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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.

A permutation cycle $(a_1{a}_2\cdots a_p)$ sends $a_1$ to $a_2$, each successive listed point to the next, and $a_p$ to $a_1$, fixing every unlisted point. Its length is $p$. For any permutation $h$,Thus conjugate group elements in the symmetric group have the same cycle length. Conversely, two cycles of the same length are conjugate: send their entries, in cyclic order, to one another and extend this bijection arbitrarily to the unused points. Length-one cycles represent the identity and cause no exception to this conclusion.

An odd-length permutation cycle is an even permutation, since its sign is $(-1{)}^{p-1}$. Let $c,d$ be such cycles on $p+2$ points and choose $h\in S_{p+2}$ with $hc{h}^{-1}=d$. If $h$ is odd, let $t$ be the transposition of the two points unused by the listed cycle $c$. Then $tc=ct$, and $ht$ is even withHence all the $p$-cycles are conjugate within $A_{p+2}$. The two spare letters supply the parity correction; they are the reason the same argument need not work with only one spare letter.

In $A_4$ the answer is no. For example, $(123)$ and $(132)$ are conjugated by the odd transposition $(23)$ in $S_4$. The centralizer of $(123)$ in $S_4$ is exactly its order-three cyclic subgroup: a commuting permutation fixes the unique fixed point $4$ and acts as a power of the cycle on the other three points. All these centralizing elements are even. Every other conjugating element differs from $(23)$ by a centralizing element, so it is also odd. There is no conjugator in $A_4$. Equivalently, the eight three-cycles split into two conjugacy classes of size $12/3=4$, illustrating the alternating conjugacy class splitting criterion.

An even permutation in the alternating group $A_6$ has one of the cycle types listed below. In the symmetric group, a type with $m_j$ cycles of length $j$ has centralizer order ${\prod }_j{j}^{m_j}{m}_j!$ and class size $6!/{\prod }_j{j}^{m_j}{m}_j!$: commuting permutations can rotate each cycle and exchange equal-length cycles.

An $S_6$ conjugacy class remains one $A_6$ class if its centralizer contains an odd permutation. Otherwise it splits into two equal classes, since the centralizer is already contained in the index-two subgroup $A_6$. This is the alternating conjugacy class splitting criterion. For the nonsplitting nonidentity types here, odd commuting permutations are respectively a constituent transposition, a transposition of fixed points, the interchange of two 3-cycles, and the constituent 4-cycle. For a 5-cycle with one fixed point, the centralizer is its cyclic group of order five and contains only even permutations.

Thus the conjugacy classes of the alternating group on six letters are:The two 5-cycle representatives lie in different classes. A relabelling that conjugates a 5-cycle to its square acts on its five positions as multiplication by two modulo five, a 4-cycle on the nonzero positions, hence is odd. Every other such conjugator differs by an even centralizer element. The sizes sum to $360=|A_6|$, so the list is exhaustive.

For simplicity of the alternating group on six letters, let $N$ be a normal subgroup. It contains the identity and is a union of full conjugacy classes; also $|N|$ divides $360$ by Lagrange's theorem. A proper subgroup has order at most $180$.

If the class of size 45 is absent, $|N|$ is odd. The odd divisors of $360$ are $1,3,5,9,15,45$. A nontrivial union has order at least 41, but no selection of the even class sizes sums to $44$, so order 45 is impossible. Thus $N=\{1\}$ in this case.

If the size-45 class is present, write $|N|=46+40a+90b+72c$ with $0\le a,c\le 2$ and $0\le b\le 1$. The possible values at most $180$ are $46,86,118,126,136,158,176$, none dividing $360$. Hence no proper nontrivial normal subgroup exists: