 Solution

Define $g\cdot (g_{1}H)=(g{g}_1)H$ on the set of left cosets. It is well-defined: if $g_{1}H=g_{2}H$, then $g_2=g_{1}h$ for some $h\in H$, so $(g{g}_2)H=(g{g}_{1}h)H=(g{g}_1)H$. The identity fixes each coset, and associativity in $G$ givesHence this is a left group action, and part (a) supplies its group homomorphism $\Phi :G\to \mathrm{Sym}(S)$.

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 $g\in H$. ThereforeThe subgroup symbol is justified because a kernel of a group homomorphism is a subgroup, and the argument proves containment.