Proof (or counter-example) of the existence of an element of order $p$ in a conjugacy class $aH$ of a subgroup $H$ of a finite group $G$

Question

Let be $G$ a finite group, $H$ a subgroup, $p$ a prime number and $a \in G \setminus H$, such that $a^p \in H$.

I know there is an element $x \in aH$ such that $x^{p^k} = e$, for some $k$ and $e$ the neutral element of $G$.

I'd like to know if we can show that $k = 1$ always works or if we can find a group $G$ (I guess non-abelian, some symmetric group) where it does not work.

I tried to poke with $S_4$ and $A_4$ around, but I'm not sure.

Answer 1

Counterexample: $a = (123456789)\in S_9$, $p=3$, $H = \langle a^3\rangle\cong \mathbb{Z}_3$. Then $aH = \{a,a^4,a^7\}$. None of elements in $aH$ has order $p = 3$.