Are you taking MATHS720 at Auck uni this semester? I am taking it :D
There are several methods to prove this problem, the first two are based on the same fact.
The first way: As the hint Steve.D already made, one can show every Sylow subgroup of H must be a Sylow subgroup of G, here the idea is to prove by contradiction. Suppose $p|gcd(|H|,|G:H|)$, then by Cauchy's theorem showed in the lecture (as a corollary of SylowE theorem) H contains an element $h$ of order p, and G has a Sylow p-subgroup P which contains $h$. Next, one needs to show $Z(P) \leq C_H(x)$ for all $x \in P\setminus H$ , notice that by $C_G(x) \leq H$ for all $x \in H$\ {$1_G$} one can show $C_H(x)$ is trivial for all$x \in P\setminus H$, so it implies Z(P) is trivial; however every non-trival p-group has non-trivial centre, contradiction.
The second way: Observe that each non-trivial class of H has order divisible by the index $|G:H|$ , so it implies $|G:H|$ divides (|H|-1), thus H is a Hall subgroup of G.