1.3 Finitely Generated Abelian Group

Theorem

Let $G$ be a finitely generated abelian group. Then $G$ is isomorphic to a finite direct sum of cyclic group, where the finite cyclic summands (if any) are of order $1<m_1\mid m_2\mid \cdots \mid m_t$.

\begin{proof} By ^d4cf25, there exists an epimorphism $\pi :{\mathbb{Z}}^n\twoheadrightarrow G$ and $G\simeq {\mathbb{Z}}^n/\ker \pi$. By ^93020f, there exists a basis $e_1,\cdots ,e_n$ of ${\mathbb{Z}}^n$ and $a_1\mid a_2\mid \cdots \mid a_q$ such that $a_1{e}_1,\cdots ,a_q{e}_q$ is a basis of $\ker \pi$. Thus $G\simeq ({\oplus }_{i=1}^n\mathbb{Z}{e}_i)/({\oplus }_{i=1}^q\mathbb{Z}{a}_i{e}_i)\simeq ({\oplus }_{i=1}^q\mathbb{Z}/a_i\mathbb{Z})\oplus ({\oplus }_{i=q+1}^n\mathbb{Z}{e}_i)$ and $\{m_1,\cdots ,m_t\}=\{a_1,\cdots ,a_q\}\setminus \{1\}$. \end{proof}

Lemma

For $m=p_1^{n_1}\cdots p_t^{n_t}\in {\mathbb{Z}}_{+}$ where $p_i$ are distinct primes and $n_i>0$. Then $\mathbb{Z}/m\mathbb{Z}\simeq {\oplus }_{i=1}^t\mathbb{Z}/p_i^{n_i}\mathbb{Z}$.

Theorem

For a finitely generated abelian group $G$, $G\simeq {\mathbb{Z}}^n\oplus ({\oplus }_{i=1}^k\mathbb{Z}/p_i^{s_i}\mathbb{Z})$ where $s_i>0$ and $p_i$ are primes. (It is possible that $p_i=p_j$.)

\begin{proof} By ^08ebc2 and ^a1db34. \end{proof}

Next, we aim to show $n,k,p_i,s_i$ in ^79ab03 are all uniquely determined. Firstly, we have the following lemma.

Lemma

If $G$ is a finitely generated abelian group and $m\in {\mathbb{Z}}_{+}$, then

• the following subsets are subgroups of $G$:
  • $mG=\{mx:x\in G\}$
  • $G[m]=\{x\in G:mx=0\}$ ($m$-torsion of $G$)
  • $G_t=\{x\in G:|x|\text{mbox}isfinite\}$ (torsion subgroup of $G$);
• if $G={\oplus }_{i\in I}{G}_i$, then $G[m]={\oplus }_{i\in I}{G}_i[m]$;
• if $f:G\stackrel{\sim }{\to }H$ is an isomorphism, then $f$ induces $f:G_t\stackrel{\sim }{\to }{H}_t$ and $f:G[m]\stackrel{\sim }{\to }H[m]$.

\begin{proof} Easy. \end{proof}

Theorem

Let $G$ be a finitely generated abelian group. By ^08ebc2, $G\simeq {\mathbb{Z}}^n\oplus ({\oplus }_{i=1}^t\mathbb{Z}/m_i\mathbb{Z})$ where $m_1\mid \cdots \mid m_t$. By ^79ab03, $G\simeq {\mathbb{Z}}^n\oplus ({\oplus }_{i=1}^k\mathbb{Z}/p_i^{s_i}\mathbb{Z})$. The numbers $(n,m_1,\cdots ,m_t)$ and $(n,p_i,s_i)$ are all uniquely determined.

\begin{proof} Part 1. $n$ is uniquely determined. By ^08ebc2, there exists some decomposition $G\simeq F\oplus T$ with $F$ free and $T$ torsion. By ^2332c5, $G_{\mathrm{tor}}\simeq F_{\mathrm{tor}}\oplus T_{\mathrm{tor}}=T$. Then $G/G_{\mathrm{tor}}\simeq (F\oplus T)/T\simeq F$. Then $\mathrm{rk}F$ is uniquely determined by ^14f54e.

Part 2. If $G\simeq {\mathbb{Z}}^n\oplus ({\oplus }_{i=1}^k\mathbb{Z}/p_i^{s_i}\mathbb{Z})$ and $G_{\mathrm{tor}}\simeq {\oplus }_{i=1}^k\mathbb{Z}/p_i^{s_i}\mathbb{Z}$, then $k,p_i,s_i$ are unique. Choose one $p$, then $G[p^{\infty }]={\cup }_{j=1}^{\infty }G[p^j]\simeq {\oplus }_{m=1}^t\mathbb{Z}/p^{a_m}\mathbb{Z}$, as $(\mathbb{Z}/q^s\mathbb{Z})[p^{\infty }]=0$ if $q\ne p$. So in the end, it suffices to prove the following statement: if $H={\oplus }_{m=1}^t\mathbb{Z}/p^{a_m}\mathbb{Z}\simeq {\oplus }_{l=1}^s\mathbb{Z}/p^{b_l}\mathbb{Z}$ where $1⩽a_1⩽\cdots ⩽a_t$ and $1⩽b_1⩽\cdots ⩽b_s$, then $t=s$ and $a_i=b_i$. First, $H[p]={\oplus }_{n=1}^t{p}^{a_m-1}\mathbb{Z}/p^{a_m}\mathbb{Z}\simeq {\oplus }_{n=1}^t\mathbb{Z}/p\mathbb{Z}$ and $H[p]{\oplus }_{l=1}^s{p}^{b_m-1}\mathbb{Z}/p^{b_m}\mathbb{Z}\simeq {\oplus }_{l=1}^s\mathbb{Z}/p\mathbb{Z}$. Thus $t=s$. Let $v(1⩽v⩽t)$ be the first integer such that $a_i=b_i$ if $i<v$ and $a_v\ne b_v$. WLOG assume that $a_v<b_v$. Then $p^{a_v}H\simeq {\oplus }_{m=x,a_x>a_v}{p}^{a_v}\mathbb{Z}/p^{a_m}\mathbb{Z}\simeq {\oplus }_{m=v}^s{p}^{a_v}\mathbb{Z}/p^{b_m}\mathbb{Z}$. Since $x>v$ and $t=s$, $v$ does not exist by the proof of Part 1. Therefore, $a_i=b_i$ for all $i\in \{1,\cdots ,t\}$.

Part 3. $G\simeq {\mathbb{Z}}^n\oplus ({\oplus }_{i=1}^t\mathbb{Z}/m_i\mathbb{Z})$ where $1<m_1\mid \cdots \mid m_t$, then $m_i$ are unique. It suffices to show, if $H\simeq {\oplus }_{i=1}^t\mathbb{Z}/m_i\mathbb{Z}\simeq {\oplus }_{j=1}^s\mathbb{Z}/n_j\mathbb{Z}$, then $t=s$ and $m_i=n_i$. Pick a prime $p\mid m_t{n}_s$. Then $H[p^{\infty }]\simeq {\oplus }_{i=1}^t\mathbb{Z}/p^{a_i}\mathbb{Z}\simeq {\oplus }_{j=1}^s\mathbb{Z}/p^{b_j}\mathbb{Z}$ where $p^{a^i}\mid \mid m_i$ and $p^{b_j}\mid \mid n_j$ and $0⩽a_1⩽\cdots ⩽a_t$ and $0⩽b_1⩽\cdots ⩽b_s$. By Part 2, non-zero $a_i,b_j$ match each other.

Claim that $m_i$ and $n_i$ have the same prime divisors. Assume that there exists a prime $p_1$ such that $p_1\nmid m_1$ and $p_1\mid n_1$. Then by the argument above, we have that $t>s$. If there exists another prime $p_2$ such that $p_2\mid m_1$ and $p_2\nmid n_1$, then $t<s$, contradiction. So WLOG we may assume that $m_1$ is a proper divisor of $n_1$ and $t>s$. It follows that $|{\oplus }_{i=1}^t\mathbb{Z}/m_i\mathbb{Z}|>|{\oplus }_{j=1}^s\mathbb{Z}/n_j\mathbb{Z}|$, which is impossible. Repeat the procedure for $i\in \{2,\cdots ,t\}$, we prove that all $m_i$ and $n_i$ have the same prime divisors.

Run this argument for all $p$, then $t=s$ and $m_i=n_i$. \end{proof}

Remark. $m_i$’s are called invariant factors of $G$, and $p_i^{s_i}$‘s are called elementary divisors.