If $G$ is a finite group and $\phi(x) = x^{p+1}$ is an automorphism of $G$ with $order(\phi) |p$ then $G$ is a $p$-group...?
If the order of $\phi$ is $1$ then $\phi(x) = x = x^{p+1} = x^px \rightarrow x^p =e$ so the order $\forall x\in G$ is $p$ therefore $G$ is a $p$-group
I'm having trouble when considering that the order of $\phi$ is $p$
If the order of $\phi$ is $p$ then
$\phi^p(x) =x= x^{(p+1)^p}$
using the binomial theorem, I get
$\forall x \in G \, \, \, order(x) | \ \displaystyle\sum_{k=1}^p \binom{p}{k}p^k = (p+1)^p-1$
At first, I thought that the only divisors of this was powers $p$ so I got this:
Suppose for contradiction that $G$ is not a $p$-group. Let $|G| = kp^n$ where $k$ is not a multiple of $p$ ($p\nmid k$). If $k>1$ then take a $q$ in $k$'s prime factorization. So we have $q|k$ and by Cauchy's Theorem $\exists y \in G$ with $y^q = e$ i.e $order(y) = q$.
Since $\forall x \in G \, \, \, order(x) | \ \displaystyle\sum_{k=1}^p \binom{p}{k}p^k$ we must have that $q|\displaystyle\sum_{k=1}^p \binom{p}{k}p^k$. A contradiction, therefore $k=1$ thus $G$ is a $p$-group.
But through examples, the divisors of $(p+1)^p-1$ also include other primes :'(. How can I show $G$ is a $p$-group?
Thanks so much :D
Update:
Taking the advice from the comments, Assume every proper subgroup of $G$ is a $p$-group, now take the Sylow $p$-subgroup of $G$, say $P$ then $P$ is normal since it is characteristic i.e $\phi(P) = P$. Since $P$ is a normal Sylow subgroup, it contains all $p$ subgroups of $G$ therefore $P$ contains all subgroups of $G$.
Can I conclude that $P=G$ since $P$ contains all subgroups of $G$?
This is false. Let $G$ be cyclic of order 7, and let $p=3$. Then $\phi:G \to G : g \mapsto g^4$ is an automorphism of order 3.
Update: The question has been changed now. You can read the original version here: https://math.stackexchange.com/revisions/823130/6