= Solution
{c}
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>, $|\operatorname{im}\Phi|=|G|/|\ker\Phi|$ also divides $75$. Hence its order divides $\gcd(6,75)=3$, giving possibilities $1$ and $3$. The action on cosets is <transitive>: $g_2g_1^{-1}$ sends $g_1H$ to $g_2H$. It cannot be the trivial action on a three-element set, so the image does not have order $1$. Therefore
$$
\boxed{|\operatorname{im}\Phi|=3.}
$$
It follows that $|\ker\Phi|=75/3=25$. Part (b)(i) gives $\ker\Phi\leq H$, and $|H|=25$; equal finite orders force equality. A <kernel of a group homomorphism> is a <normal subgroup>, so
$$
\boxed{H=\ker\Phi\mathrel{\trianglelefteq}G.}
$$
This is an instance of the fact that a <subgroup of index the smallest prime divisor is normal>.
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