An exercise from Dummit and Foote pg. $101$ ex. $9$ asks to show the following:
Let $G$ be a group of order $p^{a}n$ where $p$ does not divide $n$ and let $N\unlhd G$ so that $|N|=p^{b}m$ where $p$ does not divide $m$ and $P\leq G$ is of order $p^a$ is a Sylow p-group.
Prove that $|P\cap N|=p^{b}$.
I have seen other posts, but I found only hints or solutions that uses Sylow theorms (which have yet to be proven in the text).
I tried to use the second isomorphism theorem and I drew the subgroups lattice, I have also used the fact that $|N\cap P|=p^{r}$ for some $r$ (using Lagrange) but since I don't know what is $|PN|$ I didn't manage to find $r$.
Can someone please guide me further ?
Since $N \trianglelefteq G $, $NP$ is subgroup of $G$ and $[NP:P]=[N:N \cap P]$ (use that $NP/N \cong P/(N \cap P))$. Since $P \in Syl_p(G)$ and $[NP:P]$ divides $[G:P]$, the index $[N:N \cap P]$ is not divisible by $p$. This implies that the $p$-subgroup $N \cap P \in Syl_p(N)$.
Conversely, if $Q \in Syl_p(N)$, then the $p$-subgroup $Q$ is contained in some $P \in Syl_p(G)$ (this is a well-known lemma). Hence $Q \subseteq N \cap P$. But we just showed that $N \cap P \in Syl_p(N)$, whence $Q=N \cap P$. In other words, all Sylow $p$-subgroups of $N$ arise by intersecting those of $G$ with $N$.