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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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Hint: commutativity. –  Sanchez Nov 5 '12 at 18:44
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You should also note that you needn't send generators of $\mathbb{Z}\star\mathbb{Z}$ to generators of $\mathbb{Z}\times\mathbb{Z}$, they can be sent to any two elements. Although you should require that the two elements are different if you are to have any hope of making the homomorphism injective (not that such hope actually exists). –  Matt Pressland Nov 5 '12 at 18:45
    
@Sanchez Maybe you want to expand your comment to an answer, so that this question gets removed from the unanswered tab. If you do so, it is helpful to post it to this chat room to make people aware of it (and attract some upvotes). For further reading upon the issue of too many unanswered questions, see here, here or here. –  Julian Kuelshammer Jun 15 '13 at 10:45
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1 Answer

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.

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