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.

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