= Solution
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:
$$
\begin{array}{c|c|c}
\text{cycle type}&\text{representative}&\text{class size in }A_6\\ \hline
1^6&1&1\\
2^2\,1^2&(12)(34)&45\\
3\,1^3&(123)&40\\
3^2&(123)(456)&40\\
4\,2&(1234)(56)&90\\
5\,1&(12345)&72\\
5\,1&(12345)^2&72
\end{array}
$$
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 $0\le a,c\le2$ and $0\le b\le1$. The possible values at most $180$ are $46,86,118,126,136,158,176$, none dividing $360$. Hence no proper nontrivial <normal subgroup> exists:
$$
\boxed{A_6\text{ is simple}.}
$$
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