delta',\lambda',\omega')\mapsto(\delta',f(\lambda',\omega'))\). Specializing \(f^*\) to \(d_\omega^*\) and to \(q^*\), we see that \(T\wr_{\Lambda\times\Omega}(D\wr_{\Omega}Q)\) is generated by all \(t_{\lambda,\omega}^*,d_\omega^*\), and \(q^{**}\), where \(d_\omega^*\!:(\delta',\lambda',\omega')\mapsto(\delta',d\lambda',\omega')\) if \(\omega'=\omega\), and fixes it otherwise, and \(q^{**}\!:(\delta',\lambda',\omega')\mapsto(\delta',\lambda',q\omega')\).
A similar analysis of \((T\wr_\Lambda D)\wr_\Omega Q\) shows that it is generated by all \(q^{**},d_\omega^*\), and \((t_\lambda)_\omega^*\), where \((t_\lambda)_\omega^*\!:(\delta',\lambda',\omega')\mapsto(t\delta',\lambda',\omega')\) if \(\omega'=\omega\) and \(\lambda'=\lambda\), and fixes it otherwise. Since \((t_\lambda)_\omega^*=t_{(\lambda,\omega)}^*\), the two wreath products coincide.
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Proposition 7.4.5 If \((a,x)\in D\wr_\Omega Q\) (so that \(a\in K=\prod D_\omega\)), then
Proposition 7.4.6 If both \(D\) and \(Q\) are solvable, and \(\Omega\) is a \(Q\)-set, then \(D\wr_\Omega Q\) is solvable.
Definition 7.4.7 If \(\Omega=Q\) regarded as a \(Q\)-set acting on itself by left multiplication, then we write \(W=D\wr_r Q\), and we call \(W\) the regular wreath product. Thus, the base is the direct product of \(|Q|\) copies of \(D\), indexed by the elements of \(Q\), and \(q\in Q\) sends a \(|Q|\)-tuple \((d_x)\in\prod_{x\in Q}D_x\) into \((d_{qx})\), and \(|D\wr_r Q|=|D|^{|Q|}|Q|\). It is easy to see that the formation of regular wreath product is not associative when all groups are finite, for \(|T\wr_r(D\wr_r Q)|\neq|(T\wr_r D)\wr_r Q|\).
If \(\Omega\) is an infinite set and \(\{D_\omega:\omega\in\Omega\}\) is a family of groups, then there are two direct product constructions.
- (sometimes called the complete direct product) consists of all "vectors" \((d_\omega)\) in the cartesian product \(\prod_{\omega\in\Omega}D_\omega\) with "coordinatewise" multiplication: \((d_\omega)(d'_\omega)=(d_\omega d'_\omega)\).
- (called the restricted direct product) is the subgroup of the complete direct product consisting of all those \((d_\omega)\) with only finitely many coordinates \(d_\omega\neq\mathbf 1\).
The wreath product using the complete direct product is called the complete wreath product, while using the restricted direct product is the restricted wreath product.
Theorem 7.4.8 (Kaloujnine) If \(p\) is a prime, then a Sylow \(p\)-subgroup of \(\mathrm S_{p^n}\) is an iterated regular wreath product \(W_n=\mathbb Z_p\wr_r\mathbb Z_p\wr_r\cdots\wr_r\mathbb Z_p\) of \(n\) copies of \(\mathbb Z_p\), where \(W_{n+1}=W_n\wr_r\mathbb Z_p\).
Proof
The proof is by induction on \(n\), the case \(n=1\) holding naturally. Assume that \(n>1\), let \(\Lambda\) be a set with \(p^n\) elements and let \(D\) be a Sylow-\(p\) subgroup of \(\mathrm S_\Lambda\); thus, \(\Lambda\) is a \(D\)-set. Let \(\Omega=\{0,1,\ldots,p-1\}\), and let \(Q=\langle q\rangle\) be a cyclic group of order \(p\) acting on \(\Omega\) by \(qi=i+1\pmod p\). The permutation version of the wreath product \(P=D\wr_r\mathbb Z_p\) is a subgroup of \(\mathrm S_{\Lambda\times\Omega}\); of course, \(|\Lambda\times\Omega|=p^{n+1}\). By induction, \(D\) is a wreath product of \(n\) copies of \(\mathbb Z_p\), and so \(P\) is a wreath product of \(n+1\) copes of \(\mathbb Z_p\). To see that \(P\) is a Sylow \(p\)-subgroup, it suffices to see that its order is \(p^{\mu(n+1)}\), where \(\mu(n+1)=p^n+p^{n-1}+\cdots+p+1\). Now \(|D|=p^{\mu(n)}\), so that \(|P|=|D\wr_r\mathbb Z_p|=(p^{\mu(n)})^pp=p^{p\mu(n)+1}=p^{\mu(n+1)}\).
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Example 7.4.9 The theorem above can be used to compute the Sylow \(p\)-subgroup of \(\mathrm S_m\) for any \(m\). First write \(m\) in base \(p\):
Partition \(X=\{1,2,\ldots,m\}\) into \(a_0\) singletons, \(a_1\) \(p\)-subsets, \(a_2\) \(p^2\)-subsets, \(\ldots\), and \(a_t\) \(p^t\)-subsets. On each of these \(p^i\)-subsets \(Y\), construct a Sylow \(p\)-subgroup of \(\mathrm S_Y\). Since disjoint permutations commute, the direct product of all these Sylow subgroups is a subgroup of \(\mathrm S_X\) of order \(p^N\), where \(N=a_1+a_2\mu(2)+\cdots+a_t\mu(t)\) (recall that \(\mu(i)=p^{i-1}+p^{i-2}+\cdots+p+1\)). But it is easy to see that \(p^N\) is the highest power of \(p\) dividing \(m!\). Thus, the direct product has the right order, and so it must be a Sylow \(p\)-subgroup of \(\mathrm S_X\cong\mathrm S_m\).
Proposition 7.4.10 Let \(D\) be a (multiplicative) group. A monomial matrix \(\mu\) over \(D\) is a permutation matrix \(P\) whose nonzero entries have been replaced by elements of \(D\); we say that \(P\) is the support of \(\mu\). If \(Q\) is a group of \(n\times n\) permutation matrices, then denote
Then \(M(D,Q)\) is a group under matrix multiplication, and \(M(D,Q)\cong D\wr_\Omega Q\), where \(\Omega=\{1,2,\ldots,n\}\) is a faithful \(Q\)-set here.
Factor Sets
In discussing general extensions \(G\) of \(K\) by \(Q\), it is convenient to use the additive notation for \(G\) and its subgroup \(K\) (this is one of the rare instances in which one uses additive notation for a nonabelian group). For example, if \(k\in K\) and \(g\in G\), we shall write the conjugate of \(k\) by \(g\) as \(g+k-g\).
Definition 7.5.1 If \(K\le G\), then a (right) transversal of \(K\) in \(G\) is a subset \(T\) of \(G\) consisting of one element from each right coset of \(K\) in \(G\).
If \(T\) is a right transversal, then \(G\) is the disjoint union \(G=\bigcup_{t\in T}(K+t)\). Thus, every element \(g\in G\) has a unique factorization \(g=k+t\) for \(k\in K\) and \(t\in T\). There is a similar definition of left transversal; of course, these two notations coincide when \(K\) is normal.
If \(G\) is a semidirect product and \(Q\) is a complement of \(K\), then \(Q\) is a transversal of \(K\) in \(G\).
Definition 7.5.2 If \(\pi\!:G\rightarrow Q\) is surjective, then a lifting of \(x\in Q\) is an element \(l(x)\in G\) with \(\pi(l(x))=x\).
If one chooses a lifting \(l(x)\) for each \(x\in Q\), then the set of all such is a transversal of \(\mathrm{ker}\pi\). In this case, the function \(l\!:Q\rightarrow G\) is also called a right transversal.
Proposition 7.5.3 An extension \(G\) of \(K\) by \(Q\) is a semidirect product if and only if there is a transversal \(l\!:Q\rightarrow G\) that is a homomorphism.
Theorem 7.5.4 Let \(G\) be an extension of \(K\) by \(Q\), and let \(l\!:Q\rightarrow G\) be a transversal. If \(K\) is abelian, then there is a homomorphism \(\theta\!:Q\rightarrow\mathrm {Aut}(K)\) with
for every \(a\in K\). Moreover, if \(l_1\!:Q\rightarrow G\) is another transversal, then \(l(x)+a-l(x)=l_1(x)+a-l_1(x)\) for all \(a\in K\) and \(x\in Q\).
Proof
Since \(K\lhd G\), the restriction \(\gamma_g\big|_K\) is an automorphism of \(K\) for all \(g\in G\), where \(\gamma_g\) is conjugation by \(g\). The function \(\mu\!:G\rightarrow\mathrm{Aut}(K)\), given by \(g\mapsto\gamma_g\big|_K\), is easily seen to be a homomorphism; moreover, \(K\le\mathrm{ker}\mu\), for \(K\) being abelian implies that each conjugation by \(a\in K\) is identity. Therefore, \(\mu\) induces a homomorphism \(\mu_\#\!:G/K\rightarrow\mathrm{Aut}(K)\), namely, \(K+g\mapsto\mu(g)\).
The first isomorphism theorem says more than \(Q\cong G/K\); it gives an explicit isomorphism \(\lambda\!:Q\rightarrow G/K\): if \(l\!:Q\rightarrow G\) is a transversal, then \(\lambda(x)=K+l(x)\). If \(l_1\!:Q\rightarrow G\) is another transversal, then \(l(x)-l_1(x)\in K\), so that \(K+l(x)=K+l_1(x)\) for every \(x\in Q\). It follows that \(\lambda\) odes not depend on the choice of transversal. Let \(\theta\!:Q\rightarrow\mathrm {Aut}(K)\) be the composite: \(\theta=\mu_\#\lambda\). If \(x\in Q\), then \(\theta_x=\mu_\#\lambda(x)=\mu_\#(K+l(x))=\mu(l(x))\in\mathrm {Aut}(K)\); therefore, if \(a\in K\),
\[\theta_x(a)=\mu(l(x))(a)=l(x)+a-l(x) \]does not depend on the choice of lifting \(l(x)\).
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A homomorphism \(\theta\!:Q\rightarrow\mathrm{Aut}(K)\) makes \(K\) into a \(Q\)-set, where the action is given by \(xa=\theta_x(a)\). (For semidirect products, we denoted \( \theta_x(a)\) by \(a^x\); since we are now writing \(K\) additively, however, the notation \(xa\) is more appropriate.) The following formulas are valid for all \(x,y,\mathbf 1\in Q\) and \(a,b\in K\):
Definition 7.5.5 Call an ordered triple \((Q,K,\theta)\) data if \(K\) is an abelian group, \(Q\) is a group, and \(\theta\!:Q\rightarrow\mathrm{Aut}(K)\) is a homomorphism. We say that a group \(G\) realizes this data if \(G\) is an extension of \(K\) by \(Q\) and, for every tranversal \(l\!:Q\rightarrow G\),
for all \(x\in Q\) and \(a\in K\).
Using these terms, Theorem 7.5.3 says that when \(K\) is abelian, every extension \(G\) of \(K\) by \(Q\) determines a homomorphism \(\theta\!:Q\rightarrow\mathrm{Aut}(K)\), and \(G\) realizes the data. The intuitive meaning of \(\theta\) is that it describes how \(K\) is a normal subgroup of \(G\).
Let \(\pi\!:G\rightarrow Q\) be a surjective homomorphism with kernel \(K\), and choose a transversal \(l\!:Q\rightarrow G\) with \(l(\mathbf 1_Q)=\mathbf 0_G\). Once this transversal has been chosen, every element \(g\in G\) has a unique expression of the form
There is a formula: for all \(x,y\in Q\),
because both \(l(x)+l(y)\) and \(l(xy)\) represent the same coset of \(K\).
Definition 7.5.6 If \(\pi\!:G\rightarrow Q\) is a surjective homomorphism with kernel \(K\), and if \(l\!:Q\rightarrow G\) is a transversal with \(l(\mathbf 1_Q)=\mathbf 0_G\), then the function \(f\!:Q\times Q\rightarrow K\), defined by \((2)\) above, is called a factor set (or cocycle). Of course, the factor set \(f\) depends on the transversal \(l\).
One may think of a factor set as a "measure" of \(G\)'s deviation from being a semidirect product, for it describes the obstruction to the transversal \(l\) being a homomorphism.
Theorem 7.5.7 Let \(\pi\!:G\rightarrow Q\) be a surjective homomorphism with kernel \(K\), let \(l\!:Q\rightarrow G\) be a transversal with \(l(\mathbf 1_Q)=\mathbf 0_G\), and let \(f\!:Q\times Q\rightarrow K\) be the corresponding factor set. Then:
- for all \(x,y\in Q\), we have \(f(\mathbf 1_Q,y)=\mathbf 0_K=f(x,\mathbf 1_Q)\);
- the cocycle identity holds for every \(x,y,z\in Q\):
Proof
The definition of \(f\) gives \(l(x)+l(y)=f(x,y)+l(xy)\). In particular, \(l(\mathbf 1_Q)+l(y)=f(\mathbf 1,y)+l(y)\); since we are assuming that \(l(\mathbf 1_Q)=\mathbf 0_G\), we have \(f(\mathbf 1_Q,y)=\mathbf 0_G\). A similar calculation shows that \(f(x,\mathbf 1_Q)=\mathbf 0_G\). The cocycle identity follows from associativity:
\[\begin{aligned} [l(x)+l(y)]+l(z)&=f(x,y)+l(xy)+l(z)\\ &=f(x,y)+f(xy,z)+l(xyz). \end{aligned} \]on the other hand,
\[\begin{aligned} l(x)+[l(y)+l(z)]&=l(x)+f(y,z)+l(yz)\\ &=xf(y,z)+l(x)+l(yz)\\ &=xf(y,z)+f(x,yz)+l(xyz). \end{aligned} \]The cocycle identity follows.
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Theorem 7.5.8 Given data \((Q,K,\theta)\), a function \(f\!:Q\times Q\rightarrow K\) is a factor set if and only if it satisfies the cocycle identity
as well as \(f(\mathbf 1_Q,y)=\mathbf 0_K=f(x,\mathbf 1_Q)\) for all \(x,y,z\in Q\). More precisely, there is an extension \(G\) realizing the data and a transversal \(l\!:Q\rightarrow G\) such that \(f\) is the corresponding factor set.
Proof
To prove sufficiency, let \(G\) be the set of all ordered pairs \((a,x)\in K\times Q\) equipped with the operation
\[(a,x)+(b,y)=(a+xb+f(x,y),xy) \](note that if \(f\) is identically \(\mathbf 0_K\), then this is the semidirect product \(K\rtimes_\theta Q\)).
It is easy to check that \(G\) is indeed a group. For every transversal \(l\!:Q\rightarrow G\), that \(xa=l(x)+a-l(x)\) for all \(x\in Q\) and \(a\in K\). Now we must have \(l(x)=(b,x)\) for some \(b\in K\). Therefore,
\[\begin{aligned} l(x)+a-l(x)&=(b,x)+(a,\mathbf 1_Q)-(b,x)\\ &=(b+xa,x)+(-x^{-1}b-x^{-1}f(x,x^{-1}),x^{-1})\\ &=(b+xa+x[-x^{-1}b-x^{-1}f(x,x^{-1})]+f(x,x^{-1}),\mathbf 1_Q). \end{aligned} \]Since \(K\) is abelian, the last term simplifies to \((xa,\mathbf 1_Q)\). As any element of \(K\), we identify \(xa\) with \((xa,\mathbf 1_Q)\), and so \(G\) does realize the data.
Define a transversal \(l\!:Q\rightarrow K\) by \(l(x)=(\mathbf 0_K,x)\) for all \(x\in Q\). The factor set \(F\) corresponding to this transversal satisfies \(F(x,y)=l(x)+l(y)-l(xy)\). But a straightforward calculation shows that \(F(x,y)=(f(x,y),\mathbf 1_Q)\), and so \(f\) is a factor set, as desired.
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Notation Denote the extension \(G\) constructed in the proof of Theorem 7.5.7 by \(G_f\); it realizes \((Q,K,\theta)\) and it has \(f\) as a factor set.
Definition 7.5.9 \(\mathrm Z_\theta^2(Q,K)\) is the set of all factor sets \(f\!:Q\times Q\rightarrow K\).
Theorem 7.5.7 shows that \(\mathrm Z_\theta^2(Q,K)\) is an abelian group under pointwise addition: \(f+g\!:(x,y)\mapsto f(x,y)+g(x,y)\). If \(f\) and \(g\) are factor sets then so is \(f+g\) (for \(f+g\) also satisfies the cocycle identity and vanishes on \((\mathbf 1,y)\) and \((x,\mathbf 1)\)). If \(G_f\) and \(G_g\) are the extensions constructed from them, then \(G_{f+g}\) is also an extension; it follows that there is an abelian group structure on the family of all extensions realizing the data \((Q,K,\theta)\) whose identity element is the semidirect product (which is \(G_{\mathbf 0}\)). This group of all extensions is large, however, because the same extension occurs many times. Take a fixed extension \(G\) realizing the data, and choose two different transversals, say, \(l\) and \(l'\). Each transversal gives a factor set:
Now the factor sets \(f\) and \(f'\) are distinct, but both of them have arisen from the same extension.
Lemma 7.5.10 Let \(G\) be an extension realizing \((Q,K,\theta)\), and let \(l\) and \(l'\) be transversals with \(l(\mathbf 1_Q)=\mathbf 0_K=l'(\mathbf 1_Q)\) giving rise to factor sets \(f\) and \(f'\), respectively. Then there is a function \(h\!:Q\rightarrow K\) with \(h(\mathbf 1_Q)=\mathbf 0_K\) such that
for all \(x,y\in Q\).
Proof
For each \(x\in Q\), both \(l(x)\) and \(l'(x)\) are representatives of the same coset of \(K\) in \(G\); there is thus an element \(h(x)\in K\) with
\[l'(x)=h(x)+l(x). \]Since \(l'(\mathbf 1_Q)=\mathbf 0_K=l(\mathbf 1_Q)\), we have \(h(\mathbf 1_Q)=\mathbf 0_K\). The main formula is derived as follows.
\[\begin{aligned} l'(x)+l'(y)&=[h(x)+l(x)]+[h(y)+l(y)]\\ &=h(x)+xh(y)+l(x)+l(y)\quad(G\text{ realizes the data})\\ &=h(x)+xh(y)+f(x,y)+l(xy)\\ &=h(x)+xh(y)+f(x,y)-h(xy)+l'(xy). \end{aligned} \]Therefore, \(f'(x,y)=h(x)+xh(y)+f(x,y)-h(xy)\). The desired formula follows because each term lies in the abelian group \(K\).
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Definition 7.5.11 Given data \((Q,K,\theta)\), a coboundary is a function \(g\!:Q\times Q\rightarrow K\) for which there exists \(h\!:Q\rightarrow K\) with \(h(\mathbf 1_Q)=\mathbf 0_K\) such that
The set of all coboundaries is denoted by \(\mathrm B_\theta^2(Q,K)\).
It is easy to check that \(\mathrm B_\theta^2(Q,K)\) is a subgroup of \(\mathrm Z_\theta^2(Q,K)\); that is, every coboundary \(g\) satisfies the cocycle identity and \(g(x,\mathbf 1_Q)=\mathbf 0_K=g(\mathbf 1_Q,x)\) for all \(x\in Q\). Moreover, Lemma 7.5.9 says that factor sets \(f\) and \(f'\) arising from different transversals of the same extension satisfy \(f'-f\in\mathrm B_\theta^2(Q,K)\); that is, they lie in the same coset of \(\mathrm B_\theta^2(Q,K)\) in \(\mathrm Z_\theta^2(Q,K)\).
Proposition 7.5.12 An extension \(G\) realizing data \((Q,K,\theta)\) is a semidirect product if and only if it has a factor set \(f\in\mathrm B_\theta^2(Q,K)\).
Definition 7.5.13 Given data \((Q,K,\theta)\), then
it is called the second cohomology group of the data.
Definition 7.5.14 Two extensions \(G\) and \(G'\) realizing data \((Q,K,\theta)\) are equivalent if there are factor sets \(f\) of \(G\) and \(f'\) of \(G'\) with \(f'-f\in\mathrm B_\theta^2(Q,K)\); that is, the factor sets determine the same element of \(\mathrm H_\theta^2(Q,K)\).
Proposition 7.5.15 Any two semidirect products realizing data \((Q,K,\theta)\) are equivalent.
Theorem 7.5.16 Two extensions \(G\) and \(G'\) realizing data \((Q,K,\theta)\) are equivalent if and only if there exists an isomorphism \(\gamma\) making the following diagram commute:
where \(i\) and \(i'\) are injective, \(\pi\) and \(\pi'\) are surjective, \(\mathrm {im} i=\mathrm{ker}\pi\), and \(\mathrm{im}i'=\mathrm{ker}\pi'\).
Remark A homomorphism \(\gamma\) making the diagram commute is necessarily an isomorphism.
Proof
Assume that \(G\) and \(G'\) are equivalent. There are thus factor sets \(f,f'\!:Q\times Q\rightarrow K\), arising from lifings \(l,l'\), respectively, and a function \(h\!:Q\rightarrow K\) with \(h(\mathbf 1_Q)=\mathbf 0_K\) such that
\[\begin{align} f'(x,y)-f(x,y)=xh(y)-h(xy)+h(x) \end{align} \]for all \(x,y\in Q\). Each element of \(G\) has a unique expression of the form \(a+l(x)\), where \(a\in K\) and \(x\in Q\), and addition is given by
\[[a+l(x)]+[b+l(y)]=a+xb+f(x,y)+l(xy); \]there is a similar description of addition in \(G'\). Define \(\gamma\!:G\rightarrow G'\) by
\[\gamma(a+l(x))=a+h(x)+l'(x). \]Since \(l(\mathbf 1)=\mathbf 0\), we have \(\gamma(a)=\gamma(a+l(\mathbf 1))=a+h(\mathbf 1)+l'(\mathbf 1)=a\), for all \(a\in K\), because \(h(\mathbf 1)=\mathbf 0\); that is, \(\gamma\) fixes \(K\) pointwise. Also, \(x=\pi(a+l(x))\), while \(\pi'\gamma(a+l(x))=\pi'(a+h(x)+l'(x))=\pi'(l'(x))=x\). We have shown that the diagram commutes. It remains to show that \(\gamma\) is a homomorphism. Now
\[\begin{aligned} \gamma([a+l(x)]+[b+l(y)])&=\gamma(a+xb+f(x,y)+l(xy))\\ &= a+xb+f(x,y)+h(xy)+l'(xy), \end{aligned} \]while
\[\begin{aligned} \gamma(a+l(x))+\gamma(b+l(y))&=[a+h(x)+l'(x)]+[b+h(y)+l'(y)]\\ &= a+h(x)+xb+xh(y)+f'(x,y)+l'(xy). \end{aligned} \]The element \(l'(xy)\) is common to both expressions, and \((3)\) shows that the remaining elements of the abelian group \(K\) are equal; thus, \(\gamma\) is a homomorphism.
Conversely, assume that there exists an homomorphism \(\gamma\) as in the statement. Commutativity of the diagram gives \(\gamma(a)=a\) for all \(a\in K\). Moreover, if \(x\in Q\), then \(x=\pi(l(x))=\pi'\gamma(l(x))\); that is, \(\gamma l\!:Q\rightarrow G'\) is a lifting. Applying \(\gamma\) to the equation \(l(x)+l(y)=f(x,y)+l(xy)\) shows that \(\gamma f\) is the factor set determined by the lifting \(\gamma l\). But \(\gamma f(x,y)=f(x,y)\) for all \(x,y\in Q\), because \(f(x,y)\in K\). Therefore, \(\gamma f=f\); that is, \(f\) is a factor set of \(G'\). But \(f'\) is also a factor set of \(G'\) (arising from another lifting), and so Lemma 7.5.9 gives \(f'-f\in\mathrm B^2\); that is, \(G'\) and \(G\) are equivalent.
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Definition 7.5.17 If \(G\) is a extension of \(K\) by \(Q\), then \(\gamma\in\mathrm{Aut}(G)\) stabilizes the extension if the diagram in Theorem 7.5.15 (with \(G'\) replaced by \(G\)) commutes. The group \(A\) of all such \(\gamma\) is called the stabilizer of the extension.
Theorem 7.5.18 If \(K\) and \(Q\) are (not necessarily abelian) groups and \(G\) is an extension of \(K\) by \(Q\), then the stabilizer \(A\) of the extension is abelian.
Proof
If \(\gamma\in A\), then the hypothesis that \(\gamma\) makes the diagram commute is precisely the statement that \(\gamma\) stabilizes the series \(G> K> \mathbf 1\). It follows from Theorem 7.2.33 that the group \(A\) of all such \(\gamma\) is nilpotent of class \(\le 1\); that is, \(A\) is abelian.
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Theorem 7.5.19 (Schreier) There is a bijection from \(\mathrm H_\theta^2(Q,K)\) to the set \(E\) of all equivalence classes of extensions realizing data \((Q,K,\theta)\) taking the identity \(\mathbf 0\) to the class of the semidirect product.
Proof
Denote the equivalence class of an extension \(G\) realizing the data \((Q,K,\theta)\) by \([G]\). Define \(\varphi\!:\mathrm H_\theta^2(Q,K)\rightarrow E\) by \(\varphi(f+\mathrm B_\theta^2(Q,K))=[G_f]\), where \(G_f\) is the extension constructed in Theorem 7.5.7 from a factor set \(f\). First, \(\varphi\) is well defined, for if \(f\) and \(G\) are factor sets with \(f+\mathrm B^2=g+\mathrm B^2\), then \(f-g\in\mathrm B^2\), \(G_f\) and \(G_g\) are equivalent, and \([G_f]=[G_g]\). Conversely, \(\varphi\) is an injection: if \(\varphi(f+\mathrm B^2)=\varphi(g+\mathrm B^2)\), then \([G_f]=[G_g], f-g\in \mathrm B^2\), and \(f+\mathrm B^2=g+\mathrm B^2\). Now \(\varphi\) is a surjection: if \([G]\in E\) and \(f\) is a factor set (arising from some choice of transversal), then \([G]=[G_f]\) and \([G]=\varphi(f+\mathrm B^2)\). The last part of the theorem follows from Proposition 7.5.11.
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Thus, there is a unique group structure on \(E\) making \(\varphi\) an isomorphism, namely, \([G_f]+[G_g]=[G_{f+g}]\).
Corollary 7.5.20 \(\mathrm H_\theta^2(Q,K)=\mathbf 0\) if and only if every extension \(G\) realizing data \((Q,K,\theta)\) is a semidirect product.
Definition 7.5.21 If both \(Q\) and \(K\) are abelian and \(\theta\!:Q\rightarrow\mathrm{Aut}(K)\) is trivial, then \(\mathrm{Ext}(Q,K)\) is the set of all equivalence classes of abelian extensions \(G\) of \(K\) by \(Q\).
Corollary 7.5.22 If both \(Q\) and \(K\) are abelian and \(\theta\!:Q\rightarrow\mathrm{Aut}(K)\) is trivial, then \(\mathrm{Ext}(Q,K)\le\mathrm H_\theta^2(Q,K)\).
Proof
If \(f\!:Q\times Q\rightarrow K\) is a factor set, then the corresponding extension \(G_f\) is abelian if and only if \(f(x,y)=f(y,x)\) for all \(x,y\in Q\): since \(\theta\) is trivial,
\[\begin{aligned} (a,x)+(b,y)&=(a+b+f(x,y),xy)\\ &=(a+b+f(y,x),xy)\\ &=(b+a+f(y,x),yx)\\ &=(b,y)+(a,x). \end{aligned} \]It is easy to see that the set \(\mathrm S_\theta^2(Q,K)\) of all such "symmetric" factor sets forms a subgroup of \(\mathrm Z_\theta^2(Q,K)\), and
\[\begin{aligned} \mathrm{Ext}(Q,K)&=(\mathrm Z^2\cap\mathrm S^2)/(\mathrm B^2\cap\mathrm S^2)\\ &\cong[(\mathrm Z^2\cap\mathrm S^2)+\mathrm B^2]/\mathrm B^2\le\mathrm H_\theta^2(Q,K). \end{aligned} \]■
Remark 7.5.23 It is plain that \(|\mathrm H_\theta^2(Q,K)|\) is an upper bound for the number of nonisomorphic extensions \(G\) of \(K\) by \(Q\) realizing \(\theta\); however, this bound need not be attained.
Theorem 7.5.24 If \(D\) and \(Q\) are groups with \(Q\) finite, then the regular wreath product \(D\wr_r Q\) contains an isomorphic copy of every extension of \(D\) by \(Q\).
Corollary 7.5.25 If \(\mathscr C\) is a class of finite groups closed under subgroups and semidirect products (i.e., if \(A\in\mathscr C\) and \(S\le A\), then \(S\in\mathscr C\); if \(A,B\in\mathscr C\), then \(A\rtimes_\theta B\in\mathscr C\) for all \(\theta\)), then \(\mathscr C\) is closed under extensions.
Theorems of Schur-Zassenhaus and Gaschütz
Theorem 7.6.1 If \(K\) is an abelian normal Hall subgroup of a finite group \(G\) (i.e., \((|K|,[G:K])=1\)), then \(K\) has a complement.
Proof
Let \(|K|=m\), let \(Q=G/K\), and let \(|Q|=n\), so that \((n,m)=1\). It suffices to prove, by Corollary 7.5.19, that every factor set \(f\!:Q\times Q\rightarrow K\) is a coboundary. Define \(\sigma\:Q\rightarrow K\) by
\[\sigma(x)=\sum_{y\in Q}f(x,y); \]\(\sigma\) is well defined since \(Q\) is finite and \(K\) is abelian. Sum the cocycle identity
\[xf(y,z)-f(xy,z)+f(x,yz)=f(x,y) \]over all \(z\in Q\) to obtain
\[x\sigma(y)-\sigma(xy)+\sigma(x)=nf(x,y) \](as \(z\) ranges over all of \(Q\), so does \(yz\)). Since \((m,n)=1\), there are integers \(s\) and \(t\) with \(sm+tn=1\). Define \(h\!:Q\rightarrow K\) by \(h(x)=t\sigma(x)\). Then \(h(\mathbf 1_Q)=\mathbf 0_K\) and
\[xh(y)-h(xy)+h(x)=f(x,y)-msf(x,y). \]But \(sf(x,y)\in K\), so that \(msf(x,y)=\mathbf 0\). Therefore, \(f\) is a coboundary.
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Theorem 7.6.2 If \(K\) is an abelian normal Hall subgroup of a finite group \(G\), then any two complements of \(K\) are conjugate.
Proof
Denote \(|K|\) by \(m\) and \(|G/K|\) by \(n\), so that \((m,n)=1\). Let \(Q_1\) and \(Q_2\) be subgroups of \(G\) of order \(n\). By Proposition 7.3.12, each of these subgroups is a complement of \(K\). By Proposition 7.5.3, there are transversals \(l_i\!:G/K\rightarrow G\), for \(i=1,2\), with \(l_i(G/K)=Q_i\) and with each \(l_i\) a homomorphism. It follows that the factor sets \(f_i\) determined by \(l_i\) are identically zero. If \(h(x)\) is defined by \(l_1(x)=h(x)+l_2(x)\), then
\[\mathbf 0=f_1(x,y)-f_2(x,y)=xh(x)-h(xy)+h(x). \]Summing over all \(y\in G/K\) gives the equation in \(K\):
\[\mathbf 0=xa_0-a_0+nh(x), \]where \(a_0=\sum_{y\in G/K}h(y)\). Let \(sm+tn=1\) and define \(b_0=ta_0\). Since \(K\) has exponent \(m\),
\[h(x)=h(x)-smh(x)=-tnh(x)=xta_0-ta_0=xb_0-b_0 \]for all \(x\in G/K\). We claim that \(-b_0+Q_1+b_0=Q_2\). If \(l_1(x)\in Q_1\), then \(-b_0+l_1(x)+b_0=-b_0+xb_0+l_1(x)=-h(x)+l_1(x)=l_2(x)-l_1(x)+l_1(x)=l_2(x)\).
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Theorem 7.6.3 (Schur-Zassenhaus Lemma) A normal Hall subgroup \(K\) of a finite group \(G\) has a complement (and so \(G\) is a semidirect product of \(K\) by \(G/K\)).
Proof
Let \(|K|=m\) and let \(|G|=mn\), where \((m,n)=1\). We prove, by induction on \(m\ge 1\), that \(G\) contains a subgroup of order \(n\). The base step is trivially true. If \(K\) contains a proper subgroup \(T\) which is also normal in \(G\), then \(K/T\lhd G/T\) and \((G/T)/(K/T)\cong G/K\) has order \(n\); that is, \(K/T\) is a normal Hall subgroup of \(G/T\). If \(|K/T|=m'\), then \(m'< m\) and \([G/T:K/T]=n\). The inductive hypothesis gives a subgroup \(N/T\le G/T\) of order \(n\). Now \(|N|=n|T|\) and \((n,|T|)=1\) (for \(|T|\) divides \(m\)), so that \(T\) is a normal Hall subgroup of \(N\) (with \(|T|< m\) and with index \([N:T]=n\)). By induction, \(N\) and hence \(G\) contains a subgroup of order \(n\).
We may now assume that \(K\) is a minimal normal subgroup of \(G\). If \(p\) is a prime dividing \(m\) and if \(P\) is a Sylow \(p\)-subgroup of \(K\), then the Frattini argument gives \(G=K\mathrm N_G(P)\). By the second isomorphism theorem,
\[G/K=K\mathrm N_G(P)/K\cong \mathrm N_G(P)/(K\cap\mathrm N_G(P))=\mathrm N_G(P)/\mathrm N_K(P), \]so that \(|\mathrm N_K(P)|n=|\mathrm N_K(P)||G/K|=|\mathrm N_G(P)|\). If \(\mathrm N_G(P)\) is a proper subgroup of \(G\), then \(|\mathrm N_K(P)|<m\), and induction shows that \(\mathrm N_G(P)\) contains a subgroup of order \(n\). We may assume, therefore, that \(\mathrm N_G(P)=G\); that is, \(P\lhd G\).
Since \(K\ge P\) and \(K\) is a minimal normal subgroup of \(G\), we have \(K=P\). Since \(\mathrm Z(P)\text{ char }P\) and \(P\lhd G\), we have \(\mathrm Z(P)\lhd G\). Minimality applies again, and we get \(\mathrm Z(P)=P\). But now \(P=K\) is abelian, and the proof follows from Theorem 7.6.1.
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Theorem 7.6.4 Let \(K\) be a normal Hall subgroup of a finite group \(G\). If either \(K\) or \(G/K\) is solvable, then any two complements of \(K\) in \(G\) are conjugate.
Remark The Feit-Thompson theorem says that every group of odd order is solvable. Since \(|K|\) and \(|G/K|\) are relatively prime, at least one of them has odd order, and so complements of normal Hall subgroups are always conjugate.
Proof
Let \(|K|=m\), let \(|G/K|=n\), and let \(Q_1\) and \(Q_2\) be complements of \(K\) in \(G\); of course, \(Q_1\cong G/K\cong Q_2\).
Assume first that \(K\) is solvable. Since \(K'\text{ char }K\) we have \(K'\lhd G\); moreover, \(Q_1K'/K'\cong Q_1/(Q_1\cap K')\cong Q_1\), so that \(|Q_1K'/K'|=n\). Now \(K'< K\), because \(K\) is solvable. If \(K'=\mathbf 1\), then \(K\) is abelian and the result is Theorem 7.6.2; otherwise, \(|G/K'|<|G|\), and induction on \(|G|\) shows that the subgroups \(Q_1K'/K'\) and \(Q_2K'/K'\) are conjugate in \(G/K'\). Thus, there is \(\bar g\in G/K'\) with \(\bar g(Q_1K/K')\bar g^{-1}=Q_2K'/K'\); that is, \(gQ_1g^{-1}\le Q_2K'\) (where \(K'g=\bar g\)). But \(K'< K\) gives \(|Q_1K'|<|G|\), and so the subgroups \(gQ_1g^{-1}\) and \(Q_2\) of order \(n\) are conjugate in \(Q_2K'\), hence are conjugate in \(G\).
Assume now that \(G/K\) is solvable. We do an induction on \(|G|\) that any two complements of \(K\) are conjugate. Let \(M/K\) be a minimal normal subgroup of \(G/K\). Since \(K\le M\), the Dedekind law gives
\[\begin{align} M=M\cap G=M\cap Q_iK=(M\cap Q_i)K\qquad\text{for }i=1,2; \end{align} \]note also that \(M\cap Q_i\lhd Q_i\). Now solvability of \(G/K\) gives \(M/K\) a \(p\)-group for some prime \(p\), by Theorem 5.3.14. If \(M=G\), then \(G/K\) is a \(p\)-group (indeed, since \(M/K\) is a minimal normal subgroup of \(G/K\), we must have \(|M/K|=p\)). Therefore, \(Q_1\) and \(Q_2(\cong G/K)\) are Sylow \(p\)-subgroups of \(G\), and hence are conjugate, by the Sylow theorem.
We may assume, therefore, that \(M<G\). Now \(M\cap Q_i\) is a complement of \(K\) in \(M\), for \(i=1,2\) (because \(M=(M\cap Q_i)K\), by \((4)\), and \((M\cap Q_i)\cap K\le Q_i\cap K=\mathbf 1\)). By induction, there is \(x\in M\le G\) with \(M\cap Q_1=x(M\cap Q_2)x^{-1}=M\cap xQ_2x^{-1}\). Replacing \(Q_2\) by its conjugate \(xQ_2x^{-1}\) if necessary, we may assume that
\[M\cap Q_1=M\cap Q_2. \]If \(J=M\cap Q_1=M\cap Q_2\), then \(J\lhd Q_i\) for \(i=1,2\), and so \(Q_i\le \mathrm N_G(J)\). Two applications of the Dedekind law give
\[\mathrm N_G(J)=\mathrm N_G(J)\cap KQ_i=(\mathrm N_G(J)\cap K)Q_i \]and
\[J[\mathrm N_G(J)\cap K]\cap Q_i=J([\mathrm N_G(J)\cap K]\cap Q_i)=J \](because \((\mathrm N_G(J)\cap K)\cap Q_i\le K\cap Q_i=\mathbf 1\)). Therefore, \(Q_1/J\) and \(Q_2/J\) are complements of \(J(\mathrm N_G(J)\cap K)/J\) in \(\mathrm N_G(J)/J\). By induction, there is \(\bar y\in\mathrm N_G(J)/J\) with \(Q_1/J=\bar y(Q_2/J)\bar y^{-1}\); it follows that \(Q_1=yQ_2y^{-1}\), where \(Jy=\bar y\), as desired.
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Theorem 7.6.5 (Gaschütz) Let \(K\) be a normal abelian \(p\)-subgroup of a finite group \(G\), and let \(P\) be a Sylow \(p\)-subgroup of \(G\). Then \(K\) has a complement in \(G\) if and only if \(K\) has a complement in \(P\).
Proof
It is easy to see that if \(Q\) is a complement of \(K\) in \(G\), then \(Q\cap P\) is a complement of \(K\) in \(P\).
For the converse, assume that \(Q\) is a complement of \(K\) in \(P\), so that \(Q\) is a transversal of \(K\) in \(P\). All groups in this proof will be written additively. If \(U\) is a transversal of \(P\) in \(G\) (which need not be a subgroup), then \(P=\bigcup_{q\in Q}K+q\) and \(G=\bigcup_{u\in U}P+u=\bigcup_{q,u}K+q+u\); thus, \(Q+U=\{q+u:q\in Q,u\in U\}\) is a transversal of \(K\) in \(G\). Thus, \(-U-Q=-U+Q=\{-u+q:u\in U,q\in Q\}\) is also a transversal of \(K\) in \(G\). Let us denote \(-U\) by \(T\), so that \(|T|=[G:P]\) and \(T+Q\) is a transversal of \(K\) in \(G\).
Define \(l\!:G/K\rightarrow G\) by \(l(K+t+q)=t+q\). The corresponding factor set \(f\!:G/K\times G/K\rightarrow K\) is defined by
\[\begin{aligned} l(K+t'+q')+l(K+t+q)&=f(K+t'+q',K+t+q)+l(K+t'+q'+t+q). \end{aligned} \]In particular, if \(t=\mathbf 0\), then
\[f(x,K+q)=\mathbf 0_K\qquad\text{for all }x\in G/K,\quad\text{all }q\in Q \]Consider the cocycle identity
\[xf(y,z)-f(x+y,z)+f(x,y+z)-f(x,y)=\mathbf 0_K \]for \(x,y\in G/K\) and \(z=K+q\) with \(q\in Q\). The previous equation shows that the first two terms are zero, and so
\[\begin{align} f(x,y+z)=f(x,y)\qquad\text{for }x,y\in G/K\text{ and }z=K+q. \end{align} \]Let \(T=\{t_1,\ldots,t_n\}\), where \(n=[G:P]\). For fixed \(y=K+g\in G/K\) and for any \(t_i\), \(K+g+t_i\) lies in \(G/K\); since \(T+Q\) is a transversal of \(K\) in \(G\), there is \(t_{\pi i}\in T\) and \(q_i\in Q\) with \(K+g+t_i=K+t_{\pi i}+q_i\). We claim that \(\pi\) is a permutation. If \(K+g+t_j=K+t_{\pi i}+q_j\), then
\[g+t_i-(t_{\pi i}+q_i)\in K\quad\text{and}\quad g+t_j-(t_{\pi i}+q_j)\in K \]give
\[g+t_i-q_i-t_{\pi i}+t_{\pi i}+q_j-t_j-g\in K. \]Since \(K\lhd G\), we have \(t_i-q_i+q_j-t_j\in K\); since \(Q\) is a subgroup, \(-q_i+q_j=q\in Q\), so that \(t_i+q-t_j\in K\) and \(K+t_j=K+t_i+q\). It follows that \(t_j=t_i+q\), so that \(j=i\) (and \(q=\mathbf 0\)). Therefore, \(\pi\) is an injection and hence a permutation.
Let \(\bar T=\{K+t:t\in T\}\) for \(x\in G/K\), define
\[\sigma(x)=\sum_{x\in\bar T}f(x,z); \]\(\sigma\) is well defined because \(G/K\) is finite adn \(K\) is abelian. Summing the cocycle identity gives
\[x\sigma(y)-\sigma(x)+\sum_{x\in \bar T}f(x,y+z)=[G:P]f(x,y). \]But \(y+z=K+g+K+t_i=K+g+t_i=K+t_{\pi i}+q_i\), so that \(f(x,y+z)=f(x,K+t_{\pi i}+q_i)=f(x,K+t_{\pi i})\), by \((5)\); since \(\pi\) is a permutation, \(\sum_{x\in\bar T}f(x,y+z)=\sigma(x)\). Therefore,
\[x\sigma(y)-\sigma(x+y)+\sigma(x)=[G:P]f(x,y). \]This is an equation in \(K\le P\). Since \(([G:P],|K|)=1\), there are integers \(a\) and \(b\) with \(a|K|+b[G:P]=1\). Define \(h\!:G/K\rightarrow K\) by \(h(x)=b\sigma(x)\). Then \(h(\mathbf 1)=\mathbf 0\) and
\[xh(y)-h(x+y)+h(x)=f(x,y); \]that is, \(f\) is a coboundary and so \(G\) is a semidirect product.
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Transfer and Burnside's Theorem
Lemma 7.7.1 Let \(Q\) be a subgroup of finite index \(n\) in \(G\), and let \(\{l_1,\ldots,l_n\}\) and \(\{h_1,\ldots,h_n\}\) be left transversals of \(Q\) in \(G\). For fixed \(g\in G\) and each \(i\), there is a unique \(\sigma(i)\) (\(1\le\sigma(i)\le n\)) and a unique \(x_i\in Q\) with
Moreover, \(\sigma\) is a permutation of \(\{1,\ldots,n\}\).
Proof
Since the left cosets of \(Q\) partition \(G\), there is a unique left coset \(l_jQ\) containing \(gh_i\); the first statement follows by defining \(\sigma i=j\). Assume that \(\sigma i=\sigma k=j\). Then \(gh_i=l_j x_i\) and \(gh_k=l_jx_k\); thus \(gh_ix_i^{-1}=gh_kx_k^{-1},h_i^{-1}h_k=x_i^{-1}x_k\in Q,h_iQ=h_kQ\), and \(i=k\). Therefore, \(\sigma\) is an injection of a finite set to itself, hence is a permutation.
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Definition 7.7.2 If \(Q\) is a subgroup of finite index \(n\) in a group \(G\), then the transfer is the function \(\mathrm V\!:G\rightarrow Q/Q'\) defined by
where \(\{l_1,\ldots,l_n\}\) is a left transversal of \(Q\) in \(G\) and \(gl_i=l_j x_i\).
Remark The transver \(\mathrm V\!:G\rightarrow Q/Q'\) is often denoted by \(\mathrm V_{G\rightarrow Q}\).
Theorem 7.7.3 If \(Q\) is a subgroup of finite index in a group \(G\), then the transfer \(\mathrm V_{G\rightarrow Q}\) is a homomorphism whose definition is independent of the choice of left transversal of \(Q\) in \(G\).
Proof
Let \(\{l_1,\ldots,l_n\}\) and \(\{h_1,\ldots,h_n\}\) be left transversals of \(Q\) in \(G\). By Lemma 7.7.1, there are equations for each \(g\in G\):
\[\begin{aligned} gl_i&=l_{\sigma i}x_i\qquad &&\sigma\in\mathrm S_n,\quad&&x_i\in Q,\\ gh_i&=h_{\tau i}y_i,&&\tau\in\mathrm S_n,&&y_i\in Q,\\ h_i&=l_{\alpha i}z_i,&&\alpha\in\mathrm S_n,&&z_i\in Q. \end{aligned} \]Now
\[gh_i=gl_{\alpha i}z_i=l_{\sigma\alpha i}x_{\alpha i}z_i. \]Defining \(j\) by \(\alpha j=\sigma\alpha i\), we have \(h_j=l_{\sigma\alpha i}z_j\), where
\[gh_i=h_jz_j^{-1}x_{\alpha i}z_i. \]By Lemma 7.7.1, we have \(j=\tau i\), and
\[y_i=z_j^{-1}x_{\alpha i}z_i=z_{\alpha^{-1}\sigma\alpha i}^{-1}x_{\alpha i}z_i. \]Factors may be rearranged in the abelian group \(Q/Q'\): thus
\[\prod y_iQ'=\prod z_{\alpha^{-1}\sigma\alpha i}^{-1}x_{\alpha i}z_iQ'=\prod x_{\alpha i}Q', \]since \(\alpha^{-1}\sigma\alpha\in\mathrm S_n\), and so the inverse of each \(z_i\) occurs and cancels \(z_i\). Finally, \(\prod x_{\alpha i}Q'=\prod x_iQ'\) since \(\alpha\in\mathrm S_n\). We have shown that \(V\) is independent of the choice of transversal.
Let \(g,g'\in G\) and let \(\{l_1,\ldots,l_n\}\) be a left transversal of \(Q\) in \(G\). Thus, \(gl_i=l_{\sigma i}x_i\) and \(g'l_i=l_{\tau i}y_i\), where \(x_i,y_i\in Q\). Then
\[gg'l_i=gl_{\tau i}y_i=l_{\sigma\tau i}x_{\tau i}y_i. \]Therefore,
\[\begin{aligned} \mathrm V(gg')=\prod x_{\tau i}y_iQ'&=\left(\prod x_{\tau i}Q'\right)\left(\prod y_iQ'\right)\\&=\left(\prod x_iQ'\right)\left(\prod y_iQ'\right)=\mathrm V(g)\mathrm V(g'). \end{aligned} \]■
If a subgroup \(Q\) of finite index in a group \(G\) has a complement \(K\), then \(K=\{a_1,\ldots,a_n\}\) is a left transversal of \(Q\) in \(G\). If \(b\in K\), then \(ba_i\in K\) for all \(i: ba_i=a_{\sigma i}\). But the general formula in Lemma 7.7.1 is \(ba_i=a_{\sigma i}x_i\), so that each \(x_i=\mathbf 1\). We conclude that \(K\le\mathrm{ker}\mathrm V\). If \(Q\) is abelian, then \(Q'=\mathbf 1\) and we may identify \(Q/Q'\) with \(Q\); thus, \(\mathrm{im}\mathrm V\le Q\) in this case.
Proposition 7.7.4
- If \(\mathrm V\!:G\to Q/Q'\) is the transfer, then \(G'\le\mathrm{ker}\mathrm V\) and \(\mathrm V\) induces a homomorphism \(\bar {\mathrm V}\!:G/G'\to Q/Q'\), namely, \(G'a\mapsto \mathrm V(a)\).
- The transfer is transitive: if \(P\le Q\le G\) are subgroups of finite index, then \(\bar{\mathrm V}_{G\to P}=\bar{\mathrm V}_{Q\to P}\bar{\mathrm V}_{G\to Q}\).
Lemma 7.7.5 Let \(Q\) be a subgroup of finite index \(n\) in \(G\), and let \(\{l_1,\ldots,l_n\}\) be a left transversal of \(Q\) in \(G\). For each \(g\in G\), there exist elements \(h_1,\ldots,h_m\) of \(G\) and positive integers \(n_1,\ldots,n_m\) (all depending on \(g\)) such that:
- each \(h_i\in\{l_1,\ldots,l_n\}\);
- \(h_i^{-1}g^{n_i}h_i\in Q\);
- \(\sum n_i=n=[G:Q]\); and
- \(\mathrm V(g)=\prod(h_i^{-1}g^{n_i}h_i)Q'\).
Proof
We know that \(gl_i=l_{\sigma i}x_i\), where \(\sigma\in\mathrm S_n\) and \(x_i\in Q\). Write the complete factorization of \(\sigma\) as a product of disjoint cycles: \(\sigma=\alpha_1\ldots\alpha_m\). If \(\alpha_i=(j_1,\ldots,j_r)\), then
\[gl_{j_1}=l_{\sigma j_1}x_{j_1}=l_{j_2}=x_{j_1}, \quad gl_{j_2}=l_{j_3}x_{j_2},\quad\ldots,\quad gl_{j_r}=l_{j_1}x_{j_r}, \]and \(Q\) contains
\[x_{j_r}\cdots x_{j_1}=(l_{j_1}^{-1}gl_{j_r})\cdots(l_{j_3}^{-1}gl_{j_2})(l_{j_2}^{-1}gl_{j_1})=l_{j_1}^{-1}g^rl_{j_1}. \]Define \(h_i=l_{j_1}\) and \(n_i=r\); all the conclusions now follow.
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Theorem 7.7.6 If \(Q\) is an abelian subgroup of finite index \(n\) in a group \(G\) and if \(Q\le\mathrm Z(G)\), then \(\mathrm V(g)=g^nQ'\) for all \(g\in G\).
Proof
Since \(Q\) is abelian, we may regard the transfer as a homomorphism \(\mathrm V:G\to Q\). The condition \(Q\le \mathrm Z(G)\) implies that \(Q\) is a normal subgroup of \(G\). If \(g\in G\) and \(h^{-1}g^rh\in Q\), then normality of \(Q\) gives \(g^r=h(h^{-1}g^rh)h^{-1}\in Q\). But \(Q\le\mathrm Z(G)\) now gives \(h^{-1}g^rh=g^r\). The result now follows from Lemma 7.7.5.
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Proposition 7.7.7
- If a group \(G\) has a subgroup \(Q\) of finite index \(n\) with \(Q\le\mathrm Z(G)\), then \(g\mapsto g^n\) is a homomorphism.
- If \(Q\) has index \(n\) in \(G\), and if \(K\le G\) satisfies \(G=KQ\) and \(Q\le\mathrm C_G(K)\), then \(\mathrm V(a)=a^nQ'\) for all \(a\in G\).
Lemma 7.7.8 Let \(Q\) be a Sylow \(p\)-subgroup of a finite group \(G\) (for some prime \(p\)). If \(g,h\in\mathrm C_G(Q)\) are conjugate in \(G\), then they are conjugate in \(\mathrm N_G(Q)\).
Proof
If \(h=\gamma^{-1}g\gamma\) for some \(\gamma\in G\), then \(h\in \gamma^{-1}\mathrm C_G(Q)\gamma=\mathrm C_G(\gamma^{-1}Q\gamma)\). Since \(Q\) and \(\gamma^{-1}Q\gamma\) are contained in \(\mathrm C_G(h)\), both are Sylow subgroups of it. The Sylow theorem gives \(c\in\mathrm C_G(h)\) with \(Q=c^{-1}\gamma^{-1}Q\gamma c\). Clearly \(\gamma c\in\mathrm N_G(Q)\) and \(c^{-1}\gamma^{-1}g\gamma c=c^{-1}hc=h\).
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Theorem 7.7.9 (Burnside Normal Complement Theorem) Let \(G\) be a finite group and let \(Q\) be an abelian Sylow subgroup contained in the center of its normalizer: \(Q\le\mathrm Z(\mathrm N_G(Q))\). Then \(Q\) has a normal complement \(K\) (indeed, \(K\) is even a characteristic subgroup of \(G\)).
Proof
Since \(Q\) is abelian, we may regard the transfer \(\mathrm V\) as a homomorphism from \(G\) to \(Q\). By Lemma 7.7.5, \(\mathrm V(g)=\prod h_i^{-1}g^{n_i}h_i\). Now, if \(g\in Q\), then, for each \(i\), both \(g^{n_i}\) and \(h_i^{-1}g^{n_i}h_i\) lie in \(Q\), and so they are conjugate elements of \(\mathrm C_G(Q)\) (for \(Q\) abelian implies \(Q\le\mathrm C_G(Q)\)). By Lemma 7.7.8, there is \(c\in\mathrm N_G(Q)\) with \(h_i^{-1}g^{n_i}h_i=c^{-1}g^{n_i}c\). But \(Q\le\mathrm Z(\mathrm N_G(Q))\) implies \(c^{-1}g^{n_i}c=g^{n_i}\). Hence, if \(n=[G:Q]\), then \(\mathrm V(g)=g^n\) for all \(g\in Q\). If \(|Q|=q\), then \((n,q)=1\), and there are integers \(\alpha\) and \(\beta\) with \(\alpha n+\beta q=1\). Therefore, when \(g\in Q\), we have \(g=g^{\alpha n}g^{\beta q}=(g^\alpha)^n\), so that \(\mathrm V\!:G\to Q\) is surjective: if \(g\in Q\), then \(\mathrm V(g^{\alpha})=g^{\alpha n}=g\). The first isomorphism theorem gives \(G/K\cong Q\), where \(K=\mathrm{ker}\mathrm V\). It follows that \(G=KQ\) and \(K\cap Q=\mathbf 1\) (because \(|K|=n\) and so \((|K|,|Q|)=1\)). Therefore, \(K\) is a normal complement of \(Q\). Indeed, \(K\text{ char }G\) because it is a normal Hall subgroup.
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Proposition 7.7.10 If \(G\) is a finite group of order \(mn\), where \((m,n)=1\), and if \(Q\le\mathrm Z(G)\) is a Hall subgroup of order \(m\), then \(K=\mathrm{ker}\mathrm V\) is a normal complement of \(Q\) (where \(\mathrm V\!:G\to Q/Q'\) is the transfer), and \(G=K\times Q\).
Definition 7.7.11 If a Sylow \(p\)-subgroup of a finite group \(G\) has a normal \(p\)-completment, then \(G\) is called \(p\)-nilpotent.
Thus, Burnside's theorem says that if \(Q\le\mathrm Z(\mathrm N_G(Q))\), then \(G\) is \(p\)-nilpotent for any prime \(p\).
Theorem 7.7.12 Let \(G\) be a finite group and let \(p\) be the smallest prime divisor of \(|G|\). If a Sylow \(p\)-subgroup \(Q\) of \(G\) is cyclic, then \(G\) is \(p\)-nilpotent.
Proof
By N/C Lemma, there is an imbedding \(\mathrm N_G(Q)/\mathrm C_G(Q)\hookrightarrow\mathrm{Aut}(Q)\). Obviously, \(|\mathrm N_G(Q)/\mathrm C_G(Q)|\) divides \(|G|\). Now \(Q\) is cyclic of order \(p^m\), say, and so Proposition 7.2.9 gives \(|\mathrm{Aut}(Q)|=p^{m-1}(p-1)\). Since \(Q\) is a Sylow subgroup of \(G\) and \(Q\le \mathrm C_G(Q)\), \(p\) does not divide \(|\mathrm N_G(Q)/\mathrm C_G(Q)|\). Hence, \(|\mathrm N_G(Q)/\mathrm C_G(Q)|\) divides \(p-1\). But \((p-1,|G|)=1\) since \(p\) is the smallest prime divisor of \(|G|\), and so \(|\mathrm N_G(Q)/\mathrm C_G(Q)|=1\). Therefore, \(\mathrm N_G(Q)=\mathrm C_G(Q)\). Since \(Q\le G\) is abelian, \(Q\le\mathrm Z(\mathrm C_G(Q))=\mathrm Z(\mathrm N_G(Q))\), and thus by Burnside's theorem, \(Q\) has a normal complement.
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Theorem 7.7.13 (Hölder) If every Sylow subgroup of a finite group \(G\) is cyclic, then \(G\) is solvable.
Proof
If \(p\) is the smallest prime divisor of \(|G|\), then Theorem 7.7.12 provides a normal complement \(K\) to a Sylow \(p\)-subgroup \(Q\) of \(G\), and \(Q\cong G/K\). By induction on \(|G|\), \(K\) is solvable. Since \(Q\) is solvable (it is cyclic, hence abelian), Theorem 5.3.4 shows that \(G\) is solvable.
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Corollary 7.7.14 Every group \(G\) of squarefree order is solvable.
Theorem 7.7.15 (Tate) Let \(G\) be a finite group, and let \(P\le G\) be a Sylow \(p\)-subgroup of \(G\). If \(N\lhd G\) and \(N\cap P\le\Phi(P)\), then \(N\) is \(p\)-nilpotent.
Definition 7.7.16 Call a subgroup \(H\) of a group \(G\) \(p\)-local, for some prime \(p\), if there is some nontrivial \(p\)-subgroup \(Q\) of \(G\) with \(H=\mathrm N_G(Q)\).
Theorem 7.7.17 (Frobenius) A group \(G\) is \(p\)-nilpotent if and only if every \(p\)-local subgroup \(H\) of \(G\) is \(p\)-nilpotent.
Definition 7.7.18 If \(p\) is a prime and \(P\) is a Sylow \(p\)-subgroup of a group \(G\), define \(\mathrm E(P)=\{x\in\mathrm Z(P):x^p=\mathbf 1\}\). If \(p^{d(P)}\) is the largest order of an elementary abelian subgroup of \(P\), then the Thompson subgroup \(\mathrm J(P)\) is defined as the subgroup of \(G\) generated by all the elementary abelian \(p\)-subgroups of \(G\) of order \(p^{d(P)}\).
Theorem 7.7.19 (Thompson) Let \(p\) be an odd prime, and let \(P\) be a Sylow \(p\)-subgroup of a group \(G\). If \(\mathrm C_G(\mathrm E(P))\) and \(\mathrm N_G(\mathrm J(P))\) are \(p\)-nilpotent, then \(G\) is \(p\)-nilpotent.
Derivations
We again consider general data \((Q,K,\theta)\) consisting of a group \(Q\), an (additive) abelian group \(K\), and a not necessarily trivial homomorphism \(\theta\!:Q\to\mathrm{Aut}(K)\) giving an action of \(Q\) on \(K\).
Definition 7.8.1 If \(K\) and \(A\) are groups with \(A\) abelian, then
If \(\varphi,\psi\in\mathrm{Hom}(K,A)\), define \(\varphi\psi\!:K\to A\) by \(\varphi\psi\!:x\mapsto\varphi(x)+\psi(x)\) for all \(x\in K\); note that \(\varphi\psi\) is a homomorphism because \(A\) is abelian. It is easy to see that \(\mathrm{Hom}(K,A)\) is an abelian group under this operation.
Definition 7.8.2 Given data \((Q,K,\theta)\), a derivation (or crossed homomorphism) is a function \(d\!:Q\to K\) such that
The set \(\mathrm{Der}_\theta(Q,K)\) of all derivations is an abelian group under the following operation: if \(d,d'\in\mathrm{Der}_\theta(Q,K)\), then \(d+d'\!:x\mapsto d(x)+d'(x)\). In particular, if \(\theta\) is trivial, then \(\mathrm{Der}(Q,K)=\mathrm{Hom}(Q,K)\).
Example 7.8.3 Let data \((Q,K,\theta)\) be given. For fixed \(a\in K\), the function \(d_a\!:Q\to K\), defined by \(x\mapsto a-xa\), is easily checked to be a derivation; it is called the principal derivation determined by \(a\).
Example 7.8.4 Let \(G\) be a semidirect product \(K\rtimes Q\), so that \(G\) consists of all \((a,x)\in K\times Q\) with \((a,x)+(b,y)=(a+xb,xy)\). If \(\gamma\!:G\to G\) is a stabilizing homomorphism, then \(\gamma\!:(a,x)\mapsto (a+d(x),x)\) for some \(d(x)\in K\), and evaluating \(\gamma((\mathbf 0,x)+(\mathbf 0,y))\) in two ways shows that \(d\) is a derivation.
Theorem 7.8.5 Let \(G\) be an extension realizing data \((Q,K,\theta)\). If \(A\) is the stabilizer of this extension, then
Proof
Regard \(G\) as all \((a,x)\in K\times Q\). As in Example 7.8.4, if \(\gamma\in A\), then \(\gamma\!:(a,x)\mapsto (a+d(x),x)\), where \(d\) is a derivation. Defined \(\varphi\!:A\to\mathrm{Der}_\theta(Q,K)\) by \(\varphi(\gamma)=d\). Now \(\varphi\) is a homomorphism: if \(\gamma'\in A\), then \(\gamma'\!:(a,x)\mapsto (a+d'(x),x)\), and \(\gamma'\gamma\!:(a,x)\mapsto (a+d(x)+d'(x),x)\). We see that \(\varphi\) is an isomorphism by constructing its inverse: if \(d\) is a derivation, define \(\gamma\!:G\to G\) by \(\gamma\!:(a,x)\mapsto (a+d(x),x)\).
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Lemma 7.8.6 Let \(G\) be a semidirect product \(K\rtimes_\theta Q\), and let \(\gamma\!:(a,x)\mapsto (a+d(x),x)\) be a stabilizing automorphism of \(G\). Then \(\gamma\) is an inner automorphism if and only if \(d\) is a principal derivation.
Proof
If \(d\) is a principal derivation, then there is \(b\in K\) with \(d(x)=b-xb\). Hence, \(\gamma(a,x)=(a+b-xb,x)\). But
\[\begin{aligned} (b,\mathbf 1)+(a,x)-(b,\mathbf 1)&=(b+a,x)-(b,\mathbf 1)\\&=(b+a-xb,x), \end{aligned} \]so that \(\gamma\) is conjugation by \((b,\mathbf 1)\).
Conversely, if \(\gamma\) is conjugation by \((b,y)\), then for all \(x\in G\),
\[\begin{aligned} \gamma(\mathbf 0,x)&=(b,y)+(\mathbf 0,x)-(b,y)\\&=(b,yx)+(-y^{-1}b,y^{-1})\\&=(b-yxy^{-1}b,yxy^{-1}). \end{aligned} \]Since \(\gamma\) is stabilizing, we must have \(yxy^{-1}=x\), so that
\[\gamma(\mathbf 0,x)=(b-xb,x). \]But \(\gamma(\mathbf 0,x)=(d(x),x)\), so that \(d(x)=b-xb\) is a principal derivation.
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It is easy to see that \(\mathrm{PDer}_\theta(Q,K)\), the set of all principal derivations, is a subgroup of \(\mathrm{Der}_\theta(Q,K)\).
Definition 7.8.7 Given data \((Q,K,\theta)\), the first cohomology group \(\mathrm H^1_\theta(Q,K)=\mathrm{Der}_\theta(Q,K)/\mathrm{PDer}_\theta(Q,K)\).
In particular, if \(\theta\) is trivial, then \(\mathrm H^1_\theta(Q,K)=\mathrm{Hom}(Q,K)\).
Theorem 7.8.8 If \((Q,K,\theta)\) are data and \(G=H\rtimes_\theta Q\), then
Proof
Let \(A\) be the stabilizer of the extension and let \(\varphi\!:A\to\mathrm{Der}_\theta(Q,K)\) be the isomorphism of Theorem 7.8.5. By Theorem 7.8.6, \(\varphi(A\cap\mathrm{Inn}(G))=\mathrm{PDer}_\theta(Q,K)\). Hence
\[\begin{aligned} \mathrm H^1_\theta(Q,K)&=\mathrm{Der}_\theta(Q,K)/\mathrm{PDer}_\theta(Q,K)\\&\cong A/(A\cap\mathrm{Inn}(G))\\&\cong A\ \mathrm{Inn}(G)/\mathrm{Inn}(G)\le\mathrm{Aut}(G)/\mathrm{Inn}(G). \end{aligned} \]■
Theorem 7.8.9 Let \(G=K\rtimes_\theta Q\), where \(K\) is abelian, and let \(C\) and \(C'\) be complements of \(K\) in \(G\). If \(\mathrm H^1_\theta(Q,K)=\mathbf 0\), then \(C\) and \(C'\) are conjugate.
Proof
Since \(C\) and \(C'\) are complements of \(K\), they are isomorphic: \(C\cong G/K\cong Q\cong C'\); choose isomorphisms \(l\!:Q\to C\) and \(l'\!:Q\to C'\). Both \(C\) and \(C'\) are also transversals of \(K\) in \(G\), and so they determine factor sets \(f\) and \(f'\) (where, for example, \(l(x)+l(y)=f(x,y)+l(xy)\)). By Lemma 7.5.10, there is a function \(h\!:Q\to K\), namely, \(h(x)=l'(x)-l(x)\), with
\[f'(x,y)-f(x,y)=xh(y)-h(xy)+h(x). \]Since \(l\) and \(l'\) are homomorphisms, it follows that both \(f\) and \(f'\) are identically zero. Thus, \(f'(x,y)-f(x,y)\) is identically zero, \(h(xy)=xh(y)+h(x)\), and \(h\) is a derivation. The hypothesis \(\mathrm H^1_\theta(Q,K)=\mathbf 0\) says that \(h\) is a principal derivation; that is, there is \(b\in K\) with
\[l'(x)-l(x)=h(x)=b-xb. \]Therefore,
\[l'(x)=b+(-xb+l(x))=b+l(x)-b; \]that is, \(C'=b+C-b\) is a conjugate of \(C\).
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Definition 7.8.10 If \(d\!:Q\to K\) is a derivation, then its kernel is
For example, if \(d_a\) is a principal derivation, then \(\mathrm{ker}\ d_a\) is the set of all \(x\in Q\) which fix \(a\).
Lemma 7.8.11 Let \((Q,K,\theta)\) be data, and let \(d\!:Q\to K\) be a derivation. Then,
- \(d(\mathbf 1)=\mathbf 0\);
- \(d(x^{-1})=-x^{-1}d(x)\);
- \(\mathrm{ker}\ d\) is a subgroup of \(Q\) (however, it need not be a normal subgroup);
- \(d(x)=d(y)\) if and only if \(x^{-1}y\in\mathrm{ker}\ d\).
Given data \((Q,K,\theta)\) (it is now more convenient to write the abelian group \(K\) multiplicatively), assume that an abelian normal subgroup \(K\) has finite index \(n\) in a group \(E\), and let \(l_1,\ldots,l_n\) be a left transversal of \(K\) in \(G\); that is, \(E=\bigcup_{i=1}^nl_iK\). As in Lemma 7.7.1, if \(e\in E\), then there is a unique \(\kappa_i(e)\in K\) with \(el_i=l_{\sigma i}\kappa_i(e)\), where \(\sigma\) is a permutation (depending on \(e\)). Define \(a_i(e)=l_{\sigma i}\kappa_i(e)l_{\sigma i}^{-1}\), so that
and define a function \(d\!:E\to K\) by
Lemma 7.8.12 (Gruenberg). Let \(K\) be an abelian normal subgroup of finite index in a group \(E\), and let \(L=\{l_1,\ldots,l_n\}\) be a left transversal of \(K\) in \(E\).
- If \(\theta\!:E\to\mathrm{Aut}(K)\) is defined by \(\theta_e(k)=eke^{-1}\), then the function \(d\) defined above is a derivation.
- If \(k\in K\), then \(d(k)=k^n\).
- If \(L\) is a complement of \(K\) in \(E\), then \(L\le\mathrm{ker}\ d\).
- If \(k\in K\) and \(e\in E\), then \(d(k^{-1}ek)=d(k)^ed(e)d(k)^{-1}\).
Theorem 7.8.13 (=Theorems 7.6.2 and 7.6.3) Given data \((Q,K,\theta)\) with \(Q\) a finite group, \(K\) a finite abelian group, and \((|K|,|Q|)=1\), then every extension \(G\) of \(K\) by \(Q\) realizing the data is a semidirect product; moreover, any two complements of \(K\) in \(G\) are conjugate.
Proof
Let \(E\) be an extension of \(K\) by \(Q\) realizing the data and let \(|Q|=n\). Define \(L=\mathrm{ker}\ d\), where \(d\!:Q\to K\) is Gruenberg's derivation. By Lemma 7.8.11(iii), \(L\) is a subgroup of \(Q\). We claim that \(L\cap K=\mathbf 1\). If \(a\in K\), then \(d(a)=a^n\), by Lemma 7.8.12(ii); if \(a\in L\), then \(d(a)=\mathbf 1\), by definition of kernel; hence, \(a^n=\mathbf 1\) for all \(a\in L\cap K\). But \((|K|,n)=1\), by hypothesis, so that \(a=1\). Let us now see that \(E=KL\). If \(e\in E\), then \(d(e)\in K\). Since \((|K|, n)=1\), there is \(k\in K\) with \(d(e)=k^{-n}\). Hence \(d(e)=k^{-n}=f(k^{-1})\), by Lemma 7.8.12(ii), and so \(ke\in L=\mathrm{ker}\ d\), by Lemma 7.8.11(iv). Therefore, \(e=k^{-1}(ke)\in KL\), and \(E\) is a semidirect product.
We now show that if \(Y\) is another complement of \(K\) in \(E\), then \(Y\) and \(L\) are conjugate. Note that each \(Y\) and \(L\) is a transversal of \(K\) in \(E\). Let \(Y=\{y_1,\ldots,y_n\}\) and \(L=\{l_1,\ldots,l_n\}\), and write \(y_i=c_il_i\) for \(c_i\in K\). The derivation determined by \(L\) is \(d(e)=\prod_i a_i(e)\); the derivation determined by \(Y\) is \(\delta(e)=\prod_i\alpha_i(e)\), where \(\alpha_i(e)\in K\) and \(ey_i=\alpha_i(e)y_{\tau i}\). But
\[ey_i=ec_il_i=(ec_ie^{-1})el_i=c_i^ea_i(e)l_{\alpha i}=c_i^ea_i(e)c_{\sigma i}^{-1}y_{\sigma i}; \]it follows that \(\tau=\sigma,\,\alpha_i(e)=c_i^ea_i(e)c_{\sigma i}^{-1}\), and
\[\delta(e)=\prod_i\alpha_i(e)=\prod_ic_i^ea_i(e)c_{\sigma i}^{-1}. \]If we define \(c=\prod c_i\), then \(c\in K\) and
\[\delta(e)=c^ed(e)c^{-1}. \]Since \((|K|,n)=1\), there is \(k\in K\) with \(c=k^n\), and so \(c=d(k)\), by Lemma 7.8.12(ii). Hence, \(d(k^{-1}ek)=d(k)^ed(e)d(k)^{-1}=c^ed(e)c^{-1}\), by Lemma 7.8.12(iv). Therefore,
\[\delta(e)=d(k^{-1}ek), \]so that \(L=\mathrm{ker}\ d=k(\mathrm{ker}\ \delta)k^{-1}\ge kYk^{-1}\), by Lemma 7.8.12(iii). But \(|L|=|Y|\), for both complements of \(K\), and so \(L=kYk^{-1}\).
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