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.

  • 2
    $\begingroup$ Have you tried constructing an isomorphism between these two groups? $\endgroup$ – Dionel Jaime Apr 20 '19 at 18:01
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    $\begingroup$ Yes, direct sums preserve isomorphisms. $\endgroup$ – Bernard Apr 20 '19 at 18:05


This is because they share a presentation:

$$\begin{align} \bigoplus_{i=1}^n\langle a_i\rangle&\cong\langle a_1,\dots , a_n\mid \{[a_j, a_k]\mid j<k\}\rangle \\ &\cong\bigoplus_{i=1}^n \Bbb Z, \end{align}$$ 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.

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

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