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

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

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A start to the proof is as follows:

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

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

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

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

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

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

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

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

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

A start to the proof is as follows:

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

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

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

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

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

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

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

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