I know that $\sigma$, $\tau$ $\in$ $S_n$ are conjugate if and only if they have the same cycle structure. Is there any explicit way that we can determine whether two elements in $A_n$ are conjugates?

  • $\begingroup$ As the size of the conjugacy class of an element $x$ equals the index of its centralizer $C_G(x)$ in $G$, you have to determine if the given $x\in A_n$ is centralized by an element of $S_n\setminus A_n$ or not (that's what Mark is doing in his answer without stating this fact explicitly). $\endgroup$ – j.p. Mar 29 '15 at 9:43

For two elements $x,y \in A_n$ to be conjugate within $A_n$ they have to have the same cycle type. Suppose $axa^{-1}=y$ for some $a\in S_n$.

Suppose $x$ contains a cycle $c$ which moves an even number of points (elements of the underlying set of size $n$), and is therefore an odd permutation. We have $cxc^{-1}=x$ so that $acxc^{-1}a^{-1}=axa^{-1}=y$. One of $ac$ and $a$ must be even, so $x$ and $y$ are conjugate within $A_n$.

Suppose $x$ contains two cycles $d,e$ of the same odd length (including the case where two or more elements of the underlying set are fixed). For example, $(1 2 3)$ and $(456)$. In the example it will be found that the odd permutation $f=(14)(25)(36)$ of order $2$ exchanges the two cycles, and fixes every other element of the underlying set. Then $fxf^{-1}=x$ and we can proceed as before. This clearly works in every such case.

The remaining case has $x$ a product of cycles of distinct odd lengths. Write $y$ underneath $x$ with the same cycle order. If the permutation which takes the elements of $x$ to the corresponding elements of $y$ is even, it will conjugate one to the other within $A_n$. If it is odd, then there will be no even permutation available, and the conjugacy class will split in two.

Take $A_5$ for an example. The only elements with an even length cycle are of form $(12)(34)$ and these are all conjugate within the group. The only other elements which move at least $4$ points are the $5-$cycles - and these fit the criterion for splitting, which indeed they do. The $5-$cycles also split in $A_6$, but not in $A_7$ where they fix two points.

It is thus easy to classify the cycle types where the conjugacy classes split (or stay the same). But you have to compute explicitly to see whether two elements in a split conjugacy class are conjugate or not.

  • $\begingroup$ I understand that in $A_5$, the conjugacy class for 5-cycle splits. For instance, (1 2 3 4 5) and (2 1 3 4 5) are not conjugate in $A_5$ and they are contained in distinct conjugacy classes. Then how can we explicitly determine whether a given two 5-cycles are in the same conjugacy class in $A_n$? Just check if $\sigma \in A_n$ for ($\sigma$(1) $\sigma$(2) $\sigma$(3) $\sigma$(4) $\sigma$(5))= (2 1 3 4 5)? $\endgroup$ – Joong Mar 28 '15 at 22:39
  • $\begingroup$ @Joong That's right - that's what conjugation by $\sigma$ does $\endgroup$ – Mark Bennet Mar 29 '15 at 5:59
  • $\begingroup$ @Joong: For $n>6$ two 5-cycles are always conjugated, as thanks to their two fixed points you are in the second case of Mark's analysis: both have two cycles of the same odd length 1. Mark mentioned this also at the end of his second last paragraph. $\endgroup$ – j.p. Mar 29 '15 at 9:39

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