GTM148 抄书笔记 Part II. (Chapter IV~V)

上一篇: GTM148 抄书笔记 Part I.

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Contents

• Contents
• Chapter IV. The Sylow Theorems
  • \(p\)-Groups
  • The Sylow Theorems
• Chapter V. Normal Series
  • Some Galois Theory
  • The Jordan-Hölder Theorem
  • Solvable Groups
  • Two Theorems of P.Hall
  • Central Series and Nilpotent Groups
  • Frattini Subgroup

Chapter IV. The Sylow Theorems

\(p\)-Groups

Definition 4.1.1 If \(p\) is a prime, then a \(p\)-group is a group in which every element has order a power of \(p\).

Theorem 4.1.2 If \(G\) is a finite abelian group whose order is divisible by a prime \(p\), then \(G\) contains an element of order \(p\).

Proof

Write \(|G|=pm\), where \(m\ge 1\). We proceed by induction on \(m\) after noting that the base step is clearly true. For the inductive step, choose \(x\in G\) of order \(t>1\). If \(p|t\), then Proposition 2.2.14 shows that \(x^{t/p}\) has order \(p\), and the lemma is proved. We may, therefore, assume that the order of \(x\) is not divisible by \(p\). Since \(G\) is abelian, \(\langle x\rangle\) is a normal subgroup of \(G\), and \(G/\langle x\rangle\) is an abelian group of order \(|G|/t=pm/t\). Since \(p\nmid t\), we must have \(m/t<m\) an integer. By induction, \(G/\langle x\rangle\) contains an element \(y^*\) of order \(p\). But the natural map \(\nu\!:G\rightarrow G/\langle x\rangle\) is a surjection, and so there is \(y\in G\) with \(\nu(y)=y^*\). Thus, the order of \(y\) is a multiple of \(p\), and we have returned to the first case.
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Definition 4.1.3 Let \(G\) be a group. Partition \(G\) into its conjugacy classes and count (recall that \(\mathrm Z(G)\) consists of all the elements of \(G\) whose conjugacy class has just one element):

\[|G|=|\mathrm Z(G)|+\sum_{i}[G:\mathrm C_G(x_i)], \]

where one \(x_i\) is selected from each conjugacy class with more than one element.
The equation above is called the class equation of the finite group \(G\).

Theorem 4.1.4 (Cauchy) If \(G\) is a finite group whose order is divisible by a prime \(p\), then \(G\) contains an element of order \(p\).

Proof

If \(x\in G\), then the number of conjugates of \(x\) is \([G:\mathrm C_G(x)]\), where \(\mathrm C_G(x)\) is the centralizer of \(x\) in \(G\). If \(x\notin \mathrm Z(G)\), then its conjugacy class has more than one element, and so \(|\mathrm C_G(x)|< |G|\). If \(p\big||\mathrm C_G(x)|\) for such a noncentral \(x\), we are done, by induction. Therefore, we may assume that \(p\nmid|\mathrm C_G(x)|\) for all noncentral \(x\) in \(G\). Better, since \(|G|=[G:\mathrm C_G(x)]|\mathrm C_G(x)|\), we may assume that \(p\big|[G:\mathrm C_G(x)]\).

Consider the class equation \(|G|=|\mathrm Z(G)|+\sum_{i}[G:\mathrm C_G(x_i)]\). Since \(|G|\) and all \([G:\mathrm C_G(x_i)]\) are divisible by \(p\), it follows that \(|\mathrm Z(G)|\) is divisible by \(p\). But \(\mathrm Z(G)\) is abelian, and so it contains an element of order \(p\), by Lemma 4.2.2.
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Corollary 4.1.5 A finite group \(G\) is a \(p\)-group if and only if \(|G|\) is a power of \(p\).

Proof

If \(|G|=p^m\), then Lagrange's theorem shows that \(G\) is a \(p\)-group. Conversely, assume that there is a prime \(q\neq p\) which divides \(|G|\). By Cauchy's theorem, \(G\) contains an element of order \(q\), and this contradicts \(G\) being a \(p\)-group.
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Theorem 4.1.6 If \(G\neq \mathbf 1\) is a finite \(p\)-group, then its center \(\mathrm Z(G)\neq \mathbf 1\).

Proof

Consider the class equation \(|G|=|\mathrm Z(G)|+\sum_{i}[G:\mathrm C_G(x_i)]\). Each \(\mathrm C_G(x_i)\) is a proper subgroup of \(G\), for \(x_i\notin\mathrm Z(G)\). By Corollary 4.1.5, \([G:\mathrm C_G(x_i)]\) is a power of \(p\). Thus, \(p\) divides each \([G:\mathrm C_G(x_i)]\), and so \(p\) divides \(|\mathrm Z(G)|\).
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Corollary 4.1.7 If \(p\) is a prime, then every group \(G\) of order \(p^2\) is abelian.

Proof

If \(G\) is not abelian, then \(\mathrm Z(G)<G\); since \(\mathbf 1\neq\mathrm Z(G)\), we must have \(|\mathrm Z(G)|=p\). The quotient group \(G/\mathrm Z(G)\) is defined, since \(\mathrm Z(G)\lhd G\), and it is cyclic, because \(|G/\mathrm Z(G)|=p\); this contradicts Proposition 3.1.19.
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Theorem 4.1.8 Let \(G\) be a finite \(p\)-group.

1. If \(H\) is a proper subgroup of \(G\), then \(H< \mathrm N_G(H)\).
2. Every maximal subgroup of \(G\) is normal and has index \(p\).

Proof

1. If \(H\lhd G\), then \(\mathrm N_G(H)=G\) and the theorem is true. If \(X\) is the set of all the conjugates of \(H\), then we may assume that \(|X|=[G:\mathrm N_G(H)]\neq 1\). Now \(G\) acts on \(X\) by conjugation and, since \(G\) is a \(p\)-group, every orbit of \(X\) has size a power of \(p\). As \(\{H\}\) is an orbit of size \(1\), there must be at least \(p-1\) other orbits of size \(1\). Thus there is at least one conjugate \(gHg^{-1}\neq H\) with \(\{gHg^{-1}\}\) also an orbit of size \(1\). Now \(agHg^{-1}a^{-1}=gHg^{-1}\) for all \(a\in H\), and so \(g^{-1}ag\in \mathrm N_G(H)\) for all \(a\in H\). But \(gHg^{-1}\neq H\) gives at least one \(a\in H\) with \(g^{-1}ag\notin H\), and so \(H< \mathrm N_G(H)\).
2. If \(H\) is a maximal subgroup of \(G\), then \(H<\mathrm N_G(H)\) implies that \(\mathrm N_G(H)=H\); that is, \(H\lhd G\). By Proposition 2.7.4, \([G:H]=p\).
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Lemma 4.1.9 If \(G\) is a finite \(p\)-group and \(r_1\) is the number of subgroups of \(G\) having order \(p\), then \(r_1\equiv 1\pmod p\).

Proof

Let us first count the number of elements of order \(p\). Since \(\mathrm Z(G)\) is abelian, all its elements of order \(p\) together with \(\mathbf 1\) form a subgroup \(H\) whose order is a power of \(p\); hence the number of central elements of order \(p\) is \(|H|-1\equiv -1\pmod p\). If \(x\in G\) is of order \(p\) and not central, then its conjugacy class \(x^G\) consists of several elements of order \(p\); that is, \(|x^G|>1\) is an "honest" power of \(p\). It follows that the number of elements in \(G\) of order \(p\) is congruent to \(-1\!\!\mod p\); say, there are \(mp-1\) such elements. Since the intersection of any distinct pair of subgroups of order \(p\) is trivial, the number of elements of order \(p\) is \(r_1(p-1)\). But \(r_1(p-1)=mp-1\) implies \(r_1\equiv 1\pmod p\).
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Theorem 4.1.10 If \(G\) is a finite \(p\)-group and \(r_s\) is the number of subgroups of \(G\) having order \(p^s\), then \(r_s\equiv 1\pmod p\).

Proof

Let \(H\) be a subgroup of order \(p^s\), and let \(K_1,\ldots,K_a\) be the subgroups of \(G\) of order \(p^{s+1}\) which contain it; we claim that \(a\equiv 1\pmod p\). Every subgroup of \(G\) which normalizes \(H\) is contained in \(N=\mathrm N_G(H)\); in particular, each \(K_j\) lies in \(N\), for Theorem 4.1.8(ii) shows that \(H\lhd K\) for all \(j\). By the Correspondence Theorem, the number of subgroups of order \(p\) in \(N/H\) is equal to the number of subgroups of \(N\) containing \(H\) which have order \(p^{s+1}\). By Lemmma 4.1.9, \(a\equiv 1\pmod p\).

Now let \(K\) be a subgroup of order \(p^{s+1}\), and let \(H_1,\ldots,H_b\) be its subgroups of order \(p^s\); we claim that \(b\equiv 1\pmod p\). By Theorem 4.1.8(ii), \(H_i\lhd K\) for all \(i\). Since \(H_1H_2=K\) (for the \(H_i\) are maximal subgroups of \(K\)), the Product Formula (Theorem 2.4.2) gives \(|D|=p^{s-1}\), where \(D=H_1\cap H_2\), and \([K:D]=p^2\). By Corollary 4.1.7, the group \(K/D\) is abelian; moreover, \(K/D\) is generated by two subgroups of order \(p\), namely, \(H_i/D\) for \(i=1,2\), and so it is not cyclic. Thus \(K/D\cong \mathbb Z_p\times \mathbb Z_p\). Therefore, \(K/D\) has \(p^2-1\) elements of order \(p\) and hence has \(p+1=(p^2-1)/(p-1)\) subgroups of order \(p\). The Correspondence Theorem gives \(p+1\) subgroups of \(K\) of order \(p^s\) containing \(D\). Suppose there is some \(H_j\) with \(D\nleq H_j\). Let \(E=H_1\cap H_j\); as above, there is a new list of \(p+1\) subgroups of \(K\) of order \(p^s\) containing \(E\), one of which is \(H_1\). Indeed, \(H_1=ED\) is the only subgroup on both lists. Therefore, there are \(p\) new subgroups and \(1+2p\) subgroups counted so far. If some \(H_l\) has not yet been listed, repeat this procedure beginning with \(H_1\cap H_l\) to obtain \(p\) new subgroups. Eventually, all the \(H_i\) will be listed, and so the number of them is \(b=1+mp\) for some \(m\). Hence, \(b\equiv 1\pmod p\).

Let \(H_1,\ldots,H_{r_s}\) be all the subgroups of \(G\) of order \(p^s\), and let \(K_1,\ldots,K_{r_{s+1}}\) be all the subgroups of order \(p^{s+1}\). For each \(H_i\), let there be \(a_i\) subgroups of order \(p^{s+1}\) containing \(H_i\); for each \(K_j\), let there be \(b_j\) subgroups of order \(p^s\) contained in \(K_j\). Now

\[\sum_{i=1}^{r_s}a_i=\sum_{j=1}^{r_{s+1}}b_j, \]

for either sum counts each \(K_j\) with multiplicity the number of \(H\)'s it contains. Since \(a_i\equiv 1\pmod p\) for all \(i\) and \(b_j\equiv 1\pmod p\) for all \(j\), it follows that \(r_s\equiv r_{s+1}\pmod p\). Lemma 4.1.9 now gives the result, for \(r_1\equiv 1\pmod p\).
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Lemma 4.1.11 (Landau) Given \(n>0\) and \(q\in\mathbb Q\), there are only finitely many \(n\)-tuples \((i_1,\ldots,i_n)\) of positive integers such that \(q=\sum_{j=1}^n(1/i_j)\).

Proof

We do an induction on \(n\); the base step \(n=1\) is obviously true. Since there are only \(n!\) permutations of \(n\) objects, it suffices to prove that there are only finitely many \(n\)-tuples \((i_1,\ldots,i_n)\) with \(i_1\le i_2\le \cdots\le i_n\) which satisfy the equation \(q=\sum_{j=1}^n(1/i_j)\). For any such \(n\)-tuple, we have \(i_1\le n/q\), for

\[q=1/i_1+\cdots+1/i_n\le 1/i_1+\cdots+1/i_1=n/i_1, \]

But for each positive integer \(k\le n/q\), induction gives only finitely many \((n-1)\)-tuples \((i_2,\ldots,i_n)\) of positive integers with \(q-(1/k)=\sum_{j=2}^n(1/i_j)\). This completes the proof, for there are only finitely many such \(k\).
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Theorem 4.1.12 For every \(n\ge 1\), there are only finitely many finite groups having exactly \(n\) conjugacy classes.

Proof

Assume that \(G\) is a finite group having exactly \(n\) conjugacy classes. If \(|\mathrm Z(G)|=m\), then the class equation is

\[|G|=|\mathrm Z(G)|+\sum_{j=m+1}^n[G:\mathrm C_G(x_j)]. \]

If \(i_j=|G|\) for \(1\le j\le m\) and \(i_j=|G|/[G:\mathrm C_G(x_j)]=|\mathrm C_G(x_j)|\) for \(m+1\le j\le n\), then \(1=\sum_{j=1}^n(1/i_j)\). By Lemma 4.1.11, there are only finitely many such \(n\)-tuples, and so there is a maximum value for all possible \(i_j\)'s occurring therein, say, \(M\). It follows that a finite group \(G\) having exactly \(n\) conjugacy classes has order at most \(M\). But there are only finitely many nonisomorphic groups of any given order.
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Proposition 4.1.13 Let \(G\) be a finite \(p\)-group, and \(|G|=p^n\), where \(n\in \mathbb N^*\).

1. For any \(0\le k< n\), there exists a normal subgroup of order \(p^k\) of \(G\).
2. Let \(H\) be a nontrivial normal subgroup of \(G\), then \(H\cap \mathrm Z(G)\neq\mathbf 1\); moreover, if \(|H|=p\), then \(H\le \mathrm Z(G)\).
3. Let \(H\) be a proper subgroup of \(G\). If \(|H|=p^{s}\), then there is a subgroup of order \(p^{s+1}\) containing \(H\).

Proposition 4.1.14 The number of normal subgroups of order \(p^s\) of a finite \(p\)-group \(G\) is congruent to \(1\) mod \(p\).

Proposition 4.1.15 Let \(p\) be a prime,

1. if \(G\) is an abelian group of order \(p^2\), then \(G\) is either cyclic or isomorphic to \(\mathbb Z_p\times \mathbb Z_p\).
2. if \(G\) is a nonabelian group of order \(p^3\), then \(|\mathrm Z(G)|=p,\,G/\mathrm Z(G)\cong \mathbb Z_p\times \mathbb Z_p\), and \(\mathrm Z(G)=G'\), the commutator subgroup.

Definition 4.1.16 If \(p\) is a prime, then an elementary abelian \(p\)-group is a finite group \(G\) isomorphic to \(\mathbb Z_p\times\cdots\times\mathbb Z_p\).

Proposition 4.1.17

1. If \(G\) is an abelian group of prime exponent \(p\), then \(G\) is a vector space over \(\mathbb Z_p\), and every homomorphism \(\varphi\!:G\rightarrow G\) is a linear transformation.
2. A finite abelian \(p\)-group \(G\) is elementary if and only if it has exponent \(p\).

The Sylow Theorems

Definition 4.2.1 If \(p\) is a prime, then a Sylow \(p\)-subgroup \(P\) of a group \(G\) is a maximal \(p\)-subgroup.

Observe that every \(p\)-subgroup of \(G\) is contained in some Sylow \(p\)-subgroup; this is obvious when \(G\) is finite, and it follows from Zorn's lemma when \(G\) is infinite.

Lemma 4.2.2 Let \(P\) be a Sylow \(p\)-subgroup of a finite group \(G\).

1. \(|\mathrm N_G(P)/P|\) is prime to \(p\).
2. If \(a\in G\) has order some power \(p\) and \(aPa^{-1}=P\), then \(a\in P\).

Proof

1. If \(p\) divides \(|\mathrm N_G(P)/P|\), then Cauchy's theorem shows that \(\mathrm N_G(P)/P\) contains some element \(Pa\) of order \(p\); hence, \(S^*=\langle Pa\rangle\) has order \(p\). By the Correspondence Theorem, there is a subgroup \(S\le \mathrm N_G(P)\le G\) containing \(P\) with \(S/P\cong S^*\). Since both \(P\) and \(S^*\) are \(p\)-groups, then \(S\) is a \(p\)-group, contradicting the maximality of \(P\).
2. Replacing \(a\) by a suitable power of \(a\) if necessary, we may assume that \(a\) has order \(p\). Since \(a\) normalizes \(P\), we have \(a\in \mathrm N_G(P)\). If \(a\notin P\), then the coset \(Pa\in \mathrm N_G(P)/P\) has order \(p\), and this contradicts (i).
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Theorem 4.2.3 (Sylow)

1. If \(P\) is a Sylow \(p\)-subgroup of a finite group \(G\), then all Sylow \(p\)-subgroups of \(G\) are conjugate to \(P\).
2. If there are \(r\) Sylow \(p\)-subgroups, then \(r\) is a divisor of \(|G|\) and \(r\equiv 1\pmod p\).

Proof

Let \(X=\{P_1,\ldots,P_r\}\) be the family of all the conjugates of \(P\), where we have denoted \(P\) by \(P_1\). In Theorem 3.4.7, we saw that \(G\) acts on \(X\) by conjugation: there is a homomorphism \(\psi\!:G\rightarrow\mathrm S_X\) sending \(a\mapsto \psi_a\), where \(\psi_a(P_i)=aP_ia^{-1}\). Let \(Q\) be a Sylow \(p\)-subgroup of \(G\). Restricting \(\psi\) to \(Q\) shows that \(Q\) acts on \(X\), and we have that every orbit of \(X\) under this action has size dividing \(|Q|\); that is, every orbit has size some power of \(p\). If one of these orbits has size \(1\), then there would be an \(i\) with \(\psi_a(P_i)=P_i\) for all \(a\in Q\); that is, \(aP_ia^{-1}=P_i\) for all \(a\in Q\). By Lemma 4.2.2(ii) if \(a\in Q\), then \(a\in P_i\); that is, \(Q\le P_i\); since \(Q\) is a Sylow \(p\)-subgroup, \(Q=P_i\). If \(Q=P=P_1\), we conclude that every \(P\)-orbit of \(X\) has size an "honest" power of \(p\) except \(\{P_1\}\) which has size \(1\). Therefore, \(|X|=r\equiv 1\pmod p\).

Suppose there were a Sylow \(p\)-subgroup \(Q\) that is not a conjugate of \(P\); that is, \(Q\notin X\). If \(\{P_i\}\) is a \(Q\)-orbit of size \(1\), then we have seen that \(Q=P_i\), contradicting \(Q\in X\). Thus, every \(Q\)-orbit of \(X\) has size an honest power of \(p\), and so \(p\) divides \(|X|\); that is, \(r\equiv 0\pmod p\). The previous congruence is contradicted, and so no such subgroup \(Q\) exists. Therefore, every Sylow \(p\)-subgroup \(Q\) is conjugate to \(P\).

Finally, the number \(r\) of conjugates of \(P\) is the index of its normalizer, and so it is a divisor of \(|G|\).
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Corollary 4.2.4 A finite group \(G\) has a unique Sylow \(p\)-subgroup \(P\), for some prime \(p\), if and only if \(P\lhd G\).

Theorem 4.2.5 If \(G\) is a finite group of order \(p^em\), where \((p,m)=1\), then every Sylow \(p\)-subgroup \(P\) of \(G\) has order \(p^e\).

Proof

We claim that \([G:P]\) is prime to \(p\). Now \([G:P]=[G:N][N:P]\), where \(N=\mathrm N_G(P)\), and so it suffices to prove that each of the factors is prime to \(p\). But \([G:N]=r\), the number of conjugates of \(p\), so that \([G:N]\equiv 1\pmod p\), while \([N:P]=|N/P|\) is prime to \(p\), by Lemma 4.2.2(i). Thus by Lagrange's theorem, we have \(|P|=p^e\).
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Corollary 4.2.6 Let \(G\) be a finite group and let \(p\) be a prime. If \(p^k\) divides \(|G|\), then \(G\) contains a subgroup of order \(p^k\).

Lemma 4.2.7 If \(p\) is a prime not dividing an integer \(m\), then for all \(n\ge 1\), the binomial coefficient \(\binom{p^nm}{p^n}\) is not divisible by \(p\).

Proof

Write the binomial coefficient as follows:

\[\dfrac{p^nm(p^nm-1)\cdots(p^nm-i)\cdots(p^nm-p^n+1)}{p^n(p^n-1)\cdots(p^n-i)\cdots(p^n-p^n+1)} \]

Since \(p\) is prime, each factor equal to \(p\) of the numerator (or of the denominator) arises from a factor of \(p^nm-i\) (or of \(p^n-i\)). If \(i=0\), then the multiplicity of \(p\) in \(p^nm\) and in \(p^n\) are the same because \(p\nmid m\). If \(1\le i\le p^n\), then \(i=p^kj\), where \(0\le k< n\) and \(p\nmid j\). Now \(p^k\) is the highest power of \(p\) dividing \(p^n-i\), for \(p^n-i=p^n-p^kj=p^k(p^{n-k}-j)\) and \(p\nmid p^{n-k}-j\). A similar argument shows that the highest power of \(p\) divideing \(p^nm-i\) is also \(p^k\). Therefore, every factor of \(p\) upstairs is canceled by a factor of \(p\) downstairs, and hence the binomial coefficient has no factor equal to \(p\).
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Theorem 4.2.8 (Wielandt's Proof) If \(G\) is a finite group of order \(p^nm\), where \((p,m)=1\), then \(G\) has a subgroup of order \(p^n\).

Proof

If \(X\) is the family of all subsets of \(G\) of cardinal \(p^n\), then \(|X|\) is the binomial coefficient in the lemma, and so \(p\nmid |X|\). Let \(G\) act on \(X\) by left translation: if \(B\) is a subset of \(G\) with \(p^n\) elements, then for each \(g\in G\), define \(gB=\{gb|b\in B\}\). Now \(p\) cannot divide the size of every orbit of \(X\) lest \(p\big| |X|\); therefore, there is some \(B\in X\) with \(|\mathcal O(B)|\) not divisible by \(p\), where \(\mathcal O(B)\) is the orbit of \(B\). If \(G_B\) is the stabilizer of \(B\), then \(|G|/|G_B|=[G:G_B]=|\mathcal O(B)|\) is prime to \(p\). Hence, \(|G_B|=p^nm'\ge p^n\) (for some \(m'\) dividing \(m\)). On the other hand, if \(b_0\in B\) and \(g\in G_B\), then \(gb_0\in gB=B\); moreover, if \(g\) and \(h\) are distinct elements of \(G_B\), then \(gb_0\) and \(hb_0\) are distinct elements of \(B\). Therefore, \(|G_B|\le |B|=p^n\), and so \(|G_B|=p^n\).
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Theorem 4.2.9 (Frattini Argument) Let \(K\) be a normal subgroup of a finite group \(G\). If \(P\) is a Sylow \(p\)-subgroup of \(K\) (for some prime \(p\)), then

\[G=K\mathrm N_G(P). \]

Proof

If \(g\in G\), then \(gPg^{-1}\le gKg^{-1}=K\), because \(K\lhd G\). It follows that \(gPg^{-1}\) is a Sylow \(p\)-subgroup of \(K\), and so there exists \(k\in K\) with \(kPk^{-1}=gPg^{-1}\). Hence, \(P=(k^{-1}g)P(k^{-1}g)^{-1}\), so that \(k^{-1}g\in \mathrm N_G(P)\). The required factorization is thus \(g=k(k^{-1}g)\).
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Proposition 4.2.10 If \(Q\) is a normal \(p\)-subgroup of a finite group \(G\), then \(Q\le P\) for every Sylow \(p\)-subgroup \(P\).

Proposition 4.2.11 Let \(|G|=p^nm\), where \(p\nmid m\). If \(s\le n\) and \(r_s\) is the number of subgroups of \(G\) of order \(p^s\), then \(r_s\equiv 1\pmod p\).

Proposition 4.2.12 Let \(X\) be a finite \(G\)-set, and let \(H\le G\) act transitively on \(X\). Then \(G=HG_x\) for each \(x\in X\).

Proposition 4.2.13 Let \(P\le G\) be a Sylow subgroup. If \(\mathrm N_G(P)\le H\le G\), then \(H\) is equal to its own normalizer; that is, \(H=\mathrm N_G(H)\).

Proposition 4.2.14 Let \(G\) be a finite group and let \(P\le G\) be a Sylow subgroup. If \(H\lhd G\), then \(H\cap P\) is a Sylow subgroup of \(H\) and \(HP/H\) is a Sylow subgroup of \(G/H\).

Theorem 4.2.15 Let \(|G|=pq\), where \(p>q\) are primes. Then either \(G\) is cyclic or \(G=\langle a,b\rangle\), where \(b^p=a^q=\mathbf 1,aba^{-1}=b^m\), and \(m^q\equiv 1\pmod p\) but \(m\not\equiv 1\pmod p\). If \(q\nmid p-1\), then the second case cannot occur.

Proof

By Cauchy's theorem, \(G\) contains an element \(b\) of order \(p\): let \(S=\langle b\rangle\). Since \(S\) has order \(p\), it has index \(q\). It follows from Proposition 3.4.15 that \(S\lhd G\). Cauchy's theorem shows that \(G\) contains an element \(a\) of order \(q\); let \(T=\langle a\rangle\). Now \(T\) is a Sylow \(q\)-subgroup of \(G\), so that the number \(c\) of its conjugates is \(1+kq\) for some \(k\ge 0\). As above, either \(c=1\) or \(c=p\). If \(c=1\), then \(T\lhd G\) and \(G\cong S\times T\), and so \(G\cong \mathbb Z_p\times \mathbb Z_q\cong \mathbb Z_{pq}\). In case \(c=kq+1=p\), then \(q|p-1\), and \(T\) is not a normal subgroup of \(G\). Since \(S\lhd G\), \(aba^{-1}=b^m\) for some \(m\); furthermore, we may assume that \(m\not\equiv 1\pmod p\) lest we return to the abelian case. It is easy to prove, by induction on \(j\), that \(a^{j}ba^{-j}=b^{m^j}\). In particular, if \(j=q\), then \(m^q\equiv 1\pmod p\).
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Corollary 4.2.16 If \(p>q\) are primes, then every group \(G\) of order \(pq\) contains a normal subgroup of order \(p\). Moreover, if \(q\) does not divide \(p-1\), then \(G\) must be cyclic.

Proposition 4.2.17 If \(p\) and \(q\) are primes, then there is no simple group of order \(p^2q\).

Chapter V. Normal Series

Some Galois Theory

We are going to assume in this exposition that every field \(F\) is a subfield of an algebraically closed field \(C\). Thus, if \(f(x)\in F[x]\), the ring of all polynomials with coefficients in \(F\), and if \(f(x)\) has degree \(n\ge 1\), then there are (not necessarily distinct) elements \(\alpha_1,\ldots,\alpha_n\) in \(C\) (the roots of \(f(x)\)) and nonzero \(a\in F\) so that

\[f(x)=a(x-\alpha_1)(x-\alpha_2)\ldots(x-\alpha_n) \]

in \(C[x]\). The intersection of any family of subfields of a field is itself a subfield; define the smallest subfield of \(C\) containing a given subset \(X\) as the intersection of all those subfields of \(C\) containing \(X\). For example, if \(\alpha\in C\), the smallest subfield of \(C\) containing \(X=F\cup \{\alpha\}\) is

\[F(\alpha)=\{f(\alpha)/g(\alpha):f(x),g(x)\in F[x],g(\alpha)\neq 0\}; \]

\(F(\alpha)\) is called the subfield obtained from \(F\) by adjoining \(\alpha\). Similarly, one can define \(F(\alpha_1,\ldots,\alpha_n)\), the subfield obtained from \(F\) by adjoining \(\alpha_1,\ldots,\alpha_n\). In particular, if \(f(x)\in F[x]\) and \(f(x)=(x-\alpha_1)(x-\alpha_2)\ldots(x-\alpha_n)\in C[x]\), then \(F(\alpha_1,\ldots,\alpha_n)\), the subfield obtained from \(F\) by adjoining all the roots of \(f(x)\), is called the splitting field of \(f(x)\) over \(F\). Notice that the splitting field of \(f(x)\) does depend on \(F\).

Definition 5.1.1 Let \(f(x)\in F[x]\) have splitting field \(E\) over \(F\). Then \(f(x)\) is solvable by radicals if there is a chain of subfields

\[F=K_0\subset K_1\subset\cdots\subset K_t, \]

in which \(E\subset K_t\) and each \(K_{i+1}\) is obtained from \(K_i\) by adjoining a root of an element of \(K_i\); that is, \(K_{i+1}=K_i(\beta_{i+1})\), where \(\beta_{i+1}\in K_{i+1}\) and some power of \(\beta_{i+1}\) lies in \(K_i\).

Remark It can be shown that, for an arbitrary polynomial \(f(x)\), the concept of "\(f(x)\) is solvable by radicals" is equivalent to "there is a formula for the roots of \(f(x)\)".

Definition 5.1.2 If \(E\) and \(E'\) are fields, then a homomorphism is a function \(\sigma\!:E\rightarrow E'\) such that, for all \(\alpha,\beta\in E\),

\[\begin{aligned} \sigma(\mathbf 1)&=\mathbf 1,\\ \sigma(\alpha+\beta)&=\sigma(\alpha)+\sigma(\beta),\\ \sigma(\alpha\beta)&=\sigma(\alpha)\sigma(\beta). \end{aligned} \]

if \(\sigma\) is a bijection, then \(\sigma\) is an isomorphism; an isomorphism \(\sigma\!:E\rightarrow E\) is called an automorphism.

Lemma 5.1.3 Let \(f(x)\in F[x]\), let \(E\) be its splitting field over \(F\), and let \(\sigma\!:E\rightarrow E\) be an automorphism fixing \(F\) (i.e., \(\sigma(a)=a\) for all \(a\in F\)). If \(\alpha\in E\) is a root of \(f(x)\), then \(\sigma(a)\) is also a root of \(f(x)\).

Proof

If \(f(x)=\sum a_ix^i\), then \(\mathbf 0=\sigma(f(\alpha))=\sigma(\sum a_i\alpha^i)=\sum \sigma(a_i)\sigma(\alpha)^i=\sum a_i\sigma(\alpha)^i\), and so \(\sigma(\alpha)\) is a root of \(f(x)\).
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Lemma 5.1.4 Let \(F\) be a subfield of \(K\), let \(\{\alpha_1,\ldots,\alpha_n\}\subset K\), and let \(E=F(\alpha_1,\ldots,\alpha_n)\). If \(K'\) is a field containing \(F\) as a subfield, and if \(\sigma\!:E\rightarrow K'\) is a homomorphism fixing \(F\) with \(\sigma(\alpha_i)=\alpha_i\) for all \(i\), then \(\sigma\) is the identity.

Proof

The proof is by induction on \(n\ge 1\). If \(n=1\), then \(E\) consists of all \(g(\alpha_1)/h(\alpha_1)\), where \(g(x),h(x)\in F[x]\) and \(h(\alpha_1)\neq 0\); clearly \(\sigma\) fixes each such element. The inductive step is clear once we realizes that \(F(\alpha_1,\ldots,\alpha_n)=F^*(\alpha_n)\), where \(F^*=F(\alpha_1,\ldots,\alpha_{n-1})\).
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Definition 5.1.5 If \(F\) is a subfield of \(E\), then the Galois group, denoted by \(\mathrm{Gal}(E/F)\), is the group under composition of all those automorphisms of \(E\) which fix \(F\). If \(f(x)\in F[x]\) and \(E=F(\alpha_1,\ldots,\alpha_n)\) is the splitting field of \(f(x)\) over \(F\), then the Galois group of \(f(x)\) is \(\mathrm{Gal}(E/F)\).

Theorem 5.1.6 Let \(f(x)\in F[x]\) and let \(X=\{\alpha_1,\ldots,\alpha_n\}\) be the set of its distinct roots (in its splitting field \(E=F(\alpha_1,\ldots,\alpha_n)\) over \(F\)). Then the function \(\varphi\!:\mathrm{Gal}(E/F)\rightarrow\mathrm S_X\cong \mathrm S_n\), given by \(\varphi(\sigma)=\sigma\big|_X\), is an imbedding; that is, \(\varphi\) is completely determined by its action on \(X\).

Proof

If \(\sigma\in\mathrm{Gal}(E/F)\), then Lemma 5.1.3 shows that \(\sigma(X)\subset X\); \(\sigma\big|_X\) is a bijection because \(\sigma\) is an injection and \(X\) is finite. It is easy to see that \(\varphi\) is a homomorphism; it is an injection, by Lemma 5.1.4.
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Not every permutation of the roots of a polynomial \(f(x)\) need arise from some \(\sigma\in\mathrm{Gal}(E/F)\). For example, let \(f(x)=(x^2-2)(x^2-3)\in\mathbb Q[x]\). Then \(E=\mathbb Q(\sqrt 2,\sqrt 3)\) and there is no \(\sigma\in\mathrm{Gal}(E/\mathbb Q)\) with \(\sigma(\sqrt 2)=\sqrt 3\).

Definition 5.1.7 If \(F\) is a subfield of a field \(E\), then \(E\) is a vector space over \(F\) (if \(a\in F\) and \(\alpha\in E\), define scalar multiplication to be the given product \(a\alpha\) of two elements of \(E\)). The degree of \(E\) over \(F\), denoted by \([E:F]\), is the dimension of \(E\).

Proposition 5.1.8 Let \(p(x)\in F[x]\) be an irreducible polynomial of degree \(n\). If \(\alpha\) is a root of \(p(x)\) (in a splitting field), then \(\{\mathbf 1,\alpha,\alpha^2,\ldots,\alpha^{n-1}\}\) is a basis of \(F(\alpha)\) (viewed as a vector space over \(F\)), and so we have that \([F(\alpha):F]=n\).

Proposition 5.1.9 Let \(F\subset E\subset K\) be fields, where \([K:E]\) and \([E:F]\) are finite, then we have \([K:F]=[K:E][E:F]\).

Proposition 5.1.10 Let \(E\) be a splitting field over \(F\) of some \(f(x)\in F[x]\), and let \(K\) be a splitting field over \(E\) of some \(g(x)\in E[x]\). If \(\sigma\in\mathrm{Gal}(K/F)\), then \(\sigma\big|_E\in\mathrm{Gal}(E/F)\).

Lemma 5.1.11 Let \(p(x)\in F[x]\) be irreducible, and let \(\alpha\) and \(\beta\) be roots of \(p(x)\) in a splitting field of \(p(x)\) over \(F\). Then there exists an isomorphism \(\lambda^*\!:F(\alpha)\rightarrow F(\beta)\) which fixes \(F\) and with \(\lambda^*(\alpha)=\beta\).

Proof

By Proposition 5.1.8, every element of \(F(\alpha)\) has a unique expression of the form \(a_0+a_1\alpha+\cdots+a_{n-1}\alpha^{n-1}\). Define \(\lambda^*\) by

\[\lambda^*(a_0+a_1\alpha+\cdots+a_{n-1}\alpha^{n-1})=a_0+a_1\beta+\cdots+a_{n-1}\beta^{n-1}. \]

It is easy to see that \(\lambda^*\) is a field homomorphism; it is an isomorphism because its inverse can be constructed in the same manner.
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Remark There is a generalization of this lemma having the same proof. Let \(\lambda\!:F\rightarrow F'\) be an isomorphism of fields, let \(p(x)=a_0+a_1x+\cdots+a_nx^n\in F[x]\) be an irreducible polynomial, and let \(p'(x)=\lambda(a_0)+\lambda(a_1)x+\cdots+\lambda(a_n)x^{n}\in F'[x]\). Finally, let \(\alpha\) be a root of \(p(x)\) and let \(\beta\) be a root of \(p'(x)\) (in appropriate splitting fields). Then there is an isomorphism \(\lambda^*\!:F(\alpha)\rightarrow F'(\beta)\) with \(\lambda^*(\alpha)=\beta\) and with \(\lambda^*\big|_F=\lambda\).

Lemma 5.1.12 Let \(f(x)\in F[x]\), and let \(E\) be its splitting field over \(F\). If \(K\) is an "intermediate field", that is, \(F\subset K\subset E\), and if \(\lambda\!:K\rightarrow K\) is an automorphism fixing \(F\), then there is an automorphism \(\lambda^*\!:E\rightarrow E\) with \(\lambda^*\big|_K=\lambda\).

Proof

The proof is by induction on \(d=[E:F]\). If \(d=1\), then \(E=K\), every root \(\alpha_1,\ldots,\alpha_n\) of \(f(x)\) lies in \(K\), and we may take \(\lambda^*=\lambda\). If \(d>1\), then \(E\neq K\) and there is some root \(\alpha\) of \(f(x)\) not lying in \(K\). Now \(\alpha\) is a root of some irreducible factor \(p(x)\) of \(f(x)\); since \(\alpha\notin K\), degree \(p(x)=k>1\). By the generalized version of Lemma 5.1.11, there is \(\beta\in E\) and an isomorphism \(\lambda_1\!:K(\alpha)\rightarrow K(\beta)\) which extends \(\lambda\) and with \(\lambda_1(\alpha)=\beta\). By Proposition 5.1.9, \([E:K(\alpha)]=d/k<d\). Now \(E\) is the splitting field of \(f(x)\) over \(K(\alpha)\), for it arises from \(K(\alpha)\) by adjoining all the roots of \(f(x)\). Since all the inductive hypotheses have been verified, we conclude that \(\lambda_1\), and hence \(\lambda\), can be extended to an automorphism of \(E\).
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Remark As with the previous lemma, Lemma 5.1.12 has a more general version having the same proof. It says that if \(f(x)\in F[x]\), then any two (abstract) splitting fields of \(f(x)\) over \(F\) are isomorphic.

Theorem 5.1.13 Let \(p\) be a prime, let \(F\) be a field containing a primitive \(p\)th root of unity, say, \(\omega\), and let \(f(x)=x^{p}-a\in F[x]\).

1. If \(\alpha\) is a root of \(f(x)\) (in some splitting field), then \(f(x)\) is irreducible if and only if \(\alpha\notin F\).
2. The splitting field \(E\) of \(f(x)\) over \(F\) is \(F(\alpha)\).
3. If \(f(x)\) is irreducible, then \(\mathrm{Gal}(E/F)\cong \mathbb Z_p\).

Proof

1. If \(\alpha\in F\), then \(f(x)\) is not irreducible, for it has \(x-\alpha\) as a factor. Conversely, assume that \(f(x)=g(x)h(x)\), where degree \(g(x)=k<p\). Since the roots of \(f(x)\) are \(\alpha,\omega\alpha,\omega^2\alpha,\ldots,\omega^{p-1}\alpha\), every root of \(g(x)\) has the form \(\omega^i\alpha\) for some \(i\). If the constant term of \(g(x)\) is \(c\), then \(c=\pm\omega^r\alpha^k\) for some \(r\). As both \(c\) and \(\omega\) lie in \(F\), it follows that \(\alpha^k\in F\). But \((k,p)=1\), because \(p\) is prime, and so \(1=ks+tp\) for some integers \(s\) and \(t\). Thus

\[\alpha=\alpha^{ks+tp}=(\alpha^k)^s(\alpha^p)^t\in F. \]

2. Immediate from the observation that the roots of \(f(x)\) are of the form \(\omega^i\alpha\).
3. If \(\sigma\in\mathrm{Gal}(E/F)\), then \(\sigma(\alpha)=\omega^i\alpha\) for some \(i\), by Lemma 5.1.3. Define \(\varphi\!:\mathrm{Gal}(E/F)\rightarrow \mathbb Z_p\) by \(\varphi(\sigma)=[i]\), the congruence class of \(i\) mod \(p\). It is easy to check that \(\varphi\) is a homomorphism; it is an injection, by Lemma 5.1.4. Since \(f(x)\) is irreducible, by hypotheses, Lemma 5.1.11 shows that \(\mathrm{Gal}(E/F)\neq \mathbf 1\). Therefore, \(\varphi\) is a surjection, for \(\mathbb Z_p\) has no proper subgroups.
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Theorem 5.1.14 Let \(f(x)\in F[x]\), let \(E\) be the splitting field of \(f(x)\) over \(F\), and assume that \(f(x)\) has no repeated roots in \(E\) (i.e., \(f(x)\) has no factor of the form \((x-\alpha)^2\) in \(E[x]\)). Then \(f(x)\) is irreducible if and only if \(\mathrm{Gal}(E/F)\) acts transitively on the set \(X\) of all the roots of \(f(x)\).

Remark It can be shown that if \(F\) has characteristic \(0\) or if \(F\) is finite, then every irreducible polynomial in \(F[x]\) has no repeated roots.

Proof

Note first that Lemma 5.1.3 shows that \(\mathrm{Gal}(E/F)\) does act on \(X\). If \(f(x)\) is irreducible, then Lemma 5.1.11 shows that \(\mathrm{Gal(E/F)}\) acts transitively on \(X\). Conversely, assume that there is a factorization \(f(x)=g(x)h(x)\) in \(F[x]\). In \(E[x]\), \(g(x)=\prod(x-\alpha_i)\) and \(h(x)=\prod(x-\beta_j)\); since \(f(x)\) has no repeated roots, \(\alpha_i\neq\beta_j\) for all \(i,j\). But \(\mathrm{Gal}(E/F)\) acts transitively on the roots of \(f(x)\), so there exists \(\sigma\in\mathrm{Gal}(E/F)\) with \(\sigma(\alpha_1)=\beta_1\), and this contradicts Lemma 5.1.3.
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It is easy to see that if \(\alpha_1\) is a root of \(f(x)\), then the stabilizer of \(\alpha_1\) is \(\mathrm{Gal}(E/F(\alpha_1))\le \mathrm{Gal}(E/F)\), and \(\mathrm{Gal(E/F(\alpha_1))}\) is the Galois group of \(f(x)/(x-\alpha_1)\) over \(F(\alpha_1)\). Thus, \(f(x)/(x-\alpha_1)\) is irreducible (over \(F(\alpha_1)\)) if and only if \(\mathrm{Gal}(E/F(\alpha_1))\) acts transitively on the remaining roots.

Theorem 5.1.15 Let \(F\subset K\subset E\) be fields, where \(K\) and \(E\) are splitting fields of polynomials over \(F\). Then \(\mathrm{Gal}(E/K)\lhd\mathrm{Gal}(E/F)\) and

\[\mathrm{Gal}(E/F)/\mathrm{Gal}(E/K)\cong\mathrm{Gal}(K/F). \]

Proof

The function \(\Phi\!:\mathrm{Gal}(E/F)\rightarrow\mathrm{Gal}(K/F)\), given by \(\Phi(\sigma)=\sigma\big|_K\), is well defined (by Proposition 5.1.10, for \(K\) is a splitting field), and it is easily seen to be a homomorphism. The kernel of \(\Phi\) consists of all those automorphisms which fix \(K\); that is, \(\mathrm{ker}\Phi=\mathrm{Gal}(E/K)\), and so this subgroup is normal. We claim that \(\Phi\) is a surjection. If \(\lambda\in\mathrm{Gal}(K/F)\), that is, \(\lambda\) is an automorphism of \(K\) which fixes \(F\), then \(\lambda\) can be extended to an automorphism \(\lambda^*\) of \(E\) (by Lemma 5.1.12, for \(E\) is a splitting field). Therefore, \(\lambda^*\in\mathrm{Gal}(E/F)\) and \(\Phi(\lambda^*)=\lambda^*\big|_K=\lambda\). The first isomorphism theorem completes the proof.
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Proposition 5.1.16 Let \(f(x)=x^n-a\in F[x]\), let \(E\) be the splitting field of \(f(x)\) over \(F\), and let \(\alpha\in E\) be an \(n\)th root of \(a\). Then there are subfields

\[F=K_0\subset K_1\subset \cdots\subset K_t=F(\alpha) \]

with \(K_{i+1}=K_i(\beta_{i+1}),\beta_{i+1}^{p(i)}\in K_i\), and \(p(i)\) prime for all \(i\).

Theorem 5.1.17 (Galois) Let \(f(x)\in F[x]\) be a polynomial of degree \(n\), let \(F\) contain all \(p\)th roots of unity for every prime \(p\) dividing \(n!\), and let \(E\) be the splitting field of \(f(x)\) over \(F\). If \(f(x)\) is solvable by radicals, then there exist subgroups \(G_i\le G=\mathrm{Gal}(E/F)\) such that:

1. \(G=G_0\ge G_1\ge \cdots\ge G_t=\mathbf 1\);
2. \(G_{i+1}\lhd G_i\) for all \(i\); and
3. \(G_i/G_{i+1}\) is cyclic of prime order for all \(i\).

Remark

1. The hypothesis that \(F\) contains various roots of unity can be eliminated.
2. If \(F\) has characteristic \(0\), then the converse of this theorem is also true.

Definition 5.1.18 A normal series of a group \(G\) is a sequence of subgroups

\[G=G_0\rhd G_1\rhd\cdots\rhd G_n=\mathbf 1. \]

The factor groups of this normal series are the groups \(G_i/G_{i+1}\) for all \(i\); the length of the normal series is the number of strict inclusions; that is, the length is the number of nontrivial factor groups.

Definition 5.1.19 A finite group \(G\) is solvable if it has a normal series whose factor groups are cyclic of prime order.

In this terminology, Theorem 5.1.17 and its converse say that a polynomial is solvable by radicals if and only if its Galois group is a solvable group.

The Jordan-Hölder Theorem

Definition 5.2.1 A normal series

\[G=H_0\rhd H_1\rhd\cdots\rhd H_m=\mathbf 1 \]

is a refinement of a normal series

\[G=G_0\rhd G_1\rhd\cdots \rhd G_n=\mathbf 1 \]

if \(G_0,G_1,\ldots,G_n\) is a subsequence of \(H_0,H_1,\cdots,H_m\).

Definition 5.2.2 A composition series is a normal series

\[G=G_0\rhd G_1\rhd\cdots \rhd G_n=\mathbf 1 \]

in which, for all \(i\), either \(G_{i+1}\) is a maximal normal subgroup of \(G_i\) or \(G_{i+1}=G_i\).

Every refinement of a composition series is also a composition series; it can only repeat some of the original terms.

Proposition 5.2.3 A normal series is a composition series if and only if its factor groups are either simple or trivial.

Proposition 5.2.4

1. Every finite group has a composition series.
2. An abelian group has a composition series if and only if it is finite.

Proposition 5.2.5 If \(G\) is a finite group having a normal series with factor groups \(H_0,H_1,\ldots,H_n\), then \(|G|=\prod|H_i|\).

Definition 5.2.6 Two normal series of a group \(G\) are equivalent if there is a bijection between their nontrivial factor groups such that corresponding factor groups are isomorphic.

Lemma 5.2.7 (Zassenhaus Lemma) Let \(A\lhd A^*\) and \(B\lhd B^*\) be four subgroups of a group \(G\). Then

\[\begin{aligned} A(A^*\cap B)&\lhd A(A^*\cap B^*),\\ B(B^*\cap A)&\lhd B(B^*\cap A^*), \end{aligned} \]

and there is an isomorphism

\[\dfrac{A(A^*\cap B^*)}{A(A^*\cap B)}\cong \dfrac{B(B^*\cap A^*)}{B(B^*\cap A)}. \]

Proof

Both \(A\) and \(A^*\cap B^*\) are subgroups of \(A^*\), and \(A\lhd A^*\). By the second isomorphism theorem, \(A\cap B^*=A\cap(A^*\cap B^*)\lhd A^*\cap B^*\); similarly, \(A^*\cap B\lhd A^*\cap B^*\). It follows from Lemma 2.6.3 and Proposition 2.4.17 that \(D=(A^*\cap B)(A\cap B^*)\) is a normal subgroup of \(A^*\cap B^*\).

If \(x\in B(B^*\cap A^*)\), then \(x=bc\) for \(b\in B\) and \(c\in B^*\cap A^*\). Define \(f\!:B(B^*\cap A)\rightarrow (A^*\cap B^*)/D\) by \(f(x)=f(bc)=cD\). To see that \(f\) is well defined, assume that \(x=bc=b'c'\), where \(b'\in B\) and \(c'\in B^*\cap A\); then \(c'c^{-1}=b'^{-1}b\in (B^*\cap A^*)\cap B=B\cap A^*\le D\). It is routine to check that \(f\) is a surjective homomorphism with kernel \(B(B^*\cap A)\). The first isomorphism theorem gives \(B(B^*\cap A)\lhd B(B^*\cap A^*)\) and

\[\dfrac{B(B^*\cap A^*)}{B(B^*\cap A)}\cong \dfrac{B^*\cap A^*}{D}. \]

Transposing the symbols \(A\) and \(B\) gives \(A(A^*\cap B)\lhd A(A^*\cap B^*)\) and an isomorphism of the corresponding quotient group to \((B^*\cap A^*)/D\). It follows that the two quotient groups in the statement are isomorphic.
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Theorem 5.2.8 (Schreier Refinement Theorem) Every two normal series of an arbitrary group \(G\) have refinements that are equivalent.

Proof

Let

\[G=G_0\rhd G_1\rhd\cdots\rhd G_n=\mathbf 1 \]

and

\[G=H_0\rhd H_1\rhd\cdots\rhd H_m=\mathbf 1 \]

be normal series. Insert a "copy" of the second series between each \(G_i\) and \(G_{i+1}\). More precisely, define \(G_{i,j}=G_{i+1}(G_i\cap H_j)\) for all \(0\le j\le m\). Thus

\[G_{i,j}=G_{i+1}(G_i\cap H_j)\ge G_{i+1}(G_i\cap H_{j+1})=G_{i,j+1}. \]

Notice that \(G_{i,0}=G_i\) because \(H_0=G\) and that \(G_{i,m}=G_{i+1}\) because \(H_m=\mathbf 1\). Moreoever, setting \(A=G_{i+1}, A^*=G_i,B=H_{j+1}\), and \(B^*=H_j\) in the Zassenhaus lemma shows that \(G_{i,j+1}\lhd G_{i,j}\). It follows that the sequence

\[G_{0,0}\rhd G_{0,1}\rhd \cdots\rhd G_{0,m}\rhd G_{1,0}\rhd\cdots\rhd G_{n-1,0}\rhd\cdots\rhd G_{n-1,m}=\mathbf 1 \]

is a refinement of the first normal series (with \(mn\) terms). Similarly, if \(H_{i,j}\) is defined to be \(H_{j+1}(H_j\cap G_i)\), then \(H_{i,j}\ge H_{i+1,j}\) and

\[H_{0,0}\rhd H_{1,0}\rhd\cdots\rhd H_{n,0}\rhd H_{0,1}\rhd\cdots\rhd H_{0,m-1}\rhd\cdots\rhd H_{n,m-1}=\mathbf 1 \]

is a refinement of the second normal series (with \(mn\) terms). Finally, the function pairing \(G_{i,j}/G_{i,j+1}\) with \(H_{i,j}/H_{i+1,j}\) is a bijection, and the Zassenhaus lemma (with \(A=G_{i+1},A^*=G_i,B=G_{j+1}\), and \(B^*=H_j\)) shows that corresponding factor groups are isomorphic. Therefore, the two refinements are equivalent.
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Theorem 5.2.9 (Jordan-Hölder) Every two composition series of a group \(G\) are equivalent.

Proof

Composition series are normal series, so that every two composition series of \(G\) have equivalent refinements. But a composition series is a normal series of maximal length; a refinement of it merely repeats several of its terms, and so its new factor groups have order \(1\). Therefore, two composition series of \(G\) are equivalent.
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Definition 5.2.10 If \(G\) has a composition series, then the factor groups of this series are called the composition factors of \(G\).

Remark The Jordan-Hölder Theorem can be regarded as a unique factorization theorem.

Proposition 5.2.11 Let \(G\) and \(H\) be finite groups. If there are normal series of \(G\) and of \(H\) having the same set of factor groups, then \(G\) and \(H\) have the same composition factors.

Proposition 5.2.12 If \(n\ge 5\), then \(\mathrm S_n\) is not solvable.

Proposition 5.2.13 Assume that \(G=H_1\times \cdots \times H_n=K_1\times \cdots \times K_m\), where each \(H_i\) and \(K_j\) is simple. Then we have that \(m=n\) and there is a permutation \(\pi\) of \(\{1,2,\ldots,n\}\) with \(K_{\pi(i)}\cong H_i\) for all \(i\).

Solvable Groups

Definition 5.3.1 A solvable series of a group \(G\) is a normal series all of whose factor groups are abelian. A group \(G\) is solvable if it has a solvable series.

Remark Previously we have defined a "solvability" in Definition 5.1.18. The equivalence of these two definitions can be easily seen since we can always find in any abelian group \(G\) a cyclic subgroup of some prime order, which is also a normal subgroup of \(G\).

Theroem 5.3.2 Every subgroup \(H\) of a solvable group \(G\) is itself solvable.

Proof

If \(G=G_0\rhd G_1\rhd \cdots\rhd G_n=\mathbf 1\) is a solvable series, consider the series \(H=H_0\ge (H\cap G_1)\ge\cdots\ge(H\cap G_n)=\mathbf 1\). This is a normal series of \(H\), for the second isomorphism theorem gives \(H\cap G_{i+1}=(H\cap G_i)\cap G_{i+1}\lhd H\cap G_i\) for all \(i\). Now \((H\cap G_i)/(H\cap G_{i+1})\cong G_{i+1}(H\cap G_i)/G_{i+1}\le G_i/G_{i+1}\); as \(G_i/G_{i+1}\) is abelian, so is any of its subgroups. Therefore, \(H\) has a solvable series.
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Theroem 5.3.3 Every quotient of a solvable group is solvable.

Proof

It suffices to prove that if \(G\) is a solvable group and \(f\!:G\rightarrow H\) is a surjection, then \(H\) is a solvable group. If

\[G=G_0\rhd G_1\rhd \cdots\rhd G_t=\mathbf 1 \]

is a solvable series, then

\[H=f(G_0)\rhd f(G_1)\rhd\cdots\rhd f(G_t)=\mathbf 1 \]

is a normal series for \(H\): if \(f(x_{i+1})\in f(G_{i+1})\) and \(f(x_i)\in f(G_i)\), then \(f(x_i)f(x_{i+1})f(x_i)^{-1}=f(x_ix_{i+1}x_i^{-1})\in f(G_{i+1})\), and so \(f(G_{i+1})\lhd f(G_i)\) for every \(i\). The map \(\varphi\!:f(G_i)/f(G_{i+1})\), defined by \(x_i\mapsto f(x_i)/f(G_{i+1})\), is a surjection, for it is the composite of the surjections \(G_{i}\rightarrow f(G_i)\) and the natural map \(f(G_i)\rightarrow f(G_i)/f(G_{i+1})\). Since \(G_{i+1}\le \mathrm{ker}\varphi\), this map \(\varphi\) includes a surjection \(G_i/G_{i+1}\rightarrow f(G_i)/f(G_{i+1})\), namely, \(x_iG_{i+1}\mapsto f(x_i)f(G_{i+1})\). Now \(f(G_i)/f(G_{i+1})\) is a quotient of the abelian group \(G_i/G_{i+1}\), and so it is abelian. Therefore, \(H\) has a solvable series, and hence it is a solvable group.
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Theorem 5.3.4 If \(H\lhd G\) and if both \(H\) and \(G/H\) are solvable, then \(G\) is solvable.

Proof

Let

\[G/H\rhd K_1^*\rhd K_2^*\rhd\cdots\rhd K_n^*=\mathbf 1 \]

be a solvable series. By the correspondence theorem, we can construct a sequence looking like the beginning of a solvable series from \(G\) to \(H\); that is, there are subgroups \(K_i\) (with \(H\le K_i\) and \(K_i/H\cong K_i^*\)) such that

\[G\rhd K_1\rhd K_2\rhd\cdots\rhd K_n=H, \]

and \(K_i/K_{i+1}(\cong K_i^*/K_{i+1}^*)\) is abelian. Since \(H\) is solvable, it has a solvable series; if we splice these two sequences together at \(H\), we obtain a solvable series for \(G\).
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Corollary 5.3.5 If \(H\) and \(K\) are solvable groups, then \(H\times K\) is solvable.

Theorem 5.3.6 Every finite \(p\)-group \(G\) is solvable.

Proof

The proof is by induction on \(|G|\). By Theorem 4.1.6, \(|\mathrm Z(G)|\neq 1\). Therefore, \(G/\mathrm Z(G)\) is a \(p\)-group of order \(<|G|\), and hence it is solvable, by induction. Since every abelian group is solvable, \(\mathrm Z(G)\) is solvable. Therefore, \(G\) is solvable, by Theorem 5.3.4.
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Definition 5.3.7 The higher commutator subgroups of \(G\) are defined inductively:

\[G^{(0)}=G;\qquad G^{(i+1)}=G^{(i)\prime}; \]

that is, \(G^{(i+1)}\) is the commutator subgroup of \(G^{(i)}\). The series

\[G=G^{(0)}\rhd G^{(1)}\rhd G^{(2)}\rhd\cdots \]

is called the derived series of \(G\).

Definition 5.3.8 An automorphism of a group \(G\) is an isomorphism \(\varphi\!:G\rightarrow G\). A subgroup \(H\) of \(G\) is called characteristic in \(G\), denoted by \(H\text{ char } G\), if \(\varphi(H)=H\) for every automorphism \(\varphi\) of \(G\).

If \(\varphi(H)\le H\) for every automorphism \(\varphi\), then \(H\text{ char }G\). For each \(a\in G\), conjugation by \(a\) (i.e., \(x\mapsto axa^{-1}\)) is an automorphism of \(G\); it follows at once that every characteristic subgroup is a normal subgroup.

Lemma 5.3.9

1. If \(H\text{ char }K\) and \(K\text{ char }G\), then \(H\text{ char }G\).
2. If \(H\text{ char }K\) and \(K\lhd G\), then \(H\lhd G\).

Proof

1. If \(\varphi\) is an automorphism of \(G\), then \(\varphi(K)=K\), and so the restriction \(\varphi\big|_K\!:K\rightarrow K\) is an automorphism of \(K\); since \(H\text{ char }K\), it follows that \(\varphi(H)=\left(\varphi\big|_K\right)(H)=H\).
2. Let \(a\in G\) and let \(\varphi\!:G\rightarrow G\) be conjugation by \(a\). Since \(K\lhd G\), \(\varphi\big|_K\) is an automorphism of \(K\); since \(H\text{ char }K\), \(\left(\varphi\big|_K\right)(H)\le H\). This says that if \(h\in H\), then \(aha^{-1}=\varphi(h)\in H\).
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Theorem 5.3.10 For every group \(G\), the higher commutator subgroups are characteristic, hence normal subgroups.

Proof

The proof is by induction on \(i\ge 1\). Recall that the commutator subgroup \(G'=G^{(1)}\) is generated by all commutators; that is, by all elements of the form \(aba^{-1}b^{-1}\). If \(\varphi\) is an automorphism of \(G\), then \(\varphi(aba^{-1}b^{-1})=\varphi(a)\varphi(b)\varphi(a)^{-1}\varphi(b)^{-1}\) is also a commutator, and so \(\varphi(G')\le G'\). For the inductive step, we have just shown that \(G^{(i+1)}\text{ char }G^{(i)}\); since \(G^{(i)}\text{ char }G\), by induction, Lemma 5.3.9(i) shows that \(G^{(i+1)}\) is characteristic in \(G\).
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Lemma 5.3.11 If \(G=G_0\rhd G_1\rhd\cdots\rhd G_n=\mathbf 1\) is a solvable series, then \(G_i\ge G^{(i)}\) for all \(i\).

Proof

The proof is by induction on \(i\ge 0\). If \(i=0\), then \(G_0=G=G^{(0)}\). For the inductive step, Theorem 2.5.4 gives \(G_{i+1}\ge G_i'\), since \(G_i/G_{i+1}\) is abelian. The inductive hypothesis gives \(G_i\ge G^{(i)}\), so that \(G_i'\ge G^{(i)\prime}=G^{(i+1)}\). Therefore, \(G_{i+1}\ge G^{(i+1)}\), as desired.
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Theorem 5.3.12 A group \(G\) is solvable if and only if \(G^{(n)}=\mathbf 1\) for some \(n\).

Definition 5.3.13 A normal subgroup \(H\) of a group \(G\) is a minimal normal subgroup if \(H\neq \mathbf 1\) and there is no normal subgroup \(K\) of \(G\) with \(\mathbf 1< K< H\).

Theorem 5.3.14 If \(G\) is a finite solvable group, then every minimal normal subgroup is elementary abelian.

Proof

Let \(V\) be a minimal normal subgroup of \(G\). Now Lemma 5.3.9(ii) says that if \(H\text{ char }V\), then \(H\lhd G\); since \(V\) is minimal, either \(H=\mathbf 1\) or \(H=V\). In particular, \(V'\text{ char }V\), so that either \(V'=\mathbf 1\) or \(V'=V\). Since \(G\) is solvable, so is its subgroup \(V\). It follows from Theorem 5.3.12 that \(V'=V\) cannot occur here, so that \(V'=\mathbf 1\) and so \(V\) is abelian. Since \(V\) is abelian, a Sylow \(p\)-subgroup \(P\) of \(V\), for any prime \(p\), is characteristic in \(V\); therefore, \(V\) is an abelian \(p\)-group. But \(\{x\in V:x^{p}=\mathbf 1\}\text{ char }V\), and so \(V\) is elementary abelian.
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Theorem 5.3.15 A finite group \(G\) with no characteristic subgroups other than \(G\) and \(\mathbf 1\) is either simple or a direct product of isomorphic simple groups.

Proof

Choose a minimal normal subgroup \(H\) of \(G\) whose order is minimal among all nontrivial normal subgroups. Write \(H=H_1\), and consider all subgroups of \(G\) of the form \(H_1\times H_2\times \cdots\times H_n\), where \(n\ge 1\), \(H_i\lhd G\), and \(H_i\cong H\). Let \(M\) be such a subgroup of largest possible order. We know that \(M=G\) by showing that \(M\text{ char }G\); to see this, it suffices to show that \(\varphi(H_i)\le M\) for every \(i\) and every automorphism \(\varphi\) of \(G\). Of course, \(\varphi(H_i)\cong H=H_1\). We show that \(\varphi(H_i)\lhd G\). If \(a\in G\), then \(a=\varphi(b)\) for some \(b\in G\), and \(a\varphi(H_i)a^{-1}=\varphi(b)\varphi(H_i)\varphi(b)^{-1}=\varphi(bH_ib^{-1})\le \varphi(H_i)\), because \(H_i\lhd G\). If \(\varphi(H_i)\nleq M\), then \(\varphi(H_i)\cap M\nleq \varphi(H_i)\) and \(|\varphi(H_i)\cap M|<|\varphi(H_i)|=|H|\). But \(\varphi(H_i)\cap M\lhd G\), and so the minimality of \(|H|\) shows that \(\varphi(H_i)\cap M=\mathbf 1\). The subgroup \(\langle M,\varphi(H_i)\rangle=M\times \varphi(H_i)\) is a subgroup of the same type as \(M\) but of largere order, a contradiction. We conclude that \(M\text{ char }G\), and so \(M=G\). Finally, \(H=H_1\) must be simple: if \(N\) is a nontrivial normal subgroup of \(H\), then \(N\) is a normal subgroup of \(M=H_1\times H_2\times\cdots\times H_n=G\), and this contradicts the minimal choice of \(H\).
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Corollary 5.3.16 A minimal normal subgroup \(H\) of a finite group \(G\) is either simple or a direct product of isomorphic simple subgroups.

Proposition 5.3.17 If \(S\) and \(T\) are solvable subgroups of \(G\) with \(S\lhd G\), then \(ST\) is solvable.

Proposition 5.3.18 Every finite group \(G\) has a unique maximal normal solvable subgroup \(\mathscr S(G)\); moreover, \(G/\mathscr S(G)\) has no nontrivial normal solvable subgroups.

Proposition 5.3.19 Let \(G\) be a finite group of order \(>1\). If \(G\) is solvable, then \(G\) contains a nontrivial normal abelian subgroup; if \(G\) is not solvable, then \(G\) contains a nontrivial normal subgroup \(H\) with \(H=H'\).

Theorem 5.3.20 (Burnside's Theorem) If \(p\) and \(q\) are primes, then every group of order \(p^mq^n\) is solvable.

Proposition 5.3.21

1. If \(H\lhd G\) and \((|H|,[G:H])=1\), then \(H\text{ char }G\).
2. If \(H\text{ char }G\) and \(H\le K\le G\), then \(K/H\text{ char }G/H\) implies \(K\text{ char }G\).

Proposition 5.3.22 Let \(G\) be a group, then \(\mathrm Z(G)\text{ char }G\).

Two Theorems of P.Hall

Theorem 5.4.1 (P.Hall) If \(G\) is a solvable group of order \(ab\), where \((a,b)=1\), then \(G\) contains a subgroup of order \(a\). Moreover, any two subgroups of order \(a\) are conjugate.

Proof

The proof is by induction on \(|G|\); the base step is trivially true.

Case 1. \(G\) contains a normal subgroup \(H\) of order \(a'b'\), where \(a'|a,b'|b\), and \(b'<b\).

Existence. In this case, \(G/H\) is a solvable group of order \((a/a')(b/b')\), which is strictly less than \(ab\); by induction, \(G/H\) has a subgroup \(A/H\) of order \(a/a'\). Now \(A\) has order \((a/a')|H|=ab'<ab\); since \(A\) is solvable, it has a subgroup of order \(a\), as desired.

Conjugacy. Let \(A\) and \(A'\) be subgroups of \(G\) of order \(a\), and let \(k=|AH|\). Since \(AH\le G\), Lagrange's theorem gives \(|AH|=\alpha\beta\), where \(\alpha\big|a\) and \(\beta\big|b\), we have \(a'b'\big|\alpha\beta\), so that \(b'\big|\beta\). But the product formula gives \(k|aa'b'\), so that \(\beta|b'\). We conclude that \(|AH|=k=ab'\). A similar calculation shows that \(|A'H|=ab'\) as well. Thus, \(AH/H\) and \(A'H/H\) are subgroups of \(G/H\) of order \(a/a'\). As \(|G/H|=(a/a')(b/b')\), these subgroups are conjugate, by induction, say by \(xH\in G/H\). It is quickly checked that \(xAHx^{-1}=A'H\). Therefore, \(xAx^{-1}\) and \(A'\) are subgroups of \(A'H\) of order \(a\), and so they are conjugate, by induction.

If there is some proper normal subgroup of \(G\) whose order is not divisible by \(b\), then the theorem has been proved. We may, therefore, assume that \(b\big||H|\) for every proper normal subgroup \(H\). If \(H\) is a minimal normal subgroup, however, then Theorem 5.3.14 says that \(H\) is an (elementary) abelian \(p\)-group for some prime \(p\). It may thus be assumed that \(b=p^m\), so that \(H\) is a Sylow \(p\)-subgroup of \(G\). Normality of \(H\) forces \(H\) to be unique (for all Sylow \(p\)-subgroups are conjugate). The problem has now been reduced to the following case.

Case 2. \(|G|=ap^m\), where \(p\nmid a\), and \(G\) has a normal abelian Sylow \(p\)-subgroup \(H\), which is the unique minimal normal subgroup in \(G\).

Existence. The group \(G/H\) is a solvable group of order \(a\). If \(K/H\) is a minimal normal subgroup of \(G/H\), then \(|K/H|=q^n\) for some prime \(q\neq p\), and so \(|K|=p^mq^n\); if \(Q\) is a Sylow \(q\)-subgroup of \(K\), then \(K=HQ\). Let \(N^*=\mathrm N_G(Q)\) and let \(N=N^*\cap K=\mathrm N_K(Q)\). We claim that \(|N^*|=a\).

The Fratiini argument gives \(G=KN^*\). Since \(G/K=KN^*/K\cong N^*/(N^*\cap K)=N^*/N\), we have \(|N^*|=|G||N|/|K|\). But \(K=HQ\) and \(Q\le N\le K\) gives \(K=HN\); hence \(|K|=|HN|=|H||N|/|H\cap N|\), so that

\[\begin{aligned} |N^*|=|G||N|/|K|&=|G||N||H\cap N|/|H||N|\\&=(|G|/|H|)|H\cap N|=a|H\cap N|. \end{aligned} \]

Hence \(|N^*|=a\) if \(H\cap N=\mathbf 1\). We show that \(H\cap N=\mathbf 1\) in two stages: (i) \(H\cap N=\mathrm Z(K)\); and (ii) \(\mathrm Z(K)=\mathbf 1\).

(i). Let \(x\in H\cap N\). Every \(k\in K=HQ\) has the form \(k=hs\) for \(h\in H\) and \(s\in Q\). Now \(x\) commutes with \(h\), for \(H\) is abelian, and so it suffices to show that \(x\) commutes with \(s\). But \((xsx^{-1})s^{-1}\in Q\), because \(x\) normalizes \(Q\), and \(x(sx^{-1}s^{-1})\in H\), because \(H\) is normal; therefore, \(xsx^{-1}s^{-1}\in Q\cap H=\mathbf 1\).

(ii). By Lemma 5.3.9(ii), \(\mathrm Z(K)\lhd G\). If \(\mathrm Z(K)\neq \mathbf 1\), then it contains a minimal subgroup which must be a minimal normal subgroup of \(G\). Hence \(H\le\mathrm Z(K)\), for \(H\) is the unique minimal normal subgroup of \(G\). But since \(K=HQ\), it follows that \(Q\text{ char }K\). Thus \(Q\lhd G\), by Lemma 5.3.9(ii), and so \(H\le Q\), a contradiction. Therefore, \(\mathrm Z(K)=\mathbf 1,H\cap N=\mathbf 1\), and \(|N^*|=a\).

Conjugacy. We keep the notation of the existence proof. Recall that \(N^*=\mathrm N_G(Q)\) has order \(a\); let \(A\) be another subgroup of \(G\) of order \(a\). Since \(|AK|\) is divisible by \(a\) and by \(|K|=p^mq^n\), it follows that \(|AK|=ab=|G|, AK=G\), and \(G/K=AK/K\cong A/(A\cap K)\), and \(|A\cap K|=q^n\). By the Sylow theorem, \(A\cap K\) is conjugate to \(Q\). As conjugate subgroups have conjugate normalizers, by Proposition 3.1.9(i), \(N^*=\mathrm N_G(Q)\) is conjugate to \(\mathrm N_G(A\cap K)\), and so \(a=|N^*|=|\mathrm N_G(A\cap K)|\). Since \(A\cap K\lhd A\), we have \(A\le \mathrm N_G(A\cap K)\) and so \(A=\mathrm N_G(A\cap K)\) since they both have order \(a\). Therefore, \(A\) is conjugate to \(N^*\).
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Definition 5.4.2 If \(G\) is a finite group, then a Hall subgroup \(H\) of \(G\) is a subgroup whose order and index are relatively prime; that is, \((|H|, [G:H])=1\).

Definition 5.4.3 If \(\pi\) is a set of primes, then a \(\pi\)-number is an integer \(n\) all of whose prime factors lie in \(\pi\); the complement of \(\pi\) is denoted by \(\pi'\), and so a \(\pi'\)-number is an integer \(n\) none of whose prime factors lies in \(\pi\).

Definition 5.4.4 If \(\pi\) is a set of primes, then a group is a \(\pi\)-group if the order of each of its elements is a \(\pi\)-number; a group is a \(\pi'\)-group if the order of each of its elements is a \(\pi'\)-number.

Remark It is obvious that every Sylow \(p\)-subgroup in a finite group is a Hall \(p\)-subgroup, and Hall's theorem says that Hall \(\pi\)-subgroups always exists in finite solvable groups. However, Hall \(\pi\)-subgroups with \(|\pi|\ge 2\) of a group \(G\) need not exist.

Definition 5.4.5 If \(p\) is a prime, and \(G\) is a finite group of order \(ap^n\), where \(a\) is a \(p'\)-number, then a \(p\)-complement of \(G\) is a subgroup of order \(a\).

Hall's theorem implies that a finite solvable group has a \(p\)-complement for every prime \(p\).

Theorem 5.4.6 (P.Hall) If \(G\) is a finite group having a \(p\)-complement for every prime \(p\), then \(G\) is solvable.

Proof

We proceed by induction on \(|G|\); assume, on the contrary, that there are nonsolvable groups satisfying the hypotheses, and choose one such, say, \(G\), of smallest order. If \(G\) has nontrivial normal subgroup \(N\), and if \(H\) is any Hall \(p'\)-subgroup of \(G\), then checking orders shows that \(H\cap N\) is a Hall \(p'\)-subgroup of \(N\) and \(HN/N\) is a Hall \(p'\)-subgroup of \(G/N\). Since both \(N\) and \(G/N\) have order smaller than \(|G|\), it follows that both \(N\) and \(G/N\) are solvable. But Theorem 5.3.4 now shows that \(G\) is solvable, a contradiction.

We may assume, therefore, that \(G\) is simple, Let \(|G|=p_1^{e_1}\ldots p_n^{e_n}\), where the \(p_i\) are distinct primes and \(e_i>0\) for all \(i\). For each \(i\), let \(H_i\) be a Hall \(p_i'\)-subgroup of \(G\), so that \([G:H_i]=p_i^{e_i}\), and thus \(|H_i|=\prod_{j\neq i}p_j^{e_j}\). If \(D=H_3\cap\cdots\cap H_n\), then \([G:D]=\prod_{i=3}^np_i^{e_i}\), by Proposition 3.4.12, and so \(|D|=p_1^{e_1}p_2^{e_2}\). Now \(D\) is a solvable group, by Burnside's theorem. If \(N\) is a minimal normal subgroup of \(D\), then Theorem 5.3.14 says that \(N\) is elementary abelian; for notation, assume that \(N\) is a \(p_1\)-group, then Proposition 3.4.12 shows that \([G:D\cap H_2]=\prod_{i=2}^np_i^{e_i}\), so that \(|D\cap H_2|=p_1^{e_1}\) and \(D\cap H_2\) is a Sylow \(p_1\)-subgroup of \(D\). By Proposition 4.2.10, \(N\le D\cap H_2\) and so \(N\le H_2\). But, as above, \(|D\cap H_1|=p_2^{e_2}\), and comparison of orders gives \(G=H_2(D\cap H_1)\). If \(g\in G\), then \(g=hd\), where \(h\in H_2\) and \(d\in D\cap H_1\); if \(x\in N\), then \(gxg^{-1}=hdxd^{-1}h^{-1}=hyh^{-1}\) (where \(y=dxd^{-1}\in N\), because \(N\lhd D\)) and \(hyh^{-1}\in H_2\) (because \(N\le H_2\)). Therefore, \(N^G\le H_2\), where \(N^G\) is the normal subgroup of \(G\) generated by \(N\). Since \(H_2< G\), \(N^G\neq\mathbf 1\) is a proper normal subgroup of \(G\), and this contradicts the assumption that \(G\) is simple.
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Proposition 5.4.7 If \(G\) is a finite group and \(H\) is a normal Hall subgroup, then \(H\text{ char }G\).

Definition 5.4.8 Let \(G\) be a finite group. If \(\pi\) is a set of primes, define \(\mathrm O_\pi(G)\) to be the subgroup of \(G\) generated by all the normal \(\pi\)-subgroups of \(G\).

By definition we know that \(\mathrm O_\pi(G)\) is a characteristic subgroup of \(G\).

Proposition 5.4.9 Let \(G\) be a finite group, then \(\mathrm O_\pi(G)\) is the intersection of all the maximal \(\pi\)-subgroups of \(G\).

Central Series and Nilpotent Groups

Definition 5.5.1 If \(H,K\le G\), then

\[[H,K]=\langle[h,k]:h\in H\text{ and }k\in K\rangle, \]

where \([h,k]\) is the commutator \(hkh^{-1}k^{-1}\).

It is easily observed that \([H,K]\le G\), and \(G^{(i+1)}=[G^{(i)},G^{(i)}]\).

We say that a subgroup \(K\) normalizes \(H\) if \(K\le \mathrm N_G(H)\). It is easy to see that \(K\) normalizes \(H\) if and only if \([H,K]\le H\).

We say that a subgroup \(K\) centralizes \(H\) if \(K\le \mathrm C_G(H)\). It is easy to see that \(K\) centralizes \(H\) if and only if \([H,K]=\mathbf 1\).

If \(x,y\in G\) and \([x,y]\in K\), where \(K\lhd G\), then \(x\) and \(y\) "commute mod \(K\)"; that is, \(xKyK=yKxK\) in \(G/K\).

Lemma 5.5.2

1. If \(K\lhd G\) and \(K\le H\le G\), then \([H,G]\le K\) if and only if \(H/K\le \mathrm Z(G/K)\).
2. If \(H,K\le G\) and \(f\!:G\rightarrow L\) is a homomorphism, then \(f([H,K])=[f(H),f(K)]\).

Proof

1. If \(h\in H\) and \(g\in G\), then \(hKgK=gKhK\) if and only if \([h,g]K=K\) if and only if \([h,g]\in K\).
2. Both sides are generated by all \(f([h,k])=[f(h),f(k)]\).
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Definition 5.5.3 Define characteristic subgroups \(\mathrm C_i(G)\) of \(G\) by induction:

\[\mathrm C_0(G)=G; \qquad\mathrm C_{i+1}(G)=[\mathrm C_i(G), G]. \]

It is easy to check that \(\mathrm C_1(G)=G'=G^{(1)}\) and \(\mathrm C_{i+1}(G)\le \mathrm C_i(G)\). Moreover, Lemma 5.5.2(i) shows that \([\mathrm C_i(G), G]=\mathrm C_{i+1}(G)\) gives \(\mathrm C_i(G)/\mathrm C_{i+1}(G)\le \mathrm Z(G/\mathrm C_{i+1}(G))\).

Definition 5.5.4 The lower central series (or descending central series) of \(G\) is the series

\[G=\mathrm C_0(G)\ge \mathrm C_1(G)\ge\cdots \]

(this need not be a normal series because it may not reach \(\mathbf 1\)).

Definition 5.5.5 The higher centers \(\mathrm C^i(G)\) are the characteristic subgroups of \(G\) defined by induction:

\[\mathrm C^0(G)=\mathbf 1;\qquad\mathrm C^{i+1}(G)/\mathrm C^{i}(G)=\mathrm Z(G/\mathrm C^i(G)); \]

that is, if \(\nu_i\!:G\rightarrow G/\mathrm C^i(G)\) is the natural map, then \(\mathrm C^{i+1}(G)\) is the inverse image of the center.

It is obvious that \(\mathrm C^1(G)=\mathrm Z(G)\).

Definition 5.5.6 The upper central series (or ascending central series) of \(G\) is

\[\mathbf 1=\mathrm C^0(G)\le\mathrm C^1(G)\le\mathrm C^2(G)\le\cdots. \]

Theorem 5.5.7 If \(G\) is a group, then there is an integer \(n\) with \(\mathrm C^n(G)=G\) if and only if \(\mathrm C_n(G)=\mathbf 1\). Moreover, in this case, \(\mathrm C_{i}(G)\le \mathrm C^{n-i}(G)\) for all \(i\).

Proof

Assume that \(\mathrm C^n(G)=G\), and let us prove that the inclusion holds by induction on \(i\). If \(i=0\), then \(\mathrm C_0(G)=\mathrm C^n(G)\). If \(\mathrm C_{i}(G)\le \mathrm C_{n-i}(G)\), then

\[\mathrm C_{i+1}(G)=[\mathrm C_i(G),G]\le [\mathrm C^{n-i}(G),G]\le \mathrm C^{n-i-1}(G) \]

the last inclusion follwing from Lemma 5.5.2. In particular, if \(i=n\), then \(\mathrm C_n(G)\le \mathrm C^0(G)=\mathbf 1\).

Assume that \(\mathrm C_{n}(G)=\mathbf 1\), and let us prove that \(\mathrm C_{n-j}(G)\le \mathrm C^j(G)\) by induction on \(j\). If \(j=0\), then \(\mathrm C_n(G)=\mathbf 1=\mathrm C^0(G)\). If \(\mathrm C_{n-j}(G)\le\mathrm C^j(G)\), then the third isomorphism theorem gives a surjective homomorphism \(G/\mathrm C_{n-j}(G)\rightarrow G/\mathrm C^j(G)\). Now \([\mathrm C_{n-j-1}(G),G]=\mathrm C_{n-j}(G)\), so that Lemma 5.5.2 gives \(\mathrm C_{n-j-1}(G)/\mathrm C_{n-j}(G)\le\mathrm Z(G/\mathrm C_{n-j}(G))\). So we have

\[\mathrm C_{n-j-1}(G)\mathrm C^j(G)/\mathrm C^j(G)\le\mathrm Z(G/\mathrm C^j(G))=\mathrm C^{j+1}(G)/\mathrm C^j(G). \]

Therefore, \(\mathrm C_{n-j-1}(G)\le\mathrm C_{n-j-1}(G)\mathrm C^j(G)\le\mathrm C^{j+1}(G)\), as desired. In particular, if \(j=c\), then \(G=\mathrm C_0(G)\le \mathrm C^n(G)\).
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Theorem 5.5.8 (Schur) If \(G\) is a group with \(G/\mathrm Z(G)\) finite, then \(G'\) is also finite.

Proof

Let \(g_1,\ldots,g_n\) be representatives of the cosets of \(\mathrm Z(G)\) in \(G\); that is, each \(x\in G\) has the form \(x=g_iz\) for some \(i\) and some \(z\in \mathrm Z(G)\). For all \(x,y\in G\), we have \([x,y]=[g_iz,g_jz']=[g_i,g_j]\). Hence, every commutator has the form \([g_i,g_j]\) for some \(i,j\), so that \(G'\) has a finite number \((<n^2)\) of generators.

Each element \(g'\in G'\) can be written as a word \(c_1\cdots c_t\), where each \(c_i\) is a commutator. It suffices to prove that if a factorization of \(g'\) is chosen so that \(t=t(g')\) is minimal, then \(t(g')<n^3\) for all \(g'\in G'\).

We prove first, by induction on \(r\ge 1\), that if \(a,b\in G\), then \([a,b]^r=(aba^{-1}b^{-1})^r=(ab)^r(a^{-1}b^{-1})^ru\), where \(u\) is a product of \(r-1\) commutators. This is obvious when \(r=1\). Note, for the inductive step, that if \(x,y\in G\), then \(xy=yxx^{-1}y^{-1}xy=yx[x^{-1},y^{-1}]\); that is, \(xy=yxc\) for some commutator \(c\). Thus, if \(r>1\), then

\[\begin{aligned} (aba^{-1}b^{-1})^{r+1}&=aba^{-1}b^{-1}(aba^{-1}b^{-1})^r\\ &=ab(a^{-1}b^{-1})((ab)^r(a^{-1}b^{-1})^r)u\\ &=ab((ab)^r(a^{-1}b^{-1})^r)(a^{-1}b^{-1})cu \end{aligned} \]

for some commutator \(c\), as desired.

Since \(yx=x^{-1}(xy)x\), we have \((yx)^n=x^{-1}(xy)^nx=(xy)^n\), because \([G:\mathrm Z(G)]=n\) implies \((ab)^n\in \mathrm Z(G)\). Therefore, \((a^{-1}b^{-1})^n=((ba)^{-1})^n=((ba)^n)^{-1}=((ab)^n)^{-1}\). It follows that \([a,b]^n\) is a product of \(n-1\) commutators.

Now \(xyx=(xyx^{-1})x^2\), so that two \(x\)s can be brought together at the expense of replacing \(y\) by a conjugate of \(y\). Take an expression of an element \(g'\in G'\) as a product of commutators \(c_1,\ldots,c_t\), where \(t\) is minimal. If \(t>n^3\), then there is some commutator \(c\) occuring \(m\) times, where \(m>n\) (for there are fewer than \(n^2\) distinct commutators). But all such factors can be brought together to \(c^m\) at the harmless expense of replacing commutators by conjugates (which are still commutators); that is, the number of commutator factors in the expression is unchanged. And we have proven that the length of the minimal expression for \(g'\) is shortened, and this is a contradiction. Therefore, \(t<n^3\), and so \(G'\) is finite.
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Definition 5.5.9 A group \(G\) is nilpotent if there is an integer \(n\) such that \(\mathrm C_{n}(G)=\mathbf 1\); the least such \(n\) is called the class of the nilpotent group \(G\).

Theorem 5.5.7 shows, for nilpotent groups, that the lower and upper central series are normal series of the same length.

A group is nilpotent of class \(1\) if and only if it is abelian. By Theorem 5.5.7, a nilpotent group \(G\) of class \(2\) is described by \(\mathrm C_1(G)=G'\le \mathrm Z(G)=\mathrm C^1(G)\). Every nonabelian group of order \(p^3\) is nilpotent of class \(2\), by Proposition 4.1.15.

Theorem 5.5.10 Every finite \(p\)-group is nilpotent.

Proof

By Theorem 4.1.6, every finite \(p\)-group has a nontrivial center. If, for some \(i\), we have \(\mathrm C^i(G)<G\), then \(\mathrm Z(G/\mathrm C^i(G))\neq\mathbf 1\) and so \(\mathrm C^i(G)<\mathrm C^{i+1}(G)\). Since \(G\) is finite, there must be an integer \(i\) with \(\mathrm C^i(G)=G\); that is, \(G\) is nilpotent.
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Theorem 5.5.11

1. Every nilpotent group \(G\) is solvable.
2. If \(G\neq\mathbf 1\) is nilpotent, then \(\mathrm Z(G)\neq \mathbf 1\).

Proof

1. An easy induction shows that \(G^{(i)}\le \mathrm C_{i-1}(G)\) for all \(i\). It follows that if \(\mathrm C_{n}(G)=\mathbf 1\), then \(G^{(n+1)}=\mathbf 1\); that is, if \(G\) is nilpotent (of class \(\le n\)), then \(G\) is solvable (with derived length \(\le n+1\)).
2. Assume that \(G\neq\mathbf 1\) is nilpotent of class \(c\), so that \(\mathrm C_n(G)=\mathbf 1\) and \(\mathrm C_{n-1}(G)\neq\mathbf 1\). By Theorem 5.5.7, \(\mathbf 1\neq\mathrm C_{n-1}(G)\le\mathrm C^1(G)=\mathrm Z(G)\).
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Theorem 5.5.12 Every subgroup \(H\) of a nilpotent group \(G\) is nilpotent. Moreoever, if \(G\) is nilpotent of class \(n\), then \(H\) is nilpotent of class \(\le n\).

Proof

It is easily proved by induction that \(H\le G\) implies \(\mathrm C_i(H)\le\mathrm C_i(G)\) for all \(i\). Therefore, \(\mathrm C_n(G)=\mathbf 1\) forces \(\mathrm C_n(H)=\mathbf 1\).
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Theorem 5.5.13 If \(G\) is nilpotent of class \(n\) and \(H\lhd G\), then \(G/H\) is nilpotent of class \(\le n\).

Proof

If \(f\!:G\rightarrow L\) is a surjective homomorphism, then Lemma 5.5.2 gives \(\mathrm C_i(L)\le f(\mathrm C_i(G))\) for all \(i\). Therefore, \(\mathrm C_n(G)=\mathbf 1\) forces \(\mathrm C_n(L)=\mathbf 1\). The theorem follows by taking \(f\) to be the natural map.
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Remark If \(H\lhd G\) and both \(H\) and \(G/H\) are nilpotent, then \(G\) need not be nilpotent. For example, \(\mathrm S_3\) is not nilpotent, but both \(\mathrm A_3\cong \mathbb Z_3\) and \(\mathrm S_3/\mathrm A_3\cong \mathbb Z_2\) are abelian, hence nilpotent.

Theorem 5.5.14 If \(H\) and \(K\) are nilpotent, then their direct product \(H\times K\) is nilpotent.

Proof

An easy induction shows that \(\mathrm C_i(H\times K)\le\mathrm C_i(H)\times \mathrm C_i(K)\) for all \(i\). If \(M=\max\{n,m\}\), where \(\mathrm C_{n}(H)=\mathbf 1=\mathrm C_{m}(K)\), then \(\mathrm C_{M}(H\times K)=\mathbf 1\) and \(H\times K\) is nilpotent.
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Theorem 5.5.15 If \(G\) is nilpotent, then it satisfies the normalizer condition: if \(H< G\), then \(H<\mathrm N_G(H)\).

Proof

There exists an integer \(i\) with \(\mathrm C_{i+1}(G)\le H\) and \(\mathrm C_i(G)\nleq H\). Now \([\mathrm C_i,H]\le[\mathrm C_i,G]=\mathrm C_{i+1}\le H\), so that \(\mathrm C_i\) normalizes \(H\); that is, \(\mathrm C_i\le\mathrm N_G(H)\) Therefore, \(H\) is a proper subgroup of \(\mathrm N_G(H)\).
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Proposition 5.5.16 The following conditions on a finite group \(G\) are equivalent:

1. \(G\) is nilpotent;
2. \(G\) satisfies the normalizer condition;
3. Every maximal subgroup of \(G\) is normal.
4. Every Sylow \(p\)-subgroup of some prime \(p\) is normal.

Proposition 5.5.17

1. If \(G\) is nilpotent of class \(n\), then \(G/\mathrm Z(G)\) is nilpotent of class \(n-1\).
2. If \(H\le\mathrm Z(G)\) and if \(G/H\) is nilpotent, then \(G\) is nilpotent.
3. If \(G\) is a nilpotent group and \(H\) is a minimal normal subgroup of \(G\), then \(H\le \mathrm Z(G)\).

Theorem 5.5.18 A finite group \(G\) is nilpotent if and only if it is the direct product of its Sylow subgroups.

Proof

If \(G\) is the direct product of its Sylow subgroups, then it is nilpotent, by Theorems 5.5.10 and 5.5.14.

For the converse, let \(P\) be a Sylow \(p\)-subgroup of \(G\) for some prime \(p\). By Proposition 4.2.13, \(\mathrm N_G(P)\) is equal to its own normalizer. On the other hand, if \(\mathrm N_G(P)<G\), then Theorem 5.5.15 shows that \(\mathrm N_G(P)\) is a proper subgroup of its own normalizer. Therefore, \(\mathrm N_G(P)=G\) and \(P\lhd G\), which shows that \(G\) have a unique Sylow \(p\)-subgroup for every prime \(p\big||G|\). Since the intersection of any Sylow \(p\)-subgroup and Sylow \(q\)-subgroup is always trivial, we know that \(G\) equals to the direct product of its Sylow subgroups.
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Corollary 5.5.19 Let \(G\) be a finite nilpotent group of order \(n\). If \(m|n\), then \(G\) has a subgroup of order \(m\).

Theorem 5.5.20 If \(G\) is a nilpotent group, then every maximal subgroup \(H\) is normal and has prime index.

Proof

By Theorem 5.5.15, \(H< \mathrm N_G(H)\); since \(H\) is maximal, \(\mathrm N_G(H)=G\), and so \(H\lhd G\). Proposition 2.7.4 now shows that \(G/H\) has prime order.
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Theorem 5.5.21 Let \(G\) be a nilpotent group.

1. If \(H\) is a nontrivial normal subgroup, then \(H\cap \mathrm Z(G)\neq\mathbf 1\).
2. If \(A\) is a maximal abelian normal subgroup of \(G\), then \(A=\mathrm C_G(A)\).

Proof

1. Since \(\mathrm C^0(G)=\mathbf 1\) and \(G=\mathrm C^n(G)\) for some \(n\), there is an integer \(i\) for which \(H\cap \mathrm C^i(G)\neq\mathbf 1\); let \(m\) be the minimal such \(i\). Now \([H\cap \mathrm C^m(G),G]\le H\cap [\mathrm C^{m}(G),G]\le H\cap \mathrm C^{m-1}(G)=\mathbf 1\), because \(H\lhd G\), and this says that \(\mathbf 1\neq H\cap\mathrm C^m(G)\le H\cap \mathrm Z(G)\).
2. Since \(A\) is abelian, \(A\le \mathrm C_G(A)\). For the reverse inclusion, assume that \(g\in\mathrm C_G(A)\) and \(g\notin A\). It is easy to see, for any subgroup \(H\) (of any group \(G\)) and for all \(g\in G\), that \(g\mathrm C_G(H)g^{-1}=\mathrm C_G(g^{-1}Hg)\). Since \(A\lhd G\), it follows that \(g\mathrm C_G(A)g^{-1}=\mathrm C_G(A)\) for all \(g\in G\), and so \(\mathrm C_G(A)\lhd G\). Therefore, \(\mathrm C_G(A)/A\) is a nontrivial normal subgroup of the nilpotent group \(G/A\); by (i), there is \(x\notin A\) with \(Ax\in(\mathrm C_G(A)/A)\cap \mathrm Z(G/A)\). The correspondence theorem gives \(\langle A,x\rangle\) a normal abelian subgroup of \(G\) strictly containing \(A\), and this contradicts the maximality of \(A\).
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Definition 5.5.22 A normal series

\[G=G_0\rhd G_1\rhd\cdots\rhd G_n=\mathbf 1 \]

with each \(G_i\lhd G\) and \(G_i/G_{i+1}\le \mathrm Z(G/G_{i+1})\) is called a central series.

Proposition 5.5.23

1. If \(G\) is nilpotent, then both the upper and the lower central series of \(G\) are central series.
2. A group \(G\) is nilpotent if and only if it has a central series \(G=G_0\rhd G_1\rhd \cdots\rhd G_n=\mathbf 1\). Moreover, if \(G\) is nilpotent of class \(n\), then \(\mathrm C_{i}(G)\le G_i\le \mathrm C^{n-i}(G)\) for all \(i\).

Proposition 5.5.24 If \(H\) and \(K\) are normal nilpotent subgroups of a finite group \(G\), then \(HK\) is a normal nilpotent subgroup.

Proposition 5.5.25 Let \(G\) be a finite group, then \(G\) has a unique maximal normal nilpotent subgroup \(\mathscr F(G)\), which is called the Fitting subgroup of \(G\), and we have \(\mathscr F(G)\text{ char }G\) when \(G\) is finite.

Proposition 5.5.26 Let \(G\) be a group, then \([\mathrm C_i(G), \mathrm C_j(G)]\le \mathrm C_{i+j+1}(G)\) for all \(i,j\).

Frattini Subgroup

Definition 5.6.1 If \(G\) is a group, its Frattini subgroup \(\Phi(G)\) is defined as the intersection of all the maximal subgroups of \(G\). If \(G\) is an infinite group with no maximal subgroups, we define \(\Phi(G)=G\).

It is clear that \(\Phi(G)\text{ char }G\), and so \(\Phi(G)\lhd G\).

Definition 5.6.2 An element \(x\in G\) is called a nongenerator if it can be omitted from any generating set: if \(G=\langle x,Y\rangle\), then \(G=\langle Y\rangle\).

Theorem 5.6.3 For every group \(G\), the Frattini subgroup \(\Phi(G)\) is the set of all nongenerators.

Proof

Let \(x\) be a nongenerator of \(G\), and let \(M\) be a maximal subgroup of \(G\). If \(x\notin M\), then \(G=\langle x,M\rangle=M\), a contradiction. Therefore, \(x\in M\), for all \(M\), and so \(x\in\Phi(G)\). Conversely, if \(z\in\Phi(G)\), assume that \(G=\langle z,Y\rangle\). If \(\langle Y\rangle\neq G\), then there exists a maximal subgroup \(M\) with \(\langle Y\rangle\le M\). But \(z\in M\), and so \(G=\langle z,Y\rangle\le M\), a contradiction. Therefore, \(z\) is a nongenerator.
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Theorem 5.6.4 Let \(G\) be a finite group. Then,

1. \(\Phi(G)\) is nilpotent;
2. if \(G\) is a finite \(p\)-group, then \(\Phi(G)=G'G^{p}\), where \(G^p\) is the subgroup of \(G\) generated by all \(p\)-th powers;
3. if \(G\) is a finite \(p\)-group, then \(G/\Phi(G)\) is a vector space over \(\mathbb Z_p\).
4. if \(G\) is a finite \(p\)-group, then \(G\) is cyclic if and only if \(G/\Phi(G)\) is cyclic.

Proof

1. Let \(P\) be a Sylow \(p\)-subgroup of \(\Phi(G)\) for some \(p\). Since \(\Phi(G)\lhd G\), the Frattini argument gives \(G=\Phi(G)\mathrm N_G(P)\). But \(\Phi(G)\) consists of nongenerators, and so \(G=\mathrm N_G(P)\); that is, \(P\lhd G\) and hence \(P\lhd \Phi(G)\). Therefore, by Proposition 5.5.16, \(\Phi(G)\) is nilpotent.
2. If \(M\) is a maximal subgroup of \(G\), where \(G\) is now a \(p\)-group, then Theorem 5.5.20 gives \(M\lhd G\) and \([G:M]=p\). Thus, \(G/M\) is abelian, so that \(G'\le M\); moreover, \(G/M\) has exponent \(p\), so that \(x^p\in M\) for all \(x\in G\). Therefore, \(G'G^p\le \Phi(G)\). For the reverse inclusion, observe that \(G/G'G^p\) is an abelian group of exponent \(p\), hence is elementary abelian, and hence is a vector space over \(\mathbb Z_p\). Clearly \(\Phi(G/G'G^p)=\mathbf 1\). If \(H\lhd G\) and \(H\le\Phi(G)\), then it is easy to check that \(\Phi(G)\) is the inverse image of \(\Phi(G/H)\) under the natural map (for maximal subgroups correspond). It follows that \(\Phi(G)=G'G^p\).
3. Since \(G'G^p=\Phi(G)\), the quotient group \(G/\Phi(G)\) is an abelian group of exponent \(p\); that is, it is a vector space over \(\mathbb Z_p\).
4. Let \(|G|=p^n\). If \(G\cong \mathbb Z_{p^{n}}\), then \(G\) has only one maximal subgroup and \(\Phi(G)\cong \mathbb Z_{p^{n-1}}\), which gives that \(G/\Phi(G)\cong \mathbb Z_p\). Conversely, If \(G/\Phi(G)\) is cyclic, then \(G/\Phi(G)\cong \mathbb Z_p\) since \(G/\Phi(G)\) has exponent \(p\). So we have \(|\Phi(G)|=p^{n-1}\). But any maximal subgroup \(M\) of \(G\) has order \(p^{n-1}\), which gives that \(G\) has only one maximal subgroup. By Proposition 2.3.4, \(G\) is cyclic of order \(p^n\).
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Theorem 5.6.5 For every (possible infinite) group \(G\), one has \(G'\cap\mathrm Z(G)\le\Phi(G)\).

Proof

Denote \(G'\cap\mathrm Z(G)\) by \(D\). If \(D\nleq \Phi(G)\), there is a maximal subgroup \(M\) of \(G\) with \(D\nleq M\). Therefore, \(G=MD\), so that each \(g\in G\) has a factorization \(g=md\) with \(m\in M\) and \(d\in D\). Since \(d\in\mathrm Z(G)\), \(gMg^{-1}=mdMd^{-1}m^{-1}=mMm^{-1}=M\), and so \(M\lhd G\). By Proposition 2.7.4, \(G/M\) has prime order, hence is abelian. Therefore, \(G'\le M\). But \(D\le G'\le M\), contradicting \(D\nleq M\).
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Proposition 5.6.6 (Wielandt) A finite group \(G\) is nilpotent if and only if \(G'\le\Phi(G)\).

Proposition 5.6.7

1. A finite group \(G\) is nilpotent if and only if \(G/\Phi(G)\) is nilpotent.
2. Let \(N\) be a normal subgroup of a finite group \(G\). If \(G/\Phi(N)\) is nilpotent, then \(G\) is nilpotent.
3. Let \(G\) be a finite group. If \(\mathrm Z(G)\le\Phi(G)\), and \(G/\mathrm Z(G)\) is nilpotent, then \(G\) is nilpotent.

Definition 5.6.8 A minimal generating set of a group \(G\) is a generating set \(X\) such that no proper subset of \(X\) is a generating set of \(G\).

Theorem 5.6.9 (Burnside Basis Theorem) If \(G\) is a finite \(p\)-group, then any two minimal generating sets have the same cardinality, namely, \(\dim G/\Phi(G)\). Moreover, every \(x\notin \Phi(G)\) belongs to some minimal generating set of \(G\).

Proof

If \(\{x_1,\ldots,x_n\}\) is a minimal generating set, then the family of cosets \(\{\bar x_1,\ldots,\bar x_n\}\) spans \(G/\Phi(G)\) (where \(\bar x\) denotes the coset \(x\Phi(G)\)). If this family is dependent, then one of them, say \(\bar x_1\), lies in \(\langle\bar x_2,\ldots,\bar x_n\rangle\). There is thus \(y\in\langle x_2,\ldots, x_n\rangle\le G\) with \(x_1y^{-1}\in\Phi(G)\). Clearly, \(\{x_1y^{-1},x_2,\ldots,x_n\}\) generates \(G\), so that \(G=\langle x_2,\ldots,x_n\rangle\), by Theorem 5.6.3, and this contradicts minimality. Therefore, \(n=\dim G/\Phi(G)\), and all minimal generating sets have the same cardinality.

If \(x\notin\Phi(G)\), then \(\bar x\neq\mathbf 0\) in the vector space \(G/\Phi(G)\), and so it is part of a basis \(\{\bar x,\bar x_2,\ldots,\bar x_n\}\). If \(x_i\) represents the coset \(\bar x_i\), for \(i\ge 2\), then \(G=\langle \Phi(G),x,x_2,\ldots,x_n\rangle=\langle x,x_2,\ldots,x_n\rangle\). Moreover, \(\{x,x_2,\ldots,x_n\}\) is a minimal generating set, for the cosets of a proper subset do not generate \(G/\Phi(G)\).
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