Alternating group ($A_n$)

Every maximal proper subgroup normalizes one of its minimal normal subgroups. As $A_5$ has no proper nonabelian simple subgroup, that subgroup is elementary abelian. Its possible types are $C_2,C_2^2,C_3,C_5$. Their normalizers have orders $4,12,6,10$ respectively. The order-four normalizer lies in an order-twelve point stabilizer; the other three types are maximal and each forms a single conjugacy class. There are five, six and ten maximal subgroups of orders twelve, ten and six, respectively.

Except in degree six, the $3$-cycles are characterized among order-three elements by the largest centralizer. Their cyclic subgroups correspond to three-element supports. Two different such subgroups commute exactly for disjoint supports; with one common point their generator products have order five, and with two common points those products have order two or three. Thus an automorphism induces an automorphism of $J(n,3)$. The maximal cliques of a Johnson graph reconstruct a permutation of the points. After undoing that permutation, an automorphism fixes every cyclic $3$-subgroup. Product orders force its choices of generator inversion to be consistent on the connected graph, and inversion of every $3$-cycle would reverse a noncommuting product. Hence it fixes all $3$-cycles, which generate $A_n$.

A connected finite hypergraph of three-element supports, with a three-cycle on each support, generates the full alternating group. To merge a new triple meeting an old alternating support in two letters, conjugate over unordered pairs of old letters. For one-letter overlap compare $(abc)$ and $(a^{\prime }bc)$: their quotient is $(ab{a}^{\prime })$, reducing to two two-letter-overlap steps.

The alternating conjugacy class splitting criterion gives seven conjugacy classes in $A_6$. Their cycle types and sizes are $1^6:1$, $2^2{1}^2:45$, $31^3:40$, $3^2:40$, $42:90$, and two $51$ classes of size 72. Representatives of the split classes are a 5-cycle and its square. Their sizes sum to 360. A normal subgroup must be a union of these classes containing the identity, so these explicit sizes can prove simplicity using Lagrange's theorem.