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Define the group action as $g\cdot x:=g^{-1}xg.$ Let $G=A_5$, and $X=\{\sigma\in A_5:=\sigma=(a,b,c,d,e)\}.$ Show that the group action on X decomposes $X$ into two distinct orbits.

There are 60 elements in $A_5,$ so I assume that we need to use Burnside's theorem. But I am not sure how to use it to show that G decomposes $X$ into two distinct orbits.

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  • $\begingroup$ Does $\sigma = (a, b, c, d, e)$ mean that $\sigma$ is a transposition of 5 things? $\endgroup$ – Siddharth Bhat Apr 28 '16 at 16:12
  • $\begingroup$ It's a permutation $\endgroup$ – Anonymous Apr 28 '16 at 16:30
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Hint: show that the subset of permutations whose signature is 1 and the subset whose signature is $-1$ are preserved by the action.

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You're basically looking for the conjugacy classes of all elements of the form $\sigma = (a, b, c, d, e)$.

In any $S_n$, all elements of the same cycle length are conjugate to each other.

We know that the conjugacy class of an element of $A_n$ splits into two (from $S_n$) if the element has a cycle decomposition of odd length where every length is unique.

Here, we know that $\sigma = (a, b, c, d, e)$ which is of length 5. Hence, its conjugacy class splits into two.

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