Proof : $ P' \cap N(P) = P' \cap P $, P' and P are Sylow-P sub-groups and N(P) is the normaliser of P I am recently studying modern algebra and came across this question in my textbook. This conclusion was used directly without any proof. I just wonder why it is true.
I first searched math.stackexchange for something related to intersections of two Sylow-P subgroups and didn't get any strong statements.
Here's the question:
Proof : $ P' \cap N(P) = P' \cap P $, P' and P are Sylow-P sub-groups and N(P) is the normaliser of P.
Thanks in advance!
 A: This just comes from the following fact:


Theorem: Any $p$-subgroup of $N_G(P)$ is contained in $P$.


Indeed, we see then that $P'\cap N_G(P)$ is a $p$-subgroup of $N_G(P)$, and thus $P'\cap N_G(P)\subseteq P$ so that evidently $P'\cap N_G(P)\subseteq P'\cap P$. But, since $P'\cap N_G(P)\supseteq P\cap P'$ you get the desired equality.
To prove this theorem, we merely use the fact that if $H,K\unlhd G$ and $H\leqslant N_G(K)$ then $HK\leqslant G$. From this, we see that if $Q$ is a $p$-subgroup of $N_G(P)$ then $PQ\leqslant G$. But, 
$$|PQ|=\frac{|P||Q|}{|P\cap Q|}$$
In particular, since $|P|$ and $|Q$ are both powers of $p$, with $|P|$ being the maximal power of $p$, we must have that 
$$\frac{|P||Q|}{|P\cap Q|}=|P|$$
so that $|Q|=|P\cap Q|$ and so $Q\subseteq P$ as desired.
A: 1. If $G$ is a finite group and $H \leq G$ is a $p$-subgroup, then $H$ is contained in some Sylow $p$-subgroup of $G$
2. $P$ is normal Sylow $p$-subgroup of $N(P)$, and thus it is the only Sylow $p$-subgroup of $N(P)$
Using these two facts you can prove $P' \cap N(P) \subseteq P' \cap P$, and the other inclusion should not be difficult.
