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 $x=1$?


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

  • 6
    $\begingroup$ 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 $g$. $\endgroup$ – Arnaud Mortier Jul 3 '18 at 16:10
  • 2
    $\begingroup$ 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 $G$. $\endgroup$ – Lee Mosher Jul 3 '18 at 16:26
  • 1
    $\begingroup$ Are you assuming that $\psi$ is surjective? (Then the question is more interesting and the answer is "yes", $\psi$ is injective.) $\endgroup$ – user1729 Jul 4 '18 at 11:36
  • $\begingroup$ @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. $\endgroup$ – 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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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