let $x$ be in finite group $G$ and let order of $x$ is $p$. If $h^{-1}xh = x^{10}$ for a finite group show that $p=3$

Question

Let $G$ be a finite group, $p$ be the smallest prime divisor of $|G|$ and x $\in$ G an element of order $p$. Suppose $ h \in G $ is such that $h^{-1}xh = x^{10}$. Show that $p = 3$.

I cant solve this problem i think it should be solved without using any special theorem please firstly give me a hint .

Answer 1

Extended hints/steps (may be you want to cover the screen and reveal this one line at a time?):

• Because $x$ and $x^{10}$ are conjugates, and thus have the same order, show that $\gcd(p,10)=1$.
• By Little Fermat $10^{p-1}\equiv1\pmod p$. Why does this imply that $x$ and $h^{p-1}$ commute?
• Let $q$ be the order of $h$. Why is $\gcd(q,p-1)=1$? Why does this imply that $h$ and $x$ commute?
• Show that $p\mid 9$.