A group $G$ is residually finite if for each element $g\in G$ there exists a (surjective) homomorphism $f_g: G \rightarrow H_g$ such that $H_g$ is finite and $f_g(g)\ne 1$.

Consider the weaker condition where a finitely generated group $G = \langle x_1,\dots,x_n\rangle$ has the property that for each $x_i$ in this fixed generating set there exists a homomorphism $f$ onto a finite group $H$ such that $f(x_i)\ne 1$.

To see this is a weaker condition, assume $G$ is residually finite let $f:G\rightarrow H_{x_1}\times\cdots \times H_{x_n}$ by $f(g)=(f_{x_1}(g),...,f_{x_n}(g))$.

However, is this condition strictly weaker? That is is there a finitely generated group $G$ with a fixed generating set $X$ such that for each element of $x_i \in X$ there is a homomorphism onto a finite group $H_{x_i}$ but $G$ is not residually finite?

  • $\begingroup$ This question is possibly similar to mathoverflow.net/questions/244687/…. However, I had a different interpretation of what the question was asking then the posted anser, so I figured I would spin it off on it's own rather than edit the question when the intent was unclear. Also, I believe the answer for that question does not apply here, because this question imposes a fixed generating set. $\endgroup$ Jul 20, 2016 at 18:14
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    $\begingroup$ My answer to the other question also answers this: take any finitely generated group which has a finite quotient but is not residually finite (such as BS(n,m), n,m >1). Then choose the generating set X in the way that I described. $\endgroup$
    – Ian Agol
    Jul 20, 2016 at 19:32
  • $\begingroup$ @IanAgol: Looks like I was about half a minute ahead of your comment... $\endgroup$ Jul 20, 2016 at 19:37

1 Answer 1


Take the Baumslag-Solitar group $B(2,3) = \langle a,b\mid ba^2b^{-1}=a^3\rangle$. Take $X=\{a,b\}$.

Let $H=\langle r,s\mid r^5=s^2=1, sr=r^4s\rangle$, the dihedral group of order $10$. Since $sr^2s = (srs)^2 = r^8 = r^3$, and $s^{-1}=s$, then we get a homomorphism $f\colon B(2,3)\to H$ by $f(a)=r$, $f(b)=s$. Thus, $B(2,3)$ satisfies your weaker condition. But $B(2,3)$ is non-Hopfian, so it cannot be residually finite.


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