TITLE: Qualifying Exam Question On Elementary Group Theory
QUESTION [7 upvotes]: Question. Let $G$ be a finite group and $p$ be the smallest prime dividing $|G|$.
Let $x$ be n element of order $p$ in $G$. Assume that there exists an element $h\in G$ such that $hxh^{-1}=x^{10}$.
Show that $p=3$.

What I have done so far is use the fact that $o(hxh^{-1})=o(x)=p$ and $o(x^{10})=p/\gcd(p,10)$.
By hypothesis we have $p=p/\gcd(p,10)$, giving $\gcd(p,10)=1$.
Here I am stuck.
I have not been able to make use of the fact that $p$ is the smallest prime divisor of $|G|$.
Can somebody help?

REPLY [10 votes]: Let $n$ be the order of $h$. Then $$x = h^n x h^{-n} = x^{10^n}$$
So $10^n \equiv 1 \mod{p}$. Now apply the fact that $\gcd(p-1, n) = 1$.