# How do you calculate the number of elements conjugate to a particular permutation?

Working in $S_n$, and given a particular permutation, how would you go about calculating the number of conjugate elements to the permutation?

I guess since two elements are conjugate iff they have the same cycle type this gives some restriction on the conjugate elements. I know two elements, say $x$ and $y$, are conjugate if $g^{-1}\cdot x\cdot g=y$ for some $g$ in the group but I'm not sure how to work out the number of conjugate elements.

As an example, the permutation is $x = (12)(34)(56789)$, working in $S_{12}$. So $x$ has cycle type $[2,2,5,1,1,1]$ so clearly any conjugate elements must have cycle type $[2,2,5,1,1,1]$.

Any help would be greatly appreciated. Thanks.

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You already know the key for solving the problem. The rest is simple combinatorics: how many ways can you choose $2+2+5$ elements out of $n$? – Berci Feb 22 '13 at 12:55
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Hint: If the permutation in $S_n$ has cycle type $(\mu_1, \mu_2, \dots, \mu_k)$, start by figuring out how many ways there are to distribute $n$ numbers into boxes of size $\mu_1, \mu_2, \dots, \mu_k$. Then, once you have a box with $\mu_i$ numbers in it, how many distinct cycles can you produce? Finally, you have to account for the fact that some of the $\mu_i$'s might be equal, and you get the same permutation if you reorder the disjoint cycles.

This is worked out carefully in volume 1 of Stanley's Enumerative Combinatorics, as well as other combinatorics texts.

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Chapter 1, prop. 1.3.2, page 18 (Ed. of 1997) – DonAntonio Feb 22 '13 at 13:07

For the given example, you just have to count the possible $[2,2,5]$ groupings of $12$ elements. Note that if we exchange the two $2$-cycles, we get the same permutation.

Now it can go like: first choose $2$ elements out of $12$, then $10$ elements are left, choose $2$ again, and then choose $5$ elements out of the rest ($8$), finally divide it by $2!$ because the order of the first two chosen group doesn't matter, that is: $$\frac1{2!}\cdot\binom{12}2\cdot \binom{10}2\cdot\binom85$$

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+1 I am thinking of doing the problem using GAP for example for $S_3$ – Babak S. Feb 22 '13 at 13:23