= Solution
Conjugating a <cycle> merely relabels its entries:
$$
g(a_1\ a_2\ \cdots\ a_k)g^{-1}
=(g(a_1)\ g(a_2)\ \cdots\ g(a_k)).
$$
Thus conjugate <permutations> have the same <cycle type>. Conversely, if $\sigma$ and $\tau$ have the same cycle type, match the entries of each cycle of $\sigma$ bijectively and in cyclic order with those of a cycle of $\tau$ of the same length. Extending these matches to a permutation $g$ gives $g\sigma g^{-1}=\tau$.
Let $C_{S_n}(\sigma)$ be the <centralizer> of $\sigma$. Its $S_n$-conjugacy class remains one $A_n$-conjugacy class exactly when $C_{S_n}(\sigma)$ contains an <odd permutation>. Indeed, in that case $C_{A_n}(\sigma)$ has index two in $C_{S_n}(\sigma)$, so the <orbit-stabilizer theorem> gives
$$
|\operatorname{Cl}_{A_n}(\sigma)|
=\frac{|A_n|}{|C_{A_n}(\sigma)|}
=\frac{|S_n|}{|C_{S_n}(\sigma)|}
=|\operatorname{Cl}_{S_n}(\sigma)|.
$$
If the centralizer contains only even permutations, the $A_n$ class has half the size and the $S_n$ class splits into two $A_n$ classes.
The even cycle types in $S_5$ are
$$
1^5,\qquad 3\,1^2,\qquad 2^2 1,\qquad 5.
$$
The classes of the identity, a three-cycle, and a double transposition do not split: their centralizers contain an odd permutation. The centralizer of a five-cycle is its cyclic subgroup of order five, which lies in $A_5$, so that class splits in two. Therefore
$$
\boxed{A_5\text{ has }1+1+1+2=5\text{ conjugacy classes}}.
$$
Back to article page
