= Solution
We prove <Simplicity of the alternating group A5> by finding all its <conjugacy classes>. The <alternating group> has <group order> $|A_5|=5!/2=60$. Its even <permutations> have cycle types $1$, $(abc)$, $(ab)(cd)$ and $(abcde)$. There are $\binom53\cdot2=20$ three-cycles, $5\cdot3=15$ products of disjoint transpositions, and $4!=24$ five-cycles.
For a three-cycle, a commuting <permutation> must rotate its three-element support and may interchange the two fixed points. Its <centralizer> in $S_5$ has order six, and only the three rotations are even. Thus its <conjugacy class> in $A_5$ has size $60/3=20$ and contains every three-cycle. The <centralizer> of $(12)(34)$ in $S_5$ has order eight: one may interchange the two pairs and independently interchange the elements within each pair. It contains the odd element $(12)$, so exactly half of these commuting elements are even. The <conjugacy class> in $A_5$ consequently has size $60/4=15$. A <permutation> commuting with a five-cycle is determined by the image of one point and must be a power of that cycle. All five powers are even, so each such <conjugacy class> has size $60/5=12$. The 24 five-cycles form exactly two classes.
A <normal subgroup> contains the identity and is a union of whole <conjugacy classes>. Its possible orders therefore lie in
$$
1+\{0,12,24\}+\{0,15\}+\{0,20\}
=\{1,13,16,21,25,28,33,36,40,45,48,60\}.
$$
By <Lagrange's theorem>, its order must divide 60. Only 1 and 60 in this list do so. Hence \b[$A_5$ has no proper nontrivial <normal subgroup> and is a <simple group>].
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