I am having trouble solving the following problem:
Let $G$ be a finite group of order $p^an$, where $p$ is a prime and $p \nmid n$.
Let $P$ be a Sylow $p$-group in $G$ and let $N \unlhd G$.
It can be used without proof that the order of $N$ is $p^bm$, where $b \leq a$ and $m | n$, and that $PN \leq G$.
Show that $P$ is a Sylow $p$-group in $PN$.
I have the proof down, if only I could show that $PN$ is of order $p^kl$ for some $k,l$, where $p \nmid l$.
From there one can argue that $$P \leq PN \implies |P| \bigg| |PN| \implies p^a \bigg| p^kl \implies a \leq k.$$
And $$PN \leq G \implies |PN| \bigg| |G| \implies p^kl \bigg| p^an \implies k \leq a.$$
Which implies that $a=k$, and thus shows that $P$ indeed is a Sylow $p$-group of $PN$.
A helping hand to proof the missing link would be truly appreciated! Thank you:)