# 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 $G$

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.

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 $$p = 3$$.
• @MiKiDe I tried by letting $H$ as small as possible. If $|H| = 2$, then $o(a) = 2p$ and $a^{p+1}\in aH$ has order $p$ since $2p\mid p(p+1)$. So I tried $|H| = 3$ to make $a^{p+1}$ not of order $p$. – Hongyi Huang Jun 16 at 2:00