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I'm trying to follow a proof in group theory but I got stuck here:

In $S_{p+2}$ there are $\frac{(p+2)!}{2p}$ conjugates of any given p-cycle.

I understand that the conjugate of a p-cycle $(\ldots\;a\;b\;\ldots)$ by $\pi$ is just $(\ldots\;\pi(a)\;\pi(b)\;\ldots)$ so every p-cycle can be got this way.. then I thought there must be $\binom{p+2}{p}/p$ of them, pick any $p$ symbols for our permutation then we need to factor out the fact we can write a p-cycle in p ways (how many turns round it is written e.g. $(a\;b\;c)=(b\;c\;a)=(c\;a\;b)$)

but that gives a different number than what I should have, can you tell me how to get it right? thanks

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up vote 2 down vote accepted

You are almost right. However, $p+2\choose p$ selects $p$ elements without regarding order (and is usually not even a multiple of $p$). If you select with respecting order, the count is $\frac{(p+2)!}{2!}$ and should then be divided by $p$:

Given one of $(p+2)!$ permutations $a_1, \ldots, a_{p+2}$ we can define the $p$-cycle $(a_1\,a_2\,\ldots\, a_p)$. We obtain the same $p$-cycle if $a_{p+1}$ and $a_{p+2}$ are swapped and/or the first $p$ elements are rotated. Thus every $p$-cycle is obtained exactly $2p$ times and hence the number of $p$-cycles is $\frac{(p+2)!}{2p}$.

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As any conjugate of a p cycle is a p cycle. So the no. of conjugates of a p cycle is the no. of p cycles in $S_{p+2}$ .

Now to form a p cycle you select p no. from the nos. $\{1,2,\dots ,p+2\}$ and then consider the no. of cycles that can be formed out of those p nos.

We can select the p nos in $\binom{p+2}{p}$ ways and from each such selected p nos. no. of cycles that can be formed is $(p-1)!$ (Reason: You fix the first no. in the cycle say 1 and then permute the other nos. in the cycle in $(p-1)!$ ways .In this way you get all the cycles.)

So the no. of conjugates is $\binom{p+2}{p}\times (p-1)!$

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