This is the statement of the theorem:
Let $P$ a Sylow $p$-subgroup of $G$ and $Z$ a subgroup of $Z(P)$ that is weakly closed in $P$. Set $H=N_G(Z)$. Then $P\cap G'=P\cap H'$ and $P/(P\cap G')\simeq G/G'(p) \simeq H/H'(p)$.
I have proved the first claim.
This is the proof of the last claim, using the first part. We have that $$ P/(P\cap G') = P/(P\cap G'(p) ) \simeq PG'(p)/G'(p)=G/G'(p)$$
and $$P/(P\cap H')=P/(P\cap H'(p)) \simeq P H'(p)/H'(p)=H/H'(p). \;\; (2)$$
I don't understand why the second chain of equalities holds.
In order to say that $P\cap H'=P\cap H'(p)$ and that $PH'(p)=H$ should't I make sure that $P\subset H=N_G(Z)$? If yes, how can I show that? Are these equalities true also if the p-Sylow is not contained in $H$?
Thanks for the help!
Since $Z$ lies in the centre of $P$, certainly $P\le N_G(Z)$.