# Stabilizer in automorphism group of free group of a certain finite-index subgroup

The following question arose in my research. I'd be interested in an answer to it, but I'd also be interested in general techniques for solving this kind of problem (or, even better, pointers to computer programs that can solve it automatically).

Consider the group $G = \text{Aut}(F_2)$. Of course, $G$ acts on $F_2$. Let $x$ and $y$ be the generators for $F_2$, and let $K \subset F_2$ be the normal closure of the set $\{x^4, x^2 y^{-2}, y^{-1} x y x\}$. It is standard that $F_2/K$ is the $8$-element group of quaternions.

Question : What are generators for the subgroup $\{\text{$g \in G|g(K) = K$}\}$ of $G$?

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According to MAGMA, the stabilizer of $K$ is the whole of ${\rm Aut}(G)$.

> F<x,y>:= FreeGroup(2);
> K := ncl< F | x^4, x^2*y^-2, y^-1*x*y*x >;
> Index(F,K);
8
> G := AutomorphismGroup(F);
> G;
A group of automorphisms of GrpFP: F
Generators:
Automorphism of GrpFP: F which maps:
x |--> y
y |--> x
Automorphism of GrpFP: F which maps:
x |--> x^-1
y |--> y
Automorphism of GrpFP: F which maps:
x |--> x * y
y |--> y
> K @ G.1 eq K;
true
> K @ G.2 eq K;
true
> K @ G.3 eq K;
true


So $K$ is stabilized by all three generating automorphisms of $F$. I don't suppose it is too hard to do that calculation by hand.

More generally, for subgroups of finite index in a free group (or any other group in which you had generators for the automorphism group), you could use a standard orbit-stabilizer calculation to compute the subgroups in the orbit of $K$ under $G$, and then compute Schreier generators of the stabilizer.

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Thank you very much! –  Tina Oct 16 '12 at 2:55

Here's a short proof by hand that $K$ is characteristic in $F_2$.

The natural map $Q_8\to V=\mathbb{Z}/2\oplus\mathbb{Z}/2$ induces $\eta:F_2\to H_1(F_2,\mathbb{Z}/2)\cong V$. The kernel $N=\ker\eta$ is therefore characteristic and contains $K$ with index two.

Suppose now that $\phi$ is an automorphism of $F_2$ that does not preserve $K$. Then $\phi$ descends to an automorphism of $V$ which does not lift to an automorphism of $Q_8$. But it's easy to check that every automorphism of $V$ does, in fact, lift.

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Thank you very much! –  Tina Oct 16 '12 at 2:56