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I was wondering whether there is some injective homomorphism from $\mathbb{Z}\star\mathbb{Z}$ to $\mathbb{Z}\times\mathbb{Z}$, where with $\star$ I have denoted the free product, and with $\times$ the direct sum?

My guess is that there is no injective homomorphism, since if there were such a homomorphism then it would suffice to define it on the generators of $\mathbb{Z}\star\mathbb{Z}$ where these would be sent to the generators of $\mathbb{Z}\times\mathbb{Z}$, but cannot quite see how to argue?

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    @Sanchez Maybe you want to expand your comment to an answer, so that this question gets removed from the [unanswered tab](http://meta.math.stackexchange.com/q/3138). If you do so, it is helpful to post it to [this chat room](http://chat.stackexchange.com/rooms/9141) to make people aware of it (and attract some upvotes). For further reading upon the issue of too many unanswered questions, see [here](http://meta.stackexchange.com/q/143113), [here](http://meta.math.stackexchange.com/q/1148) or [here](http://meta.math.stackexchange.com/a/9868).2013-06-15

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As $\mathbb{Z} \times \mathbb{Z}$ is abelian, any (group) homomorphism $\mathbb{Z} * \mathbb{Z} \to \mathbb{Z} \times \mathbb{Z}$ has kernel containing the commutator subgroup of $\mathbb{Z} * \mathbb{Z}$. In particular, it is not injective.