I have a question that I feel I am going about in a roundabout way, and would like some help on. I am preparing for an exam.

Problem: Let G be a group with |G|=150. Let H be a non-normal subgroup in G with |H|=25.

(a) How many elements of order 5 does G have?

(b) How many elements of order 25 does G have?

My attempt:

G has 6 subgroups of order 25 because the number of subgroups of order 25 has to divide . So there are 6 such subgroups: $ $. Let $G$ act on this set by conjugation. The permutation representation of the group action of conjugation on this set is . Because has at most one factor of $5$, $5$ divides $ $. Suppose $25$ divides $ $. Then $ $ has a Sylow 5-subgroup $H_i$. So $ $. By the Second Sylow Theorem, $ . So $\ker(\phi)$ has a subgroup $K$ with $|K|=5$. So for some $ $

What next? Surely this shouldn't be so long-winded.