# Finite group

• Oct 7th 2011, 01:27 PM
dwsmith
Finite group
Let G be a finite group which possesses an automorphism $\displaystyle \sigma$ such that $\displaystyle \sigma(g)=g$ iff. g = 1.

$\displaystyle (\Rightarrow)$
Suppose |G| = n and $\displaystyle \sigma(g)=g$.

First, I don't get why this would only work if g is 1.
• Oct 7th 2011, 01:30 PM
girdav
Re: Finite group
I don't understand what we have to show.
• Oct 7th 2011, 01:34 PM
dwsmith
Re: Finite group
Quote:

Originally Posted by girdav
I don't understand what we have to show.

Then you aren't alone.
• Oct 7th 2011, 01:41 PM
girdav
Re: Finite group
In your first post, you say that we have a finite group which possesses an automorphism $\displaystyle \sigma$ such that $\displaystyle \sigma (g)=g$ if and only if $\displaystyle g=1$.
Then, what do we have to do?
• Oct 7th 2011, 01:50 PM
dwsmith
Re: Finite group
Quote:

Originally Posted by girdav
In your first post, you say that we have a finite group which possesses an automorphism $\displaystyle \sigma$ such that $\displaystyle \sigma (g)=g$ if and only if $\displaystyle g=1$.
Then, what do we have to do?

Show it is a homomorphism, monic, and epi.
• Oct 7th 2011, 02:48 PM
Drexel28
Re: Finite group
Quote:

Originally Posted by dwsmith
Let G be a finite group which possesses an automorphism $\displaystyle \sigma$ such that $\displaystyle \sigma(g)=g$ iff. g = 1.

$\displaystyle (\Rightarrow)$
Suppose |G| = n and $\displaystyle \sigma(g)=g$.

First, I don't get why this would only work if g is 1.

If I had to guess, you are asking about the common problem: "If $\displaystyle G$ is a finite group and $\displaystyle \sigma\in\text{Aut}(G)$ such that $\displaystyle \sigma^2=\text{id}$ and $\displaystyle \sigma$ possesses no non-identity fixed points, then $\displaystyle \sigma$ is the inverse map and $\displaystyle G$ is abelian."