Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 3.3 Exercise 3.3.9
Let $p$ be a prime and let $G$ be a finite group of order $p^am$, where $p$ does not divide $m$. Let $P \leq G$ be a subgroup of order $p^a$ and $N \leq G$ a normal subgroup of order $p^bn$ where $p$ does not divide $n$. Prove that $|P \cap N| = p^b$ and that $|PN/N| = p^{a-b}$.
Solution: By the Second Isomorphism Theorem, we have $PN \leq G$, $N \leq PN$ normal, $P \cap N \leq P$ normal, and $P/(P \cap N) \cong PN/N$.
Now $|PN|$ divides $|G|$ by Lagrange, so that $|PN| = p^k\ell$ for some $k$ where $p$ does not divide $\ell$; then $\ell|m$. Because $P \leq PN$, we have $k = a$, and because $N \leq PN$, $n|\ell$. Thus $|PN/N| = p^{a-b}q$, where $p$ does not divide $q$.
Note that $|P/(P \cap N)| = p^k$ for some $k$, and we have $p^k = p^{a-b}q$. Thus $q = 1$ and we have $|PN/N| = p^{a-b}$. Finally, we have $|P|/|P \cap N| = p^a / p^b$, so that $|P \cap N| = p^b$.
