This is mostly a continuation on the group I gave in the last post, which is given by the presentation

$$G = \langle a,t ~|~ t^{-1}a^2t = a^3\rangle.$$ At the risk of beating a dead horse, I proved that the homomorphism $f:G\to G$ given on generators by $f(t) = t$ and $f(a) = a^2$ is surjective but not injective. Groups for which surjective homomorphisms are isomorphisms are called **Hopfian**, and so our group $G$ is not Hopfian.

As I've been talking about frequently in the past little while, a group $G$ is called **residually finite** if for every nontrivial $x\in G$ there exists a homomorphism $\varphi:G\to F$ such that $F$ is finite and $\varphi(x)$ is not the identity of $F$. In the post on residually finite groups, I explained the classic proof that a finitely-generated, residually finite group is Hopfian.

Now the particular group $G$ here that is not Hopfian is finitely-generated, and so of course it can't be residually finite. I was wondering, can we find an explicit nontrivial element $x\in G$ such that for every homomorphism $\varphi:G\to F$ where $F$ is finite, the element $\varphi(x)\in F$ is the identity of $F$?

Yes, that's actually quite easy. It is because in the last post we already proved that the commutator $[t^{-1}at,a]$ is sent to the identity under the endomorphism $f:G\to G$ (recall, which was given by $f(t) = t$ and $f(a) = a^2$). But if we carefully examine the proof of the statement "every finitely-generated residually finite group is Hopfian", we see that the kernel of $f$ is actually contained in every finite-index normal subgroup of $G$. Therefore, in particular, the commutator $[t^{-1}at,a]$, which is nontrivial by Britton's lemma, is mapped to the identity under every homomorphism $G\to F$ where $F$ is a finite group!