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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.

Shaun
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Alex Doe
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    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)" – sTertooy Jan 02 '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 02 '17 at 16:25
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    Just edited with my meaning for extension – Alex Doe Jan 02 '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 02 '17 at 16:43
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    The result appears to be false. On http://www.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 02 '17 at 17:22
  • Thank you! I see that it must be a splitting extension, otherwise one could run into Gupta's example. – Alex Doe Jan 02 '17 at 17:53
  • @DerekHolt The first example is actually due to John Millson, Real vector bundles with discrete structure group, Topology, 1979. – Moishe Kohan Jan 02 '17 at 23:21
  • So, the example shows that my claim does not hold for soluble groups, but what about nilpotent groups? Do you think it holds for them? – Alex Doe Jan 04 '17 at 10:17
  • Could somebody sketch the proof of Mal'cev in "On homomorphisms onto finite groups (1958)" ? because I can't find the article. – 1123581321 Oct 22 '20 at 15:40

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As pointed out in the comment the claim is false. This was proven by John Millson in the paper Real vector bundles with discrete structure group, Topology, 1979 doi.

"We also obtain the following theorem in group theory. Let $\mathcal{O}$ be the ring of integers in the real quadratic field $\mathbb{Q}(\sqrt{p})$, where $p$ is a rational prime $p = 7 \pmod 8$. Then by choosing an embedding of $\mathbb{Q}(\sqrt{p})$ into $\mathbb{R}$ we obtain $p : \operatorname{SL}(n, \mathcal{O})\to\operatorname{SL}(n, \mathbb{R})$ and consequently an extension $\mathbb{Z}/2\mathbb{Z} \to \overline{\operatorname{SL}(n, \mathcal{O})}\to \operatorname{SL}(n, \mathcal{O})$. We prove that this extension does not become trivial when restricted to any subgroup of finite index of $\operatorname{SL}(n, \mathcal{O})$, consequently the group $\overline{\operatorname{SL}(n, \mathcal{O})}$ is not residually finite." - Bottom of first page.

The "correct", stronger result of Mal'cev is:

Theorem (Mal'cev). A split extension $G$ of a finitely generated, residually finite group $N$ by a residually finite group $K$ is residually finite.

The proof is surprisingly straightforward.

Proof. Write $G=N\rtimes K$. As $N$ is finitely generated, the intersection $N_i$ of all subgroups of a fixed finite $i$ in $N$, $N_i=\cap_{|N_i:H|=i}H$, is a characteristic subgroup of $N_i$ which still has finite index. Hence, $N_i$ is normal in $G$. Therefore, if $K'$ is a finite index subgroup of $K$ then $N_iK'$ is a finite index subgroup of $G$. The result now holds, as if $g=(n, k)$ is some element of $G$ then we may take $N_i$ and $K'$ such that $n\not\in N_i$ and $k\not\in K'$, so $g\not\in N_iK$ as required. QED

Note that we assume finite generation of the kernel. This is required to ensure that $N_iK'$ is a subgroup, and without it the result also does not hold (for example, take the wreath product of a nonabelian finite group $H$ by the integers, which splits as $H'\rtimes\mathbb{Z}$ where $H'$ denotes the countable direct product of $H$ with itself).

A powerful result which does not assume finite generation of the kernel is the following:

Theorem (Baumslag). A finitely generated cyclic extension of a free group is residually finite.

Here we are assuming finite generation of the group, rather than of the kernel. The proof here is much more complicated than the proof of Mal'cev's theorem given above!

user1729
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