Past exam of the mathematics course of the University of Cambridge / 2004 / ib / Paper 2 / 2F (Groups, Rings and Modules)

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 $\left(\genfrac{}{}{0ex}{}{5}{3}\right)\cdot 2=20$ three-cycles, $5\cdot 3=15$ 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 inBy Lagrange's theorem, its order must divide 60. Only 1 and 60 in this list do so. Hence $A_5$ has no proper nontrivial normal subgroup and is a simple group.