= Solution
Act by left multiplication on the five <cosets> of $H$. The kernel is the core $\bigcap_{g\in G}gHg^{-1}$, a <normal subgroup> contained in $H$, so the hypothesis makes it trivial. Thus $G$ embeds in $S_5$, and its image has order $60$ and index two. An index-two <subgroup> is normal, so the quotient is of order two. Every three-cycle maps trivially to that quotient, since its order divides three. Three-cycles generate $A_5$, so the image contains $A_5$; their equal order gives $\boxed{G\cong A_5}$.
For <completeness>, prove simplicity rather than merely invoking it. The nonidentity even cycle types in $S_5$ are three-cycles, products of two disjoint transpositions and five-cycles. In $A_5$, a three-cycle has <centralizer> of order three, so its <conjugacy class> has size $20$. A double transposition has <centralizer> of order four in $A_5$, giving class size $15$. A five-cycle has <centralizer> its cyclic <subgroup> of order five, giving class size $12$; the $24$ such elements therefore split into two classes of size $12$. These account for all $60$ elements with the <identity element>[identity].
A <normal subgroup> is a union of whole <conjugacy classes>, containing the <identity element>[identity], and its order must divide $60$. No nonempty proper sum of the class sizes $15,20,12,12$, plus one, is a divisor of $60$ greater than one: the possible proper nontrivial sums are $13,16,21,25,28,33,36,40,45,48$. Only the full union has order $60$. Hence $\boxed{A_5\text{ is simple}}$, and so is $G$.
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