# Residually finite extension of a finite group

Do you know how to prove that the extension of a finte group by a residually finite group is residually finite? I know that there is a result of Mal'cev proving something stronger, but I cannot find it either.

I say that a group $$G$$ is an extension of a finte group by a residually finite group, in this case, if there exists a finite normal subgroup $$N$$ of $$G$$ such that $$G/N$$ is residually finite.

• Mal'cev proved in "On homomorphisms onto finite groups (1958)" that a split extension of a finitely generated residually finite group by a residually finite group is residually finite. Miller improved on this result in "On Group-Theoretic Decision Problems and their Classification (1971)" – TastyRomeo Jan 2 '17 at 16:25
• What do you mean by extension of group A by group B? There are two possible meanings, and they are both used roughly equally often, so you cannot use this terminology without explaining which you mean. – Derek Holt Jan 2 '17 at 16:25
• Just edited with my meaning for extension – Alex Doe Jan 2 '17 at 16:31
• It may be helpful to do a few reductions. Since $C_G(N)$ has finite index in $G$, you can assume that $N \le Z(G)$. You could then assume by induction on $|N|$ that $|N|$ is prime. I think the result is easy for abelian groups, and so you can assume that $N \le [G,G]$, since otherwise it follows from looking at $G/[G,G]$. So $N$ is a quotient of the Schur Multiplier $H_2(G)$. That's as far as I have got! – Derek Holt Jan 2 '17 at 16:43
• The result appears to be false. On math.umbc.edu/~campbell/CombGpThy/RF_Thesis/2_RF_Results.html in the section on extensions of residually finite groups, there is an example attributed to Kanta Gupta of a group $G$ that is not residually finite with a normal subgroup $K$ of order $2$ such that $Q \cong G/K$ is residually finite. – Derek Holt Jan 2 '17 at 17:22