Past exam of the Probability, statistics and options II course of the University of Oxford / 2009 / B / B2 / ii

Act by left multiplication on the five cosets of $H$. The kernel is the core ${\bigcap }_{g\in G}gH{g}^{-1}$, 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.

A normal subgroup is a union of whole conjugacy classes, containing the 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 $G$.