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