1. ## Automorphisms and Conjugation

Exercise 1 in Dummit and Foote section 4.4 Automorphisms reads as follows:

If $\sigma$ $\in$ 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).

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 ${\phi}_g$ is h $\longrightarrow$ gh $g^{-1}$

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

2. ## Re: Automorphisms and Conjugation

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

If $\sigma$ $\in$ 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).

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 ${\phi}_g$ is h $\longrightarrow$ gh $g^{-1}$

(3) I am not sure what Dummit and Foote mean by ${\sigma{(g)}$ - unless they mean the set of automorphisms that map g into g???
What they mean is this. There is a natural map $\Phi:G\to \text{Aut}(G)$ defined by the rule $(\Phi(g))(x)=gxg^{-1}$ (where $x\in G$--note that since $\Phi(g)$ is supposed to be a map that this makes sense). They then define $\text{im }\Phi$ to be $\text{Inn}(G)$. They then ask you to show that $\sigma\circ\Phi(g)\circ\sigma^{-1}=\Phi(\sigma(g))$ (where, note again, that $\sigma(g)\in G$) and so one can deduce that $\text{Inn}(G)\unlhd \text{Aut}(G)$.

3. ## Re: Automorphisms and Conjugation

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

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

Peter

4. ## Re: Automorphisms and Conjugation

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

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

Peter

Think about this. Take $G=\mathbb{Z}$, $\sigma(x)=2x$, and $g=3$. Then, they're trying to have you prove that $\sigma\circ\Phi(3)\circ\sigma^{-1}=\Phi(\sigma(3))=\Phi(6)$.

5. ## 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:

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

and

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

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

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

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

Peter

6. ## 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 $\sigma$. this is a bijective homomorphism, and let's say $\sigma:x \to y$. clearly, $\sigma^{-1}:y \to x$.

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

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

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

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

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

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

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

this shows that for any automorphism $\sigma$, that $\sigma\mathrm{Inn}(G)\sigma^{-1} \subseteq \mathrm{Inn}(G)$

7. ## Re: Automorphisms and Conjugation

Thanks for that really helpful post

I will now work through the proof

Peter

8. ## Re: Automorphisms and Conjugation

note: i just noticed i use $\sigma_g$ where you use $\phi_g$, which is probably preferrable to avoid confusion.