I'm trying to understand the following proof from Kurzweil-Stellmacher:
How does $PgP=Pg$ imply $gPg^{-1}=P$?
Transcription of image:
3.1.10 Let $P$ be a $p$-subgroup of $G$ and $p$ be a divisor of $|G:P|$. Then $P<N_G(P)$.
Proof. By 3.1.1 (c) $P$ acts on the set $\Omega$ of right cosets $Pg, g\in G$, by right multiplication, and $$|\Omega| = |G:P|\equiv 0\pmod p.$$ From 3.1.7 we get (with $P$ in place of $G$): $$|C_\Omega(P)|\equiv |\Omega|\equiv 0\pmod p.$$ Moreover $C_\Omega(P)\neq\emptyset$ since $P\in C_\Omega(P)$. Hence there exists $Pg \in C_\Omega(P)$ such that $P\neq Pg$. This implies $g\notin P$ and $PgP=Pg$. Thus $gPg^{-1}=P$ and $g\in N_G(P)\setminus P$.