Note on Finitely Generated Abelian Groups

We state and prove a sufficient condition for finitely generated Abelian Groups to be the direct product of its generators, and state a counterexample to the conclusion when the condition is not satisfied.

Theorem

Let $G$ be an abelian group and $G=\langle g_1,\dots, g_n\rangle$ .

Suppose the generators $g_1,\dots,g_n$ are linearly independent over $\mathbb{Z}$ , that is, whenever $c_1g_1+\dots+c_ng_n=0$ for some integers $c_i\in\mathbb{Z}$ , we have $c_1=\dots=c_n=0$ .

(Here we are using additive notation for $(G,+)$ , where the identity of $G$ is written as 0, the inverse of $g$ is written as $-g$ ).

Then $\displaystyle G\cong\langle g_1\rangle\times\dots\times\langle g_n\rangle.$

Proof

Define the following map $\psi:\langle g_1\rangle\times\dots\times\langle g_n\rangle\to\langle g_1,\dots,g_n\rangle$ by $\displaystyle \psi((c_1g_1,\dots,c_ng_n))=c_1g_1+\dots+c_ng_n.$

We can check that $\psi$ is a group homomorphism.

We have that $\psi$ is surjective since any element $x\in\langle g_1,\dots,g_n\rangle$ is by definition a combination of finitely many elements of the generating set and their inverses. Since $G$ is abelian, $x=c_1g_1+\dots+c_ng_n$ for some $c_i\in\mathbb{Z}$ .

Also, $\psi$ is injective since if $c_1g_1+\dots+c_ng_n=0$ , then all the coefficients $c_i$ are zero (by the linear independence condition). Thus $\ker\psi$ is trivial.

Hence $\psi$ is an isomorphism.

Remark

Note that without the linear independence condition, the conclusion may not be true. Consider $G=\mathbb{Z}_2\times\mathbb{Z}_3\times\mathbb{Z}_5$ which is abelian with order 30. Consider $g_1=(1,1,0)$ , $g_2=(0,1,1)$ .

We can see that $G=\langle g_1,g_2\rangle$ , by observing that $3g_1=(1,0,0)$ , $4g_1=(0,1,0)$ , $2g_1+g_2=(0,0,1)$ . However $\langle g_1\rangle\times\langle g_2\rangle=\mathbb{Z}_6\times\mathbb{Z}_{15}$ has order 90. Thus $\langle g_1,g_2\rangle\not\cong\langle g_1\rangle\times\langle g_2\rangle$ .