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The cycle type records the lengths of the disjoint cycles in a permutation, including fixed points as length-one cycles. The possibilities for three objects areThese exhaust the six elements of the symmetric group $S_3$, denoted ${\Sigma }_3$ in the source. They are also the conjugacy classes of the symmetric group on three letters.

In a symmetric group, two permutations are in the same conjugacy class exactly when they have the same cycle type. For $S_4$ the five possibilities and their sizes areThere are $\left(\genfrac{}{}{0ex}{}{4}{2}\right)=6$ transpositions, three pairings into two unordered pairs, $\left(\genfrac{}{}{0ex}{}{4}{3}\right)(3-1)!=8$ three-cycles, and $(4-1)!=6$ four-cycles. Together with the identity these counts sum to $24=|S_4|$.

The sign homomorphism $\mathrm{sgn}:S_n\to \{+1,-1\}$ records parity of a permutation. The kernel of a group homomorphism in this case is the alternating group $A_n$, consisting of even permutations. For $n\ge 2$ it has index two in the symmetric group and order $n!/2$.

The cycle type is the multiset of lengths of the disjoint cycles, including fixed points if desired. For $A_4$ the complete count isA single transposition and a four-cycle are odd and are excluded. Thus

For a finite group over the complex numbers, the sum of the squared dimensions of its inequivalent irreducible representations equals the group order. There is always a one-dimensional trivial representation, and part (b) supplies an irreducible representation of dimension three. Thus the remaining squared dimensions sum toThey must be two further dimensions equal to one. HenceThe number of inequivalent irreducible representations equals the number of conjugacy classes, so $A_4$ has four classes. There are only three cycle types in part (a); consequently cycle type cannot classify conjugacy within $A_4$.

More explicitly, the identity is a singleton class, and conjugation by three-cycles permutes the three double transpositions transitively. For a three-cycle, the centralizer in $A_4$ consists of its three powers, so its conjugacy class has $12/3=4$ elements. The eight three-cycles therefore split into two classes of four:For example, $(123)$ and $(132)$ have the same cycle type but are not conjugate in $A_4$. Any permutation conjugating the first to its inverse must fix its unique fixed vertex and reverse the orientation of the remaining three labels, so it is odd. They are conjugate in $S_4$, illustrating exactly why restricting the ambient group changes the answer.

Squaring an odd-length permutation cycle retains its length, whereas an even-length cycle splits into two cycles of half its length. Thus if any even cycle is present, the number of cycles strictly increases, so the cycle type changes. Since conjugate permutations in a symmetric group have the same cycle type, a permutation is conjugate to its square precisely when all its cycle lengths are odd, equivalently when its order of a group element is odd.

Conjugation preserves order of a group element, because $(gx{g}^{-1}{)}^k=g{x}^k{g}^{-1}$, and preserves the sign of a permutation, because sign is a group homomorphism into $\{1,-1\}$. These invariants do not determine cycle type. In $S_6$, $(123)$ and $(123)(456)$ both have order three and positive sign, but have three and zero fixed points respectively. Conjugation relabels cycles, so these permutations are not conjugate.