# Is this surjective homomorphism necessarily injective? (Free Product of Cyclic groups)

Let $G=\langle g_1,\dots,g_m\mid g_1^{k_1}=\dots=g_m^{k_m}=1\rangle$. (Where $k_1,\dots,k_m$ are integers.)

Let $\psi:G\to G$ be a surjective homomorphism.

Is $\psi$ necessarily injective?

In other words, if $x\in\ker\psi$, then is $x=1$?

Thanks.

"Intuitively" it seems true that $\psi$ is injective, but I can't seem to find a proof for it.

• No. The trivial homomorphism is never injective if $G$ is not the trivial group. And no matter what $G$ is, there is always a trivial homomorphism, $\Psi(g)=e$ for all $g$. – Arnaud Mortier Jul 3 '18 at 16:10
• And you can easily find lots of other noninjective homomorphisms: for each $i=1,...,m$, you are free to pick the value of $\psi(g_i)$ as the identity or as any element of $G$ of order dividing $k_i$. So, for example, if $k_i$ and $k_j$ are both even then you can pick $\psi(g_i)=\psi(k_j)$ to be the same order 2 element of $G$. – Lee Mosher Jul 3 '18 at 16:26
• Are you assuming that $\psi$ is surjective? (Then the question is more interesting and the answer is "yes", $\psi$ is injective.) – user1729 Jul 4 '18 at 11:36
• @user1729 Yes, you are right. I wish to require that $\psi$ is surjective. Otherwise as Arnaud mention, the trivial homomorphism is already a counter-example. – yoyostein Jul 4 '18 at 11:58

Yes. In this setting, $\psi$ is injective if it is also surjective. That is, free products of finitely many cyclic groups (in fact, free products of finitely many finite groups) are Hopfian.*

One reason for this is the following two results:

Lemma 1. Free products of finitely many finite groups are virtually free, and hence residually finite.

Lemma 2. Finitely generated, residually finite groups are Hopfian.

I once wrote out the proof of Lemma 2 in this old answer.

You can find a direct proof that your groups are Hopfian, which does not apply Lemma 2, on MathOverflow.

*Examples of finitely presented, non-Hopifan groups can be found here.