Let $G$ be a finite group, let $P$ be one of its $2$-sylow subgroups. Let $H$ be a proper subgroup of $P$, namely $H<P$ with $H\neq P$. Is it possible that $$\bigcup_{g\in G}g^{-1}Hg=\bigcup_{g\in G}g^{-1}Pg?$$

  • 3
    $\begingroup$ What about $G=A_4$ and $H$ a subgroup of order $2$. Then $P \lhd G$ and $\cup_{g \in G} g^{-1}Hg = P$. $\endgroup$ – Derek Holt Jan 6 at 20:54
  • $\begingroup$ @Derek Right, thank you! $\endgroup$ – Qixiao Ma Jan 7 at 1:26
  • $\begingroup$ Just to build on Derek Holt's example, there are many simple examples. For example, let $G$ be any of the simple groups ${\rm SL}(2,2^{n})$ where $n >1.$ Then $G$ has a unique conjugacy class of involutions, so f $P$ is a Sylow $2$-subgroup of $G$ and $T$ is a (proper) subgroup of $P$ of order $2,$ we have $\bigcup_{g \in G} g^{-1}Tg = \bigcup_{g \in G} g^{-1}Pg.$ $\endgroup$ – Geoff Robinson Jan 7 at 12:54

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Browse other questions tagged or ask your own question.