In 'Finite Groups' by Gorenstein, it is stated that if $P$ is a Sylow $p$-subgroup of $G$ then there exists a normal subgroup $K$ such that $G/K$ is isomorphic to $P/P \cap G'$′. The proof is the following:

$P \cap G'$ is a Sylow $p$-subgroup of $G'$, as $G/G'$ is abelian and $G' \triangleleft G$. If $K$ denotes the inverse image of $O_{P'}(G/G')$ in $G$, $P \cap G' = P \cap K$, $K \triangleleft G$ and $G/K$ is an abelian $p$-group isomorphic to $P/P \cap G'$.

I understand the first line, and that $O_{P'}(G/G')$ represents the maximal normal $p'$-subgroup of $G/G'$ but don't quite understand what the inverse image is, or how this immediately yields $P \cap G' = P \cap K$, could anyone explain this in better terms?

  • $\begingroup$ Please state the theorem with an edit. $\endgroup$ – Shaun Dec 29 '18 at 18:48
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    $\begingroup$ The theorem was in the title, but I put it in the text body if thats what you were referring to :) $\endgroup$ – whereismymind96 Dec 29 '18 at 19:34
  • $\begingroup$ Thank you. Of course, yes; I'm sorry :) $\endgroup$ – Shaun Dec 29 '18 at 19:36
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    $\begingroup$ The inverse image is, by definition, the group $K$ with $G' \le K$ such that $K/G' = O_{p'}(G/G')$. Clearly $P \cap G' \le P \cap K$, and $|K:G'|$ is not divisible by $p$, so we get equality. $\endgroup$ – Derek Holt Dec 29 '18 at 20:13
  • $\begingroup$ That makes sense, thank you! $\endgroup$ – whereismymind96 Dec 29 '18 at 21:14

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