Let $G$ be a finite group and $H$ be a normal subgroup of $G$.Moreover,$P$ be a sylow subgroup of $H$.
I have to show that,
i) $\forall g\in G\ \ \exists h\in H $ such that $gPg^{-1}=hPh^{-1}$
ii)Let $N=\{g\in G|gPg^{-1}=P\}$.And $HN$ is defined as usual by $HN=\{hn|h\in H;n\in N \}$.Prove that $G=HN.$
My attempt: I can do the first part as follows.
$\because H\trianglelefteq G$ $\therefore$ $gPg^{-1}\subseteq H$,So $P$ and $gPg^{-1}$ both are sylow subgroups inside $H$.($\because |P|=|gPg^{-1}$).And we know that any two sylow subgroups inside a group are conjugate.Therefore,i can get a $h\in H$ such that $gPg^{-1}=hPh^{-1}$
But i am having a bit of problem to show the second part.What would a conceptual solution to that one(without involving large computations)?
Best Answer
Intuition / skeleton solution:
Clearly $HN\le G$, so you need to show $G\le HN$.
So, let $g\in G$. You need $g=hn$ for some $h\in H$ and $n\in N$.
You can use the first part to obtain an interesting $h$ and you would need $n=h^{-1}g$, so you'd need to show $h^{-1}g\in N$.