# Finite index of automorphisms that fix normal subgroup

does anyone have some thoughts on this:

Let G be a group and K a normal subgroup in G of finite index. Then the set X of all automorphisms phi of G that fix K (meaning phi(K)=K, so not necessarily pointwise) has finite index in Aut G.

Is this even true? I am trying to work out a larger problem and this would be the final step but I am not sure this is even correct.

I tried explaining it by using that the inner automorphisms of G (Int G) are a subset of X (since K is normal). Therefore, if [Aut G:Int G] is finite, so must be [Aut G:X]. But again, I'm not sure this holds...

Thanks very much for your help. I really appreciate it!

## migrated from mathoverflow.netFeb 5 '15 at 1:01

This question came from our site for professional mathematicians.

• Sadly, it is not homework... if it was, I would soon find out the solution. Could you maybe give me a pointer as to how to solve it? – LostInMath Feb 5 '15 at 0:31
• Homework or not, it is very uncool to double post it simultaneously to multiple sites (in this case matheplanet.com) ! Now three people spend time writing virtually identical answers to this question, when one over there or two here would have sufficed. – Johannes Hahn Feb 5 '15 at 1:57

## 2 Answers

If $\alpha$ and $\beta$ are automorphisms of $G$, then the cosets $\alpha X$ and $\beta X$ are the same iff $\alpha(K)=\beta(K)$. Thus $X$ will fail to have finite index if there are infinitely many different subgroups of $G$ that are conjugate to $K$ under automorphisms of $G$. For instance, if $G$ is an infinite-dimensional vector space over a finite field and $K$ is subspace such that $G/K$ is finite-dimensional, then automorphisms of $G$ can send $K$ to any subspace $K'$ such that $G/K'$ has the same dimension, and there are infinitely many such subspaces.

I don't think this is true without some further finiteness condition that limits the number of images of $K$:

Take $G$ the (two-sided) infinite sequences with entries in 0,1 and as operation component-wise addition (so its an infinite direct product of $C_2$ with itself), and as $K$ the kernel of the projection on one particular component.

Clearly right/left shift are automorphisms of $G$, and under these $K$ has infinitely many images, so the stabilizer of $K$ in the automorphism group must have infinite index.