Number of m-Cycles in Symmetric Group

Theorem

Let $S_n$ be the symmetric group on $n$ letters.

Let $m \in \N: 1 \le m \le n$.

The number of $m$-cycles in $S_n$ is:

$\dfrac {n \left({n-1}\right) \left({n-2}\right) \ldots \left({n-m+1}\right)} m$

Proof

Suppose $n \ge m$, and consider the number of $m$-cycles in $S_n$.

An $m$-cycle can be represented by a selection of $m$ elements from $n$ without any repeats.

From Number of Permutations, the number of permutations of $m$ elements from $n$ possible elements is $\dfrac {n!} {\left({n - m}\right)!}$.

However, each such string is merely a representation of an $m$-cycle; the cycle does not depend on the starting element in the string.

Since there are $m$ possible starting numbers, we must divide this number by $m$.

Hence, the number of $m$-cycles is

$\dfrac {n!}{m \left({n - m}\right)!}$

which can be expressed as

$\dfrac {n \left({n-1}\right) \left({n-2}\right) \ldots \left({n-m+1}\right)} m$

$\blacksquare$

Sources

• Allan Clark: Elements of Abstract Algebra (1971)... (previous)... (next): $\S 79 \alpha$