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.

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    $\begingroup$ 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)" $\endgroup$ – sTertooy Jan 2 '17 at 16:25
  • $\begingroup$ 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. $\endgroup$ – Derek Holt Jan 2 '17 at 16:25
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    $\begingroup$ Just edited with my meaning for extension $\endgroup$ – Alex Doe Jan 2 '17 at 16:31
  • $\begingroup$ 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! $\endgroup$ – Derek Holt Jan 2 '17 at 16:43
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    $\begingroup$ 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. $\endgroup$ – Derek Holt Jan 2 '17 at 17:22

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