Suppose $n$ is a non negative integer $\geq 4$ and $\sigma\in S_n$ a permutation. Conjugacy classes in $S_n$ are completley determined by the cycle structure of $\sigma$. If we let the alternating group $A_n$.act on $S_n$ by conjugation, the orbits coincide with the conjugacy classes in $S_n$ except for those (even) permutations whose cycle structure is composed of even cycles (which corresponds to cycles of uneven length) of distinct order (fixed points are treated as cycles of length $1$.) See http://groupprops.subwiki.org/wiki/Splitting_criterion_for_conjugacy_classes_in_the_alternating_group

For instance a $3$-cycle in $A_4$ has cycle structure $[3,1]$ which fits the bill, and a $5$-cycle in $A_5$ or $A_6$ works too, since the cycle structure is $[5]$ and $[5,1]$ respectively. On the other hand, a $3$-cycle in $A_5$ will does not satisfy the criteria as it has cycle structure $[3,1,1]$ with two fixed points.

In this case, the conjugacy class of $\sigma$ inside $S_n$ splits in $2$ orbits of the conjugation action of $A_n$, and if $\tau$ is any odd permutation, for instance $(12)$, then we have $$\mathrm{Conj}_{S_n}(\sigma)=\mathcal{O}(\sigma)\coprod\mathcal{O}(\tau\sigma\tau^{-1})$$

My question is How can we differentiate between the two orbits? Does it have geometric meaning?

I can see that this information can be useful when looking at representations of the alternating groups, since there are as many irreducible representations as conjugacy classes. As for some geometric interpretation, I ask because there is one for the three cycles in $A_4$. If $S_4$ is understood as the group of symmetries of a regular tetrahedron, then $A_4$ is the subgroup of direct symmetries, and a $3$-cycle corresponds to a rotation of angle $\frac{2\pi}{3}$ or $\frac{4\pi}{3}$ with axis passing through one of the vertices. Conjugacy classes inside $A_n$ preserve the angle, and that's how we can tell them apart. In this case the computations are so simple we don't really need this picture, but I think it is a nice way to understand the fact $3$-cycles in are not all conjugate inside $A_4$...

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    $\begingroup$ Those two classes are interchanged under the action of the automorphisms of $A_n$, so the answer to your question is probably no. $\endgroup$ – Mariano Suárez-Álvarez May 13 '12 at 19:13
  • $\begingroup$ If you instead whant to identify that class given, say, the pair $(A_n,V)$ with $V$ some representation, then you can tell them apart by looking, for example, the values of the character at the two classes (for some $V$s; those that appear as proper summands of reps of $S_n$—as the other characters do not distinguish the classes) $\endgroup$ – Mariano Suárez-Álvarez May 13 '12 at 19:20

The statement you make about tetrahedra generalizes to arbitrary $n$. Specifically, the symmetric group $S_n$ is the group of symmetries of a regular $n$-simplex, and the alternating group $A_n$ acts on this simplex by rotations. (In fact, $A_n$ is precisely the set of rotational symmetries.) Elements of different conjugacy classes of $A_n$ are geometrically distinguishable, in the sense that they would "look different" to an observer in $\mathbb{R}^n$.

One way of making the notion of "look different" precise is that non-conjugate elements of $A_n$ correspond to non-conjugate elements of the rotation group $SO(n)$. Thus, two non-conjugate elements of $A_n$ do not look the same up to rotation of the simplex.

Incidentally, the simplest algorithm to distinguish conjugacy classes in $A_n$ is essentially to check the sign of the conjugator. For example, the elements $(5)(2\;6\;3)(1\;9\;4\;8\;7)$ and $(2)(1\;4\;8)(3\;7\;5\;6\;9)$ are conjugate in $S_9$, and their conjugacy class in $S_9$ splits into two conjugacy classes in $A_9$. To check whether the two elements are conjugate in $A_9$, we consider a permutation that maps between corresponding numbers: $$ \begin{bmatrix} 5 & 2 & 6 & 3 & 1 & 9 & 4 & 8 & 7\\ 2 & 1 & 4 & 8 & 3 & 7 & 5 & 6 & 9 \end{bmatrix} \;=\; (1\;3\;8\;6\;4\;5\;2)(7\;9) $$ This permutation is odd, so the two elements are not in the same conjugacy class in $A_9$. It is possible to construe this algorithm geometrically, since the difference between odd and even permutations is the same as the difference between right-handed and left-handed coordinate systems.

  • $\begingroup$ It is likely I'm not understanding your argument, but what you showed is that the standard construction of the conjugating element is odd, not that any conjugating element is odd. $\endgroup$ – user217285 Nov 21 '15 at 4:32
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    $\begingroup$ @Nitin We already know that this conjugacy class splits into two orbits under $A_n$ --- the only question is whether the two permutations are in the same orbit. Conjugating by an even permutation preserves the orbits, while conjugating by any odd permutation switches them. $\endgroup$ – Jim Belk Nov 21 '15 at 6:17

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