Automorphisms and Conjugation

**Bernhard** · November 25th 2011, 06:53 PM

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???

**Drexel28** · November 25th 2011, 07:51 PM

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 $\text{Inn}(G)\unlhd \text{Aut}(G)$ .

**Bernhard** · November 25th 2011, 09:34 PM

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

**Drexel28** · November 25th 2011, 09:52 PM

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 $\sigma\circ\Phi(3)\circ\sigma^{-1}=\Phi(\sigma(3))=\Phi(6)$ .

**Bernhard** · November 26th 2011, 12:13 AM

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

**Deveno** · November 26th 2011, 05:35 AM

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 $\sigma\mathrm{Inn}(G)\sigma^{-1} \subseteq \mathrm{Inn}(G)$

**Bernhard** · November 26th 2011, 02:40 PM

Thanks for that really helpful post

I will now work through the proof

Peter

**Deveno** · November 26th 2011, 02:53 PM

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

**Bernhard** · November 26th 2011, 03:18 PM

Thanks for that helpful clarification

Peter

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