5
$\begingroup$

Recall that a group $G$ is called residually finite if for any nontrivial element $g\in G$ there exists a finite group $H$ and a homomorphism $f$ from $G$ to $H$ such that $f(g)\neq1$.

My question is

for which kind of finitely presented group $G=\{x_1, \cdots, x_n|r_1, \cdots, r_m\}$, there exists a finite group $H$ and a homomorphism $f$ from $G$ to $H$, such that $f(x_i)\neq 1$ $(1\leq i\leq n)$.

Finite groups and free groups are naive examples.

$\endgroup$

1 Answer 1

13
$\begingroup$

I'm a bit unclear on what you're asking, but I'll assume that you mean that given $G$, is there a presentation $G=\{x_1, \ldots, x_n|r_1, \ldots, r_m\}$ for which the property holds (note that your question only concerns the generators $x_1,\ldots x_n$, so the relators $r_j$ are irrelevant)? Otherwise, if the presentation is given, then the property will hold iff none of the $x_i$ lie in the kernel of the map $G\to \hat{G}$, the profinite completion.

Given this interpretation, then the answer is that as long has $G$ has a non-trivial homomorphism to a finite group $H$, then you can always find a generating set with this property. Starting with some generators $G=\langle x_1, \ldots, x_n\rangle$, and a non-trivial homomorphism $f:G\to H$, then $f$ must be non-trivial on some generator, say $f(x_1)\neq 1$. Now, consider all the generators $x_i, \ldots, x_n$ (by relabeling) such that $f(x_j)=1$, $i\leq j\leq n$. We obtain a different generating set $G=\langle x_1,\ldots, x_{i-1}, x_ix_1^{-1}, \ldots, x_nx_1^{-1}\rangle$. Clearly now $f$ is non-trivial on each one of these generators.

$\endgroup$

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.