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 $\displaystyle $150/25=6$ and $6 \cong 1 (\mod 5)$$. So there are 6 such subgroups: $$\displaystyle \{H_1,H_2,H_3,H_4,H_5,H_6\}$$. Let $G$ act on this set by conjugation. The permutation representation of the group action of conjugation on this set is $\displaystyle $\phi:G \rightarrow S_6$, and $|S_6|=720$. Also, $|\ker(\phi)||im(\phi)|=150$$. Because $\displaystyle $|im(\phi)|$$ has at most one factor of $5$, $5$ divides $$\displaystyle |\ker(\phi)|$$. Suppose $25$ divides $$\displaystyle |\ker(\phi)|$$. Then $$\displaystyle \ker(\phi)$$ has a Sylow 5-subgroup $H_i$. So $$\displaystyle H_i \subset \ker(\phi) \implies aH_ia^{-1} \subset a(\ker(\phi))a^{-1} = \ker(\phi)$$. By the Second Sylow Theorem, $$\displaystyle H_1 \cup \ldots \cup H_6 \subset \ker(\phi) \implies G = \ker(\phi) \implies \phi$ trivial$. So $\ker(\phi)$ has a subgroup $K$ with $|K|=5$. So for some $$\displaystyle H_i, \ker(\phi) \cap H_i=K \implies a(\ker(\phi))a^{-1} \cap aH_ia^{-1} = aKa^{-1}$$

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