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Let $H$ be a finite index subgroup of a finitely generated group $G$. Assume that $Out(H)$ is finite. Can $Out(G)$ be infinite?

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follow-up:… – YCor Jul 2 '15 at 20:25

I think the answer is yes. Let $A$ be a f.g. group. Assume that $\text{Out}(A)=1$ ($\text{Out}(A)$ finite would probably be enough) and that $A$ contains a central copy of $C^{(\infty)}$, the direct sum of countably many copies of the cyclic group $C$ of prime order $p$. Then $\text{Out}(A\times C)$ is infinite: indeed if $f$ is a homomorphism from $C$ to the center of $A$, then if we set $u_f(a,t)=(af(t),t)$, then $f\mapsto u_f$ injects $\text{Hom}(C,Z(A))$ into $\text{Out}(A\times C)$. Actually all these automorphisms are identity on the finite index subgroup $A$, and this is essentially the reason I don't expect Mark's argument to work.

It remains to find an example of such a group $A$. I expect some complicated nilpotent-by-abelian group over $\mathbf{F}_p[t]$ to work (or Kac-Moody groups??), but I'm not prone to go into computations and maybe somebody else has one example.

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There is a nice result by Ould Houcine: every countable abelian group injects into the centre of some finitely presented group: – Alain Valette May 1 '11 at 20:47
Yes but there are many classical ways to inject an infinite-dimensional $\mathbf{F}_p$-vector space into the center of a f.p. group (e.g. into Abels' group over $\mathbf{F}_p$). But we need it to have finite Out. I think it should work by picking some suitable f.g. $\mathbf{F}_p$-algebra $A$ with a finite automorphism group, with an infinite $K_2$. Probably some localization of $\mathbf{F}_p[u,v]$ should work. Then the universal central extension of $\mathbf{SL}_d(A)$ should be the example. --YC – YCor May 2 '11 at 11:34

No, it is not possible. Here is a rough argument (I do not have too much time to give more details) $Aut(G)$ acts on the set of all subgroups of $G$ of the same index as $H$. Then a finite index subgroup $T$ of it fixes $H$. Then $T/H$ is infinite and $TG/G$ is infinite too.

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This argument cannot work (unless its validity is determined by the number of votes :) ); the point is that an infinite group of automorphisms can act as the identity on a finite index subgroup. The answer I give provides an example of a group T of automorphisms of $G$ that are trivial on $H$ and has infinite image in Out$(G)$. This is not enough to obtain a negative answer to the question (because there may be other automorphisms) but it is enough to show that your argument fails. – YCor Dec 30 '12 at 16:43

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