1. ## 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.

2. ## Re: Finite group

I don't understand what we have to show.

3. ## Re: Finite group

Originally Posted by girdav
I don't understand what we have to show.
Then you aren't alone.

4. ## 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?

5. ## Re: Finite group

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.

6. ## Re: Finite group

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."