Finite group

**dwsmith** · October 7th 2011, 01:27 PM

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

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

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

**girdav** · October 7th 2011, 01:30 PM

I don't understand what we have to show.

**dwsmith** · October 7th 2011, 01:34 PM

Originally Posted by girdav

I don't understand what we have to show.

Then you aren't alone.

**girdav** · October 7th 2011, 01:41 PM

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

**dwsmith** · October 7th 2011, 01:50 PM

Originally Posted by girdav

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

Show it is a homomorphism, monic, and epi.

**Drexel28** · October 7th 2011, 02:48 PM

Originally Posted by dwsmith

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

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

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 $G$ is a finite group and $\sigma\in\text{Aut}(G)$ such that $\sigma^2=\text{id}$ and $\sigma$ possesses no non-identity fixed points, then $\sigma$ is the inverse map and $G$ is abelian."

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