Number of subgroups and elements 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 $150/25=6$ and $6 \cong 1 (\mod 5)$. So there are 6 such subgroups: $\{H_1,H_2,H_3,H_4,H_5,H_6\}$. Let $G$ act on this set by conjugation. The permutation representation of the group action of conjugation on this set is $\phi:G \rightarrow S_6$, and $|S_6|=720$. Also, $|\ker(\phi)||im(\phi)|=150$. Because $|im(\phi)|$ has at most one factor of $5$, $5$ divides $|\ker(\phi)|$. Suppose $25$ divides $|\ker(\phi)|$. Then $\ker(\phi)$ has a Sylow 5-subgroup $H_i$. So $H_i \subset \ker(\phi) \implies aH_ia^{-1} \subset a(\ker(\phi))a^{-1} = \ker(\phi)$. By the Second Sylow Theorem, $H_1 \cup \ldots \cup H_6 \subset \ker(\phi) \implies G = \ker(\phi) \implies \phi$ trivial. So $\ker(\phi)$ has a subgroup $K$ with $|K|=5$. So for some $H_i, \ker(\phi) \cap H_i=K \implies a(\ker(\phi))a^{-1} \cap aH_ia^{-1} = aKa^{-1}$
What next? Surely this shouldn't be so long-winded.

Your first part is correct $G$ has 6 Sylow 5-subgroups. $H$ is abelian and either cyclic or isomorphic to $C_5 \times C_5$. So ...
