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 $p = 3$.

  • $\begingroup$ How did you find this? I mean, what was the reasoning? $\endgroup$ – MiKiDe Jun 15 at 15:27
  • $\begingroup$ @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 $p$. $\endgroup$ – Hongyi Huang Jun 16 at 2:00

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