Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
Hint: commutativity. – user27126 Nov 5 '12 at 18:44
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). – Matthew 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

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.