Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 3.3 Exercise 3.3.10
Solution: We have $[HN:H \cap N] = [HN: H \cap N]$, so that $$[HN:N][N:H \cap N] = [HN:H][H:H \cap N].$$ By the Second Isomorphism Theorem, we have $[HN:N] = [H:H \cap N]$. Thus $[N:H \cap N] = [HN:H]$; because $HN \leq G$, we have that $[HN:H]$ divides $[G:H]$. Moreover, $|H \cap N|$ divides $|H|$. Thus $|H \cap N|$ and $[N:H \cap N]$ are relatively prime.
Now $[G:H] = [G:HN][HN:H]$, so that $[G:HN]$ divides $[G:H]$. By the Third Isomorphism Theorem, $[G:HN] = [G/N : HN/N]$, so that $[G:HN] = [G/N : HN/N]$ divides $[G:H]$. Now $$|HN| = |H| \cdot |N|/|H \cap N|,$$ so that $|HN/N| = |H|/|H \cap N|$. Thus $|HN/N|$ divides $|H|$, so that $|HN/N|$ and $[G/N : HN/N]$ are relatively prime.
