= Solution
Let $G$ act by left multiplication on its five left cosets of $H$. This gives a <group homomorphism> $\varphi:G\to S_5$. Its <kernel> is the <normal core of a subgroup>[core of a subgroup], $\bigcap_{g\in G}gHg^{-1}$: an element fixes every coset precisely when all its conjugates lie in $H$. It is a <normal subgroup> contained in $H$, so the hypothesis forces the <kernel> to be trivial. Thus $G$ embeds in $S_5$.
The image has order $60$ and index two, hence is the <kernel> of a surjection $S_5\to C_2$. All transpositions are conjugate and generate $S_5$, so under any such surjection they must all have the same nonidentity image. This determines the map uniquely as the sign <homomorphism>. Its <kernel> is $A_5$, proving $G\cong A_5$.
To prove simplicity directly, the nonidentity <conjugacy classes> in $A_5$ have sizes $15,20,12,12$. A double transposition has centralizer of order four in $A_5$; a three-cycle has centralizer of order three; a five-cycle has centralizer of order five, yielding the two classes of size twelve. The class sizes together with the identity account for all sixty elements. A <normal subgroup> must be a <union> of these classes containing the identity, and its order must divide sixty. None of the proper nontrivial sums of these sizes with one is a divisor of sixty. Therefore \b[$A_5$, and hence $G$, is simple].
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