GTM148 抄书笔记 Part III. (Chapter VI)

上一篇: GTM148 抄书笔记 Part II.

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Contents

• Contents
• Chapter VI. Finite Direct Products
  • The Basic Theorem
  • The Fundamental Theorem of Finite Abelian Groups
  • The Krull-Schmidt Theorem
  • Operator Groups
• Appendix A
  • Groups of Small Order

Chapter VI. Finite Direct Products

The Basic Theorem

Definition 6.1.1 If a nonabelian group \(G=H\times K\) is a direct product, then \(H\) is called a direct factor of \(G\); in additive notation, one writes \(G=H\oplus K\), and calls \(H\) a (direct) summand of \(G\).

It is easily observed that if \(X\) is a nonempty subset of an abelian group \(G\), then \(\langle X\rangle\) is the set of all linear combinations of elements in \(X\) having coefficients in \(\mathbb Z\).

Definition 6.1.2 If \(G\) is an abelian \(p\)-group for some prime \(p\), then \(G\) is also called a \(p\)-primary group.

Theorem 6.1.3 (Primary Decomposition) Every finite abelian group \(G\) is a direct sum of \(p\)-primary groups.

Remark The theorem is obvious since every abelian group is nilpotent. But here we have a more elementary approach.

Proof

Since \(G\) is finite, it has exponent \(n\) for some \(n\): we have \(nx=\mathbf 0\) for all \(x\in G\). For each prime divisor \(p\) of \(n\), define

\[G_p=\{x\in G:p^ex=\mathbf 0\text{ for some }e\}. \]

Now \(G_p\) is a subgroup of \(G\), for if \(p^nx=\mathbf 0\) and \(p^my=\mathbf 0\), where \(m\le n\), then \(p^n(x-y)=\mathbf 0\). We claim that \(G=\sum G_p\), and we use the criterion in Proposition 2.8.6.

Let \(n=p_1^{e_1}\ldots p_t^{e_t}\), where the \(p_i\) are distinct primes and \(e_i>0\) for all \(i\). Set \(n_i=n/p_i^{e_i}\), and observe that the \(\gcd(n_1,\ldots,n_t)=1\). Then there are integers \(s_i\) with \(\sum s_in_i=1\), and so \(x=\sum(s_in_ix)\). But \(s_in_ix\in G_{p_i}\), because \(p_i^{e_i}s_in_ix=s_inx=\mathbf 0\). Therefore, \(G\) is generated by the family of \(G_p\)'s.

Assume that \(x\in G_p\cap\langle\bigcup_{q\neq p}G_q\rangle\). On the one hand, \(p^ex=\mathbf 0\) for some \(e\ge 0\); on the other hand, \(x=\sum x_q\), where \(q^{e_q}x_q=\mathbf 0\) for exponents \(e_q\). If \(m=\prod q^{e_q}\), then \(m\) and \(p^e\) are relatively prime, and there are integers \(r\) and \(s\) with \(1=rm+sp^{e}\). Therefore, \(x=rmx+sp^{e}x=\mathbf 0\), and so \(G_p\cap\langle U_{q\neq p}G_q\rangle=\mathbf 0\).
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Definition 6.1.4 The subgroups \(G_p\) are called the \(p\)-primary components of \(G\).

Definition 6.1.5 A set \(\{x_1,\ldots,x_r\}\) of nonzero elements in an abelian group is independent if, whenever there are integers \(m_1,\ldots,m_r\) with \(\sum_{i=1}^rm_ix_i=\mathbf 0\), then each \(m_ix_i=\mathbf 0\).

Lemma 6.1.6 If \(G\) is an abelian group, then a subset \(\{x_1,\ldots,x_r\}\) of nonzero elements of \(G\) is independent if and only if \(\langle x_1,\ldots,x_r\rangle=\langle x_1\rangle\oplus\cdots\oplus\langle x_r\rangle\).

Proof

Assume independence: if \(y\in\langle x_i\rangle\cap \langle \{x_j:j\neq i\}\rangle\), then there are integers \(m_1,\ldots,m_r\) with \(y=-m_ix_i=\sum_{j\neq i}m_jx_j\), and so \(\sum_{k=1}^rm_kx_k=\mathbf 0\). By independence, \(m_kx_k=\mathbf 0\) for all \(k\); in particular, \(m_ix_i=\mathbf 0\) and so \(y=-m_ix_i=\mathbf 0\). Proposition 2.8.6 now shows that \(\langle x_1,\ldots,x_r\rangle=\langle x_1\rangle\oplus\cdots\oplus\langle x_r\rangle\).

For the converse, assume that \(\sum m_ix_i=\mathbf 0\). For each \(j\), we have \(-m_jx_j=\sum_{k\neq j}m_kx_k\in\langle x_j\rangle\cap\langle\{x_k:k\neq j\}\rangle=\mathbf 0\). Therefore, each \(m_jx_j=\mathbf 0\) and \(\{x_1,\ldots,x_r\}\) is independent.
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Corollary 6.1.7 Every finite abelian group \(G\) of prime exponent \(p\) is an elementary abelian \(p\)-group.

Lemma 6.1.8 Let \(\{x_1,\ldots,x_r\}\) be an independent subset of a \(p\)-primary abelian group \(G\).

1. If \(\{z_1,\ldots,z_r\}\subset G\), where \(pz_i=x_i\) for all \(i\), then \(\{z_1,\ldots,z_r\}\) is independent.
2. If \(k_1,\ldots,k_r\) are integers with \(k_ix_i\neq\mathbf 0\) for all \(i\), then \(\{k_1x_1,\ldots,k_rx_r\}\) is also independent.

Definition 6.1.9 If \(G\) is an abelian group and \(m>0\) is an integer, then

\[mG=\{mx:x\in G\}. \]

It is easy to see that \(mG\) is a subgroup of \(G\); indeed, since \(G\) is abelian, the function \(\mu_m:G\rightarrow G\), defined by \(x\mapsto mx\), is a homomorphism (called multiplication by \(m\)), and \(mG\) = \(\mathrm{im}\mu_m\). We denote \(\mathrm{ker}\mu_m\) by \(G[m]\); that is,

\[G[m]=\{x\in G:mx=\mathbf 0\}. \]

Theorem 6.1.11 (Basic Theorem) Every finite abelian group \(G\) is a direct sum of primary cyclic groups.

Proof

By Theorem 6.1.3, we may assume that \(G\) is \(p\)-primary for some prime \(p\). We prove the theorem by induction on \(n\), where \(p^nG=\mathbf 0\). If \(n=1\), then the theorem is Corollary 6.1.7.

Suppose that \(p^{n+1}G=\mathbf 0\). If \(H=pG\), then \(p^{n}H=\mathbf 0\), so that induction gives \(H=\sum_{i=1}^r\langle y_i\rangle\). Since \(y_i\in H=pG\), there are \(z_i\in G\) with \(pz_i=y_i\). By Lemma 6.1.6, \(\{y_1,\ldots,y_r\}\) is independent; by Lemma 6.1.8(i) \(\{z_1,\ldots,z_r\}\) is independent, and so \(L=\langle z_1,\ldots,z_r\rangle\) is a direct sum: \(L=\sum_{i=1}^r\langle z_i\rangle\).

For each \(i\), let \(k_i\) be the order of \(y_i\), so that \(k_iz_i\) has order \(p\). The linearly independent subset \(\{k_1z_1,\ldots,k_rz_r\}\) of the vector space \(G[p]\) can be extended to a basis: there are elements \(\{x_1,\ldots,x_s\}\) so that \(\{k_1z_1,\ldots,k_rz_r,x_1,\ldots,x_s\}\) is a basis of \(G[p]\). If \(M=\langle x_1,\ldots,x_s\rangle\), then independence gives \(M=\sum\langle x_j\rangle\). We now show that \(M\) consists of the resurrected summands of order \(p\); that is, \(G=L\oplus M\), and this will complete the proof.

1. \(L\cap M=\mathbf 0.\quad\) If \(g\in L\cap M\), then \(g=\sum b_iz_i=\sum a_jx_j\). Now \(pg=\mathbf 0\), because \(g\in M\), and so \(\sum pb_iz_i=\mathbf 0\). By independence, \(pb_iz_i=b_iy_i=\mathbf 0\) for all \(i\). It follows that \(b_i=b_i'k_i\) for some \(b_i'\). Therefore, \(\mathbf 0=\sum b_i'k_iz_i-\sum a_jx_j\), and so independence of \(\{k_1z_1,\ldots,k_rz_r,x_1,\ldots,x_s\}\) gives each term \(\mathbf 0\); hence \(g=\sum a_jx_j=\mathbf 0\).
2. \(L+M=G.\quad\) If \(g\in G\), then \(pg\in pG=H\), and so \(pg=\sum c_iy_i=\sum pc_iz_i\). Hence, \(p(g-\sum c_iz_i)=\mathbf 0\) and \(g-\sum c_iz_i\in G[p]\). Therefore, \(g-\sum c_iz_i=\sum b_ik_iz_i+\sum a_jx_j\), so that \(g=\sum(c_i+b_ik_i)z_i+\sum a_jx_j\in L+M\).
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Corollary 6.1.12 Every finite abelian group \(G\) is a direct sum of cyclic groups: \(G=\sum_{i=1}^t\langle x_i\rangle\), where \(x_i\) has order \(m_i\), and

\[m_1\mid m_2\mid\ldots\mid m_t. \]

Proof

Let the primary decomposition of \(G\) be \(G=\sum_{i=1}^r G_{p_i}\). By the basis theorem, we may assume that each \(G_{p_i}\) is a direct sum of cyclic groups; let \(C_i\) be a cyclic summand of \(G_{p_i}\) of largest order, say, \(p_i^{e_i}\). It follows that \(G=K\oplus(C_1\oplus\cdots\oplus C_r)\), where \(K\) is the direct sum of the remaining cyclic summands. But \(C_1\oplus\cdots\oplus C_t\) is cyclic of order \(m=\prod p_i^{e_i}\). Now repeat this construction: let \(K=H\oplus D\), where \(D\) is cyclic of order \(n\), say. If there is a cyclic summand \(S_i\) in \(D\) arising from \(G_{p_i}\), that is, if \(G_{p_i}\neq C_i\), then \(S_i\) has order \(p_i^{f_i}\le p_i^{e_i}\), so that \(p_i^{f_i}\mid p_i^{e_i}\), for all \(i\), and \(n\mid m\). This process ends in a finite number of steps.
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Definition 6.1.13 If \(G\) has a decomposition as a direct sum \(G=\sum C_i\), where \(C_i\) is cyclic of order \(m_i\) and \(m_1\mid m_2\mid \ldots\mid m_t\), then one says that \(G\) has invariant factors \((m_1,\ldots,m_t)\).

It is obvious that if \(G\) is an abelian group with invariant factors \((m_1,\ldots,m_t)\), then the order of \(G\) is \(\prod m_i\) and the minimal exponent of \(G\) is \(m_t\).

Proposition 6.1.14 If \(G\) is a finite \(p\)-primary abelian group, and if \(x\in G\) has largest order, then \(\langle x\rangle\) is a direct summand of \(G\).

Proposition 6.1.15 If \(p\) is an odd prime, the multiplicative group

\[\mathrm U(\mathbb Z_{p^n})=\{[a]\in\mathbb Z_{p^n}:(a,p)=1\}, \]

is cyclic of order \((p-1)p^{n-1}\).

Proposition 6.1.16 Let \(G\) be a finite \(p\)-primary group and \(\mathrm d(G)\) be the minimal number of generators of \(G\), then:

1. \(\Phi(G)=pG\), and \(\mathrm d(G)=\dim G/pG\);
2. If \(G\) and \(H\) are elementary abelian \(p\)-groups, then \(\mathrm d(G\oplus H)=\mathrm d(G)+\mathrm d(H)\);
3. Let \(G\) be a direct sum of \(b\) cyclic groups of order \(p^m\). If \(n< m\), then \(p^nG/p^{n+1}G\) is elementary and \(\mathrm d(p^nG/p^{n+1}G)=b\).

The Fundamental Theorem of Finite Abelian Groups

Lemma 6.2.1 If a \(p\)-primary abelian group \(G\) has a decomposition \(G=\sum C_i\) into a direct sum of cyclic groups, then the number of \(C_i\) having order \(\ge p^{n+1}\) is \(\mathrm d(p^nG/p^{n+1}G)\), the minimal number of generators of \(p^nG/p^{n+1}G\).

Proof

Let \(B_k\) be the direct sum of all \(C_i\), if any, of order \(p^k\); say, there are \(b_k\ge 0\) such summands in \(B_k\). Thus,

\[G=B_1\oplus\cdots\oplus B_t. \]

Now \(p^nG=p^nB_{n+1}\oplus\cdots\oplus p^nB_t\), because \(p^nB_1=\cdots=p^nB_n=\mathbf 0\), and \(p^{n+1}G=p^{n+1}B_{n+2}\oplus\cdots\oplus p^{n+1}B_t\). Therefore, \(p^nG/p^{n+1}G\cong p^nB_{n+1}\oplus(p^nB_{n+2}/p^{n+1}B_{n+2})\oplus\cdots\oplus(p^nB_t/p^{n+1}B_t)\), and so Proposition 6.1.16 gives \(\mathrm d(p^nG/p^{n+1}G)=b_{n+1}+b_{n+2}+\cdots+b_t\).
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Definition 6.2.2 If \(G\) is a finite \(p\)-primary abelian group and \(n\ge 0\), then

\[\mathrm U_p(n,G)=\mathrm d(p^{n}G/p^{n+1}G)-\mathrm d(p^{n+1}G/p^{n+2}G). \]

Theorem 6.2.3 If \(G\) is a finite \(p\)-primary abelian group, then any two decompositions of \(G\) into direct sums of cyclic groups have the same number of summands of each kind. More precisely, for every \(n\ge 0\), the number of cyclic summands of order \(p^{n+1}\) is \(\mathrm U_p(n,G)\).

Proof

For any decomposition of \(G\) into a direct sum of cyclic groups, the lemma shows that there are exactly \(\mathrm U_p(n,G)\) cyclic summands of order \(p^{n+1}\). The result follows, for \(\mathrm U_p(n,G)\) does not depend on the choice of decomposition.
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Corollary 6.2.4 If \(G\) and \(H\) are finite \(p\)-primary abelian groups, then \(G\cong H\) if and only if \(\mathrm U_p(n,G)=\mathrm U_p(n,H)\) for all \(n\ge 0\).

If \(\varphi\!:G\rightarrow H\) is an isomorphism, then \(\varphi(p^nG)=p^nH\) for all \(n\ge 0\), and so \(\varphi\) includes isomorphisms \(p^nG/p^{n+1}G\cong p^nH/p^{n+1}H\) for all \(n\). Therefore, \(\mathrm U_p(n,G)=\mathrm U_p(n,H)\) for all \(n\).

Conversely, if \(G\) and \(H\) each have direct sum decompositions into cyclic groups with the same number of summands of each kind, then it is easy to construct an isomorphism \(G\rightarrow H\).
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Proposition 6.2.5 If \(G\) and \(H\) are finite abelian groups, then

\[\mathrm U_p(n,G\oplus H)=\mathrm U_p(n,G)+\mathrm U_p(n,H). \]

Proposition 6.2.6

1. If \(A,B\) and \(C\) are finite abelian groups with \(A\oplus C\cong B\oplus C\), then \(A\cong B\).
2. If \(A\) and \(B\) are finite abelian groups for which \(A\oplus A\cong B\oplus B\), then \(A\cong B\).

Definition 6.2.7 The orders of the primary cyclic summands of \(G\), that is, the numbers \(p^{n+1}\) with multiplicity \(\mathrm U_p(n,G)>0\) for all primes \(p\) and all \(n\ge 0\), are called the elementary divisors of \(G\).

Example 6.2.8 Let \(G=\mathbb Z_2\oplus \mathbb Z_2\oplus\mathbb Z_2\oplus \mathbb Z_4\oplus \mathbb Z_3\oplus \mathbb Z_9\), then the elementary divisors of \(G\) are \(\{2,2,2,4;3,9\}\), while the invariant factors of \(G\) are \(\{2,2,6,36\}\).

Theorem 6.2.9 (Fundamental Theorem of Finite Abelian Groups) If \(G\) and \(H\) are finite abelian groups, then \(G\cong H\) if and only if, for all primes \(p\), they have the same elementary divisors.

Corollary 6.2.10 Let \(G\) be a finite abelian group.

1. If \(G\) has invariant factors \((m_1,\ldots,m_t)\) and invariant factors \((k_1,\ldots,k_s)\), then \(s=t\) and \(k_i=m_i\) for all \(i\).
2. Two finite abelian groups \(G\) and \(H\) are isomorphic if and only if they have the same invariant factors.

Proposition 6.2.11 The number of nonisomorphic abelian groups of order \(n=\prod p_i^{e_i}\) is \(\prod_i\mathscr P(e_i)\), where the \(p_i\) are distinct primes and the \(e_i\) are positive integers, and \(\mathscr P(e)\) denotes the number of partitions of \(e\).

Proposition 6.2.12 If \(G\) is a finite abelian group and \(H\le G\), then \(G\) contains a subgroups isomorphic to \(G/H\).

The Krull-Schmidt Theorem

Definition 6.3.1 An endomorphism of a group \(G\) is a homomorphism \(\varphi\!:G\rightarrow G\).

Definition 6.3.2 If \(G=H_1\times\cdots\times H_m\), then the maps \(\pi_i\!:G\rightarrow H_i\), defined by \(\pi_i(h_1\ldots,h_m)=h_i\), are called projections.

If the inclusion \(H_i\hookrightarrow G\) is denoted by \(\lambda_i\), then the maps \(\lambda_i\pi_i\) are endomorphisms of \(G\). Indeed, they are idempotent: \(\lambda_i\pi_i\circ\lambda_i\pi_i=\lambda_i\pi_i\).

Definition 6.3.3 An endomorphism \(\varphi\) of a group \(G\) is normal if \(\varphi(axa^{-1})=a\varphi(x)a^{-1}\) for all \(a,x\in G\).

It is easy to see that if \(G\) is a direct product, then the maps \(\lambda_i\pi_i\) are normal endomorphisms of \(G\).

Proposition 6.3.4

1. If \(\varphi\) and \(\psi\) are normal endomorphisms of a group \(G\), then so is their composite \(\varphi\circ\psi\).
2. If \(\varphi\) is a normal endomorphism of \(G\) and if \(H\lhd G\), then \(\varphi(H)\lhd G\).
3. If \(\varphi\) is a normal automorphism of a group \(G\), then \(\varphi^{-1}\) is also normal.

Definition 6.3.5 If \(\varphi\) and \(\psi\) are endomorphisms of a group \(G\), then \(\varphi+\psi\!:G\rightarrow G\) is the function defined by \(x\mapsto \varphi(x)\psi(x)\).

If \(G\) is abelian, then \(\varphi+\psi\) is always an endomorphism. It is easy to see that if \(\varphi\) and \(\psi\) are normal endomorphisms and if \(\varphi+\psi\) is an endomorphism, then it is normal.

Lemma 6.3.6 Let \(G=H_1\times\cdots\times H_m\) have projections \(\pi_i\!:G\rightarrow H_i\) and inclusions \(\lambda_i\!:H_i\hookrightarrow G\). Then the sum of any \(k\) distinct \(\lambda_i\pi_i\) is a normal endomorphism of \(G\). Moreover, the sum of all the \(\lambda_i\pi_i\) is the identity function on \(G\).

Proof

Note that \(\lambda_i\pi_i(h_1\ldots h_m)=h_i\). If \(\varphi=\sum_{i=1}^k\lambda_i\pi_i\), then \(\varphi(h_1\ldots h_m)=h_1\ldots h_k\); that is, \(\varphi=\lambda\pi\), where \(\pi\) is the projection of \(G\) onto the direct factor \(H_1\times\cdots\times H_k\) and \(\lambda\) is the inclusion of \(H_1\times\cdots\times H_k\) into \(G\). It follows that \(\varphi\) is a normal endomorphism of \(G\) and, if \(k=m\), that \(\varphi=\mathbf 1_G\).
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Definition 6.3.7 A group \(G\) is indecomposable if \(G\neq\mathbf 1\) and if \(G=H\times K\), then either \(H=\mathbf 1\) or \(K=\mathbf 1\).

Definition 6.3.8 A group \(G\) has ACC (ascending chain condition) if every increasing chain of normal subgroups stops; that is, if

\[K_1\le K_2\le K_3\le\cdots \]

is a chain of normal subgroups of \(G\), then there is an integer \(t\) for which \(K_t=K_{t+1}=K_{t+2}=\cdots\).

A group \(G\) has DCC (descending chain condition) if every decreasing chain of normal subgroups stops; that is, if

\[H_1\ge H_2\ge H_3\ge\cdots \]

is a chain of normal subgroups of \(G\), then there is an integer \(s\) for which \(H_s=H_{s+1}=H_{s+2}=\cdots\).

Every finite group has both chain conditions. The group \(\mathbb Z\) has ACC but not DCC; the additive group of rationals \(\mathbb Q\) has neither ACC nor DCC.

Lemma 6.3.9

1. If \(H\lhd G\) and both \(H\) and \(G/H\) have both hain conditions, then \(G\) has both chain conditions. In particular, if \(H\) and \(K\) have both chain conditions, then so does \(H\times K\).
2. If \(G=H\times K\) and \(G\) has both chain conditions, then each of \(H\) and \(K\) has both chain conditions.

Proof

1. If \(G_1\ge G_2\ge\cdots\) is a chain of normal subgroups of \(G\), then \(H\cap G_1\ge H\cap G_2\ge\cdots\) is a chain of normal subgroups of \(H\) and \(HG_1/H\ge HG_2/H\ge\cdots\) is a chain of normal subgroups of \(G/H\). By hypothesis, there is an integer \(t\) with \(H\cap G_t=H\cap G_{t+1}=\cdots\), and there is an integer \(s\) with \(HG_s/H=HG_{s+1}/H=\cdots\); that is, \(HG_s=HG_{s+1}=\cdots\). Let \(l=\max\{s,t\}\). By the Dedekind law, for all \(i\ge l\),

\[\begin{aligned} G_i&=G_iH\cap G_i=G_{i+1}H\cap G_i\\ &=G_{i+1}(H\cap G_i)=G_{i+1}(H\cap G_{i+1})\le G_{i+1}, \end{aligned} \]

and so \(G_{i}=G_{i+1}=\cdots\). A similar argument holds for ascending chains.

2. If \(G=H\times K\), then every normal subgroup of \(H\) is also a normal subgroup of \(G\). Therefore, every (ascending or descending) chain of normal subgroups of \(H\) is a chain of normal subgroups of \(G\), hence must stop.
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Lemma 6.3.10 If \(G\) has either chain condition, then \(G\) is a direct product of a finite number of indecomposition groups.

Proof

Call a group good if it satisfies the conclusion of this lemma; call it bad otherwise. An indecomposable group is good and, if both \(A\) and \(B\) are good groups, then so is \(A\times B\). Therefore, a bad group \(G\) is a direct product, say, \(G=U\times V\), with both \(U\) and \(V\) proper subgroups, and with \(U\) or \(V\) bad.

Suppose there is a bad group \(G\). Define \(H_0=G\). By induction, for every \(n\), there are bad subgroups \(H_0,H_1,\ldots,H_n\) such that each \(H_i\) is a proper bad direct factor of \(H_{i-1}\). There is thus a strictly decreasing chain of normal subgroups of \(G\)

\[G=H_0>H_1>H_2>\cdots; \]

if \(G\) has DCC, we have reached a contradiction.

Suppose that \(G\) has ACC. Since each \(H_i\) is a direct factor of \(H_{i-1}\), there are normal subgroups \(K_i\) with \(H_{i-1}=H_i\times K_i\). There is thus an ascending chain of normal subgroups

\[K_1<K_1\times K_2<K_1\times K_2\times K_3<\cdots, \]

and we reach a contradiction in this case, too.
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Lemma 6.3.11 Let \(G\) have both chain conditions. If \(\varphi\) is a normal endomorphism of \(G\), then \(\varphi\) is an injection if and only if it is a surjection. (Thus, either property ensures that \(\varphi\) is an automorphism.)

Proof

Suppose that \(\varphi\) is an injection and that \(g\notin \varphi(G)\). We prove, by induction, that \(\varphi^n(g)\notin\varphi^{n+1}(g)\). Otherwise, there is an element \(h\in G\) with \(\varphi^n(g)=\varphi^{n+1}(h)\), so that \(\varphi(\varphi^{n-1}(g))=\varphi(\varphi^n(h))\). Since \(\varphi\) is an injection, \(\varphi^{n-1}(g)=\varphi^n(h)\), contradicting the inductive hypothesis. There is thus a strictly decreasing chain of subgroups

\[G>\varphi(G) >\varphi^2(G)>\cdots. \]

Now \(\varphi\) normal implies \(\varphi^n\) is normal; by Lemma 6.3.9 \(\varphi^n(G)\lhd G\) for all \(n\), and so the DCC is violated. Therefore, \(\varphi\) is a surjection.

Assume that \(\varphi\) is a surjection. Define \(K_n=\mathrm{ker} \varphi^n\); each \(K_n\) is a normal subgroup of \(G\) because \(\varphi^n\) is a homomorphism (the normality of \(\varphi^n\) is here irrelevant). There is an ascending chain of normal subgroups

\[\mathbf 1=K_0\le K_1\le K_2\le \cdots. \]

This chain stops because \(G\) has ACC; let \(t\) be the smallest integer for which \(K_t=K_{t+1}=K_{t+2}=\cdots\). We claim that \(t=0\), which will prove the result. If \(t\ge 1\), then there is \(x\in K_t\) with \(x\notin K_{t-1}\); that is, \(\varphi^t(x)=\mathbf 1\) and \(\varphi^{t-1}(x)\neq \mathbf 1\). Since \(\varphi\) is a surjection, there is \(g\in G\) with \(x=\varphi(g)\). Hence, \(\mathbf 1=\varphi^t(x)=\varphi^{t+1}(g)\), so that \(g\in K_{t+1}=K_t\). Therefore, \(\mathbf 1=\varphi^t(g)=\varphi^{t-1}(\varphi(g))=\varphi^{t-1}(x)\), a contradiction. Thus, \(\varphi\) is an injection.
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Definition 6.3.12 An endomorphism \(\varphi\) of \(G\) is nilpotent if there is a positive integer \(k\) such that \(\varphi^k=\mathbf 0\), where \(\mathbf 0\) denotes the endomorphism which sends every element of \(G\) into \(\mathbf 1\).

Theorem 6.3.13 (Fitting's Lemma) Let \(G\) have both chain conditions and let \(\varphi\) be a normal endomorphism of \(G\). Then \(G=K\times H\), where \(K\) and \(H\) are each invariant under \(\varphi\) (i.e., \(\varphi(K)\le K\) and \(\varphi(H)\le H\)), \(\varphi\big|_K\) is nilpotent, and \(\varphi\big|_H\) is a surjection.

Proof

Let \(K_n=\mathrm {ker}\varphi^n\) and let \(H_n=\mathrm{im}\varphi^n\). As in the proof of Lemma 6.3.11, there are two chains of normal subgrouups of \(G\):

\[G\ge H_1\ge H_2\ge\cdots\qquad\text{and}\qquad \mathbf 1\le K_1\le K_2\le\cdots. \]

Since \(G\) has both chain conditions, each of these chains stops: the \(H_n\) after \(t\) steps, the \(K_n\) after \(s\) steps. Let \(l\) be the larger of \(t\) and \(s\), so that \(H_l=H_{l+1}=H_{l+2}=\cdots\) and \(K_l=K_{l+1}=K_{l+2}=\cdots\). Define \(H=H_l\) and \(K=K_l\); it is easy to check that both \(H\) and \(K\) are invariant under \(\varphi\).

Let \(x\in H\cap K\). Since \(x\in H\), there is \(g\in G\) with \(x=\varphi^l(g)\); since \(x\in K\), \(\varphi^l(x)=\mathbf 1\). Therefore, \(\varphi^{2l}(g)=\varphi^l(x)=\mathbf 1\), so that \(g\in K_{2l}=K_l\). Hence, \(x=\varphi^l(g)=\mathbf 1\), and so \(H\cap K=\mathbf 1\).

If \(g\in G\), then \(\varphi^l(g)\in H_l=H_{2l}\), so there is \(y\in G\) with \(\varphi^l(g)=\varphi^{2l}(y)\). Applpying \(\varphi^l\) to \(g\varphi^l(y^{-1})\) gives \(\mathbf 1\), so that \(g\varphi^l(y^{-1})\in K_l=K\). Therefore, \(g=[g\varphi^l(y^{-1})]\varphi^l(y)\in KH\), and so \(G=K\times H\).

Now \(\varphi(H)=\varphi(H_l)=\varphi(\varphi^l(G))=\varphi^{l+1}(G)=H_{l+1}=H_l=H\), so that \(\varphi\) is a surjection. Finally, if \(x\in K\), then \(\varphi^l(x)\in K\cap H=\mathbf 1\), and so \(\varphi\big|_K\) is nilpotent.
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Corollary 6.3.14 If \(G\) is an indecomposable group having both chain conditions, then every normal endomorphism \(\varphi\) of \(G\) is either nilpotent or an automorphism.

Proof

By Fitting's lemma, \(G=K\times H\) with \(\varphi\big|_K\) nilpotent and \(\varphi\big|_H\) surjective. Since \(G\) is indecomposable, either \(G=K\) or \(G=H\). In the first case, \(\varphi\) is nilpotent. In the second case, \(\varphi\) is surjective and, by Lemma 6.3.11, \(\varphi\) is an automorphism.
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Lemma 6.3.15 Let \(G\) be an indecomposable group with both chain conditions, and let \(\varphi\) and \(\psi\) be normal nilpotent endomorphisms of \(G\). If \(\varphi+\psi\) is an endomorphism of \(G\), then it is nilpotent.

Proof

We have already observed that if \(\varphi+\psi\) is an endomorphism, then it is normal, and so Corollary 6.3.14 says that it is either nilpotent or an automorphism. If \(\varphi+\psi\) is an automorphism, then Proposition 6.3.4 says that its inverse \(\gamma\) is also normal. For each \(x\in G\), \(x=(\varphi+\psi)\gamma x=\varphi\gamma(x)\psi\gamma(x)\), so that if we define \(\lambda=\varphi\gamma\) and \(\mu=\psi\gamma\), then \(\mathbf 1_G=\lambda+\mu\). In particular, \(x^{-1}=\lambda(x^{-1})\mu(x^{-1})\) and, taking inverses, \(x=\mu(x)\lambda(x)\); that is, \(\lambda+\mu=\mu+\lambda\). The equation \(\lambda(\lambda+\mu)=(\lambda+\mu)\lambda\) (which holds because \(\lambda+\mu=\mathbf 1_G\)) implies that \(\lambda\mu=\mu\lambda\). It follows that the set of all endomorphisms of \(G\) obtained from \(\lambda\) and \(\mu\) forms an algebraic system in which the binomial theorem holds: for every integer \(m>0\),

\[(\lambda+\mu)^m=\sum_i\binom{m}{i} \varphi^i\psi^{m-i}. \]

Nilpotence of \(\varphi\) and \(\psi\) implies nilpotence of \(\lambda=\varphi\gamma\) and \(\mu=\psi\gamma\) (they cannot be automorphisms because they have notrivial kernels); there are thus positive integers \(r\) and \(s\) with \(\lambda^{r}=\mathbf 0\) and \(\mu^s=\mathbf 0\). If \(m=r+s-1\), then either \(i\ge r\) or \(m-i\ge s\). It follows that each term \(\lambda^i\mu^{m-i}\) in the binomial expansion of \((\lambda+\mu)^m\) is \(\mathbf 0\): if \(i\ge r\), then \(\lambda^i=\mathbf 0\); if \(m-i\ge s\), then \(\mu^{m-i}=\mathbf 0\). Hence, \((\mathbf 1_G)^m=(\lambda+\mu)^m=\mathbf 0\), and \(\mathbf 1_G=\mathbf 0\), forcing \(G=\mathbf 1\). This is a contradiction, for every indecomposable group is nontrivial.
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Corollary 6.3.16 Let \(G\) be an indecomposable group having both chain conditions. If \(\varphi_1,\ldots,\varphi_n\) is a set of normal nilpotent endomorphisms of \(G\) such that every sum of distinct \(\varphi\)'s is an endomorphism, then \(\varphi_1+\cdots+\varphi_n\) is nilpotent.

Theorem 6.3.17 (Krull-Schmidt) Let \(G\) be a group having both chain conditions. If

\[G=H_1\times\cdots\times H_s=K_1\times\cdots\times K_t \]

are two decompositions of \(G\) into indecomposable factors, then \(s=t\) and there is a reindexing so that \(H_i\cong K_i\) for all \(i\). Moreover, given any \(r\) between \(1\) and \(s\), the reindexing may be chosen so that

\[G=H_1\times\cdots\times H_r\times K_{r+1}\times\cdots\times K_s. \]

Proof

We shall give the proof when \(r=1\); The rest for general \(r\) can be completed by induction. Given the first decomposition, we must find a reindexing of the \(K\)'s so that \(H_i\cong K_i\) for all \(i\) and \(G=H_1\times K_2\times\cdots\times K_t\). Let \(\pi_i\!:G\rightarrow H_i\) and \(\lambda_i\!:H_i\hookrightarrow G\) be the projections and inclusions from the first decomposition, and let \(\sigma_j\!:G\rightarrow K_j\) and \(\mu_j\!:K_j\hookrightarrow G\) be the projections and inclusions from the second decomposition. The maps \(\lambda_i\pi_i\) and \(\mu_j\sigma_j\) are normal endomorphisms of \(G\).

By Lemma 6.3.6, every partial sum \(\sum \mu_j\sigma_j\) is a normal endomorphism of \(G\). Hence, every partial sum of

\[\mathbf 1_{H_1}=\pi_1\lambda_1=\pi_1\circ\mathbf 1_G\circ\lambda_1=\pi_1\circ\left(\sum\mu_j\sigma_j\right)\circ\lambda_1=\sum\pi_1\mu_j\sigma_j\lambda_1 \]

is a normal endomorphism of \(H_1\). Since \(\mathbf 1_H=\pi_1\lambda_1\) is not nilpotent, Lemma 6.3.9 and Corollary 6.3.16 give an index \(j\) with \(\pi_1\mu_j\sigma_j\lambda_1\) an automorphism. We reindex so that \(\pi_1\mu_1\sigma_1\lambda_1\) is an automorphism of \(H_1\); let \(\gamma\) be its inverse.

We claim that \(\sigma_1\lambda_1\!:H_1\rightarrow K_1\) is an isomorphism. The definition of \(\gamma\) gives \((\gamma\pi_1\mu_1)(\sigma_1\lambda_1)=\mathbf 1_H\). To compute the composite in the reverse order, let \(\theta=\sigma_1\lambda_1\gamma_1\pi_1\mu_1\!:K_1\rightarrow K_1\), and note that \(\theta^2=\theta\):

\[\theta\circ\theta=\sigma_1\lambda_1[\gamma\pi_1\mu_1\sigma_1\lambda_1]\gamma\pi_1\mu_1=\theta. \]

Now \(\mathbf 1_{H_1}=\mathbf 1_{H_1}\circ\mathbf 1_{H_1}=\gamma\pi_1\mu_1\sigma_1\lambda_1\gamma\pi_1\mu_1\sigma_1\lambda_1=\gamma\pi_1\mu_1\theta\sigma_1\lambda_1\); it follows that \(\theta\neq\mathbf 0\) (lest \(\mathbf 1_{H_1}=\mathbf 0\)). Were \(\theta\) nilpotent, then \(\theta^2=\theta\) would force \(\theta=\mathbf 0\). Therefore, \(\theta\) is an automorphism of \(K_1\), and so \(\theta^2=\theta\) gives \(\theta=\mathbf 1\). It follows that \(\sigma_1\lambda_1\!:H_1\rightarrow K_1\) is an isomorphism (with inverse \(\gamma\pi_1\mu_1\)).

Now \(\sigma_1\) sends \(K_2\times \cdots\times K_t\) into \(\mathbf 1\) while \(\sigma_1\lambda_1\) restricts to an isomorphism on \(H_1\). Therefore

\[H_1\cap(K_2\times\cdots\times K_t)=\mathbf 1. \]

If we define \(G^*=\langle H_1,K_2\times\cdots\times K_t\rangle\le G\), then

\[G^*=H_1\times K_2\times\cdots\times K_t. \]

If \(x\in G\), then \(x=k_1k_2\ldots k_t\), where \(k_j\in K_j\). Since \(\pi_1\mu_1\) is an isomorphism, the map \(\beta\!:G\rightarrow G\), defined by \(x\mapsto \pi_1\mu_1(k_1)k_2\ldots k_t\), is an injection with image \(G^*\). By Lemma 6.3.11, \(\beta\) is a surjection; that is, \(G=G^*=H_1\times K_2\times\cdots\times K_t\). Finally,

\[K_2\times\cdots\times K_t\cong G/H_1\cong H_2\times\cdots\times H_s, \]

so that the remaining uniqueness assertions follow by induction on \(\max\{s,t\}\).
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Proposition 6.3.18 Let \(G\) be a group with both chain conditions. If there is a group \(H\) with \(G\times G\cong H\times H\), then \(G\cong H\).

Definition 6.3.19 A subgroup \(H\le G\) is subnormal if there is a normal series

\[G\rhd G_1\rhd G_2\cdots\rhd H\rhd \mathbf 1. \]

Proposition 6.3.20

1. If \(G\) has a composition series and if \(H\) is subnormal in \(G\), then \(G\) has a composition series one of whose terms is \(H\).
2. A group \(G\) has a composition series if and only if it has both chain conditions on subnormal subgroups.

Operator Groups

Definition 6.4.1 Let \(\Omega\) be a set and let \(G\) be a group. Then \(\Omega\) is a set of operators on \(G\) and \(G\) is an \(\Omega\)-group if there is a function \(\Omega\times G\rightarrow G\), denoted by \((\omega,g)\mapsto\omega g\), such that

\[\omega(gh)=(\omega g)(\omega h) \]

for all \(\omega\in\Omega\) and \(g,h\in G\).

Definition 6.4.2 If \(G\) and \(H\) are \(\Omega\)-groups, then a function \(\varphi\!:G\rightarrow H\) is an \(\Omega\)-map if \(\varphi\) is a homomorphism such that \(\varphi(\omega g)=\omega\varphi(g)\) for all \(\omega\in\Omega\) and \(g\in G\).

If \(G\) is an \(\Omega\)-group, then a subgroup \(H\le G\) is an admissible subgroup if \(\omega h\in H\) for all \(\omega\in\Omega\) and \(h\in H\).

Example 6.4.3

1. If \(\Omega=\varnothing\), then an \(\Omega\)-group is just a group, every homomorphism is an \(\Omega\)-map, and every subgroup is admissible.
2. If \(G\) is an abelian group and \(\Omega\) is a ring, then every \(\Omega\)-module is an \(\Omega\)-group, every \(\Omega\)-homomorphism is an \(\Omega\)-map, and every submodule is admissible.
3. If \(\Omega\) is the set of all conjugations of a group \(G\), then \(G\) is an \(\Omega\)-group whose admissible subgroups are the normal subgroups. An \(\Omega\)-map to \(G\) to itself is a normal endomorphism. An \(\Omega\)-isomorphism between \(\Omega\)-groups is called a central isomorphism.
4. If \(\Omega\) is the set of all automorphisms of a group \(G\), then \(G\) is an \(\Omega\)-group whose admissible subgroups are the characteristic subgroups.
5. If \(\Omega\) is the set of all endomorphisms of a group \(G\), then \(G\) is an \(\Omega\)-group whose admissible subgroups are the fully invariant subgroups.

Proposition 6.4.4

1. The intersection of admissible subgroups is admissible.
2. If \(H\) and \(K\) are admissible subgroups at least one of which is normal, then \(HK\) is admissible.
3. If \(H\) is a normal admissible subgroup of an \(\Omega\)-group \(G\), then the quotient group \(G/H\) is an \(\Omega\)-group (where one defines \(\omega(gH)=(\omega g)H\)).
4. The kernel of an \(\Omega\)-map is an admissible normal subgroup, and the isomorphism theorems hold, as does the correspondence theorem.
5. The direct product of \(\Omega\)-groups \(H\) and \(K\) becomes an \(\Omega\)-group if one defines \(\omega(h,k)=(\omega h,\omega k)\).

Definition 6.4.5 An \(\Omega\)-group \(G\) is \(\Omega\)-simple if has no admissible subgroups other than \(\mathbf 1\) and \(G\).

An admissible normal subgroup \(H\) of an \(\Omega\)-group \(G\) is maximal such if and only if \(G/H\) is \(\Omega\)-simple.

Definition 6.4.6 Let \(G\) be an \(\Omega\)-group. An \(\Omega\)-series for \(G\) is a normal series

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

with each \(G_i\) admissible; an \(\Omega\)-composition series is an \(\Omega\)-series whose factor groups are \(\Omega\)-simple.

An \(\Omega\)-group is \(\Omega\)-indecomposable if it is not the direct product of nontrivial admissible subgroups.

Example 6.4.7

1. If \(\Omega\) is the set of all conjugations of \(G\), then an \(\Omega\)-series is a normal series \(G=G_0\rhd G_1\rhd\cdots\rhd G_n=\mathbf 1\) in which each \(G_i\) is a normal subgroup of \(G\), and such a normal series is called a chief series or a principal series.
2. If \(\Omega\) is the set of all automorphisms of a finite group \(G\), then \(G\) is \(\Omega\)-simple (or characteristically simple) if and only if it is a direct product of isomorphic simple groups (Theorem 5.3.15).
3. If \(\Omega\) is a ring, then an \(\Omega\)-module \(V\) with no submodules other than \(\mathbf 0\) and \(V\) is an \(\Omega\)-simple group. In particular, when \(\Omega\) is a field, a one-dimensional \(\Omega\)-module is \(\Omega\)-simple.

Theorem 6.4.8 (Jordan-Hölder) Every two \(\Omega\)-composition series of an \(\Omega\)-group are equivalent.

Corollary 6.4.9

1. (Let \(\Omega=\) conjugations) Any two chief composition series of a group have centrally isomorphic factor groups.
2. (Let \(\Omega=\) automorphisms) Any two characteristic series in which the factor groups are products of isomorphic simple groups have isomorphic factor groups.

Theorem 6.4.10 (Fitting's Lemma) Let \(G\) be an \(\Omega\)-group having both chain conditions on admissible subgroups, and let \(\varphi\) be an \(\Omega\)-endomorphism of \(G\). Then \(G=H\times K\), where \(H\) and \(K\) are admissible subgroups, \(\varphi\big|_H\) is nilpotent, and \(\varphi\big|_K\) is a surjection.

Theorem 6.4.11 (Krull-Schmidt) Let \(G\) be an \(\Omega\)-group having both chain conditions on admissible subgroups. If

\[G=H_1\times\cdots\times H_s=K_1\times\cdots\times K_t \]

are two decompositions of \(G\) into \(\Omega\)-indecomposable factors, then \(s=t\) and there is a reindexing so that \(H_i\cong K_i\) for all \(i\). Moreover, given any \(r\) between \(1\) and \(s\), the reindexing may be chosen so that \(G=H_1\times\cdots\times H_r\times K_{r+1}\times\cdots\times K_s\).

Appendix A

Groups of Small Order

Definition A.1.1 The following four permutations form a group \(\mathbf V\) (which is called the 4-group):

\[1;\qquad(1\ \ 2)(3\ \ 4);\qquad(1\ \ 3)(2\ \ 4);\qquad(1\ \ 4)(2\ \ 3). \]

Definition A.1.2 The dihedral group \(\mathrm D_{2n}\), for \(2n\ge 4\), is a group of order \(2n\) which is generated by two elements \(s\) and \(t\) such that

\[s^n=\mathbf 1,\qquad t^2=\mathbf 1,\qquad\text{and}\qquad tst=s^{-1}. \]

It is easily observed that \(\mathrm D_4\cong\mathbf V\), \(\mathrm D_6\cong \mathrm S_3\), and \(\mathrm D_{12}\cong\mathrm S_3\times\mathbb Z_2\).

Theorem A.1.3 If \(G\) is a finite group and if \(a,b\in G\) have order \(2\), then \(\langle a,b\rangle\cong \mathrm D_{2n}\) for some \(n\).

Proof

Since \(G\) is finite, the element \(ab\) has finite order \(n\), say. If \(s=ab\), then \(asa=a(ab)a=ba=(ab)^{-1}=s^{-1}\), because both \(a\) and \(b\) have order \(2\).

It remains to show that \(|\langle a,b\rangle|=2n\). We claim that \(as^{i}\neq\mathbf 1\) for all \(i\ge 0\). Otherwise, choose \(i\ge 0\) minimal with \(as^{i}=\mathbf 1\). Now \(i\neq 0\) (for \(a\neq\mathbf 1\)) and \(i\neq 1\) (lest \(\mathbf 1=as=aab=b\)), so that \(i\ge 2\). But \(\mathbf 1=as^{i}=aabs^{i-1}=bs^{i-1}\); conjugating by \(b\) gives \(\mathbf 1=s^{i-1}b=s^{i-2}abb=s^{i-2}a\), and conjugating by \(a\) now gives \(as^{i-2}=\mathbf 1\), contradicting the minimal choice of \(i\). It follows that \(as^{i}\neq s^j\) for all \(i,j\); hence \(\langle a,b\rangle\) contains the disjoint union \(\langle s\rangle\cup a\langle s\rangle\), and so \(|\langle a,b\rangle|=|\langle a,s\rangle|\ge 2n\). For the reverse inequality, it suffices to show that

\[H=\{a^js^i:0\le j<2,0\le i<n\} \]

is a subgroup. It only needs to check four cases: \(as^ias^k=s^{-i}s^k=s^{k-i}\in H\); \(as^{i}s^k=as^{i+k}\in H\); \(s^is^k=s^{i+k}\in H\); \(s^ias^k=a(as^ia)s^k=as^{-i}s^k=as^{k-i}\in H\).
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Theorem A.1.4 If \(p\) is a prime, then every group \(G\) of order \(2p\) is either cyclic or dihedrdal.

Proof

If \(p=2\), then \(|G|=4\), and the result is trivial. If \(p\) is an odd prime, then Cauchy's theorem shows that \(G\) contains an element \(s\) of order \(p\) and an element \(t\) of order \(2\). If \(H=\langle s\rangle\), then \(H\) has index \(2\) in \(G\), and so \(H\lhd G\). Therefore, \(tst=s^{i}\) for some \(i\). Now \(s=t^2st^2=t(tst)t=ts^it=s^{i^2}\); hence, \(i^2\equiv 1\pmod p\) and, because \(p\) is prime, we have \(i\equiv\pm 1\pmod p\). Thus, either \(tst=s\) or \(tst=s^{-1}\). In the first case, \(s\) and \(t\) commute, \(G\) is abelian, and we have \(G\cong \mathbb Z_p\times\mathbb Z_2\cong \mathbb Z_{2p}\); in the second case, \(G\cong\mathrm D_{2p}\).
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Definition A.1.5 The quaternions is a group \(\mathbf Q=\langle a,b\rangle\) of order \(8\) with \(a^4=\mathbf 1,b^2=a^2\), and \(bab^{-1}=a^{-1}\).

Proposition A.1.6

1. Every subgroup of \(\mathbf Q\) is normal.
2. \(\mathbf Q\) contains no subgroup isomorphic to \(\mathbf Q/\mathrm Z(\mathbf Q)\).
3. If \(G\) is a group each of whose subgroups is normal, we call \(G\) hamiltonian. Let \(G\) be finite, then \(G\) is hamiltonian if and only if \(G=\mathbf Q\times A\times B\), where \(A\) is a (necessarily abelian) group of exponent \(2\) and \(B\) is an abelian group in which every element has odd order.

Definition A.1.7 If \(n\ge 3\), a generalized quaternion group (or dicyclic group) is a group \(\mathbf Q_n\) of order \(2^n\) generated by elements \(a\) and \(b\) such that

\[a^{2^{n-1}}=\mathbf 1,\qquad bab^{-1}=a^{-1},\qquad\text{and}\qquad b^2=a^{2^{n-2}}. \]

Proposition A.1.8

1. If \(G\) is a group of ordrer \(2^n\) which is generated by elements \(a\) and \(b\) such that \(a^{2^{n-2}}=b^2=(ab)^2\), then \(G\cong\mathbf Q_n\).
2. \(\mathbf Q_n\) has a unique involution \(z\), and \(\mathrm Z(\mathbf Q_n)=\langle z\rangle\).
3. \(\mathbf Q_n/\mathrm Z(\mathbf Q_n)\cong\mathrm D_{2^{n-1}}\).

Theorem A.1.9 \(\mathbf Q\) and \(\mathrm D_8\) are the only nonabelian groups of order \(8\).

Proof

A nonabelian group \(G\) of order \(8\) has no element of order \(8\) (lest it be cyclic), and not every nonidentity element has order \(2\); thus, \(G\) has an element \(a\) of order \(4\). Now \(\langle a\rangle\lhd G\), for it has index \(2\), and \(G/\langle a\rangle\cong\mathbb Z_2\). If \(b\in G\) and \(b\notin\langle a\rangle\), then \(b^2\in\langle a\rangle\). If \(b^2=a\) or \(b^2=a^3\), then \(b\) has order \(8\), a contradiction; therefore, either

\[b^2=a^2\qquad\text{or}\qquad b^2=\mathbf 1. \]

Furthermore, \(bab^{-1}\in\langle a\rangle\), because \(\langle a\rangle\) is normal, so that

\[bab^{-1}=a\qquad\text{or}\qquad bab^{-1}=a^3 \]

(these are the only possibilities because \(a\) and \(bab^{-1}\) have the same order). The first case is ruled out, for \(G=\langle a,b\rangle\) and \(G\) is abelian if \(a\) and \(b\) commute. The following case remain:

1. \(a^4=\mathbf 1,\,b^2=a^2\), and \(bab^{-1}=a^3\); and
2. \(a^4=\mathbf 1,\,b^2=\mathbf 1\), and \(bab^{-1}=a^3\).

Since \(a^3=a^{-1}\), (i) describes \(\mathbf Q\) and (ii) describes \(\mathrm D_8\).
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Proposition A.1.10 If \(G\) is a group of order \(8\) having only one involution, then either \(G\cong\mathbb Z_8\) or \(G\cong\mathbf Q\).

Theorem A.1.11 If \(G\) is a \(p\)-group having a unique subgroup of order \(p\) and more than one cyclic subgroup of index \(p\), then \(G\cong\mathbf Q\).

Proof

If \(A\) is a subgroup of \(G\) of index \(p\), then \(A\lhd G\), by Theorem 5.5.20. Thus, if \(x\in G\), then \(Ax\in G/A\), a group of order \(p\), and so \(x^p\in A\). Let \(A=\langle a\rangle\) and \(B=\langle b\rangle\) be distinct subgroups of index \(p\), and let \(D=A\cap B\); note that \(D\lhd G\), for it is the intersection of normal subgroups. Our initial remarks show that the subset

\[G^p=\{x^p:x\in G\} \]

is contained in \(D\). Since \(A\) and \(B\) are distinct maximal subgroups, it follows that \(AB=G\), and so the product formula gives \([G:D]=p^2\). Hence, \(G/D\) is abelian and \(G'\le D\). As \(G=AB\), each \(x\in G\) is a product of a power of \(a\) and a power of \(b\); but every element of \(D\) is simultaneously a power of \(a\) and a power of \(b\), and so it commutes with each \(x\in G\); that is, \(D\le\mathrm Z(G)\). We have seen that

\[G'\le D\le\mathrm Z(G) \]

so that the hypothesis of Lemma 2.5.6(i) holds. Hence, for every \(x,y\in G\), \([y,x]^p=[y^p,x]\). But \(y^p\in D\le\mathrm Z(G)\), and so \([y,x]^p=\mathbf 1\). Now Lemma 2.5.6(ii) gives \((xy)^p=[y,x]^{p(p-1)/2}x^py^p\). If \(p\) is odd, then \(p|(p-1)/2\), and \((xy)^p=x^py^p\). By Proposition 2.4.9, if \(G[p]=\{x\in G:x^p=\mathbf 1\}\) and \(G^p=\{x^p:x\in G\}\), then both these subsets are subgroups and \([G:G[p]]=|G^p|\). Thus,

\[|G[p]|=[G:G^p]=[G:D][D:G^p]\ge p^2, \]

and \(G[p]\) contains a subgroup \(E\) of order \(p^2\); but \(E\) must be elementary abelian, so that \(G[p]\), hence \(G\), contains more than one subgroup of order \(p\). We conclude that \(p=2\).

When \(p=2\), the commutator identity gives

\[(xy)^4=[y,x]^6x^4y^4 \]

for all \(x,y\in G\). Since \([y,x]^2=\mathbf 1\), the map \(\varphi\!:x\mapsto x^4\) is a homomorphism, and so \(G[4]=\{x\in G:x^4=\mathbf 1\}=\mathrm{ker}\varphi\) and \(G^4=\{x^4:x\in G\}=\mathrm{im}\varphi\) are subgroups of \(G\). The first isomorphism theorem gives

\[|G[4]|=[G:G^4]=[G:D][D:G^4]=4[D:G^4]. \]

Now \(A/D=A/(A\cap B)\cong AB/B=G/B\), for \(A\) and \(B\) are distinct maximal subgroups of \(G\). Therefore, \([A:D]=2\) and \(D=\langle a^2\rangle\). Now \(G^2\le\langle a\rangle\cap\langle b\rangle=D=\langle a^2\rangle\), so that \(G^4\le\langle a^4\rangle\). It follows that \(G^4=\langle a^4\rangle\), for the reverse inclusion \(\langle a^4\rangle\le G^4\) is obvious. Therefore, \([D:G^4]=[\langle a^2\rangle:\langle a^4\rangle]=2\), and \(|G[4]|=8\). It follows that \(G[4]\cong\mathbf Q\), for it has order \(8\), exponent \(4\), and a unique subgroup of order \(2\). If \(a^4\neq\mathbf 1\), then \(D=\langle a^2\rangle\) has order \(2^m\ge 4\), and hence it contains a cyclic subgroup \(\langle u\rangle\) of order \(4\); of course, \(\langle u\rangle\le G[4]\). There is a second cyclic subgroup \(\langle v\rangle\le G[4]\cong \mathbf Q\). Since \(\langle u\rangle\le D\le\mathrm Z(G)\), however, the subgroup \(H=\langle u\rangle\langle v\rangle\) is abelian. But \(H\) contains more than one involution, for either \(u^2\neq v^2\) or \(u^2\neq uv^{-1}\). Thus, \(a^4=\mathbf 1, |D|=|\langle a^2\rangle|=2\), and \(|G|=8\). Therefore, \(G=G[4]\cong\mathbf Q\).
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Lemma A.1.12 Let \(\mathrm U(\mathbb Z_{2^m})\) be the multiplicative group

\[\mathrm U(\mathbb Z_{2^m})=\{[a]\in\mathbb Z_{2^m}:a\text{ is odd}\}. \]

If \(m\ge 3\), then

\[\mathrm U(\mathbb Z_{2^m})=\langle[-1],[5]\rangle\cong\mathbb Z_2\times\mathbb Z_{2^{m-2}}. \]

Proof

It is easily known that \(|\mathrm U(\mathbb Z_{2^m})|=\varphi(2^m)=2^{m-1}\), and \(5^{2^{m-3}}=1+2^{m-1}\pmod{2^m}\).

Since \(\mathrm U(\mathbb Z_{2^m})\) is a \(2\)-group, \([5]\) has order \(2^s\), for some \(s\ge m-2\). Of course, \([-1]\) has order \(2\). We claim that \(\langle[5]\rangle\cap\langle[-1]\rangle=1\). If not, then \([5^{t}]=[-1]\) for some \(t\); that is, \(5^t\equiv -1\pmod{2^m}\). Since \(m\ge 3\), this congruence implies \(5^{t}\equiv -1\pmod 4\), a contradiction. It follows that these two cyclic subgroups generate their direct product, which is a subgroup of order at least \(2\times 2^s\ge 2\times 2^{m-2}=2^{m-1}=\varphi(2^m)\). This subgroup is thus all of \(\mathrm U(\mathbb Z_{2^m})\).
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Corollary A.1.13 Let \(G\) be a group containing elements \(x\) and \(y\) such that \(x\) has order \(2^m\) (where \(m\ge 3\)), \(y^2=x^{2^r}\), and \(yxy^{-1}=x^t\). Then

\[t=\pm 1\qquad\text{or}\qquad t=\pm 1+2^{m-1}. \]

In the latter two cases, \(G\) contains at least two involutions.

Proof

Since \(y^2=x^{2^r}\) commutes with \(x\), we have

\[x=y^2xy^{-2}=yx^ty^{-1}=x^{t^2}, \]

so that \(t^2\equiv 1\pmod {2^m}\), and the congruence class \([t]\) is an element of order \(2\) in \(\mathrm U(\mathbb Z_{2^m})\). If \(m\ge 3\), the lemma exhibits the only four such elements, and this gives the first statement.

One involution in \(G\) is \(x^{2^{m-1}}\). Suppose \(t=1+2^{m-1}\). For any integer \(k\),

\[(x^ky)^2=x^k(yx^ky^{-1})y^2=x^{k+kt+2^r}=x^{2s}, \]

where \(s=k(1+2^{m-2})+2^{r-1}\). Since \(m\ge 3\), \(1+2^{m-2}\) is odd, and we can solve the congruence \(s=k(1+2^{m-2})+2^{r-1}\equiv 0\pmod {2^{m-1}}\). For this choice of \(k\), we have \((x^ky)^2=x^{2s}=x^{2^m}=\mathbf 1\), so that \(x^ky\) is a second involution (lest \(y\in\langle x\rangle\)).

Suppose that \(t=-1+2^{m-1}\). As above, for any integer \(k\),

\[\begin{align} (x^ky)^2=x^{k+kt+2^r}=x^{k2^{m-1}+2^r}. \end{align} \]

Now

\[yx^{2^r}y^{-1}=(yxy^{-1})2^r=(x^{-1+2^{m-1}})^{2^r}=x^{-2^r+2^{m+r-1}}=x^{-2^r}, \]

because \(r\ge 1\) implies \(m+r-1\ge m\), and so \(x^{2^{m+r-1}}=\mathbf 1\). Hence

\[\mathbf 1=(yx^{2^r}y^{-1})x^{2^r}=yy^2y^{-1}y^2=y^4=(x^{2^r})^2=x^{2^{r+1}}; \]

we conclude that \(r+1=m\) (because \(x\) has order \(2^m\)) and, also, that \(y^4=\mathbf 1\); hence, \(y^2\) is an involution. If we set \(k=1\) in equation (1), then \((xy)^2=\mathbf 1\); that is, \(xy\) is an involution. But \(y^2\neq xy\), and so we have displayed two distinct involutions in \(G\).
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Theorem A.1.14 A finite \(p\)-group \(G\) having a unique subgroup of order \(p\) is either cyclic or generalized quaternion.

Proof

The proof is by induction on \(n\), where \(|G|=p^n\); of course, the theorem is true when \(n=0\).

Assume first that \(p\) is odd. If \(n>0\), then \(G\) has a subgroup \(H\) of index \(p\), by Proposition 4.1.13(i), and \(H\) is cyclic, by induction. But Theroem A.1.11 shows that there can be no other subgroup of index \(p\), lest \(G\) be the quaternions, which is a \(2\)-group. Therefore, \(H\) is the unique maximal subgroup of \(G\), and so it contains every proper subgroup of \(G\). But if \(G\) is not cyclic, then \(\langle x\rangle\) is a proper subgroup of \(G\) for every \(x\in G\), and so \(G\le H\), which is absurd.

Assume now that \(G\) is a \(2\)-group. If \(G\) is abelian, then Theorem 2.3.3 shows that \(G\) is cyclic; therefore, we may assume that \(G\) is not abelian. Let \(A\) be a maximal normal abelian subgroup of \(G\). Since \(A\) has a unique involution, \(A\) is cyclic, by Theorem 2.3.3, say, \(A=\langle a\rangle\). We claim that \(A\) has index \(2\). Assume, on the contrary, that \(|G/A|\ge 4\). If \(G/A\) does not have exponent \(2\), then there is \(Ab\in G/A\) with \(b^2\notin A\). Consider \(H=\langle a,b^2\rangle< \langle a,b\rangle\le G\). If \(H\) is abelian, then \(b^2\) centralizes \(A\), contradicting Theorem 5.5.21(ii). As \(H\) is not abelian, it must be generalized quaternion, by induction. We may thus assume that \(b^2ab^{-2}=a^{-1}\). Now \(\langle a\rangle\lhd G\) gives \(bab^{-1}=a^{i}\) for some \(i\), so that

\[a^{-1}=b^2ab^{-2}=b(bab^{-1})b^{-1}=ba^{i}b^{-i}=a^{i^2}, \]

and \(i^2\equiv -1\pmod{2^e}\), where \(2^e\) is the order of \(a\). Note that \(e\ge 2\), for \(A\) properly contains \(\mathrm Z(G)\). But there is no such congruence: if \(e\ge 3\), then Lemma A.1.12 shows that this congruence never holds; if \(e=2\), then \(-1\) is not a square \(\mod 4\). It follows that \(G/A\) must have exponent \(2\). Since \(|G/A|\ge 4\), \(G/A\) contains a copy of \(V\). Therefore, there are elements \(c\) and \(d\) with \(c,d,c^{-1}d\notin A\) and with \(\langle a,c\rangle\), \(\langle a,d\rangle\), and \(\langle a,c^{-1}d\rangle\) properly subgroups of \(G\). Now none of these can be abelian, lest \(c,d\) or \(c^{-1}d\) centralize \(A\), so that all three are generalized quaternion. But there are equations \(cac^{-1}=a^{-1}=dad^{-1}\), giving \(c^{-1}d\in\mathrm C_G(A)\), a contradiction. We conclude that \(A=\langle a\rangle\) must have index \(2\) in \(G\).

Choose \(b\in G\) with \(b^2\in\langle a\rangle\). Replacing \(a\) by another generator of \(A\) if necessary, we may assume, that there is some \(r\le n-2\) with \(b^2=a^{2^r}\). Now \(bab^{-1}=a^t\) for some \(t\), because \(\langle a\rangle\lhd G\). Since \(G\) has only one involution, Corollary A.1.13 gives \(t=\pm 1\). But \(t=1\) says that \(a\) and \(b\) commute, so that \(G\) is abelian, hence cyclic. Therefore, we may assume that \(t=-1\) and \(G=\langle a,b\rangle\), where

\[a^{2^{n-1}}=\mathbf 1,\qquad bab^{-1}=a^{-1},\qquad b^2=a^{2^r}. \]

To complete the proof, we need only show that \(r=n-2\). This follows from Lemma A.1.12: since \(t=-1\), we have \(2^r\equiv -2^r\pmod{2^{n-1}}\), so that \(2^{r+1}\equiv 0\pmod{2^{n-1}}\), and \(r=n-2\).
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Theorem A.1.15 \(\mathrm A_4\) is a group of order \(12\) having no subgroup of order \(6\).

Proof

If such a subgroup \(H\) exists, then it has index \(2\), and so \(\alpha^2\in H\) for every \(\alpha\in\mathrm A_4\). If \(\alpha\) is a \(3\)-cycle, however, then \(\alpha=\alpha^4=(\alpha^2)^2\), and this gives \(8\) elements in \(H\), a contradiction.
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Proposition A.1.16 \(\mathrm A_4\) is the only subgroup of \(\mathrm S_4\) having order \(12\).

Lemma A.1.17 If \(G\) has order \(12\) and \(G\not\cong\mathrm A_4\), then \(G\) contains an element of order \(6\); moreover, \(G\) has a normal Sylow \(3\)-subgroup, hence has exactly two elements of order \(3\).

Proof

If \(P\) is a Sylow \(3\)-subgroup of \(G\), then \(|P|=3\) and so \(P=\langle b\rangle\) for some \(b\) of order \(3\). Since \([G:P]=4\), there is a homomorphism \(\psi\!:G\rightarrow\mathrm S_4\) whose kernel \(K\) is a subgroup of \(P\); as \(|P|=3\), either \(K=\mathbf 1\) or \(K=P\). If \(K=\mathbf 1\), then \(\psi\) is an injection and \(G\) is isomorphic to a subgroup of \(\mathrm S_4\) of order \(12\); thus, \(G\cong A_4\). Therefore, \(K=P\) and so \(P\lhd G\); it follows that \(P\) is the unique Sylow \(3\)-subgroup, and so the only elements in \(G\) of order \(3\) are \(b\) and \(b^2\). Now \([G:\mathrm C_G(b)]\) is the number of conjugates of \(b\). Since every conjugate of \(b\) has order \(3\), \([G:\mathrm C_G(b)]\le 2\) and \(|\mathrm C_G(b)|=6\) or \(12\); in either case, \(\mathrm C_G(b)\) contains an element \(a\) of order \(2\). But \(a\) commutes with \(b\), and so \(ab\) has order \(6\).
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Definition A.1.18 \(\mathbf T\) is a group of order \(12\) which is generated by two elements \(a\) and \(b\) such that \(a^6=\mathbf 1\) and \(b^2=a^3=(ab)^2\).

Theorem A.1.19 Every nonabelian group \(G\) of order \(12\) is isomorphic to either \(\mathrm A_4,\mathrm D_{12}\), or \(\mathbf T\).

Proof

It suffices to show that if \(G\) is a nonabelian group of order \(12\) which is not isomorphic to \(\mathrm A_4\), then either \(G\cong \mathrm D_{12}\) or \(G\cong\mathbf T\).

Let \(K\) be a Sylow \(3\)-subgroup of \(G\) (so that \(K=\langle k\rangle\) is cyclic of order \(3\) and, by Lemma A.1.17, \(K\lhd G\)), and let \(P\) be a Sylow \(2\)-subgroup of \(G\) (so that \(P\) has order \(4\)). Notice that \(G=KP\), for \(P\) is a maximal subgroup because it has prime index. Now either \(P\cong\mathbf V\) or \(P\cong \mathbb Z_4\).

In the first case, \(P=\{\mathbf 1,x,y,z\}\), where \(x,y\), and \(z\) are involutions. Not every element of \(P\) commutes with \(k\) lest \(G\) be abelian; therefore, there exists an involution in \(P\), say \(x\), with \(xkx\neq k\). But \(xkx\) is a conjugate of \(k\), hence has order \(3\); as \(G\) has only two elements of order \(3\), \(xkx=k^{-1}\), and \(\langle x,k\rangle\cong\mathrm D_6\). We claim that either \(y\) or \(z\) commutes with \(k\). If \(y\) does not commute with \(k\), then, as above, \(yky=k^{-1}\), so that \(xkx=yky\) and \(z=xy\) commutes with \(k\). If \(a=zk\), then \(G=\langle a,x\rangle\), \(a\) has order \(6\), and \(xax=xzkx=zxkx=zk^{-1}=a^{-1}\), so that \(G\cong\mathrm D_{12}\).

In the second case, \(P=\langle x\rangle\cong\mathbb Z_4\). Now \(x\) and \(k\) do not commute, because \(G=\langle x,k\rangle\) is not abelian, and so \(xkx^{-1}=k^{-1}\). But \(x^2\) does commute with \(k\), for \(x^2kx^{-2}=x(xkx^{-1})x^{-1}=xk^{-1}x^{-1}=k\); hence, \(a=x^{2}k\) has order \(6\). Since \(x^2\) and \(k\) commute, \(a^3=(x^2k)^3=x^6k^3=x^2\). Finally, \((ax)^2=axax=kx^3kx^3=(kx^{-1})kx^{-1}=(x^{-1}k^{-1})kx^{-1}=x^{-2}=x^2\). We have shown that \(G\cong \mathbf T\) in this case.
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Proposition A.1.20 There is no nonabelian simple group of order less than \(60\), and any simple group \(G\) of order \(60\) is isomorphic to \(\mathrm A_5\).

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