# If $P$ is a Sylow $p$-subgroup of $G$ then there exists a normal subgroup $K$ such that $G/K$ is isomorphic to $P/P \cap G'$

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?

• Please state the theorem with an edit. – Shaun Dec 29 '18 at 18:48
• The theorem was in the title, but I put it in the text body if thats what you were referring to :) – whereismymind96 Dec 29 '18 at 19:34
• Thank you. Of course, yes; I'm sorry :) – Shaun Dec 29 '18 at 19:36
• 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. – Derek Holt Dec 29 '18 at 20:13
• That makes sense, thank you! – whereismymind96 Dec 29 '18 at 21:14