# Thread: Automorphisms of a group G - Inn(G) and Aut(G)

1. ## Automorphisms of a group G - Inn(G) and Aut(G)

Dummit and Foote Section 4.4 Automorphisms Exercise 1 reads as follows:

Let $\displaystyle \sigma \in Aut(G)$ and $\displaystyle \phi_g$ is conjugation by g, prove that $\displaystyle \sigma \phi_g \sigma^{-1}$ = $\displaystyle \phi_{\sigma (g)}$.

Deduce that Inn(G) is a normal subgroup of Aut(G)

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A start to the proof is as follows:

$\displaystyle \sigma \phi_g \sigma^{-1}$

= $\displaystyle \sigma (\phi_g (\sigma^{-1} (x)))$

= $\displaystyle \sigma (g. \sigma^{-1} (x). g^{-1} )$

= $\displaystyle \sigma (g). x . \sigma (g^{-1} )$

= $\displaystyle \sigma (g). x . {[\sigma (g)]}^{-1}$

Thus $\displaystyle \sigma \phi_g \sigma^{-1}$ = $\displaystyle \phi_{\sigma (g)}$.

My question is - how do we use this result to show Inn(G) is a normal subgroup of Aut(G)?

Peter

2. ## Re: Automorphisms of a group G - Inn(G) and Aut(G)

Originally Posted by Bernhard
Dummit and Foote Section 4.4 Automorphisms Exercise 1 reads as follows:

Let $\displaystyle \sigma \in Aut(G)$ and $\displaystyle \phi_g$ is conjugation by g, prove that $\displaystyle \sigma \phi_g \sigma^{-1}$ = $\displaystyle \phi_{\sigma (g)}$.

Deduce that Inn(G) is a normal subgroup of Aut(G)

================================================== ====

A start to the proof is as follows:

$\displaystyle \sigma \phi_g \sigma^{-1}$

= $\displaystyle \sigma (\phi_g (\sigma^{-1} (x)))$

= $\displaystyle \sigma (g. \sigma^{-1} (x). g^{-1} )$

= $\displaystyle \sigma (g). x . \sigma (g^{-1} )$

= $\displaystyle \sigma (g). x . {[\sigma (g)]}^{-1}$

Thus $\displaystyle \sigma \phi_g \sigma^{-1}$ = $\displaystyle \phi_{\sigma (g)}$.

My question is - how do we use this result to show Inn(G) is a normal subgroup of Aut(G)?

Peter
Well, you first need to show that Inn is a subgroup. I am assuming you have done this?

Now, a normal subgroup, $\displaystyle N\unlhd G$, is a subgroup whhere $\displaystyle xyx^{-1}\in N$ for all $\displaystyle y\in N$ and $\displaystyle x\in G$. So...you took an arbitrary automorphism, $\displaystyle \sigma$, and an arbitrary inner automorphism, $\displaystyle \phi_g$, and showed that $\displaystyle \sigma\phi_g\sigma^{-1}=\phi_{\simga(g)}\in \operatorname{Inn}(G)$ which is precisely what you want. Here, $\displaystyle x:=\sigma$, $\displaystyle y:=\phi_g$, $\displaystyle G:=\operatorname{Aut}(G)$ and $\displaystyle N:=\operatorname{Inn}(G)$.