= Solution
{c}
Define $g\cdot(g_1H)=(gg_1)H$ on the set of <left cosets>. It is well-defined: if $g_1H=g_2H$, then $g_2=g_1h$ for some $h\in H$, so $(gg_2)H=(gg_1h)H=(gg_1)H$. The identity fixes each coset, and associativity in $G$ gives
$$
g\cdot(h\cdot(g_1H))=(ghg_1)H=(gh)\cdot(g_1H).
$$
Hence this is a left <group action>, and part (a) supplies its <group homomorphism> $\Phi:G\to\operatorname{Sym}(S)$.
An element of the <kernel of a group homomorphism> $\Phi$ fixes every coset, in particular $H$ itself. Thus $g\in\ker\Phi$ implies $gH=H$, which implies $g\in H$. Therefore
$$
\boxed{\ker\Phi\leq H.}
$$
The subgroup symbol is justified because a <kernel of a group homomorphism> is a <subgroup>, and the argument proves containment.
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