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Let $G$ be a finite group and let $H$ be a subgroup of $G$. Suppose that every element of $G$ is conjugate to some element of $H$ that is for all $y \in G$, there exists a $g \in G$ and $x \in H$ such that $g^{-1}xg = y)$. Prove that $H = G$.

I'm really not quite sure how to approach this problem. It seems like there might be some way to construct a normal subgroup, or some kind of argument using conjugacy classes/ normalizers.

Shaun
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Important_man74
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    Basically : prove that $G \subset H$ and you're done ($H \subset G$ follows directly from the fact that $H$ is a subgroup of $G$). – Abezhiko Aug 16 '23 at 07:06
  • Of course, hence that is not of great help ;-) – Anne Bauval Aug 16 '23 at 07:27
  • Right, that hint is useless. Better hint: count. Can $G$ be covered by conjugates of a subgroup $H$? How many elements are in a conjugate $H^g$? How many conjugates are there? – Sean Eberhard Aug 16 '23 at 11:55

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