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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 $$\bigcup_{g\in G}g^{-1}Hg=\bigcup_{g\in G}g^{-1}Pg?$$

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    $\begingroup$ 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

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