Number of conjugates of an $m$-cycle $\sigma$ in $S_n$ I am currently reading the following from Dummit and Foote can someone explain to me why the number of conjugates of $\sigma$ is:
$$\frac{n(n\ -\ 1)...(n\ -\ m\ + \ 1)}{m}\ ?$$ 
 A: Two permutations in $S_{n}$ are conjugate if and only if they have the same cycle type, so really the question is how many $m$-cycles there are in $S_{n}$. Well, choose your $m$ elements from $n$ in $n\choose{m}$ ways. But we can rearrange those $m$ elements in $m!$ ways. Finally, each of our cycles has been represented $m$ times by a cyclic rearrangement, so we divide by $m$. So our final count is $$\frac{n!}{(n-m)!m}$$
as required.
A: A permutation is  conjugate if it has the same cycle structure. 
In this case you want to find the no. conjugates of a m cycle, so your question boils down to find the no. of m cycles in Sn.
Which can be done by n(n-1)(n-2)...(n-m+1)/m ways as for first entry you have n options, for next entry you have (n-1) options and finally you divide n(n-1)...(n-m+1) by m as each m cycle can be represented in m ways.
A: Denote
$$
\sigma=(a_{1}a_{2}\ldots a_{m})
$$
There is a theorem that states that for every $\tau\in S_{n}$ 
$$
\tau^{-1}\sigma\tau=(\tau(a_{1}),\tau(a_{2}),\ldots,\tau(a_{m}))
$$
thus, given any $m$ elements of $\{1,...,n\}$ there is some $\tau$
s.t $\tau^{-1}\sigma\tau$ will result in a cycle of those exactly
$m$ elements.
For the above formula recall two things:


*

*$\binom{n}{m}$ is the number of ways choosing $m$ elements from
a set of $n$ elements

*Any cyclic shift of the $m$ elements results in the same cycle
e.g $(123)=(231)=(312)$ so to cancel those repetitions we need to
divide by $m$ which is the number of such cyclic shifts 
A: The number of combinations of $n$ distinct things taken $m$ at a time is your starting point, for which the binomial coefficient comes in handy: $n!/m!(n-m)!$ Since each combination will represent more than just one $m$-cycle, for every choice of the 1st spot of an $m$-cycle there will be $m-1$ choices for the 2nd spot, $m-2$ choices for the 3rd spot, etc. You then arrive at $(m-1)! \times n!/m!(n-m)!$
