**Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 4.5 Exercise 4.5.1**

Solution: If $P \leq H \leq G$ is a Sylow $p$-subgroup of $G$, then $p$ does not divide $[G : P]$. Now $[G : P] = [G : H] [H : P]$, so that $p$ does not divide $[H : P]$; hence $P$ is a Sylow $p$-subgroup of $H$.

A trivial counterexample to the converse statement is the subgroup $\langle r \rangle \leq D_8; \langle r \rangle$ is a Sylow 2-subgroup of itself, but clearly not a Sylow 2-subgroup of $D_8$ since $|D_8| = 2^3$.