I am stuck on the question related to Sylow p-subgroups. I tried to read through the Sylow Theorems but I could not relate it back to the questions and can't see the way to apply it. It would be great if anyone can guide me through the questions with explanation.
The question is as below:
Let $G$ be a finite group with subgroup $H$. Show that
a) $H$ contains a Sylow $p$-subgroup of $G$ if and only if |$G$:$H$| is coprime to $p$
b) if $N_G(P) \leq H$ for some Sylow $p$-subgroup of $H$, show that $p$ does not divide |$G$:$H$|
Hint: Normalisers grow in nilpotent groups
For a)
Note that if $|G| =p^e m $ with $(p,m)=1$ then the order of any Sylow $p $-subgroup is $p^e$ then, if a Sylow $p $-subgroup is in $H$ then the order of $H $ order must be at least $p^e $ let's say it is $p^e k$ with $k$ $|$ $m $ now we get $[G:H]= |G| / |H|=p^e m / p^e k= m/k$ which obviously is coprime to p.
The other direction is easy taking in account what I wrote above.
For b)
Check that the fact that $N_G (P) \le H$ means $P \le H $ and a).