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Given $$G=\mathbb{Z}_2*\mathbb{Z}_2=P(a,b\mid a^2,b^2)$$ among other things I wanted to show that this group is infinite, what I did is consider the words of the form $$abababa\ldots$$ they are all diferent from each other since they do not contain any of the relations described below, and also there is an infinite amount of them.

You can do the same trick for any 2 finite-presented groups, so my question is, is the free product of two arbitrary (non trivial) groups infinite always? If so, how do you show it?

I also had to calculate $\text{Ab}(G)$ and I get $$\text{Ab}(G)=\mathbb{Z}_2\times\mathbb{Z}_2$$ is it true that, for any two abelian groups $G_1,G_2$ the following holds? $$\text{Ab}(G_1*G_2)=G_1\times G_2$$

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  • $\begingroup$ "they are all diferent from each other since they do not contain any of the relations described below" This argument doesn't work without more explanation. For instance, in the group $P(a,b\mid ab)$, $a=ba^2$ even though neither side of the equation contains the relation $ab$. $\endgroup$ – Eric Wofsey Feb 5 '16 at 9:53
  • $\begingroup$ You're right, but in your example you can see that $ba\in\langle\langle ab\rangle\rangle$. Then in mine would be enough to show that $ababab..\notin\langle\langle a^2,b^2\rangle\rangle$ right? Which seems fairly obvious. $\endgroup$ – Smurf Feb 5 '16 at 10:29
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  1. Yes, the free product of two nontrivial groups is infinite. Let $a\in G_1$ and $b\in G_2$ be non-identity elements. Then the words $(ab)^n$ are all distinct. This is because adjacent elements come from different groups, so the words are not reducible. (This is using the word-based model for free groups.)

  2. Yes, and in fact $\operatorname{Ab}(G_1*G_2)=\operatorname{Ab}(G_1)\times\operatorname{Ab}(G_2)$. See Abelianization of free product is the direct sum of abelianizations

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