Transitive group action acting on a finite group

Question

I've faced this problem earlier today and I think I've found a proof, but I'm not so sure of it. The problem is the following (it's translated so maybe the terminology may be a little bit off)
Let $F$ be a finite group, and $G=Aut(F)$. Let's consider the natural action of $G$ on $F-\{1\}$. If the action is transitive, show that $F$ is an abelian elementary $p$-group for some prime $p$.

Now, we just have to show that every element have the same order, which is a prime, and that $F$ is abelian; the result will thus be implied by the classification theorem. For the first result we have for $f_1,f_2\in F$, $f_2=gf_1$, but as an element and it's image have the same order, we have that all the elements have the same order; furthermore, if this order has a non trivial divisor $m$, then we have $f_1=g_1f_1^m$ for some $g_1 \in G$, but that raises a clear contradiction, so that the order of each element is a prime. On the other hand, let's take $z \in Z=Z(F)$; we can do so as a $p$-group has always a non-trivial centre. For every $f$ we can find a $g$ so that $f=gz$, but for a second $f_0$ we can find $f_g$ so that $f_0=gf_g$: that's because the orbit of $O(z)=F$ for transitivity; for every two $f,f_0$ we thus have $ff_0=(gz)(gf_g)=g(zf_g)=g(f_gz)=(gf_g)(gz)=f_0f$, so that we're done (I think).

Answer 1

Two major issues:

• You need to show $F$ is a $p$-group. By Cauchy's theorem, if a prime number $q$ divides $|F|$, then $F$ contains an element of order $q$. But all non-identity elements have the same order, hence $q$ is unique. (I would also prefer Cauchy to show all non-identity elemnts have the same prime order, but what you did is fine.)
• $f_0=gf_g$ is not because $O(z)=F\setminus\{1\}$ (not $F$, which is a notation error), but becuase $g$ is an automorphism hence surjective. If you apply the transitivity, you would establish $f_0=g'z$ which is not very useful.

The overall readibility can be improved by using e.g. $T\in G=\operatorname{Aut}(F)$ and $a, b\in F$, instead of $g\in G, f\in F$. Also it's better to use the proper quantifier, e.g. "for a second $f_0$" is better paraphrased as "for any $f_0\in F$".