# Direct sum of $n$ (infinite) cyclic groups isomorphic to direct sum of $n$ copies of $\mathbb{Z}$?

I'm currently selfstudying some algebra and i am currently covering the various equivalent definitions of free abelian groups. However, in order to understand why these definitions are indeed equivalent, this question came up and i havent managed to figure it out on my own. The actual question is:

Assume I have $$n$$ infinite cyclic groups $$\langle a_1 \rangle,\langle a_2 \rangle,...,\langle a_n \rangle.$$

Does it hold that $$\bigoplus_{i=1}^n \langle a_i \rangle \cong \bigoplus_{i}^n \mathbb{Z} \quad (n\ \text{copies of}\ \mathbb{Z})?$$

I know that every infinite cyclic group is isomorphic to $$\mathbb{Z}$$. But I couldn't figure out whether direct sums preserve isomorphisms.

If someone has a recommendation for a good book regarding this topic, I'd also appreciate any recommendation.

Thanks for any help.

• Have you tried constructing an isomorphism between these two groups? – Dionel Jaime Apr 20 '19 at 18:01
• Yes, direct sums preserve isomorphisms. – Bernard Apr 20 '19 at 18:05

Yes.

This is because they share a presentation:

\begin{align} \bigoplus_{i=1}^n\langle a_i\rangle&\cong\langle a_1,\dots , a_n\mid \{[a_j, a_k]\mid j where $$[x,y]=x^{-1}y^{-1}xy$$ is the commutator of $$x$$ and $$y$$.

They're also both free abelian groups of rank $$n$$.

I recommend reading Magnus et al.'s "Combinatorial Group Theory [. . .]".

Another way to see the isomorphism is to construct one out of the $$n$$ isomorphisms between the cyclic groups and $$\Bbb Z$$, like so: if $$\varphi_i:\langle a_i\rangle\to\Bbb Z$$ is given by $$\varphi_i(a_i)=1$$, then define

\begin{align} \varphi: \bigoplus_{i=1}^n\langle a_i\rangle &\to \bigoplus_{i=1}^n \Bbb Z,\\ (a_1^{\alpha_1},\dots ,a_n^{\alpha_n})&\mapsto (\alpha_1, \dots , \alpha_n). \end{align} Then $$\varphi$$ is an isomorphism.

• beautiful, thank you very much for your detailed response. highly appreciating it. – Zest Apr 20 '19 at 18:32
• You're very welcome, @Zest :) – Shaun Apr 20 '19 at 18:41