Does every sequence of group epimorphisms (between finitely generated groups) contain a stable subsequence?

Question

I have a question that is related to the topic of limit groups:

Let $G$ and $H$ be finitely generated groups and let $(\varphi_n: G \to H)_{n \in \mathbb{N}}$ be a sequence of group epimorphisms. Does there exist a stable subsequence of $(\varphi_n)_{n \in \mathbb{N}}$?

This is the definition of "stable sequence":

Let $G$ be a group and $(\varphi_n) _{n \in \mathbb{N}}⊂ \mathrm{Hom}(G, H)$. The sequence $(\varphi_n) _{n \in \mathbb{N}}$ is stable if for any $g ∈ G$ either $\varphi_n(g) = 1$ for almost all $n$ or $\varphi_n(g)\neq 1$ for almost all $n$.

I conjecture that the answer to the question is positive, because the statement of the question is used on page 27 of the following article (in the article we have a sequence of epimorphisms $g_n: F_l \to L$ to which the statement of my question is applied to):

https://arxiv.org/pdf/2002.10278.pdf

Edit: In the above article we have the situation that G is a free group of rank at least two and H is a non-abelian limit group. So it would be sufficient if somebody can answer my question for this situation. Of course you can still say something to the general situation.

Edit: "Almost all" means that the statement ist true for all natural numbers with possibly a finite number of exceptions.

Answer 1

Yes. This is just (metrizable) compactness in the space of normal subgroups of $G$. It is enough to assume that $G$ is countable (finitely generated plays no role).

Namely, let $N(G)\subset 2^G$ be the set of normal subgroups. It is closed in the compact space $2^G$, hence is compact.

Let $M_n$ be the kernel of $\varphi_n:G\to H$. Then there is a converging subsequence $M_{i(n)}\to M$. Then for $g\notin M$ (resp. $g\in M$), for each $n$ large enough (bound depending on $g$) we have $\phi_{i(n)}(g)\neq 1$ (resp. $\phi_{i(n)}(g)=1$).