GTM148 抄书笔记 Part III. (不定期更新)

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

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 \(G_p\cap\langle U_{q\neq p}G_q\rangle=\mathbf 0\).

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.

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 \(g=\sum(c_i+b_ik_i)z_i+\sum a_jx_j\in L+M\).

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.

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 \(\mathrm d(p^nG/p^{n+1}G)=b_{n+1}+b_{n+2}+\cdots+b_t\).

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.

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 \(G\rightarrow H\).

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 \(G/H\).