1
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Neil Hoffman Jul 20 '16 at 18:14
  • 2
    $\begingroup$ 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 '16 at 19:32
  • $\begingroup$ @IanAgol: Looks like I was about half a minute ahead of your comment... $\endgroup$ – Arturo Magidin Jul 20 '16 at 19:37
5
$\begingroup$

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.

$\endgroup$

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.