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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?

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1 Answer 1

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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.

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  • $\begingroup$ 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? $\endgroup$
    – Sandy
    Commented Jun 1, 2013 at 19:18
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    $\begingroup$ @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 ;-) $\sigma\tau = \tau \sigma$. $\endgroup$
    – martini
    Commented Jun 1, 2013 at 23:07
  • $\begingroup$ Ok, thank you a lot :) $\endgroup$
    – Sandy
    Commented Jun 2, 2013 at 6:00
  • $\begingroup$ $\tau$ 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? $\endgroup$
    – roi_saumon
    Commented Aug 1, 2020 at 11:17
  • $\begingroup$ @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? $\endgroup$ Commented Aug 18, 2022 at 19:11

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