Proving $P$ is a Sylow $p$-group of $PN$

Question

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:)

Answer 1

In general, if $H \leq G$, $P \in Syl_p(G)$ with $P \subseteq H$, then $P \in Syl_p(H)$. Proof: $|H:P|$ divides the $p'$-number $|G:P|$.

Answer 2

Update: Woke up with the following possible solution on mind.

Since $P,N \leq PN$, we know that $|P| \bigg| |PN|$ and $|N| \bigg| |PN|$ i.e. $p^a \bigg| |PN|$ and $p^bm \bigg| |PN|$.

So $|PN|$ must be of the form $p^kmq$ for some $k,q$, where $k \geq a \geq b$, and it can be assumed without loss of generality that $p \nmid q$ (since any $p$ dependence can be absorbed in $p^k$).

Further, since $m|n$, by definition $n=mj$ for some $j$. Which implies that $m=n\frac{1}{j}$, but since $\frac{1}{j} \leq 1$ and we already know that $p \nmid n$, we thus have that $p \nmid m$.

Finally, we can define $l=mq$ to get $|PN|$ in the desired form: $|PN|=p^kl$, where $p \nmid l$.