# Automorphisms and Conjugation

• Nov 25th 2011, 06:53 PM
Bernhard
Automorphisms and Conjugation
Exercise 1 in Dummit and Foote section 4.4 Automorphisms reads as follows:

If $\displaystyle \sigma$$\displaystyle \in Aut(G) and \displaystyle {\phi}_g is conjugation by g prove that \displaystyle \sigma$$\displaystyle {\phi}_g$$\displaystyle {\sigma}^{-1} = \displaystyle {\phi}_{\sigma{(g)}. Deduce that Inn(G) is a normal subgroup of Aut(G). Can anyone help me get started on this problem? Notes: (1) I have attached Dummit and Foote section 4.4 in case readers need to view the terminology and notation used. This upload includes the text of the exercise. (2) I am assuming that the mapping \displaystyle {\phi}_g is h \displaystyle \longrightarrow gh\displaystyle g^{-1} (3) I am not sure what Dummit and Foote mean by \displaystyle {\sigma{(g)} - unless they mean the set of automorphisms that map g into g??? • Nov 25th 2011, 07:51 PM Drexel28 Re: Automorphisms and Conjugation Quote: Originally Posted by Bernhard Exercise 1 in Dummit and Foote section 4.4 Automorphisms reads as follows: If \displaystyle \sigma$$\displaystyle \in$ Aut(G) and $\displaystyle {\phi}_g$ is conjugation by g prove that $\displaystyle \sigma$$\displaystyle {\phi}_g$$\displaystyle {\sigma}^{-1}$ = $\displaystyle {\phi}_{\sigma{(g)}$.

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

Can anyone help me get started on this problem?

Notes:

(1) I have attached Dummit and Foote section 4.4 in case readers need to view the terminology and notation used. This upload includes the text of the exercise.

(2) I am assuming that the mapping $\displaystyle {\phi}_g$ is h $\displaystyle \longrightarrow$ gh$\displaystyle g^{-1}$

(3) I am not sure what Dummit and Foote mean by $\displaystyle {\sigma{(g)}$ - unless they mean the set of automorphisms that map g into g???

What they mean is this. There is a natural map $\displaystyle \Phi:G\to \text{Aut}(G)$ defined by the rule $\displaystyle (\Phi(g))(x)=gxg^{-1}$ (where $\displaystyle x\in G$--note that since $\displaystyle \Phi(g)$ is supposed to be a map that this makes sense). They then define $\displaystyle \text{im }\Phi$ to be $\displaystyle \text{Inn}(G)$. They then ask you to show that $\displaystyle \sigma\circ\Phi(g)\circ\sigma^{-1}=\Phi(\sigma(g))$ (where, note again, that $\displaystyle \sigma(g)\in G$) and so one can deduce that $\displaystyle \text{Inn}(G)\unlhd \text{Aut}(G)$.
• Nov 25th 2011, 09:34 PM
Bernhard
Re: Automorphisms and Conjugation
I am still struggling with both the meaning and the difference between $\displaystyle \phi_g$ and $\displaystyle \phi_{\sigma{(g)}}$?

Can anyone clarify this for me and go a bit further with the proof?

Peter
• Nov 25th 2011, 09:52 PM
Drexel28
Re: Automorphisms and Conjugation
Quote:

Originally Posted by Bernhard
I am still struggling with both the meaning and the difference between $\displaystyle \phi_g$ and $\displaystyle \phi_{\sigma{(g)}}$?

Can anyone clarify this for me and go a bit further with the proof?

Peter

Think about this. Take $\displaystyle G=\mathbb{Z}$, $\displaystyle \sigma(x)=2x$, and $\displaystyle g=3$. Then, they're trying to have you prove that $\displaystyle \sigma\circ\Phi(3)\circ\sigma^{-1}=\Phi(\sigma(3))=\Phi(6)$.
• Nov 26th 2011, 12:13 AM
Bernhard
Re: Automorphisms and Conjugation
Thanks

The example was very helpful in getting an understanding of the problem.

Basically he equivalence between your notation and D&Fs is as follows:

$\displaystyle \phi_g$ $\displaystyle \equiv$ $\displaystyle \Phi (g)$

and

$\displaystyle \phi_{\sigma (g)}$$\displaystyle \equiv$$\displaystyle \Phi (\sigma (g))$ where $\displaystyle \sigma (g)$ (as is usual) is the map of g under $\displaystyle \sigma$

I am still struggling with the proof of $\displaystyle \sigma$$\displaystyle \phi_g$$\displaystyle \sigma^{-1}$ = $\displaystyle \phi_{\phi (g)}$ and would appreciate some further help.

Regarding Inn(G) $\displaystyle \unlhd$ Aut(G) it appears to me that $\displaystyle \phi_g$ is essentially conjugation of G by g and hence $\displaystyle \phi_g$ is an inner automorphism of G and hence $\displaystyle \phi_g$ $\displaystyle \in$ Inn(G).

But where to from here to show Inn(G) $\displaystyle \unlhd$ Aut(G)

Peter
• Nov 26th 2011, 05:35 AM
Deveno
Re: Automorphisms and Conjugation
two maps are equal if they take on equal values for every element in their domain.

let's say we have any old automorphism of G, which we call $\displaystyle \sigma$. this is a bijective homomorphism, and let's say $\displaystyle \sigma:x \to y$. clearly, $\displaystyle \sigma^{-1}:y \to x$.

now by $\displaystyle \sigma_g$, we mean the particular automorphism that is conjugation by g: $\displaystyle \sigma_g:x \to gxg^{-1}$ for any x. so:

$\displaystyle \sigma \circ \sigma_g \circ \sigma^{-1}(y) = \sigma(\sigma_g(\sigma^{-1}(y)))$

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

$\displaystyle = \sigma(g)\sigma(x)\sigma(g^{-1})$ (since $\displaystyle \sigma$ is a homomorphism)

$\displaystyle = \sigma(g)y(\sigma(g))^{-1}$ (again, since $\displaystyle \sigma$ is a homomorphism, and by our definition of y)

$\displaystyle = \sigma_{\sigma(g)}(y)$.

so when we conjugate the inner automorphism corresponding to conjugation by g, by another automorphism $\displaystyle \sigma$, we get the same map as conjugation by the element $\displaystyle \sigma(g)$.

this shows that for any automorphism $\displaystyle \sigma$, that $\displaystyle \sigma\mathrm{Inn}(G)\sigma^{-1} \subseteq \mathrm{Inn}(G)$
• Nov 26th 2011, 02:40 PM
Bernhard
Re: Automorphisms and Conjugation
Thanks for that really helpful post

I will now work through the proof

Peter
• Nov 26th 2011, 02:53 PM
Deveno
Re: Automorphisms and Conjugation
note: i just noticed i use $\displaystyle \sigma_g$ where you use $\displaystyle \phi_g$, which is probably preferrable to avoid confusion.
• Nov 26th 2011, 03:18 PM
Bernhard
Re: Automorphisms and Conjugation