Splitting of conjugacy class in alternating group

Browsing the web I came across this:

The conjugacy class of an element $$g\in A_{n}$$:

1. splits if the cycle decomposition of $$g\in A_{n}$$ comprises cycles of distinct odd length. Note that the fixed points are here treated as cycles of length $$1$$, so it cannot have more than one fixed point; and
2. does not split if the cycle decomposition of $$g$$ contains an even cycle or contains two cycles of the same length.

Anybody with a proof?

• Commented May 28, 2013 at 12:29

Note the following: (1) The conjugacy class in $$S_n$$ of an element $$\sigma \in A_n$$ splits, iff there is no element $$\tau \in S_n\setminus A_n$$ commuting with $$\sigma$$. For if there is one, for each $$\tau' \in S_n \setminus A_n$$ we have $$\tau'\sigma{\tau'}^{-1} = \tau'\sigma\tau\tau^{-1}\tau'{}^{-1} = (\tau'\tau)\sigma(\tau'\tau)^{-1}$$ and $$\tau\tau' \in A_n$$. On the other hand, if $$\tau\sigma\tau^{-1}$$ and $$\sigma$$ with $$\tau \in S_n\setminus A_n$$ are conjugate in $$A_n$$, then for some $$\tau' \in A_n$$, we have $$\tau\sigma\tau^{-1} = \tau'\sigma\tau'^{-1}$$, giving $$\tau'{}^{-1}\tau \sigma = \sigma\tau'{}^{-1}\tau$$ and hence $$\tau'{}^{-1}\tau \in S_n\setminus A_n$$ commutes with $$\sigma$$.
Now suppose, $$\sigma$$ has a cycle $$c_i$$ of even length. A cycle of even length is an element of $$S_n \setminus A_n$$, and as $$\sigma$$ commutes with its cycles, we are done by the above. If $$\sigma$$ has two cycles $$(a_1\ldots a_\ell)$$ and $$(b_1 \ldots b_\ell)$$ of the same odd length $$\ell$$, then $$(a_1b_1) \ldots (a_\ell b_\ell)$$ is a product of $$\ell$$ transpositions (hence odd, so an element of $$S_n \setminus A_n$$) commuting with $$\sigma$$.
Now suppose $$\sigma = c_1 \cdots c_s$$ is a product of odd cycles $$c_i$$ of distinct length $$d_i$$. Let $$\tau \in S_n$$ be a permutation commuting with $$\sigma$$. Then $$\tau$$ must fix each of the $$c_i$$, that is, $$\tau$$ must be of the form $$\tau = c_1^{a_1} \cdots c_s^{a_s}$$ for some $$a_i \in \mathbb Z$$. But as the $$c_i$$ are even permutations (as cycles of odd length), we have $$\tau \in A_n$$. So no $$\tau \in S_n \setminus A_n$$ commutes with $$\sigma$$ and we are done.
• It is not my question, but could you explain to me what the product $(a_1 b_1)...(a_l b_l)$ means? The two cycles $(a_1...a_l), \ (b_1 ... b_l)$ are supposed to be disjoint, aren't they? Commented Jun 1, 2013 at 19:18
• @Sandy Yes, they are. With $\sigma = (a_1, \ldots, a_\ell)(b_1, \ldots, b_\ell)$ we let $\tau = (a_1, b_1) \cdots (a_\ell b_\ell)$ the product of $\ell$ (disjoint) permutations $(a_i, b_i)$ [the 2-cycle exchanging $a_i$ and $b_i$, their product as permutations]. We have, as computation shows ;-) $\sigma\tau = \tau \sigma$. Commented Jun 1, 2013 at 23:07
• $\tau$ fixe each $c_i$ mean $\tau c_i \tau^{-1} = c_i$ right? Also in the criterion "comprise cycle of different length" mean actually all the cycles in the decomposition have different length right? Last thing : in fact this criterion cover all cases because if we decompose a permutation into disjoint cycles and there is no even permutations and no two permutations of same cycle length then the decomposition is composed of odd cycles with distinct length right? Commented Aug 1, 2020 at 11:17
• @martini I don't understand why $\tau$ must be in that form. There could be some cycle of even lenght, moving some numbers that $\sigma$ doesn't move, i.e an element of the centralizer of $\sigma$ which doesn't move the elements moved by $\sigma$ if there is enough space. Is this case not possible? Commented Aug 18, 2022 at 19:11