GTM148 抄书笔记 Part I. (Chapter I~III)

写在最前面的一点闲话

GTM148是我最开始想学习抽象代数时找到的；然而当时的我过于地菜，看不懂一点便丢掉了。直到最近想开始复习抽代，我又重新发现了这本书。然后发现它的内容挺有深度，而且不是我想象的那么难读懂(雾)，于是开坑。
个人认为GTM148不适合初学抽象代数的人看；相比于研究生教材，这本书更像是一本群论字典。想要入门抽象代数的可以左转知乎0003大佬的抽象代数专栏(也是我入门抽象代数的专栏)。
这个系列几乎原封不动地把GTM148上的内容搬了过来(也正所谓“抄书笔记”)。(求轻喷)

Contents

Chapter I. Groups and Homomorphisms

Permutations

Definition 1.1.1 If \(X\) is a nonempty set, a permutation of \(X\) is a bijection \(\alpha\!:X\rightarrow X\). We denote the set of all permutations of \(X\) by \(\mathrm S_X\).

In the important special case when \(X=\{1,2,\ldots,n\}\), we write \(\mathrm S_n\) instead of \(\mathrm S_X\). Note that \(|\mathrm S_n|=n!\), where \(|Y|\) denotes the number of elements in a finite set \(Y\).

Remark Given a rearrangement \(i_1,i_2,\ldots,i_n\) define a function \(\alpha\!:X\rightarrow X\) by \(\alpha(j)=i_j\) for all \(j\in X\). This function \(\alpha\) is an injection because the list has no repetitions; it is a surjection because all of the elements of \(X\) appear on the list. Thus, every rearrangement gives a bijection. Conversely, any bijection \(\alpha\) can be denoted by two rows:

\[\alpha=\begin{pmatrix}1&2&\ldots&n\\\alpha 1&\alpha 2&\ldots&\alpha n\end{pmatrix}, \]

and the bottom row is a rearrangement of \(\{1,2,\ldots,n\}\). Thus, the two versions of permutation, rearrangement and bijection, are equivalent. The advantage of the new viewpoint is that two permutations in \(\mathrm S_X\) can be "multiplied," for the composite of two bijections is again a bijection.

Cycles

Definition 1.2.1 If \(x\in X\) and \(\alpha\in\mathrm S_{X}\), then \(\alpha\) fixes \(x\) if \(\alpha(x)=x\) and \(\alpha\) moves \(x\) if \(\alpha(x)\neq x\).

Definition 1.2.2 Let \(i_1,i_2,\ldots,i_r\) be distinct integers between \(1\) and \(n\). If \(\alpha\in\mathrm S_{n}\) fixes the remaining \(n-r\) integers and if

\[\alpha(i_1)=i_2,\alpha(i_2)=i_3,\ldots,\alpha(i_{r-1})=i_r,\alpha(i_r)=i_1, \]

then \(\alpha\) is an \(r\)-cycle; one also says that \(\alpha\) is a cycle of length \(r\). Denote \(\alpha\) by \(\begin{pmatrix}i_1&i_2&\cdots&i_r\end{pmatrix}\).

Every \(1\)-cycle fixes every element of \(X\), and so all \(1\)-cycles are equal to the identity. A \(2\)-cycle, which merely interchanges a pair of elements, is called a transposition.

Definition 1.2.3 Two permutations \(\alpha,\beta\in\mathrm S_X\) are disjoint if every \(x\) moved by one is fixed by the other. In symbols, if \(\alpha(x)\neq x\), then \(\beta(x)=x\) and if \(\beta(y)\neq y\), then \(\alpha(y)=y\). A family of permutations \(\alpha_1,\alpha_2,\ldots,\alpha_m\) is disjoint if each pair of them is disjoint.

Proposition 1.2.4(the cancellation law for permutations) If either \(\alpha\beta=\alpha\gamma\) or \(\beta\alpha=\gamma\alpha\), then \(\beta=\gamma\).

Proposition 1.2.5 If \(\alpha\) and \(\beta\) are disjoint permutations, then \(\alpha\beta=\beta\alpha\); that is \(\alpha\) and \(\beta\) commute.

Definition 1.2.6 A permutation \(\alpha\in\mathrm S_n\) is regular if either \(\alpha\) has no fixed points and it is the product of disjoint cycles of the same length or \(\alpha=1\).

Proposition 1.2.7 A permutation \(\alpha\in\mathrm S_n\) is regular if and only if \(\alpha\) is a power of an \(n\)-cycle \(\beta\); that is, \(\alpha=\beta^m\) for some \(m\).

Proposition 1.2.8 If \(\alpha\) is an \(n\)-cycle, then \(a^k\) is a product of \((n,k)\) disjoint cycles, each of length \(n/(n,k)\).

Factorization into Disjoint Cycles

Theorem 1.3.1 Every permutation \(\alpha\in\mathrm S_n\) is either a cycle or a product of disjoint cycles.

Proof

The proof is by induction on the number \(k\) of points moved by \(\alpha\). The base step \(k=0\) is true, for then \(\alpha\) is the identity. If \(k>0\), let \(i_1\) be a point moved by \(\alpha\). Define \(i_2=\alpha(i_1),i_3=\alpha(i_2),\ldots,i_{r+1}=\alpha(i_r)\), where \(r\) is the smallest integer for which \(i_{r+1}\in\{i_1,i_2,\ldots,i_r\}\). We claim that \(\alpha(i_r)=i_1\). Otherwise, \(\alpha(i_r)=i_j\) for some \(j\ge 2\); but \(\alpha(i_{j-1})=i_j\), and this contradicts the hypothesis that \(\alpha\) is an injection. Let \(\sigma\) be the \(r\)-cycle \(\begin{pmatrix}i_1&i_2&\cdots&i_r\end{pmatrix}\). If \(r=n\), then \(\alpha\) is the cycle \(\sigma\). If \(r<n\) and \(Y\) consists of the remaining \(n-r\) points, then \(\alpha(Y)=Y\) and \(\sigma\) fixes the points in \(Y\). Now \(\sigma\big|_{\{i_1,i_2,\ldots,i_r\}}=\alpha\big|_{\{i_1,i_2,\ldots,i_r\}}\). If \(\alpha'\) is the permutation with \(\alpha'\big|_Y=\alpha\big|_Y\) and which fixes \(\{i_1,i_2,\ldots,i_r\}\), then \(\sigma\) and \(\alpha'\) are disjoint and \(\alpha=\sigma\alpha'\). Since \(\alpha'\) moves fewer points than does \(\alpha\), the inductive hypothesis shows that \(\alpha'\), and hence \(\alpha\), is a product of disjoint cycles.
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Definition 1.3.2 A complete factorization of a permutation \(\alpha\) is a factorization of \(\alpha\) as a product of disjoint cycles which contains one \(1\)-cycle \((i)\) for every \(i\) fixed by \(\alpha\).

Theorem 1.3.3 Let \(\alpha\in\mathrm S_n\) and let \(\alpha=\beta_1\ldots\beta_t\) be a complete factorization into disjoint cycles. This factorization is unique except for the order in which the factors occur.

Proof

Disjoint cycles commute, so that the order of the factors in a complete facatorization is not uniquely determined; however, we shall see that the factors themselves are uniquely determined. Since there is exactly one \(1\)-cycle \((i)\) for every \(i\) fixed by \(\alpha\), it suffices to prove uniqueness of the cycles of length at least \(2\). Suppose \(\alpha=\gamma_1\ldots\gamma_s\) is a second complete factorization into disjoint cycles. If \(\beta_t\) moves \(i_1\), then \(\beta_1^k(i_1)=\alpha^k(i_1)\) for all \(k\). Now some \(\gamma_j\) must move \(i_1\); since disjoint cycles commute, we may assume that \(\gamma_j=\gamma_s\). But \(\gamma_s^k(i_1)=\alpha^k(i_1)\) for all \(k\), and so we have \(\beta_t=\gamma_s\). The cancellation law gives \(\beta_1\ldots\beta_{t-1}=\gamma_1\ldots\gamma_{s-1}\), and the proof is completed by an induction on \(\max\{s,t\}\).
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Proposition 1.3.4 Let \(p\) be a prime and let \(\alpha\in\mathrm S_n\). If \(\alpha^p=1\), then either \(\alpha=1\), \(\alpha\) is a \(p\)-cycle, or \(\alpha\) is a product of disjoint \(p\)-cycles.

Even and Odd Permutations

Theorem 1.4.1 Every permutation \(\alpha\in\mathrm S_n\) is a product of transpositions.

Proof

By Theorem 1.3.1
it suffices to factor cycles, and \(\begin{aligned}\begin{pmatrix}1&2&\ldots&r\end{pmatrix}=\begin{pmatrix}1&r\end{pmatrix}\begin{pmatrix}1&r-1\end{pmatrix}\ldots\begin{pmatrix}1&2\end{pmatrix}\end{aligned}.\)
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Definition 1.4.2 A permutation \(\alpha\in \mathrm S\) is even if it is a product of an even number of transpositions; otherwise, \(\alpha\) is odd.

Lemma 1.4.3 If \(k,l\ge 0\), then

\[\begin{pmatrix}a&b\end{pmatrix}\begin{pmatrix}a&c&\ldots&c_k&b&d_1&\ldots&d_l\end{pmatrix}=\begin{pmatrix}a&c_1&\ldots&c_k\end{pmatrix}\begin{pmatrix}b&d_1&\ldots&d_l\end{pmatrix} \]

and

\[\begin{pmatrix}a&b\end{pmatrix}\begin{pmatrix}a&c_1&\ldots&c_k\end{pmatrix}\begin{pmatrix}b&d_1&\ldots&d_l\end{pmatrix}=\begin{pmatrix}a&c&\ldots&c_k&b&d_1&\ldots&d_l\end{pmatrix} \]

Definition 1.4.4 If \(\alpha\in\mathrm S_n\) and \(\alpha=\beta_1\ldots\beta_t\) is a complete factorization into disjoint cycles, then signum \(\alpha\) is defined by

\[\operatorname{sgn}(\alpha)=(-1)^{n-t}. \]

It is easy to prove that \(\operatorname{sgn}\) is well defined.

Lemma 1.4.5 If \(\beta\in\mathrm S_n\) and \(\tau\) is a transposition, then

\[\operatorname{sgn}{\tau\beta}=-\operatorname{sgn}(\beta). \]

Proof

Let \(\tau=\begin{pmatrix}a&b\end{pmatrix}\) and let \(\beta=\gamma _1\cdots\gamma _t\) be a complete factorization of \(\beta\) into disjoint cycles. If \(a\) and \(b\) occur in the same \(\gamma\), say, in \(\gamma _1\), then \(\gamma _1=\begin{pmatrix}a&c_1&\ldots&c_k&b&d_1&\cdots&d_l\end{pmatrix}\), where \(k\ge 0\) and \(l\ge 0\). By Lemma 1.4.3,

\[\tau\gamma_1=\begin{pmatrix}a&c_1&\cdots&c_k\end{pmatrix}\begin{pmatrix}b&d_1&\cdots&d_l\end{pmatrix}, \]

and so \(\tau\beta=(\tau\gamma _1)\gamma _2\ldots\gamma _t\) is a complete factorization with an extra cycle. Therefore, \(\operatorname{sgn}(\tau\beta)=(-1)^{n-(t+1)}=-\operatorname{sgn}(\beta)\).

The other possibility is that \(a\) and \(b\) occur in different cycles, say, \(\gamma _1=\begin{pmatrix}a&c_1&\ldots&c_k\end{pmatrix}\) and \(\gamma _2=\begin{pmatrix}b&d_1&\cdots&d_l\end{pmatrix}\), where \(k\ge 0\) and \(l\ge 0\). But now \(\tau\beta=(\tau\gamma _1\gamma _2)\gamma _3\ldots\gamma _t\), and Lemma 1.4.3 gives

\[\tau\gamma_1\gamma_2=\begin{pmatrix}a&c_1&\cdots&c_k&b&d_1&\cdots&d_l\end{pmatrix}. \]

Therefore, the complete factorization of \(\tau\beta\) has one fewer cycle than does \(\beta\), and so \(\operatorname{sgn}(\tau\beta)=(-1)^{n-(t-1)}=-\operatorname{sgn}(\beta)\).
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Theorem 1.4.6 For all \(\alpha,\beta\in\mathrm S_n\), \(\operatorname{sgn}(\alpha\beta)=\operatorname{sgn}(\alpha)\operatorname{sgn}(\beta)\).

Proof

Assume that \(\alpha\in\mathrm S_n\) is given and that \(\alpha=\tau_1\ldots\tau_m\) is a factorization of \(\alpha\) into transpositions with \(m\) minimal. We prove, by induction on \(m\), that \(\operatorname{sgn}(\alpha\beta)=\operatorname{sgn}(\alpha)\operatorname{sgn}(\beta)\) for every \(\beta\in\mathrm S_n\). The base step is precisely Lemma 1.4.5. If \(m>1\), then the factorization \(\tau_2\ldots\tau_m\) is also minimal: if \(\tau_2\ldots\tau_m=\sigma_1\ldots\sigma_q\) with each \(\sigma_j\) a transposition and \(q<m-1\), then the factorization \(\alpha=\tau_1\sigma_1\ldots\sigma_q\) violates the minimality of \(m\). Therefore, we have

\[\begin{aligned}\operatorname{sgn}&=\operatorname{sgn}(\tau_1\cdots\tau_m\beta)=-\operatorname{sgn}(\tau_2\cdots\tau_m\beta)\qquad&&(\textit{Lemma 1.4.5})\\ &=-\operatorname{sgn}(\tau_2\cdots\tau_m)\operatorname{sgn}(\beta)&&(\textit{by induction})\\ &=\operatorname{sgn}(\tau_1\cdots\tau_m)\operatorname{sgn}(\beta)&&(\textit{Lemma 1.4.5})\\ &=\operatorname{sgn}(\alpha)\operatorname{sgn}(\beta).\end{aligned} \]

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Theorem 1.4.7

  1. A permutation \(\alpha\in\mathrm S_n\) is even if and only if \(\operatorname{sgn}(\alpha)=1\);
  2. A permutation \(\alpha\) is odd if and only if it is a product of an odd number of transpositions.

Proposition 1.4.8 \(\mathrm S_n\) has the same number of even permutations as of odd permutations.

Semigroups

Definition 1.5.1 A (binary) operation on a nonempty set \(G\) is a function \(\mu\!: G\times G\rightarrow G\).

Definition 1.5.2 An operation \(*\) on a set \(G\) is associative if

\[(a*b)*c=a*(b*c) \]

for every \(a,b,c\in G\).

Definition 1.5.3 An expression \(a_1*a_2*\cdots*a_n\) needs no parentheses if, no matter what choices of multiplications of adjacent factors are made, the resulting elements of \(G\) are all equal.

Theorem 1.5.4(Generalized Associativity) If \(*\) is an associative operation on a set \(G\), then every expression \(a_1*a_2*\cdots*a_n\) needs no parentheses.

Definition 1.5.5 A semigroup \((G,*)\) is a nonempty set \(G\) equipped with an associative operation \(*\).

Definition 1.5.6 Let \(G\) be a semigroup and let \(\alpha\in G\). Define \(a^1=a\) and, for \(n\ge 1\), define \(a^{n+1}=a*a^n\).

Corollary 1.5.7 Let \(G\) be a semigroup, let \(a\in G\), and let \(m\) and \(n\) be positive integers. Then \(a^m*a^n=a^{m+n}=a^n*a^m\) and \((a^m)^n=a^{mn}=(a^n)^m\).

Groups

Definition 1.6.1 A group is a semigroup \(G\) containing an element \(e\) such that:

  1. \(e*a=a=a*e\) for all \(a\in G\);
  2. for every \(a\in G\), there is an element \(b\in G\) with

\[a*b=e=b*a. \]

It is easy to show that \(\mathrm S_X\) is a group with composition as operation; it is called the symmetric group on \(X\). When \(X=\{1,2,\ldots,n\}\), then \(\mathrm S_X\) is denoted by \(\mathrm S_n\) and is called the symmetric group on \(n\) letters.

Definition 1.6.2 A pair of elements \(a\) and \(b\) in a semigroup commutes if \(a*b=b*a\). A group (or a semigroup) is abelian if every pair of its elements commutes.

Examples 1.6.3

  1. The set \(\mathbb Z\) of all integers is an abelian group with ordinary addition as operation: \(a*b=a+b;e=0;-a+a=0\). Some other additive abelian groups are the rational numbers \(\mathbb Q\), the real numbers \(\mathbb R\), and the complex numbers \(\mathbb C\).
  2. Recall that if \(n\ge 2\) and \(a\) and \(b\) are integers, then \(a\equiv b\pmod n\) means that \(n\) is a divisor of \(a-b\). Denote the congruence class of an integer \(a\text{ mod }n\) by \([a]\); that is,
    \([a]=\{b\in\mathbb Z:b\equiv a\pmod n\}\). The set \(\mathbb Z_n\) of all the congruence classes \(\text{mod }n\) is called the integers modulo \(n\); it is an abelian group when equipped with the operation: \([a]+[b]=[a+b]\); here \(e=[0]\) and \([-a]+[a]=[0]\). It is easy to prove that these operations are well defined.
  3. If \(k\) is a field, then the set of all \(n\times n\) nonsingular matrices with entries in \(k\) is a group, denoted by \(\mathrm{GL}(n,k)\), called the general linear group; here the operation is matrix multiplication, \(e\) is the identity matrix \(E\), and if \(A^{-1}\) is the inverse of the matrix \(A\), then \(AA^{-1}=E=A^{-1}A\). If \(n\ge 2\), then \(\mathrm{GL}(n,k)\) is not abelian; if \(n=1\), then \(\mathrm{GL}(1, k)\) is abelian: it is the multiplicative group \(k^{\times }\) of all the nonzero elements in \(k\).

Theorem 1.6.4 If \(G\) is a group, there is a unique element \(e\) with \(e*a=a=a*e\) for all \(a\in G\). Moreover, for each \(a\in G\), there is a unique \(b\in G\) with \(a*b=e=b*a\).

Proof

Suppose that \(e'*a=a=a*e'\) for all \(a\in G\). In particular, if \(a=e\), then \(e'*e=e\). On the other hand, the defining property of \(e\) gives \(e'*e=e'\), and so \(e'=e\).
Suppose that \(a*c=e=c*a\), then \(c=c*e=c*(a*b)=(c*a)*b=e*b=b\), as desired.
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As a result of the uniqueness assertions of the theorem, we may now give names to \(e\) and to \(b\). We call \(e\) the identity of \(G\) and, if \(a*b=e=b*a\), then we call \(b\) the inverse of \(a\) and denote it by \(a^{-1}\).

Corollary 1.6.5 If \(G\) is a group and \(a\in G\), then \((a^{-1})^{-1}=a\).

Definition 1.6.6 If \(G\) is a group and \(a\in G\), define the powers of \(a\) as follows: if \(n\) is a positive integer, then \(a^n\) is defined as in any semigroup; define \(a^0=e\); define \(a^{-n}=(a^{-1})^n\).

Theorem 1.6.7 If \(G\) is a semigroup with an element \(e\) such that:

  1. \(e*a=a\) for all \(a\in G\);
  2. for each \(a\in G\) there is an element \(b\in G\) with \(b*a=e\),

then \(G\) is a group.

Proof

We claim that if \(x*x=x\) in \(G\), then \(x=e\). There is an element \(y\in G\) with \(y*x=e\), and \(y*(x*x)=y*x=e\). On the other hand, \(y*(x*x)=(y*x)*x=e*x=x\). Therefore, \(x=e\).

If \(b*a=e\), let us show that \(a*b=e\). Now \((a*b)*(a*b)=a*[(b*a)*b]=a*[e*b]=a*b\), and so our claim gives \(a*b=e\).

If \(a\in G\), we must show that \(a*e=a\). Choose \(b\in G\) with \(b*a=e=a*b\), then we have \(a*e=a*(b*a)=(a*b)*a=e*a=a\), as desired.
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Proposition 1.6.8 Let \(G\) be a group, let \(a\in G\), and let \(m\) and \(n\) be relatively prime integers. If \(a^m=e\), then there exists \(b\in G\) with \(a=b^n\).

Proposition 1.6.9 Let \(G\) be a group. For each \(a\in G\), the functions \(L_a\!:G\rightarrow G\), defined by \(x\mapsto a*x\) (called left translation by \(a\)), and \(R_a\!:G\rightarrow G\), defined by \(x\mapsto x*a^{-1}\) (called right translation by \(a\)), are bijections.

Proposition 1.6.10 Let \(G\) be a group. For all \(a,b\in G\), we have \(L_{a*b}=L_a\circ L_b\), \(R_{a*b}=R_a\circ R_b\), and \(L_a\circ R_b=R_b\circ L_a\).

Homomorphisms

Definition 1.7.1
Let \((G,*)\) and \((H,\circ)\) be groups. A function \(f\!:G\rightarrow H\) is a homomorphism if, for all \(a, b\in G\), we have \(f(a*b)=f(a)\circ f(b)\).

An isomorphism is a homomorphism that is also a bijection. We say that \(G\) is isomorphic to \(H\), denoted by \(G\cong H\), if there exists an isomorphism \(f\!:G\rightarrow H\).

Theorem 1.7.2 Let \(f\!:(G,*)\rightarrow(G',\circ)\) be a homomorphism.

  1. \(f(e)=e'\), where \(e'\) is the identity in \(G'\).
  2. If \(a\in G\), then \(f(a^{-1})=f(a)^{-1}\).
  3. If \(a\in G\) and \(n\in\mathbb Z\), then \(f(a^n)=f(a)^n\).
Proof
  1. Applying \(f\) to the equation \(e=e*e\) gives \(f(e)=f(e*e)=f(e)\circ f(e)\). Now multiply each side of the equation by \(f(e)^{-1}\) to obtain \(e'=f(e)\).
  2. Applying \(f\) to the equations \(a*a^{-1}=e=a^{-1}*a\) gives \(f(a)\circ f(a^{-1})=e'=f(a^{-1})\circ f(a)\). It follows from Theorem 1.6.4, the uniqueness of the inverse, that \(f(a^{-1})=f(a)^{-1}\).
  3. An easy induction proves \(f(a^n)=f(a)^n\) for all \(n\ge 0\), and then \(f(a^{-n})=f((a^{-1})^n)=f(a^{-1})^n=f(a)^{-n}\).
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Proposition 1.7.3 Let \(f\!:X\rightarrow Y\) be a bijection between sets \(X\) and \(Y\), then \(\alpha\mapsto f\circ \alpha\circ f^{-1}\) is an isomorphism \(\mathrm S_X\rightarrow \mathrm S_Y\).

Proposition 1.7.4

  1. If \(f\!:G\rightarrow H\) and \(g\!:H\rightarrow K\) are homomorphisms, then so is the composite \(g\circ f: G\rightarrow K\).
  2. If \(f\!:G\rightarrow H\) is an isomorphism, then its inverse \(f^{-1}\!:H\rightarrow G\) is also an isomorphism.

Proposition 1.7.5 Let \(a\) be a fixed element of a group \(G\), define the conjugation by \(a\) as \(\gamma_a\!:G\rightarrow G\) by \(\gamma_a(x)=a*x*a^{-1}\). Then

  1. \(\gamma_a\) is an isomorphism.
  2. If \(a,b\in G\), then \(\gamma_a\gamma_b=\gamma_{a*b}\).

Proposition 1.7.6

  1. A group \(G\) is abelian if and only if the function \(f\!:G\rightarrow G\), defined by \(f(a)=a^{-1}\), is a homomorphism.
  2. Let \(f\!:G\rightarrow G\) be an isomorphism from a finite group \(G\) to itself. If \(f\) has no nontrivial fixed points and if \(f\circ f\) is the identity function, then \(f(x)=x^{-1}\) for all \(x\in G\) and \(G\) is abelian.

Chapter II. The Isomorphism Theorems

Subgroups

Notation We now drop the \(*\) notation for the operation in a group. Henceforth, we shall write \(ab\) instead of \(a*b\), and we shall denote the identity element by \(\mathbf 1\) instead of by \(e\).

Definition 2.1.1 A nonempty subset \(S\) of a group \(G\) is a subgroup of \(G\) if \(s\in S\) implies \(s^{-1}\in S\) and \(s,t\in S\) imply \(st \in S\).

If \(X\) is a subset of group \(G\), we write \(X\subset G\); if \(X\) is a subgroup of \(G\), we write \(X\le G\).

Theorem 2.1.2 If \(S\le G\), then \(S\) is a group in its own right.

Theorem 2.1.3 A subset \(S\) of a group \(G\) is a subgroup if and only if \(1\in S\) and \(s,t\in S\) imply \(st^{-1}\in S\).

Definition 2.1.4 If \(G\) is a group and \(a\in G\), then the cyclic subgroup generated by \(a\), denoted by \(\langle a\rangle\), is the set of all the powers of \(a\). A group \(G\) is called cyclic if there is \(a\in G\) with \(G=\langle a\rangle\); that is, \(G\) consists of all the powers of \(a\).

Definition 2.1.5 If \(G\) is a group and \(a\in G\), then the order of \(a\) is \(|\langle a\rangle|\), the number of elements in \(\langle a\rangle\).

Theorem 2.1.6 If \(G\) is a group and \(a\in G\) has finite order \(m\), then \(m\) is the smallest positive integer such that \(a^m=\mathbf 1\).

Proof

If \(a=\mathbf 1\), then \(m=1\). If \(a\neq \mathbf 1\), there is an integer \(k>1\) so that \(\mathbf 1,a,a^2,\ldots,a^{k-1}\) are distinct elements of \(G\) while \(a^k=a^i\) for some \(i\) with \(0\le i\le k-1\). We claim that \(a^k=\mathbf 1=a^0\). If \(a^k=a^i\) for some \(i\ge 1\), then \(k-i\le k-1\) and \(a^{k-i}=\mathbf 1\), contradicting the original list \(\mathbf 1, a, a^2, \ldots, a^{k-1}\) having no repetitions. It follows that \(k\) is the smallest positive integer with \(a^k=\mathbf 1\).

It now suffices to prove that \(k=m\); that is, that \(\langle a\rangle=\{\mathbf 1,a,a^2,\ldots,a^{k-1}\}\). Clearly \(\langle a\rangle\supset\{\mathbf 1,a,a^2,\ldots,a^{k-1}\}\). For the reverse inclusion, let \(a^l\) be a power of \(a\). By the division algorithm, \(l=qk+r\), where \(0\le r<k\). Hence, \(a^l=a^{qk+r}=a^{qk}a^{r}=a^{r}\), and so \(a^{l}=a^{r}\in\{\mathbf 1,a,a^2,\ldots,a^{k-1}\}\).
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Corollary 2.1.7 If \(G\) is a finite group, then a nonempty subset \(S\) of \(G\) is a subgroup if and only if \(s,t\in S\) imply \(st\in S\).

Proof

Necessity is obvious. For sufficiency, we must show that \(s\in S\) implies \(s^{-1}\in S\). It follows easily by induction that \(S\) contains all the powers of \(s\). Since \(G\) is finite, \(s\) has finite order, say, \(m\). Therefore, \(\mathbf 1=s^m\in S\) and \(s^{-1}=s^{m-1}\in S\).
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Example 2.1.8 If \(G\) is a group, then \(G\) itself and \(\{\mathbf 1\}\) are always subgroups. Any subgroup \(H\) other than \(G\) is called proper, and we denote this by \(H<G\); the subgroup \(\mathbf 1\) is often called the trivial group.

Definition 2.1.9 Let \(f\!:G\rightarrow H\) be a homomorphism, and define
the kernel of \(f\) as \(\mathrm {ker}f=\{a\in G:f(a)=\mathbf 1\}\) and the image of \(f\) as \(\mathrm {im} f=\{h\in H:h=f(a)\text{ for some }a\in G\}\).

Proposition 2.1.10 Let \(f\!:G\rightarrow H\) be a homomorphism, then \(K=\mathrm{ker}f\) is a subgroup of \(G\) and \(\mathrm{im}f\) is subgroup of \(H\).

Theorem 2.1.11 The intersection of any family of subgroups of a group \(G\) is again a subgroup of \(G\).

Proof

Let \(\{S_i:i\in I\}\) be a family of subgroups of \(G\). Now \(\mathbf 1\in S_i\) for every \(i\), and so \(\mathbf 1\in\bigcap S_i\). If \(a,b\in\bigcap S_i\), then \(a,b\in S_i\) for every \(i\), and so \(ab^{-1}\in S_i\) for every \(i\); hence, \(ab^{-1}\in\bigcap S_i\), and \(\bigcap S_i\le G\).
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Corollary 2.1.12 If \(X\) is a subset of a group \(G\), then there is a smallest subgroup \(H\) of \(G\) containing \(X\); that is, if \(X\subset S\) and \(S\le G\), then \(H\le S\).

Definition 2.1.13 If \(X\) is a subset of a group \(G\), then the smallest subgroup of \(G\) containing \(X\), denoted by \(\langle X\rangle\), is called the subgroup generated by \(X\). One also says that \(X\) generates \(\langle X\rangle\).

In particular, if \(H\) and \(K\) are subgroups of \(G\), then the subgroup \(\langle H\cup K\rangle\) is denoted by \(H\lor K\).

If \(X\) consists of a single element \(a\), then \(\langle X\rangle=\langle a\rangle\), the cyclic subgroup generated by \(a\). If \(X\) is a finite set, say, \(X=\{a_1,a_2,\ldots,a_n\}\) then we write \(\langle X\rangle=\langle a_1,a_2,\ldots,a_n\rangle\) instead of \(\langle X\rangle=\langle\{ a_1,a_2,\ldots,a_n\}\rangle\).

Definition 2.1.14 If \(X\) is a nonempty subset of a group \(G\), then a word on \(X\) is an element \(w\in G\) of the form

\[w=x_1^{e_1}x_2^{e_2}\ldots x_n^{e_n}, \]

where \(x_i\in X,e_i=\pm 1\), and \(n\ge 1\).

Theorem 2.1.15 Let \(X\) be a subset of a group \(G\). If \(X\neq\varnothing\), then \(\langle X\rangle=\mathbf 1\); if \(X\) is nonempty, then \(\langle X\rangle\) is the set of all the words on \(X\).

Proof

If \(X=\varnothing\), then the subgroup \(\mathbf 1=\{\mathbf 1\}\) contains \(X\), and so \(\langle X\rangle=\mathbf{1}\). If \(X\) is nonempty, let \(W\) denote the set of all the words on \(X\). It is easy to see that \(W\) is a subgroup of \(G\) containing \(X\) : \(\mathbf 1=x_1^{-1}x_1\in W\); the inverse of a word is a word; the product of two words is a word. Since \(\langle X\rangle\) is the smallest subgroup containing \(X\), we have \(\langle X\rangle\subset W\). The reverse inclusion also holds, for every subgroup \(H\) containing \(X\) must contain every word on \(X\). Therefore, \(W\le H\), and \(W\) is the smallest subgroup containing \(X\).
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Proposition 2.1.16 If \(k\) is a field, then the set of all \(n\times n\) matrices over \(k\) having determinant \(1\), denoted by \(\mathrm SL(n,k)\), is a subgroup of \(\mathrm GL(n,k)\). We call \(\mathrm SL(n,k)\) the special linear group over \(k\).

Proposition 2.1.17 Let \(\mathrm A_n\) be the set of all even permutations in \(\mathrm S_n\) (and is also called the alternating group on \(n\) letters), then \(\mathrm A_n\) is a subgroup with \(n!/2\) elements.

Proposition 2.1.18 Suppose that \(X\) is a nonempty subset of a set \(Y\), then \(\mathrm S_X\) can be imbedded in \(\mathrm S_Y\); that is, \(\mathrm S_X\) is isomorphic to a subgroup of \(\mathrm S_Y\).

Proposition 2.1.19 For any \(n\ge 2\), \(\mathrm S_n\) can be imbedded in \(\mathrm A_{n+2}\), but not in \(\mathrm A_{n+1}\).

Proposition 2.1.20

  1. For any \(n\ge 2\), \(\mathrm S_n\) can be generated by \((1\ 2),(1\ 3),\ldots,(1\ n)\).
  2. For any \(n\ge 2\), \(\mathrm S_n\) can be generated by \((1\ 2),(2\ 3),\ldots,(i\ i+1),\ldots,(n-1\ n)\).
  3. For any \(n\ge 2\), \(\mathrm S_n\) can be generated by the two elements \((1\ 2)\) and \((1\ 2\ \ldots\ n)\).
  4. For any \(n> 2\), \(\mathrm A_n\) can be generated by all the \(3\)-cycles.

Langrange's Theorem

Definition 2.2.1 If \(S\) is a subgroup of \(G\) and if \(t\in G\), then a right coset of \(S\) in \(G\) is the subset of \(G\)

\[St=\{st:s\in S\} \]

(a left coset is \(tS=\{ts:s\in S\}\)). One calls \(t\) a representative of \(St\) (and also of \(tS\)).

Example 2.2.2 If \(G\) is the additive group \(\mathbb Z\) of all integers, if \(S\) is the set of all multiples of an integer \(n\) (\(S=\langle n\rangle\), the cyclic subgroup generated by \(n\)), and if \(a\in\mathbb Z\), then the coset \(a+S=\{a+qn:q\in\mathbb Z\}=\{k\in\mathbb Z:k\equiv a\pmod n\}\); that is, the coset \(a+\langle n\rangle\) is precisely the congruence class \([a]\) of \(a\!\!\mod n\).

Lemma 2.2.3 If \(S\le G\), then \(Sa=Sb\) if and only if \(ab^{-1}\in S\) (\(aS=bS\) if and only if \(b^{-1}a\in S\)).

Proof

If \(Sa=Sb\), then \(a=\mathbf 1a\in Sa=Sb\), and so there is \(s\in S\) with \(a=sb\); hence, \(ab^{-1}=s\in S\). Conversely, assume that \(ab^{-1}=\sigma\in S\); hence \(a=\sigma b\). To prove that \(Sa=Sb\), we prove two inclusions. If \(x\in Sa\), then \(x=sa\) for some \(s\in S\), and so \(x=s\sigma b\in Sb\); similarly, if \(y\in Sb\), then \(y=s'b\) for some \(s'\in S\), and \(y=s'\sigma^{-1}a\in Sa\). Therefore, \(Sa=Sb\).
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Theorem 2.2.4 If \(S\le G\), then any two right (or any two left) cosets of \(S\) in \(G\) are either identical or disjoint.

Proof

We show that if there exists an element \(x\in Sa\cap Sb\), then \(Sa=Sb\). Such an \(x\) has the form \(sb=x=ta\), where \(s,t\in S\). Hence \(ab^{-1}=t^{-1}s\in S\), and so Lemma 2.2.3 gives \(Sa=Sb\).
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Theorem 2.2.4 may be paraphrased to say that the right cosets of a subgroup \(S\) comprise a partition of \(G\) (each such coset is nonempty, and \(G\) is their disjoint union). This being true, their must be an equivalence relation on \(G\) lurking somewhere in the background: it is given, for \(a,b\in G\), by \(a\equiv b\) if \(ab^{-1}\in S\), and its equivalence classes are the right cosets of \(S\).

Theorem 2.2.5 If \(S\le G\), then the number of right cosets of \(S\) in \(G\) is equal to the number of left cosets of \(S\) in \(G\).

Proof

We give a bijection \(f\!:\mathscr R\rightarrow\mathscr L\), where \(\mathscr R\) is the family of right cosets of \(S\) in \(G\) and \(\mathscr L\) is the family of left cosets. If \(Sa\in\mathscr R\), define \(f(Sa)=a^{-1}S\), and it can be verified that \(f\) is well defined; that is, if \(Sa=Sb\), then \(a^{-1}S=b^{-1}S\).
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Definition 2.2.6 If \(S\le G\), then the index of \(S\) in \(G\), denoted by \([G:S]\), is the number of right cosets of \(S\) in \(G\).

Proposition 2.2.7 In a finite group \(G\), one can always choose a common system of representatives for the right and left cosets of a subgroup \(S\); if \([G:S]=n\), there exists elements \(t_1,\ldots,t_n\in G\) so that \(t_1S,\ldots,t_nS\) is the family of all left cosets and \(St_1,\ldots,St_n\) is the family of all right cosets.

Definition 2.2.8 If \(G\) is a group, then the order of \(G\), denoted by \(|G|\), is the number of elements in \(G\).

Theorem 2.2.9 (Langrange) If \(G\) is a finite group and \(S\le G\), then \(|S|\) divides \(|G|\) and \([G:S]=|G|/|S|\).

Proof

By Theorem 2.2.4, \(G\) is partitioned into its right cosets

\[G=St_1\cup St_2\cup \ldots \cup St_n, \]

and so \(|G|=\sum_{i=1}^n|St_i|\). But it is easy to see that \(f_i\!:S\rightarrow St_i\), defined by \(f_i(s)=st_i\), is a bijection, and so \(|St_i|=|S|\) for all \(i\). Thus, \(|G|=n|S|\), where \(n=[G:S]\).
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Corollary 2.2.10 If \(G\) is a finite group and \(a\in G\), then the order of \(a\) divides \(|G|\).

Definition 2.2.11 A group \(G\) has exponent \(n\) if \(x^n=\mathbf 1\) for all \(x\in G\).

Corollary 2.2.12 If \(p\) is a prime and \(|G|=p\), then \(G\) is a cyclic group.

Corollary 2.2.13 (Fermat) If \(p\) is a prime and \(a\) is an integer, then \(a^p\equiv a\pmod p\).

Proof

Let \(G=\mathrm U(\mathbb Z_p)\), the multiplicative group of nonzero elements of \(\mathbb Z_p\); since \(p\) is prime, \(\mathbb Z_p\) is a field and \(G\) is a group of order \(p-1\).

Recall that for integers \(a\) and \(b\), one has \(a\equiv b\pmod p\) if and only if \([a]=[b]\) in \(\mathbb Z_p\). If \(a\in \mathbb Z\) and \([a]=[0]\) in \(\mathbb Z_p\), then it is clear that \([a]^p=[0]=[a]\). If \([a]\neq [0]\), then \([a]\in G\) and so \([a]^{p-1}=[1]\), by Corollary 2.2.10; multiplying by \([a]\) now gives the desired result.
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Proposition 2.2.14 Let \(a\in G\) have order \(n=mk\), where \(m,k\ge 1\), then \(a^k\) has order \(m\).

Proposition 2.2.15 If \(a\in G\) has order \(n\) and \(k\) is an integer with \(a^k=\mathbf 1\), then \(n\) divides \(k\). Indeed, \(\{k\in\mathbb Z:a^{k}=\mathbf 1\}\) consists of all the multiplies of \(n\).

Proposition 2.2.16 If \(G\) is a finite group and \(K\le H\le G\), then \([G:K]=[G:H][H:K]\).

Proposition 2.2.17 If \(a\in G\) has finite order and \(f\!:G\rightarrow H\) is a homomorphism, then the order of \(f(a)\) divides the order of \(a\).

Proposition 2.2.18 Every subgroup of a cyclic group is cyclic.

Proposition 2.2.19 Two cyclic groups are isomorphic if and only if they have the same order.

Proposition 2.2.20 If \(G=\langle a\rangle\) is cyclic of order \(n\), then \(a^k\) is also a generator of \(G\) if and only if \((k,n)=1\).

Cyclic Groups

Lemma 2.3.1 If \(G\) is a cyclic group of order \(n\), then there exists a unique subgroup of order \(d\) for every divisor \(d\) of \(n\).

Proof

If \(G=\langle a\rangle\), then \(\langle a^{n/d}\rangle\) is a subgroup of order \(d\), by Proposition 2.2.14. Assume that \(S=\langle b\rangle\) is a subgroup of order \(d\). Now \(b^d=\mathbf 1\); moreover, \(b=a^m\) for some \(m\), by Proposition 2.2.15. Then we have \(md=nk\) for some integer \(k\), and \(b=a^m=(a^{n/d})^k\). Therefore, \(\langle b\rangle\le\langle a^{n/d}\rangle\), and this inclusion is equality because both subgroups have order \(d\).
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Theorem 2.3.2 A group \(G\) of order \(n\) is cyclic if and only if, for each divisor \(d\) of \(n\), there is at most one cyclic subgroup \(G\) having order \(d\).

Proof

If \(G\) is cyclic, then the result is Lemma 2.3.1. For the converse, recall from the previous proof that \(G\) is the disjoint union \(\bigcup \mathrm {gen}(C)\), where \(C\) ranges over all the cyclic subgroups of \(G\) and \(\mathrm {gen}(C)\) denote the set of all the generators of \(C\). Hence, \(n=|G|=\sum|\mathrm{gen}(C)|\le\sum_{d|n}\varphi(d)=n\). We conclude that \(G\) must have a cyclic subgroup of order \(d\) for every divisor \(d\) of \(n\); in particular, \(G\) has a cyclic subgroup of order \(d=n\), and so \(G\) is cyclic.
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Theorem 2.3.3 Let \(p\) be a prime. A group \(G\) of order \(p^n\) is cyclic if and only if it is an abelian group having a unique subgroup of order \(p\).

Proof

Necessity follows at once from Lemma 2.3.1. For the converse, let \(a\in G\) have largest order, say \(p^k\) (it follows that \(g^{p^k}=\mathbf 1\) for all \(g\in G\)). Of course, the unique subgroup \(H\) of order \(p\) is a subgroup of \(\langle a\rangle\). If \(\langle a\rangle\) is a proper subgroup of \(G\), then there is \(x\in G\) with \(x\notin \langle a\rangle\) but with \(x^p\in\langle a\rangle\); let \(x^p=a^l\). If \(k=1\), then \(x^p=\mathbf 1\) and \(x\in H\le\langle a\rangle\), a contradiction; we may, therefore, assume that \(k>1\). Now

\[\mathbf 1=x^{p^{k}}=(x^p)^{p^{k-1}}=a^{lp^{k-1}}, \]

so that \(l=pm\) for some integer \(m\), by Proposition 2.2.15. Hence, \(x^p=a^{mp}\), and so \(\mathbf 1=x^{-p}a^{mp}\). Since \(G\) is abelian, \(x^{-p}a^{mp}=(x^{-1}a^m)^p\), and so \(x^{-1}a^m\in H\le\langle a\rangle\). This gives \(x\in\langle a\rangle\), a contradiction. Therefore, \(G=\langle a\rangle\) and hence is cyclic.
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Proposition 2.3.4 A subgroup \(H< G\) is a maximal subgroup of \(G\) if there is no subgroup \(N\) of \(G\) with \(H< N< G\). If \(G\) is a finite group with only one maximal subgroup, then \(G\) is cyclic.

Normal Subgroups

Definition 2.4.1 If \(S\) and \(T\) are nonempty subsets of a group \(G\), then

\[ST=\{st:s\in S\text{ and }t\in T\}. \]

If \(S\le G,t\in G\), and \(T=\{t\}\), then \(ST\) is the right coset \(St\). Notice that the family of all the nonempty subsets of \(G\) is a semigroup under this operation: if \(S,T\) and \(U\) are nonempty subsets of \(G\), then \((ST)U=S(TU)\), for either side consists of all the elements of \(G\) of the form \((st)u=s(tu)\) with \(s\in S,t\in T\), and \(u\in U\).

Theorem 2.4.2 If \(S\) and \(T\) are subgroups of a finite group \(G\), then \(|ST||S\cap T|=|S||T|\).

Proof

Define a function \(\varphi\!:S\times T\rightarrow ST\) by \((s,t)\mapsto st\). Since \(\varphi\) is a surjection, it suffices to show that if \(x\in ST\), then \(|\varphi^{-1}(x)|=|S\cap T|\). We show that \(\varphi^{-1}(x)=\{(sd,d^{-1}t)|d\in S\cap T\}\). It is clear that \(\varphi^{-1}(x)\) contains the right side. For the reverse inclusion, let \((s,t),(\sigma,\tau)\in\varphi^{-1}(x)\); that is, \(s,\sigma\in S\), \(t,\tau\in T\), and \(st=x=\sigma\tau\). Thus, \(s^{-1}\sigma=t\tau^{-1}\in S\cap T\); let \(d=s^{-1}\sigma=t\tau^{-1}\) denote their common value. Then \(\sigma=s(s^{-1}\sigma)=sd\) and \(d^{-1}t=\tau t^{-1}t=\tau\), as desired.
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Definition 2.4.3 A subgroup \(K\le G\) is a normal subgroup, denoted by \(K\lhd G\), if \(gKg^{-1}=K\) for every \(g\in G\).

If \(K\le G\) and there are inclusions \(gKg^{-1}\le K\) for every \(g\in G\), then \(K\lhd G\): replacing \(g\) by \(g^{-1}\), we have the inclusion \(g^{-1}Kg\le K\), and this gives the reverse inclusion \(K\le gKg^{-1}\).

Proposition 2.4.4 If \(K\le H\le G\) and \(K\lhd G\), then \(K\lhd H\).

Proposition 2.4.5 Let \(G\) be a group, then the intersection of any family of normal subgroups of \(G\) is itself a normal subgroup of \(G\).

Proposition 2.4.6 Let \(f\!:G\rightarrow H\) be a homomorphism, then the kernel \(\mathrm{ker} f\) is a normal subgroup of \(G\).

Definition 2.4.7 If \(x\in G\), then a conjugate of \(x\) in \(G\) is an element of the form \(axa^{-1}\) for some \(a\in G\); equivalently, \(x\) and \(y\) are conjugate if \(y=\gamma_a(x)=axa^{-1}\) for some \(a\in G\).

Proposition 2.4.8 If \(S\le G\), then \(S\lhd G\) if and only if \(\gamma(S)\le S\) for every conjugation \(\gamma\).

Proposition 2.4.9 If \(S\) is a finite nonempty subset of \(G\) with \(SS=S\), then \(S\) is a subgroup; however, when \(S\) is infinite, the statement may be false.

Definition 2.4.10 If \(S\) and \(T\) are (not necessarily distinct) subgroups of \(G\), then an (\(S\)-\(T\))-double coset is a subset of \(G\) of the form \(SgT\), where \(g\in G\).

Proposition 2.4.11 Let \(S\) and \(T\) be subgroups of \(G\), then the family of all (\(S\)-\(T\))-double coset partitions \(G\).

Proposition 2.4.12 Let \(S\) and \(T\) be subgroups of a finite group \(G\), and suppose that \(G\) is the disjoint union

\[G=\bigcup_{i=1}^nSg_iT, \]

then we have \([G:T]=\sum_{i=1}^n[S:S\cap g_iTg_i^{-1}]\).

Proposition 2.4.13 Let \(G\) be a finite group, and let \(S\) and \(T\) be (not necessarily distinct) nonempty subsets, then either \(G=ST\) or \(|G|\ge |S|+|T|\) is true.

Proposition 2.4.14 If \(S\le G\) and \([G:S]=2\), then \(S\lhd G\).

Definition 2.4.15 If \(X\) is a subset of \(G\), then there is a smallest normal subgroup of \(G\) which contains \(X\); it is called the normal subgroup generated by \(X\), often denoted by \(\langle X\rangle^G\).

Proposition 2.4.16 If \(X=\varnothing\), then \(\langle X\rangle^G=\mathbf 1\); if \(X\neq\varnothing\), then \(\langle X\rangle^G\) is the set of all words on the conjugates of elements in \(X\).

Proposition 2.4.17 If \(H\) and \(K\) are normal subgroups of \(G\), then \(H\lor K\lhd G\).

Proposition 2.4.18 If a normal subgroup \(H\) of \(G\) has index \(n\), then \(g^n\in H\) for all \(g\in G\).

Quotient Groups

Theorem 2.5.1 If \(N\lhd G\), then the cosets of \(N\) in \(G\) form a group, denoted by \(G/N\), of order \([G:N]\).

Proof

As operation, we propose the multiplication of nonempty subsets of \(G\) defined earlier. We have already observed that this operation is associative. Now

\[\begin{aligned} NaNb&=Na(a^{-1}Na)b\\ &=N(aa^{-1})Nab=NNab=Nab \end{aligned} \]

Thus, \(NaNb=Nab\), and so the product of two cosets is a coset. It is easy to prove that the identity is the coset \(N=N\mathbf 1\) and that the inverse of \(Na\) is \(N(a^{-1})\).
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Corollary 2.5.2 If \(N\lhd G\), then the natural map (i.e., the function \(\nu\!:G\rightarrow G/N\) defined by \(\nu(a)=Na\)) is a surjective homomorphism with kernel \(N\).

As for the integer group \(\mathbb Z\), the quotient group \(\mathbb Z/\langle n\rangle\) is equal to \(\mathbb Z_n\), the group of integers modulo \(n\). An arbitrary quotient group \(G/N\) is often called \(G\!\!\mod N\) because of this example.

Proposition 2.5.3 Let \(H\lhd G\), let \(\nu\!:G\rightarrow G/H\) be the natural map, and let \(X\subset G\) be a subset such that \(\nu(X)\) generates \(G/H\), then \(G=\langle H\cup X\rangle\).

Definition 2.5.4 If \(a,b\in G\), the commutator of \(a\) and \(b\), denoted by \([a,b]\), is

\[[a,b]=aba^{-1}b^{-1}. \]

The commutator subgroup (or derived subgroup) of \(G\), denoted by \(G'\), is the subgroup of \(G\) generated by all the commutators.

Theorem 2.5.5 The commutator subgroup \(G'\) is a normal subgroup of \(G\). Moreover, if \(H\lhd G\), then \(G/H\) is abelian if and only if \(G'\le H\).

Proof

If \(f\!:G\rightarrow G\) is a homomorphism, then \(f(G')\le G'\) because \(f([a,b])=[fa,fb]\). It follows from Proposition 2.4.8 that \(G'\lhd G\).

Let \(H\lhd G\). If \(G/H\) is abelian, then \(HaHb=HbHa\) for all \(a,b\in G\); that is, \(Hab=Hba\), and so \(ab(ba)^{-1}=aba^{-1}b^{-1}=[a,b]\in H\). Thus we have \(G'\le H\). Conversely, suppose that \(G'\le H\). For every \(a,b\in G\), we have \(ab(ba)^{-1}=[a,b]\in G'\le H\), and so \(Hab=Hba\); that is, \(G/H\) is abelian.
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Proposition 2.5.6 Let \(x,y\in G\) and assume that both \(x\) and \(y\) commute with \([x,y]\). Then:

  1. \([x,y]^n=[x^n,y]=[x,y^n]\) for all \(n\in\mathbb Z\);
  2. \((xy)^n=[y,x]^{n(n-1)/2}x^ny^n\) for all \(n\ge 0\).

Proposition 2.5.7 If \(x,y,z\in G\), denote \(yxy^{-1}\) by \(x^y\) and \([x,[y,z]]\) by \([x,y,z]\).

  1. we have \([x,yz]=[x,y][x,z]^y\) and \([xy,z]=[y,z]^x[x,z]\).
  2. Jacobi identity \([x,y^{-1},z]^y[y,z^{-1},x]^z[z,x^{-1},y]^{x}=\mathbf 1\).

Proposition 2.5.8 Let \(H,K,L\) be subgroups of \(G\) and let \([H,K,L]=\langle[h,k,l]:h\in H,k\in K,l\in L\rangle\).

  1. Three subgroups lemma If \(N\lhd G\) and \([H,K,L][K,L,H]\le N\), then \([L,H,K]\le N\).
  2. If \(H,K\) and \(L\) are all normal subgroups of \(G\), then \([L,H,K]\le [H,K,L][K,L,H]\).

The Isomorphism Theorems

Theorem 2.6.1 (First Isomorphism Theorem) Let \(f\!:G\rightarrow H\) be a homomorphism with kernel \(K\). Then \(K\) is a normal subgroup of \(G\) and \(G/K\cong \mathrm{im}f\).

Proof

We have already noted that \(K\lhd G\). Define \(\varphi: G/K\rightarrow H\) by \(\varphi(Ka)=f(a)\). To see that \(\varphi\) is well defined, assume that \(Ka=Kb\); that is, \(ab^{-1}\in K\). Then \(\mathbf 1=f(ab^{-1})=f(a)f(b)^{-1}\), and \(f(a)=f(b)\); it follows that \(\varphi(Ka)=\varphi(Kb)\), as desired. Now \(\varphi\) is a homomorphism:

\[\varphi(KaKb)=\varphi(Kab)=f(ab)=f(a)f(b)=\varphi(Ka)\varphi(Kb). \]

It is plain that \(\mathrm {im}\varphi=\mathrm {im} f\). Finally, we show that \(\varphi\) is an injection. If \(\varphi(Ka)=\varphi(Kb)\), then \(f(a)=f(b)\); hence \(f(ab^{-1})=\mathbf 1,\ ab^{-1}\in K\), and \(Ka=Kb\) (note that \(\varphi\) being an injection is the converse of \(\varphi\) being well defined). Thus we have shown that \(\varphi\) is an isomorphism.
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It follows that there is no significant difference between a quotient group and a homomorphic image.

If \(\nu\!:G\rightarrow G/K\) is the natural map, then the following "commutative diagram" (i.e., \(f=\varphi\circ\nu\)) with surjection \(\nu\) and injection \(\varphi\) describes the theorem:

It is easy to describe \(\varphi^{-1}\!:\mathrm{im}f\rightarrow G/K\): if \(x\in\mathrm{im}f\), then there exists \(a\in G\) with \(f(a)=x\), and \(\varphi^{-1}(x)=Ka\).

Proposition 2.6.2 A homomorphism \(f\!:G\rightarrow H\) is an injection if and only if \(\mathrm{ker}f=1\).

Lemma 2.6.3 If \(S\) and \(T\) are subgroups of \(G\) and if one of them is normal, then \(ST=S\lor T=TS\).

Proof

Recall that \(ST\) and \(TS\) are subsets of \(S\lor T\) containing \(S\cup T\). If \(ST\) and \(TS\) are subgroups, then the reverse inclusion will follow from Corollary 2.1.12. Assume that \(T\lhd G\). If \(s_1t_1\) and \(s_2t_2\in ST\), then

\[\begin{aligned} (s_1t_1)(s_2t_2)^{-1}&=s_1t_1t_2^{-1}s_2^{-1}=s_1(s_2^{-1}s_2)t_1t_2^{-1}s_2^{-1}\\ &=s_1s_2^{-1}t_3=(s_1s_2^{-1})t_3\in ST, \end{aligned} \]

where \(t_3=s_2(t_1t_2^{-1})s_2^{-1}\in T\) because \(T\lhd G\). Therefore, \(ST=S\lor T\). A similar proof shows that \(TS\) is a subgroup, and so \(TS=S\lor T=ST\).
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Suppose that \(S\le H\le G\) are subgroups with \(S\lhd G\). Then \(S\lhd H\) and the quotient \(H/S\) is defined; it is the subgroup of \(G/S\) consisting of all those cosets \(Sh\) with \(h\in H\). In particular, if \(S\lhd G\) and \(T\) is any subgroup of \(G\), then \(S\le ST\le G\) and \(ST/S\) is the subgroup of \(G/S\) consisting of all those cosets \(Sst\), where \(st\in ST\). Since \(Sst=St\), it follows that \(ST/S\) consists precisely of all those cosets of \(S\) having a representative in \(T\).

Theorem 2.6.4 (Second Isomorphism Theorem) Let \(N\) and \(T\) be subgroups of \(G\) with \(N\) normal. Then \(N\cap T\) is normal in \(T\) and \(T/(N\cap T)\cong NT/N\).

Remark The following diagram is a mnemonic for this theorem:

Proof

Let \(\nu\!:G\rightarrow G/N\) be the natural map, and let \(\nu'=\nu\big|_T\), the restriction of \(\nu\) to \(T\). Since \(\nu'\) is a homomorphism whose kernel is \(N\cap T\), Theorem 2.6.1 gives \(N\cap T\lhd T\) and \(T/(N\cap T)\cong \mathrm{im}\nu'\). However, \(\mathrm{im}\nu'\) is just the family of all those cosets of \(N\) having a representative in \(T\); that is, \(\mathrm{im}\nu'\) consists of all the cosets in \(NT/N\).
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Theorem 2.6.5 (Third Isomorphism Theorem) Let \(K\le H\le G\), where both \(K\) and \(H\) are normal subgroups of \(G\). Then \(H/K\) is a normal subgroup of \(G/K\) and \((G/K)/(H/K)\cong G/H\).

Proof

Define \(f\!:G/K\rightarrow G/H\) by \(f(Ka)=Ha\) (this "enlargement of coset" map \(f\) is well defined because \(K\le H\)). Then it is easily observed that \(f\) is a surjective homomorphism with kernel \(H/K\). By Theorem 2.6.1 we have \((G/K)/(H/K)\cong G/H\).
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Proposition 2.6.6 (Modular Law) Let \(A,B\) and \(C\) be subgroups of \(G\) with \(A\le B\). If \(A\cap C=B\cap C\) and \(AC=BC\), then \(A=B\).

Proposition 2.6.7 (Dedekind Law) Let \(H,K\) and \(L\) be subgroups of \(G\) with \(H\le L\). Then \(HK\cap L=H(K\cap L)\).

Proposition 2.6.8 Let \(f\!:G\rightarrow G^*\) be a homomorphism and let \(S^*\) be a subgroup of \(G^*\). Then \(f^{-1}(S^*)=\{x\in G:f(x)\in S^*\}\) is a subgroup of \(G\) containing \(\mathrm{ker}f\).

Correspondence Theorem

Let \(X\) and \(X^*\) be sets. A function \(f\!:X\rightarrow X^*\) induces a "forward motion" and a "backward motion" between subsets of \(X\) and subsets of \(X^*\). The forward motion assigns to each subset \(S\subset X\) the subset \(f(S)=\{f(s):s\in S\}\) of \(X^*\); the backward motion assigns to each subset \(S^*\) of \(X^*\) the subset \(f^{-1}(S^*)=\{x\in X:f(x)\in S^*\}\) of \(X\). Morerover, if \(f\) is a surjection, these motions define a bijection between all the subsets of \(X^*\) and certain subsets of \(X\).

Theorem 2.7.1 (Correspondence Theorem) Let \(K\lhd G\) and let \(\nu\!:G\rightarrow G/K\) be the natural map. Then \(S\mapsto \nu(S)=S/K\) is a bijection from the family of all those subgroups \(S\) of \(G\) which contain \(K\) to the family of all the subgroups of \(G/K\).

Moreover, if we denote \(S/K\) by \(S^*\), then:

  1. \(T\le S\) if and only if \(T^*\le S^*\), and then \([S:T]=[S^*:T^*]\);
  2. \(T\lhd S\) if and only if \(T^*\lhd S^*\), and then \(S/T\cong S^*/T^*\).

Remark The following diagram is a mnemonic for this theorem:

Proof

We show first that \(S\mapsto S/K\) is an injection: if \(S\) and \(T\) are subgroups containing \(K\), and if \(S/K=T/K\), then \(S=T\). To see this, let \(s\in S\); since \(S/K=T/K\), there exists \(t\in T\) with \(Ks=Kt\). Hence, \(s=kt\) for some \(k\in K\le T\) and \(s\in T\). The reverse inclusion is proved similarly. To see that the correspondence \(S\mapsto S/K\) is a surjection, we must show that if \(A\le G/K\), then there is a subgroup \(S\) of \(G\) containing \(K\) with \(A=S/K\). By Proposition 2.6.8 \(S=\nu^{-1}(A)\) is a subgroup of \(G\) containing \(K\); moreover, that \(\nu\) is a surjection implies that \(S/K=\nu(S)=\nu\nu^{-1}(A)=A\).

It is plain that if \(K\le T\le S\), then \(T/K\le S/K\). To prove that \([S:T]=[S^*:T^*]\), it suffices to show that there is a bijection from the family of all cosets of the form \(Ts\), where \(s\in S\), to the family of all cosets \(T^*s^*\), where \(s^*\in S^*\). It is easy to check that \(\alpha\), defined by \(\alpha\!:Ts\mapsto T^*\nu(s)\), is such a bijection.

If \(T\lhd S\), then the third isomorphism theorem gives \(T/K\lhd S/K\) and \((S/K)/(T/K)\cong S/K\); that is, \(T^*\lhd S^*\) and \(S^*/T^*\cong S/T\). It remains to show that if \(T^*\lhd S^*\), then \(T\lhd S\). However, it is easily verified that \(T=\mathrm{ker}\mu\nu_0\!:S\rightarrow S^*/T^*\), where \(\nu_0=\nu\big|_S\) and \(\mu\!:S^*\rightarrow S^*/T^*\) is the natural map.
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Proposition 2.7.2 If \(G'\le H\le G\), where \(G'\) is the commutator subgroup of \(G\), then \(H\lhd G\) and \(G/H\) is abelian.

Definition 2.7.3 A group \(G\neq \mathbf 1\) is simple if it has no normal subgroups other than \(G\) and \(\mathbf 1\).

Proposition 2.7.4 A subgroup \(H< G\) is a maximal subgroup of \(G\) if there is no subgroup \(N\) of \(G\) with \(H< N< G\). If \(H\lhd G\) is a maximal subgroup, then \([G:H]\) is finite and equal to a prime.

Proposition 2.7.5 A subgroup \(H< G\) is a maximal normal subgroup of \(G\) if there is no normal subgroup \(N\) of \(G\) with \(H< N< G\). Then \(H\) is a maximal normal subgroup of \(G\) if and only if \(G/H\) is simple.

Proposition 2.7.6 (Schur) Let \(f\!:G\rightarrow H\) be a homomorphism that does not send every element of \(G\) into \(\mathbf 1\). If \(G\) is simple, then \(f\) must be an injection.

Direct Products

Definition 2.8.1 If \(H\) and \(K\) are groups, then their direct product, denoted by \(H\times K\), is the group with elements all ordered pairs \((h,k)\), where \(h\in H\) and \(k\in K\), and with operation

\[(h,k)(h',k')=(hh',kk'). \]

It is easy to check that \(H\times K\) is a group: the identity is \((\mathbf 1,\mathbf 1)\); the inverse \((h,k)^{-1}\) is \((h^{-1},k^{-1})\). Notice that neigher \(H\) nor \(K\) is a subgroup of \(H\times K\), but \(H\times K\) does contain isomorphic replicas of each, namely, \(H\times \mathbf 1=\{(h,\mathbf 1):h\in H\}\) and \(\mathbf 1\times K=\{(\mathbf 1,k):k\in K\}\).

Theorem 2.8.2 Let \(G\) be a group with normal subgroups \(H\) and \(K\). If \(HK=G\) and \(H\cap K=\mathbf 1\), then \(G\cong H\times K\).

Proof

If \(a\in G\), then \(a=hk\) for some \(h\in H\) and \(k\in K\) (because \(G=HK\)). We claim that \(h\) and \(k\) are uniquely determined by \(a\). If \(a=h_1k_1\) for \(h_1\in H\) and \(k_1\in K\), then \(hk=h_1k_1\) and \(h^{-1}h_1=kk_1^{-1}\in H\cap K=\mathbf 1\); hence \(h=h_1\) and \(k=k_1\).

Define \(f\!:G\rightarrow H\times K\) by \(f(a)=(h,k)\), where \(a=hk\). If \(a=hk\) and \(a'=h'k'\), then \(aa'=hkh'k'\) which is not in the proper form for evaluating \(f\). However, we could consider the commutator \(h'kh'^{-1}k^{-1}\). Now \((h'kh'^{-1})k^{-1}\in K\) for \(K\lhd G\), and, similarly, \(h'(kh'^{-1}k^{-1})\in H\); therefore, \(h'kh'^{-1}k^{-1}\in H\cap K=\mathbf 1\) and \(h'\) and \(k\) commute. Now it is easily observed that \(f\) is a homomorphism and a bijection; that is, \(f\) is an isomorphism.
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Theorem 2.8.3 If \(A\lhd H\) and \(B\lhd K\), then \(A\times B\lhd H\times K\) and

\[(H\times K)/(A\times B)\cong (H/A)\times (K/B). \]

Proof

The homomorphism \(\varphi\!:H\times K\rightarrow (H/A)\times (K/B)\), defined by \(\varphi(h,k)=(Ah,Bk)\), is surjective and \(\mathrm{ker}\varphi=A\times B\). The first isomorphism theorem now gives the result.
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Corollary 2.8.4 If \(G=H\times K\), then \(G/(H\times \mathbf 1)\cong K\).

There are two versions of the direct product \(H\times K\): the external version, whose elements are ordered pairs and which contains isomorphic copies of \(H\) and \(K\) (namely, \(H\times \mathbf 1\) and \(\mathbf 1\times K\)); the internal version which does contain \(H\) and \(K\) as normal subgroups and in which \(HK=G\) and \(H\cap K=\mathbf 1\). By Theorem 2.8.2, the two versions are isomorphic. In the future, we shall not distinguish between external and internal; in almost all cases, however, our point of view is internal. For example, we shall write Corollary 2.8.4 as \((H\times K)/H\cong K\).

Proposition 2.8.5 The operation of direct product is commutative and associative in the following sense: for groups \(H,K\) and \(L\), we have

\[H\times K\cong K\times H\text{ and }H\times (K\times L)\cong (H\times K)\times L. \]

This means that the notations \(H_1\times\cdots\times H_n\) and \(\prod_{i=1}^nH_i\) are unambiguous.

Proposition 2.8.6 Let \(G\) be a group having normal subgroups \(H_1,\ldots,H_n\).

  1. If \(G=\left\langle\bigcup_{i=1}^nH_i\right\rangle\) and, for all \(j\), \(\mathbf 1=H_j\cap\left\langle\bigcup_{i\neq j}H_i\right\rangle\), then \(G\cong H_1\times\cdots\times H_n\).
  2. If each \(a\in G\) has a unique expression of the form \(a=h_1\ldots h_n\), where each \(h_i\in H_i\), then \(G\cong H_1\times\cdots\times H_n\).

Proposition 2.8.7 Let \(H_1,\ldots,H_n\) be normal subgroups of a group \(G\), and define \(\varphi\!:G\rightarrow G/H_1\times\cdots\times G/H_n\) by \(\varphi(x)=(H_1x,\ldots,H_nx)\), then we have

  1. \(\mathrm{ker}\varphi=H_1\cap\cdots\cap H_n\);
  2. if each \(H_i\) has finite index in \(G\) and if \((|G/H_i|,|G/H_j|)=1\) for all \(i\neq j\), then \(\varphi\) is a surjection and

\[[G:H_1\cap\cdots\cap H_n]=\prod_{i=1}^n|G/H_i|. \]

Chapter III. Symmetric Groups and \(G\)-Sets

Conjugates

Lemma 3.1.1 If \(G\) is a group, then the relation "\(y\) is a conjugate of \(x\) in \(G\)," that is, \(y=gxg^{-1}\) for some \(g\in G\), is an equivalence relation.

Definition 3.1.2 If \(G\) is a group, then the equivalence class of \(a\in G\) under the relation "\(y\) is a conjugate of \(x\) in \(G\)" is called the conjugacy class of \(a\); it is denoted by \(a^G\).

If \(a\) and \(b\) are conjugate in \(G\), say, \(b=gag^{-1}\), then there is an isomorphism \(\gamma_g\!:G\rightarrow G\), namely, conjugation by \(g\), with \(\gamma_g(a)=b\). It follows that all the elements in the same conjugacy class have the same order.

If \(a\in G\) is the sole resident of its conjugacy class, then \(a=gag^{-1}\) for all \(g\in G\); that is, \(a\) commutes with every element of \(G\).

Definition 3.1.3 The center of a group \(G\), denoted by \(\mathrm Z(G)\), is the set of all \(a\in G\) that commute with every element of \(G\).

It is easy to check that \(\mathrm Z(G)\) is a normal abelian subgroup of \(G\).

Definition 3.1.4 If \(a\in G\), then the centralizer of \(a\) in \(G\), denoted by \(\mathrm C_G(a)\), is the set of all \(x\in G\) which commute with \(a\).

It is easy to check that \(\mathrm C_G(a)\) is a subgroup of \(G\).

Theorem 3.1.5 If \(a\in G\), the number of conjugates of \(a\) is equal to the index of its centralizer:

\[\left|a^{G}\right|=[G:\mathrm C_G(a)], \]

and this number is a divisor of \(|G|\) when \(G\) is finite.

Proof

Denote the family of all left cosets of \(C=\mathrm C_G(a)\) in \(G\) by \(G/C\), and define \(f\!:a^G\rightarrow G/C\) by \(f(gag^{-1})=gC\). Now \(f\) is well defined: if \(gag^{-1}=hah^{-1}\) for some \(h\in G\), then \(h^{-1}gag^{-1}h=a\) and \(h^{-1}g\) commutes with \(a\); that is, \(h^{-1}g\in C\), and so \(hC=gC\). The function \(f\) is an injection: if \(gC=f(gag^{-1})=f(kak^{-1})=kC\) for some \(k\in G\), then \(k^{-1}g\in C\), \(k^{-1}g\) commutes with \(a\), \(k^{-1}gag^{-1}k=a\), and \(gag^{-1}=kak^{-1}\); the function \(f\) is a surjection: if \(g\in G\), then \(gC=f(gag^{-1})\). Therefore, \(f\) is a bijection and \(\left|a^G\right|=|G/C|=[G:\mathrm C_G(a)]\). When \(G\) is finite, Lagrange's theorem applies.
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Definition 3.1.6 If \(H\le G\) and \(g\in G\), then the conjugate \(gHg^{-1}\) is \(\{ghg^{-1}:h\in H\}\). The conjugate \(gHg^{-1}\) is often denoted by \(H^g\).

The conjugate \(H^g\) is a subgroup of \(G\) isomorphic to \(H\): if \(\gamma_g\!:G\rightarrow G\) is conjugation by \(g\), then \(\gamma_g\big|_H\) is an isomorphism from \(H\) to \(H^g\).

Note that a subgroup \(H\) is a normal subgroup if and only if it has only one conjugate.

Definition 3.1.7 If \(H\le G\), then the normalizer of \(H\) in \(G\), denoted by \(\mathrm N_G(H)\), is

\[\mathrm N_G(H)=\{a\in G:aHa^{-1}=H\}. \]

It is immediate that \(\mathrm N_G(H)\) is a subgroup of \(G\). Notice that \(H\lhd\mathrm N_G(H)\); indeed, \(\mathrm N_G(H)\) is the largest subgroup of \(G\) in which \(H\) is normal.

Theorem 3.1.8 If \(H\le G\), then the number \(c\) of conjugates of \(H\) in \(G\) is equal to the index of its normalizer: \(c=[G:\mathrm N_G(H)]\), and \(c\) divides \(|G|\) when \(G\) is finite. Moreover, \(H^a=H^b\) if and only if \(b^{-1}a\in \mathrm N_G(H)\).

Proof

Let \([H]\) denote the family of all the conjugates of \(H\), and let \(G/N\) denote the family of all left cosets of \(N=\mathrm N_G(H)\) in \(G\). Define \(f\!:[H]\rightarrow G/N\) by \(f(H^a)=aN\). Now \(f\) is well defined: if \(H^a=H^b\) for some \(b\in G\), then \(H^{b^{-1}a}=H\) and \(b^{-1}a\) normalizes \(H\); that is, \(b^{-1}a\in N\), and so \(bN=aN\). The function \(f\) is an injection: if \(aN=f(H^a)=f(H^c)=cN\) for some \(c\in G\), then \(c^{-1}a\in N\), \(c^{-1}a\) normalizes \(H\), \(H^{c^{-1}a}=H\), and \(H^a=H^c\); the function \(f\) is a surjection: if \(a\in G\), then \(aN=f(H^a)\). Therefore, \(f\) is a bijection and \(|[H]|=|G/N|=[G:\mathrm N_G(H)]\). When \(G\) is finite, Lagrange's theorem applies.
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Proposition 3.1.9 Let \(H\) be a subgroup of \(G\), and \(a\in G\), then we have

  1. \(\mathrm N_G(H^{a})=\mathrm N_G(H)^{a}\); that is, \(\mathrm N_G(aHa^{-1})=a\mathrm N_G(H)a^{-1}\);
  2. If \(H\le K\le G\), then \(\mathrm N_K(H)=\mathrm N_G(H)\cap K\).
  3. If \(H,K\le G\), then \(\mathrm N_G(H)\cap \mathrm N_G(K)\le \mathrm N_G(H\cap K)\).

Definition 3.1.10 If \(H\) is a subset of \(G\), then the centralizer of \(H\) in \(G\), denoted by \(\mathrm C_G(H)\), is the set of all \(x\in G\) which commute with any element in \(H\); that is, \(\mathrm C_G(H)=\{x\in G:xy=yx\text{ for every }y\in H\}\).

Proposition 3.1.11 Let \(H\) be a subset of \(G\), then we have

  1. \(\mathrm C_G(H)\) is a subgroup of \(G\); that is, \(\mathrm C_G(H)\le G\);
  2. \(\mathrm C_G(H)\lhd \mathrm N_G(H)\) when \(H\le G\).

Proposition 3.1.12 If \(H\le G\), then \(\mathrm N_G(H)\le\{a\in G:aHa^{-1}\le H\}\).

Remark When \(H\le G\) is an infinite subgroup, there may exist \(a\in G\) such that \(H^a=aHa^{-1}< H\).

Proposition 3.1.13 If \(H\) is a proper subgroup of a finite group \(G\), then \(G\) is not the union of all the conjugates of \(H\).

Proposition 3.1.14 If \(G\) is a finite group with conjugacy classes \(C_1,\ldots,C_m\), and if \(g_i\in C_i\), then \(G=\langle g_1,\ldots,g_m\rangle\).

Proposition 3.1.15 Let \(G\) be a finite group, let \(H\) be a normal subgroup of prime index, and let \(x\in H\) satisfy \(\mathrm C_H(x)<\mathrm C_G(x)\). If \(y\in H\) is conjugate to \(x\) in \(G\), then \(y\) is conjugate to \(x\) in \(H\).

Definition 3.1.16 Let \(G\) be a group. If \(\mathrm Z(G)=\mathbf 1\), we call \(G\) centerless.

Proposition 3.1.17 If \(n\ge 3\), then \(\mathrm S_n\) is centerless.

Proposition 3.1.18 If \(\alpha\in \mathrm S_n\) is an \(n\)-cycle, then its centralizer is \(\langle \alpha\rangle\).

Proposition 3.1.19

  1. Let \(G\) be a group. If \(G\) is not abelian, then \(G/\mathrm Z(G)\) is not cyclic.
  2. If \(G\) is a group with \(G=G'\), then \(G/\mathrm Z(G)\) is centerless.
  3. If \(H\lhd G\) and \(H\cap G'=\mathbf 1\), then \(H\le \mathrm Z(G)\).

Symmetric Group

Definition 3.2.1 Two permutations \(\alpha,\beta\in\mathrm S_n\) have the same cycle structure if their complete factorizations into disjoint cycles have the same number of \(r\)-cycles for each \(r\).

Lemma 3.2.2 If \(\alpha,\beta\in \mathrm S_n\), then \(\alpha\beta\alpha^{-1}\) is the permutation with the same cycle structure as \(\beta\) which is obtained by applying \(\alpha\) to the symbols in \(\beta\).

Proof

Let \(\pi\) be the permutation defined in the lemma. If \(\beta\) fixes a symbol \(i\), then \(\pi\) fixes \(\alpha(i)\), for \(\alpha(i)\) resides in a \(1\)-cycle; but \(\alpha\beta\alpha^{-1}(\alpha(i))=\alpha\beta(i)=\alpha(i)\), and so \(\alpha\beta\alpha^{-1}\) fixes \(\alpha(i)\) as well. Assume that \(\beta\) moves \(i\); say, \(\beta(i)=j\). Let the complete factorization of \(\beta\) be

\[\beta=\gamma_1\gamma_2\cdots(\cdots\ \ i\ \ j\ \ \cdots)\cdots\gamma_t. \]

If \(\alpha(i)=k\) and \(\alpha(j)=l\), then \(\pi\!:k\mapsto l\). But \(\alpha\beta\alpha^{-1}\!:k\mapsto i\mapsto j\mapsto l\), and so \(\alpha\beta\alpha^{-1}=\pi(k)\). Therefore, \(\pi\) and \(\alpha\beta\alpha^{-1}\) agree on all symbols of the form \(k=\alpha(i)\); since \(\alpha\) is a surjection, it follows that \(\pi=\alpha\beta\alpha^{-1}\).
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Theorem 3.2.3 Permutations \(\alpha,\beta\in\mathrm S_n\) are conjugate if and only if they have the same cycle structure.

Proof

The lemma shows that conjugate permutations do have the same cycle structure. For the converse, define \(\gamma\in \mathrm S_n\) as follows: place the complete factorization of \(\alpha\) over that of \(\beta\) so that cycles of the same length correspond, and let \(\gamma\) be the function sending the top to the bottom. For example, if

\[\begin{aligned} \alpha&=\gamma_1\gamma_2\cdots(\cdots\ \ i\ \ j\ \ \cdots)\cdots\gamma_t,\\ \beta&=\delta_1\delta_2\cdots(\cdots\ \ k\ \ l\ \ \cdots)\cdots\delta_t, \end{aligned} \]

then \(\gamma(i)=k,\gamma(j)=l\), etc. Notice that \(\gamma\) is a permutation, for every \(i\) between \(1\) and \(n\) occurs exactly once in a complete factorization. The lemma gives \(\gamma\alpha\gamma^{-1}=\beta\), and so \(\alpha\) and \(\beta\) are conjugate.
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Corollary 3.2.4 A subgroup \(H\) of \(\mathrm S_n\) is a normal subgroup if and only if, whenever \(\alpha\in H\), then every \(\beta\) having the same cycle structure as \(\alpha\) also lies in \(H\).

Definition 3.2.5 If \(n\) is a positive integer, then a partition of \(n\) is a sequence of integers \(1\le i_1\le i_2\le \cdots\le i_r\) with \(\sum{i_j}=n\).

Proposition 3.2.6 The number of conjugacy classes in \(\mathrm S_n\) is the number of partitions of \(n\).

The Simplicity of Alternating Groups

Lemma 3.3.1 \(\mathrm A_5\) is simple.

Proof

(i). All \(3\)-cycles are conjugate in \(\mathrm A_5\).
If, for example, \(\alpha=(1\ \ 2\ \ 3)\), then the odd permutation \((4\ \ 5)\) commutes with \(\alpha\). Since \(\mathrm A_5\) has index \(2\) in \(\mathrm S_5\), it is a normal subgroup of prime index, and so Proposition 3.1.15 says that \(\alpha\) has the same number of conjugates in \(\mathrm A_5\) as it does in \(\mathrm S_5\) because \(\mathrm C_{\mathrm A_5}(\alpha)< \mathrm C_{\mathrm S_5}(\alpha)\).

(ii). All products of distinct transpositions are conjugate in \(\mathrm A_5\).
If, for example, \(\alpha=(1\ \ 2)(3\ \ 4)\), then the odd permutation \((1\ \ 2)\) commutes with \(\alpha\). Since \(\mathrm A_5\) has index \(2\) in \(\mathrm S_5\), Proposition 3.1.15 says that \(\alpha\) has the same number of conjugates in \(\mathrm A_5\) as it does in \(\mathrm S_5\).

(iii). There are two conjugacy classes of \(5\)-cycles in \(\mathrm A_5\), each of which has \(12\) elements.
In \(\mathrm S_5\), \(\alpha=(1\ \ 2\ \ 3\ \ 4\ \ 5)\) has \(24\) conjugates, so that \(\mathrm C_{\mathrm S_5}(\alpha)\) has \(5\) elements; these must be the powers of \(\alpha\). By Proposition 3.1.18, \(\mathrm C_{\mathrm A_5}(\alpha)\) has order \(5\), hence, index \(60/5=12\).

We have now surveyed all the conjugacy classes occuring in \(\mathrm A_5\). Since every normal subgroup \(H\) is a union of conjugacy classes, \(|H|\) is a sum of \(1\) and certain of the numbers: \(12\), \(12\), \(15\), and \(20\). It is easily checked that no such sum is a proper divisor of \(60\), so that \(|H|=60\) and \(\mathrm A_5\) is simple.
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Lemma 3.3.2 Let \(H\lhd \mathrm A_n\), where \(n\ge 5\). If \(H\) contains a \(3\)-cycle, then \(H=\mathrm A_n\).

Proof

We show that \((1\ \ 2\ \ 3)\) and \((i\ \ j\ \ k)\) are conjugate in \(\mathrm A_n\) (and thus that all \(3\)-cycles are conjugate in \(\mathrm A_n\)). If these cycles are not disjoint, then each fixes all the symbols outside of \(\{1,2,3,i,j\}\), say, and the two \(3\)-cycles lie in \(A^*\), the group of all even permutations on these \(5\) symbols. Of course, \(A^*\cong \mathrm A_5\), and, as in part (i) of the previous proof, \((1\ \ 2\ \ 3)\) and \((i\ \ j\ \ k)\) are conjugate in \(A^*\); a fortiori, they are conjugate in \(\mathrm A_n\). If the cycles are disjoint, then we have just seen that \((1\ \ 2\ \ 3)\) is conjugate to \((3\ \ j\ \ k)\) and that \((3\ \ j\ \ k)\) is conjugate to \((i\ \ j\ \ k)\), so that \((1\ \ 2\ \ 3)\) is conjugate to \((i\ \ j\ \ k)\) in this case as well.

A normal subgroup \(H\) containing a \(3\)-cycle \(\alpha\) must contain every conjugate of \(\alpha\); as all \(3\)-cycles are conjugate, \(H\) contains every \(3\)-cycle. But Proposition 2.1.20 shows that \(\mathrm A_n\) is generated by the \(3\)-cycles, and so \(H=\mathrm A_n\).
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Lemma 3.3.3 \(\mathrm A_6\) is simple.

Proof

Let \(H\neq \mathbf 1\) be a normal subgroup of \(\mathrm A_6\), and let \(\alpha\in H\) be distinct from \(\mathbf 1\). If \(\alpha\) fixes some \(i\), define

\[F=\{\beta\in\mathrm A_6:\beta(i)=i\}. \]

Now \(F\cong \mathrm A_5\) and \(\alpha\in H\cap F\). But \(H\cap F\lhd F\), by the second isomorphism theorem, so that \(F\) simple and \(H\cap F\neq \mathbf 1\) give \(H\cap F=F\); that is, \(F\le H\). Therefore, \(H\) contains a \(3\)-cycle, \(H=\mathrm A_6\) (by the lemma), and we are done.

We may now assume that no \(\alpha\in H\) with \(\alpha\neq\mathbf 1\) fixes any \(i\), for \(1\le i\le 6\). It could be easily shown that the cycle structure of \(\alpha\) is either \((1\ \ 2)(3\ \ 4\ \ 5\ \ 6)\) or \((1\ \ 2\ \ 3)(4\ \ 5\ \ 6)\). In the first case, \(\alpha^2\in H,\,\alpha^2\neq\mathbf 1\), and \(\alpha^2\) fixes \(1\) (and \(2\)), a contradiction. In the second case, \(H\) contains \(\alpha(\beta\alpha^{-1}\beta^{-1})\), where \(\beta=(2\ \ 3\ \ 4)\), and it is easily checked that this element is not the identity and it fixes \(1\), a contradiction. Therefore, no such normal subgroup \(H\) can exist.
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Theorem 3.3.4 \(\mathrm A_n\) is simple for all \(n\ge 5\).

Proof

Let \(n\ge 5\) and let \(H\neq 1\) be a normal subgroup of \(\mathrm A_n\). If \(\beta\in H\) and \(\beta \neq \mathbf 1\), then there is an \(i\) with \(\beta(i)=j\neq i\). If \(\alpha\) is a \(3\)-cycle fixing \(i\) and moving \(j\), then \(\alpha\) and \(\beta\) do not commute: \(\beta\alpha(i)=\beta(i)=j\) and \(\alpha\beta(i)\alpha(j)\neq j\); therefore, their commutator is not the identity. Furthermore, \(\alpha(\beta\alpha^{-1}\beta^{-1})\) lies in the normal subgroup \(H\), and, by Lemma 3.2.2, it is a product of two \(3\)-cycles \((\alpha\beta\alpha^{-1})\beta^{-1}\); thus it moves at most \(6\) symbols, say, \(i_1,\ldots,i_6\). If \(F=\{\gamma\in\mathrm A_m:\gamma\text{ fixes the other symbols}\}\), then \(F\cong \mathrm A_6\) and \(\alpha\beta\alpha^{-1}\beta^{-1}\in H\cap F\lhd F\). Since \(\mathrm A_6\) is simple, \(H\cap F=F\) and \(F\le H\). Therefore \(H\) contains a \(3\)-cycle, \(H=\mathrm A_n\) (Lemma 3.3.2), and the proof is complete.
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Proposition 3.3.5 If \(n\neq 4\), then \(\mathrm A_n\) is the only proper nontrivial normal subgroup of \(\mathrm S_n\).

Proposition 3.3.6 If \(X=\{1,2,\ldots\}\) is the set of all positive integers, then the infinite alternating group \(\mathrm A_{\infty}\) is the subgroup of \(\mathrm S_X\) generated by all the \(3\)-cycles, and we have that \(\mathrm A_{\infty}\) is an infinite simple group.

Some Representation Theorems

Theorem 3.4.1 (Cayley) Every group \(G\) can be imbedded as a subgroup of \(\mathrm S_G\). In particular, if \(|G|=n\), then \(G\) can be imbedded in \(\mathrm S_n\).

Proof

Recall that for each \(a\in G\), left translation \(L_a\!:G\rightarrow G\), defined by \(x\mapsto ax\), is a bijection; that is, \(L_a\in\mathrm S_G\). The theorem is proved if the function \(L\!:G\rightarrow \mathrm S_G\), given by \(a\mapsto L_a\) is an injection and a homomorphism, for then \(G\cong \mathrm{im}L\). If \(a\neq b\), then \(L_a(\mathbf 1)=a\neq b=L_b(\mathbf 1)\), and so \(L_a\neq L_b\). Finally, we show that \(L_{ab}=L_a\circ L_b\). If \(x\in G\), then \(L_{ab}(x)=(ab)x\), while \((L_a\circ L_b)(x)=L_a(L_b(x))=L_a(bx)=a(bx)\); associativity shows that these are the same.
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Definition 3.4.2 The homomorphism \(L\!:G\rightarrow \mathrm S_G\), given by \(a\mapsto L_a\), is called the left regular representation of \(G\).

Corollary 3.4.3 If \(k\) is a field and \(G\) is a finite group of order \(n\), then \(G\) can be imbedded in \(\mathrm {GL}(n,k)\).

Proof

The group \(P(n,k)\) of all \(n\times n\) permutation matrices is a subgroup of \(\mathrm {GL}(n,k)\) that is isomorphic to \(\mathrm S_n\). Now apply Cayley's theorem to imbed \(G\) into \(P(n,k)\).
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Theorem 3.4.4 If \(H\le G\) and \([G:H]=n\), then there is a homomorphism \(\rho\!:G\rightarrow \mathrm S_n\) with \(\mathrm{ker}\rho\le H\).

Proof

If \(a\in G\) and \(X\) is the family of all the left cosets of \(H\) in \(G\), define a function \(\rho_a\!:X\rightarrow X\) by \(gH\mapsto agH\) for all \(g\in G\). It is easy to check that each \(\rho_a\) is a permutation of \(X\) (its inverse is \(\rho_{a^{-1}}\)) and that \(a\mapsto \rho_a\) is a homomorphism \(\rho\!:G\rightarrow \mathrm S_X\cong \mathrm S_n\). If \(a\in\mathrm{ker}\rho\), then \(agH=gH\) for all \(g\in G\); in particular, \(aH=H\), and so \(a\in H\); therefore, \(\mathrm{ker}\rho\le H\).
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Definition 3.4.5 The homomorphism \(\rho\) in Theorem 3.4.4 is called the representation of \(G\) on the cosets of \(H\).

Corollary 3.4.6 A simple group \(G\) which contains a subgroup \(H\) of index \(n\) can be imbedded in \(\mathrm S_n\).

Proof

There is a homomorphism \(\rho\!:G\rightarrow \mathrm S_n\) with \(\mathrm {ker}\rho\le H< G\). Since \(G\) is simple, \(\mathrm{ker}\rho=\mathbf 1\), and so \(\rho\) is an injection.
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Theorem 3.4.7 Let \(H\le G\) and let \(X\) be the family of all the conjugates of \(H\) in \(G\). There is a homomorphism \(\psi\!:G\rightarrow \mathrm S_X\) with \(\mathrm{ker}\psi\le \mathrm N_G(H)\).

Proof

If \(a\in G\), define \(\psi_a\!:X\rightarrow X\) by \(\psi_a(gHg^{-1})=agHg^{-1}a^{-1}\). If \(b\in G\), then

\[\psi_a\psi_b(gHg^{-1})=\psi_a(bgHg^{-1}b^{-1})=abgHg^{-1}b^{-1}a^{-1}=\psi_{ab}(gHg^{-1}). \]

We conclude that \(\psi_a\) has inverse \(\psi_{a^{-1}}\), so that \(\psi_a\in\mathrm S_X\) and \(\psi\!:G\rightarrow \mathrm S_X\) is a homomorphism.

If \(a\in\mathrm{ker}\psi\), then \(agHg^{-1}a^{-1}=gHg^{-1}\) for all \(g\in G\). In particular, \(aHa^{-1}=H\), and so \(a\in \mathrm N_G(H)\); hence \(\mathrm{ker}\psi\le\mathrm N_G(H)\).
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Definition 3.4.8 The homomorphism \(\psi\) in Theorem 3.4.7 is called the representation of \(G\) on the conjugates of \(H\).

Proposition 3.4.9 If \(\rho\) is the representation of a group \(G\) on the cosets of a subgroup \(H\), then \(\mathrm{ker}\rho=\bigcap_{x\in G}H^x\).

Proposition 3.4.10 If \(\psi\) is the representation of a group \(G\) on the conjugates of a subgroup \(H\), then \(\mathrm{ker}\psi=\bigcap_{x\in G}\mathrm N_G(H)^x\).

Proposition 3.4.11 Let \(a\in G\), where \(G\) is finite. If \(a^n\) has \(m\) conjugates and \(a\) has \(k\) conjugates, then \(m|k\).

Proposition 3.4.12 If \(H\) and \(K\) are subgroups of \(G\) having finite index, then \([G:H\cap K]\le [G:H][G:K]\). Moreover, if \(([G:H],[G:K])=1\), then \([G:H\cap K]=[G:H][G:K]\).

Proposition 3.4.13 Let \(G\) be an infinite simple group, then

  1. every \(x\in G\) with \(x\neq \mathbf 1\) has infinitely many conjugates;
  2. every proper subgroup \(H\neq \mathbf 1\) has infinitely many conjugates.

Proposition 3.4.14 If \(A\) and \(G\) are groups, then a homomorphism \(h\!:A\rightarrow G\) is a surjection if and only if it is right cancellable: for every group \(L\) and every pair of homomorphisms \(f,g\!:G\rightarrow L\), the equation \(f\circ h=g\circ h\) implies \(f=g\).

Proposition 3.4.15 Let \(G\) be a finite group containing a subgroup \(H\) of index \(p\), where \(p\) is the smallest prime divisor of \(|G|\). Then \(H\) is a normal subgroup of \(G\).

\(G\)-Sets

Definition 3.5.1 If \(X\) is a set and \(G\) is a group, then \(X\) is a \(G\)-set if there is a function \(\alpha\!:G\times X\rightarrow X\) (called an action), denoted by \(\alpha\!:(g,x)\mapsto gx\), such that:

  1. \(\mathbf 1x=x\) for all \(x\in X\);
  2. \(g(hx)=(gh)x\) for all \(g,h\in G\) and \(x\in X\).

One also says that \(G\) acts on \(X\). If \(|X|=n\), then \(n\) is called the degree of the \(G\)-set \(X\).

Theorem 3.5.2 If \(X\) is a \(G\)-set with action \(\alpha\), then there is a homomorphism \(\tilde\alpha\!:G\rightarrow \mathrm S_X\) given by \(\tilde\alpha(g)\!:x\mapsto gx=\alpha(g,x)\). Conversely, every homomorphism \(\varphi\!:G\rightarrow\mathrm S_X\) defines an action, namely, \(gx=\varphi(g)x\), which makes \(X\) into a \(G\)-set.

Proof

If \(X\) is a \(G\)-set, \(g\in G\), and \(x\in X\), then

\[\tilde\alpha(g^{-1})\tilde\alpha(g)\!:x\mapsto \tilde\alpha(g^{-1})(gx)=g^{-1}(gx)=(g^{-1}g)x=\mathbf 1x=x; \]

it follows that each \(\tilde\alpha(g)\) is a permutation of \(X\) with the inverse \(\tilde\alpha(g^{-1})\). That \(\tilde\alpha\) is a homomorphism is immediate from (ii) of the definition of \(G\)-set. The converse is also routine.
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If \(\sigma\in\mathrm S_n\), define

\[g^{\sigma}(x_1,\ldots,x_n)=g(x_{\sigma 1},\ldots,x_{\sigma n}); \]

If \(g^{\sigma}=g\) for all \(\sigma\in\mathrm S_n\), then \(g\) is called a symmetric function. If a polynomial \(f(x)=\sum_{i=0}^na_ix^i\) has roots \(r_1,\ldots,r_n\), then each of the coefficients \(a_i\) of \(f(x)=a_n\prod_{i=0}^n(x-r_i)\) is a symmetric function of \(r_1,\ldots,r_n\). Other interesting functions of the roots may not be symmetric. For example, the discriminant of \(f(x)\) is defined to be the numer \(d^2\), where \(d=\prod_{i< j}(r_i-r_j)\). If \(D(x_1,\ldots,x_n)=\prod_{i< j}(x_i-x_j)\), then it is easy to see, for every \(\sigma\in\mathrm S_n\), that \(D^{\sigma}=\pm D\). Indeed, \(D\) is an alternating function of the roots: \(D^{\sigma}=D\) if and only if \(\sigma\in\mathrm A_n\).

Given \(g(x_1,\ldots,x_n)\), find

\[\mathscr S(g)=\{\sigma\in\mathrm S_n:g^{\sigma}=g\}; \]

it is easy to see that \(\mathscr S(g)\le \mathrm S_n\); moreover, \(g\) is symmetric if and only if \(\mathscr S(g)=\mathrm S_n\), while \(\mathscr S(D)=\mathrm A_n\).

Definition 3.5.3 If \(X\) is a \(G\)-set and \(x\in X\), then the \(G\)-orbit of \(x\) is

\[Gx=\mathcal O(x)=\{gx:g\in G\}\subset X \]

Usually, we will say orbit instead of \(G\)-orbit. The orbits of \(X\) form a partition; indeed, the relation \(x\equiv y\) defined by "\(y=gx\) for some \(g\in G\)" is an equivalence relation whose equivalence classes are the orbits.

Definition 3.5.4 If \(X\) is a \(G\)-set and \(x\in X\), then the stabilizer of \(x\), denoted by \(G_x\), is the subgroup

\[G_x=\{g\in G:gx=x\}\le G. \]

Example 3.5.5 If \(G\) acts on itself by conjugation and \(x\in G\), then \(\mathcal O(x)\) is the conjugacy class of \(x\) and \(G_x=\mathrm C_G(x)\).

Example 3.5.6 If \(G\) acts by conjugation on the family of all its subgroups and if \(H\le G\), then \(\mathcal O(H)=\{\text{all the conjugates of }H\}\) and \(G_H=\mathrm N_G(H)\).

Theorem 3.5.7 If \(X\) is a \(G\)-set and \(x\in X\), then

\[|\mathcal O(x)|=[G:G_x]. \]

Proof

If \(x\in X\), let \(G/G_x\) denote the family of all left cosets of \(G_x\) in \(G\). Define \(f\!:\mathcal O(x)\rightarrow G/G_x\) by \(f(ax)=aG_x\). Now \(f\) is well defined: if \(ax=bx\) for some \(b\in G\), then \(b^{-1}ax=x\), \(b^{-1}a\in G_x\), and \(aG_x=bG_x\). The function \(f\) is an injection: if \(aG_x=f(ax)=f(cx)=cG_x\) for some \(c\in G\), then \(c^{-1}a\in G_x\), \(c^{-1}ax=x\), and \(ax=cx\); the function \(f\) is a surjection: if \(a\in G\), then \(aG_x=f(ax)\). Therefore, \(f\) is a bijection and \(|\mathcal O(x)|=|G/G_x|=[G:G_x]\).
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Corollary 3.5.8 If a finite group \(G\) acts on a set \(X\), then the number of elements in any orbit is a divisor of \(|G|\).

Proposition 3.5.9 Let \(X\) be a \(G\)-set, let \(x,y\in X\), and let \(y=gx\) for some \(g\in G\). Then we have \(G_y=gG_xg^{-1}\), and also \(|G_y|=|G_x|\).

Definition 3.5.10 A \(G\)-set \(X\) is transitive if it has only one orbit; that is, for every \(x,y\in X\), there exists \(\sigma\in G\) with \(y=\sigma x\).

Proposition 3.5.11 If \(X\) is a \(G\)-set, then each of its orbits is a transitive \(G\)-set.

Proposition 3.5.12 If \(H\le G\), then \(G\) acts transitively on the set of all left cosets of \(H\), and \(G\) acts transitively on the set of all conjugates of \(H\).

Proposition 3.5.13 Let \(X\) be a \(G\)-set with action \(\alpha\!:G\times X\rightarrow X\), and let \(\tilde\alpha\!:G\rightarrow \mathrm S_X\) send \(g\in G\) into the permutation \(x\mapsto gx\).

  1. If \(K=\mathrm{ker}\tilde\alpha\), then \(X\) is a \((G/K)\)-set if one defines \((gK)x=gx\).
  2. If \(X\) is a transitive \(G\)-set, then \(X\) is a transitive \((G/K)\)-set.
  3. If \(X\) is a transitive \(G\)-set, then \(|\mathrm{ker}\tilde\alpha|\le |G|/|X|\).

Proposition 3.5.14 If \(G\le \mathrm S_n\), then \(G\) acts on \(X=\{1,2,\ldots,n\}\). In particular, \(\langle\alpha\rangle\) acts on \(X\) for every \(\alpha\in\mathrm S_n\). If the complete factorization of \(\alpha\) into disjoint cycles is \(\alpha=\beta_1\ldots\beta_t\) and if \(i\) is a symbol appearing in \(\beta_j\), then \(\mathcal O(i)=\{\alpha^{k}(i):k\in\mathbb Z\}\) consists of all the symbols appearing in \(\beta_j\).

Counting Orbits

Let us call a \(G\)-set \(X\) finite if both \(X\) and \(G\) are finite.

Theorem 3.6.1 (Burnside's Lemma) If \(X\) is a finite \(G\)-set and \(N\) is the number of \(G\)-orbits of \(X\), then

\[N=(1/|G|)\sum_{\tau\in G}F(\tau), \]

where, for \(\tau\in G\), \(F(\tau)\) is the number of \(x\in X\) fixed by \(\tau\).

Proof

In the sum \(\sum_{\tau\in G}F(\tau)\), each \(x\in X\) is counted \(|G_x|\) times (for \(G_x\) consists of all those \(\tau\in G\) which fix \(x\)). If \(x\) and \(y\) lie in the same orbit, then Proposition 3.5.9 gives \(|G_y|=|G_x|\), and so the \([G:G_x]\) elements constituting the orbit of \(x\) are, in the above sum, collectively counted \([G:G_x]|G_x|=|G|\) times. Each orbit thus contributes \(|G|\) to the sum, and so \(\sum_{\tau\in G}F(\tau)=N|G|\).
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Corollary 3.6.2 If \(X\) is a finite transitive \(G\)-set with \(|X|>1\), then there exists \(\tau\in G\) having no fixed points.

Proof

Since \(X\) is transitive, the number \(N\) of orbits of \(X\) is \(1\), and so Burnside's lemma gives

\[1=(1/|G|)\sum_{\tau\in G}F(\tau). \]

Now \(F(\mathbf 1)=|X|>1\); if \(F(\tau)>0\) for every \(\tau\in G\), then the right hand side is too large.
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Definition 3.6.3 If \(G\le \mathrm S_X\), where \(X=\{1,\ldots,n\}\), and if \(\mathscr C\) is a set of colors, then \(\mathscr C^n\) is a \(G\)-set if we define \(\tau(c_1,\ldots,c_n)=(c_{\tau 1},\ldots,c_{\tau n})\) for all \(\tau\in G\). If \(|\mathscr C|=q\), then an orbit of \(\mathscr C^n\) is called a \((q,G)\)-coloring of \(X\).

Lemma 3.6.4 Let \(\mathscr C\) be a set of \(q\) colors, and let \(G\le \mathrm S_X\cong \mathrm S_n\). If \(\tau\in G\), then \(F(\tau)=q^{t(\tau)}\), where \(t(\tau)\) is the number of cycles occurring in the complete factorization of \(\tau\).

Proof

Since \(\tau(c_1,\ldots,c_n)=(c_{\tau 1},\ldots,c_{\tau n})=(c_1,\ldots,c_n)\), we see that \(c_{\tau i}=c_i\) for all \(i\), and so \(\tau i\) has the same color as \(i\). It follows that \(\tau^ki\) has the same color as \(i\), for all \(k\); that is, all \(i\) in the \(\langle\tau\rangle\)-orbit of \(X\) have the same color. But Proposition 3.5.14 shows that if the complete factorization of \(\tau\) is \(\tau=\beta_1\ldots\beta_{t(\tau)}\), and if \(i\) occurs in \(\beta_j\), then the set of symbols occuring in \(\beta_j\) is the \(\langle\tau\rangle\)-orbit containing \(i\). Since there are \(t(\tau)\) orbits and \(q\) colors, there are \(q^{t(\tau)}\) \(n\)-tuples fixed by \(\tau\) in its action on \(\mathscr C^n\).
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Definition 3.6.5 If the complete factorization of \(\tau\in\mathrm S_n\) has \(e_r(\tau)\ge 0\) \(r\)-cycles, then the index of \(\tau\) is

\[\mathrm{ind}(\tau)=x_1^{e_1(\tau)}x_2^{e_2(\tau)}\ldots x_n^{e_n(\tau)}. \]

If \(G\le \mathrm S_n\), then the cycle index of \(G\) is the polynomial

\[P_G(x_1,\ldots,x_n)=(1/|G|)\sum_{\tau\in G}\mathrm{ind}(\tau)\in \mathbb Q[x_1,\ldots,x_n]. \]

Corollary 3.6.6 If \(|X|=n\) and \(G\le \mathrm S_n\), then the number of \((q,G)\)-colorings of \(X\) is \(P_G(q,\ldots,q)\).

Proof

By Burnside's lemma for the \(G\)-set \(\mathscr C^n\), the number of \((q,G)\)-colorings of \(X\) is

\[(1/|G|)\sum_{\tau\in G}F(\tau). \]

By Lemma 3.6.4, this number is

\[(1/|G|)\sum_{\tau\in G}q^{t(\tau)}, \]

where \(t(\tau)\) is the number of cycles in the complete factorization of \(\tau\). On the other hand,

\[\begin{aligned} P_G(x_1,\ldots,x_n)&=(1/|G|)\sum_{\tau\in G}\mathrm{ind}(\tau)\\ &=(1/|G|)\sum_{\tau\in G}x_1^{e_1(\tau)}x_2^{e_2(\tau)}\ldots x_n^{e_n(\tau)}, \end{aligned} \]

and so

\[\begin{aligned} P_G(q,\ldots,q)&=(1/|G|)\sum_{\tau\in G}q^{e_1(\tau)+e_2(\tau)+\cdots+e_n(\tau)}\\ &=(1/|G|)\sum_{\tau\in G}q^{t(\tau)}. \end{aligned} \]

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Theorem 3.6.7 (Pólya) Let \(G\le \mathrm S_X\), where \(|X|=n\), let \(|\mathscr C|=q\), and, for each \(i\ge 1\), define \(\sigma_i=c_1^i+\cdots+c_q^i\). Then the number of \((q,G)\)-colorings of \(X\) with \(f_r\) elements of color \(c_r\), for every \(r\), is the coefficient of \(c_1^{f_1}c_2^{f_2}\cdots c_q^{f_q}\) in \(P_G(\sigma_1,\ldots,\sigma_n)\).

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