 Solution

By Lagrange's theorem, the index of a subgroup is $[G:H]=75/25=3$. Thus the set of left cosets has three elements, and the preceding group action yields a group homomorphism $\Phi :G\to S_3$ after labelling them.

Its image is a subgroup of $S_3$, so its order divides $6$. By the first isomorphism theorem, $|\mathrm{im}\Phi |=|G|/|\ker \Phi |$ also divides $75$. Hence its order divides $\gcd (6,75)=3$, giving possibilities $1$ and $3$. The action on cosets is transitive: $g_2{g}_1^{-1}$ sends $g_{1}H$ to $g_{2}H$. It cannot be the trivial action on a three-element set, so the image does not have order $1$. ThereforeIt follows that $|\ker \Phi |=75/3=25$. Part (b)(i) gives $\ker \Phi \le H$, and $|H|=25$; equal finite orders force equality. A kernel of a group homomorphism is a normal subgroup, soThis is an instance of the fact that a subgroup of index the smallest prime divisor is normal.