## Inner Automorphisms as a Normal Subgroup

1. The problem statement, all variables and given/known data

Let G be a group. We showed in class that the permutations of G which send products to products form a subgroup Aut(G) inside all the permutations. Furthermore, the mappings of the form $\sigma_b(g)=bgb^{-1}$ form a subgroup inside Aut(G) called the inner automorphisms and denoted Inn(g).

Prove that the inner automorphisms form a normal subgroup of Aut(G).

2. Relevant equations

None

3. The attempt at a solution

I attempted this problem one way, and my professor said I was going about it the wrong way - I did multiplications of the permutations and apparently I'm supposed to use function composition of the permutations. So this is what I have: I must show that, given a permutation $\tau$ that $\tau \sigma_b \tau^{-1}$ is in the set of inner automorphisms (at least, that's how we've been showing normality in class and my professor told me that this is at least correct).

So, I have
$$\tau\left(\sigma_b\left(\tau^{-1}\left(g\right)\right)\right)&=&\tau\left(b\tau^{-1}(g)b^{-1}\right)\\ &=&\tau(b)\tau\left(\tau^{-1}(g)\right)\tau(b^{-1})\\ &=&\tau(b)g\tau(b)^{-1}\\ &=&\sigma_{\tau(b)}(g)$$

My question is, is $\tau$ equal to $\tau^{-1}(g)$ or is it $\left(\tau(g)\right)^{-1}$, or are these the same thing? I've done the problem assuming that $\tau^{-1}$ means $\tau^{-1}(g)$ and it seems to work, but I'm a bit uneasy about this. Can anybody confirm whether I've done this properly or not?

 Quote by PingPong So, I have $$\tau\left(\sigma_b\left(\tau^{-1}\left(g\right)\right)\right)&=&\tau\left(b\tau^{-1}(g)b^{-1}\right)\\ &=&\tau(b)\tau\left(\tau^{-1}(g)\right)\tau(b^{-1})\\ &=&\tau(b)g\tau(b)^{-1}\\ &=&\sigma_{\tau(b)}(g)$$
This is good
 Quote by PingPong My question is, is $\tau$ equal to $\tau^{-1}(g)$ or is it $\left(\tau(g)\right)^{-1}$, or are these the same thing?
$\tau$ is a mapping and $\tau^{-1}(g)$ is an element of the group on which the mapping is defined. Your question is not making sense. You have successfully shown that that the inner automorphisms are normal in the group of automorphisms. You can stop there.

 Quote by Mathdope This is good $\tau$ is a mapping and $\tau^{-1}(g)$ is an element of the group on which the mapping is defined. Your question is not making sense. You have successfully shown that that the inner automorphisms are normal in the group of automorphisms. You can stop there.
Okay, so I worded my question a bit strangely. Sorry

What I meant to ask was whether $\tau^{-1}(g)=\left(\tau(g)\right)^{-1}$ in general. I have a feeling that it's not and that's what I was uneasy about.

Recognitions:
Gold Member
 Quote by PingPong What I meant to ask was whether $\tau^{-1}(g)=\left(\tau(g)\right)^{-1}$ in general. I have a feeling that it's not and that's what I was uneasy about.