# Show that if $p$ divides $\lvert\operatorname{Aut}(G)\rvert$, then $p=4k+3$

1. Give an example of some odd prime number $p$ and some group $G$ such that $|G|=p+1$ and $p$ divides $\lvert\operatorname{Aut}(G)\rvert$.

2. Let $p$ is an odd prime number and $G$ is a group where $|G|=p+1$. Show that if $p$ divides $\lvert\operatorname{Aut}(G)\rvert$, then $p=4k+3$, for some integer $k$.

I assume $G$ is a Klein Four-Group, then $p$ must be $3$. Since an automorphism send identity element to itself, then the mapping of the rest elements is as same as rearranges them to theirself. Hence, $\lvert\operatorname{Aut}(G)\rvert=3!=6$ and $p$ divides $\lvert\operatorname{Aut}(G)\rvert$.

But for the second problem, I didn't know how can I deal with it.

If $p$ divides $|{\rm Aut}(G)|$, then an element of order $p$ in ${\rm Aut}(G)$ must act transitively on $G \setminus \{1\}$. That implies that $G$ is an elementary abelian $q$-subgroup for some prime $q$.
But $|G|$ is even, so $q=2$ and $p = 2^k-1$ for some $k$.
• I wonder if there is a weaker argument actually leading to $4k+3$ instead of $2^k-1$ ... – Hagen von Eitzen Mar 2 '18 at 8:15
• Possibly, but my argument is reasonably elementary. You don't even need to show that $G$ is elementary abelian. The fact that all of its elements have the same order proves that it has prime power order by Cauchy's theorem. – Derek Holt Mar 2 '18 at 8:23