Number of $(n-1)-$cycles in $S_n$

Question

In a solution to an old exam, it is claimed that for $n$ even, the number of $(n-1)$-cycles in $S_n$ is precisely $\frac{n!}{n-1} = n(n-2)!$. I understand the rest of the solution, but this part is the only part that is not explained, and of course the only part that eludes me..

Any thoughts?

Answer 1

I would do this using the orbit-stabiliser theorem and the action of $S_n$ on itself by conjugacy. For $c \in S_n$ the size of the conjugacy class $c^{S_n}$ is $|S_n|/|\mathrm{Stab}_{S_n} (c)|$ where $\mathrm{Stab}_{S_n}(c) = \{h \in S_n : hc = ch \}$.

Picking the particular $(n-1)$-cycle $c = (1,2,\ldots, n-1)$, we have $hc = ch$ if and only if $c^h = c$, so if and only if

$$(1,2,\ldots,n-1) = (1^h, 2^h, \ldots, (n-1)^h).$$

If $1^h = k$ then we must have $2^h = k+1$, $3^h = k+2$, and so on, wrapping around at $n-1$. So $h = (1,2,\ldots, n-1)^{k-1}$ and $h$ is a power of $c$. Since $c$ clearly commutes with itself, we get that $\mathrm{Stab}_{S_n} (c)$ is generated by $c$. Hence $|\mathrm{Stab}_{S_n}(c)| = |\langle c \rangle| = n-1$.

By the first paragraph it follows that the number of $(n-1)$-cycles is $n! /(n-1) = n(n-2)!$. Note this doesn't need that $n$ is even, only that $n \ge 2$.

Answer 2

Since we want a $n-1$ cycle that means one element among $1,2,3, \ldots ,n$ will be fixed. That element can be chosen in $n$ ways. The remaining $n-1$ elements can be permuted in $(n-1)!$ ways, however each cyclical shift produces the same permutation function (i.e. if $(a_1 \, a_2 \ldots \,a_{n-1})$ is a permutation function, then $(a_2 \, a_3 \ldots \,a_{n-1}\, \color{blue}{a_1})$ is the same permutation function) so we have to factor out $n-1$ to avoid overcounting. This means the number of $n-1$ cycles are $$n\frac{(n-1)!}{n-1}=n(n-2)!$$

Answer 3

To see why the number of $(n-1)$-cycles in $S_n$ is $n(n-2)!$, observe that you have $n$ choices for the number in $\{1,\dots, n\}$ which will not appear in the $(n-1)$-cycle, and that if you write the $(n-1)$-cycle so that the smallest number appears first (i.e., for $n=4$ you would write $(132)$ rather than $(213)$) then there are $(n-2)!$ possible orderings of the remaining numbers which will appear in the cycle.

Answer 4

Consider the symmetric group $S_n$ and let $1 \leq m \leq n$. Then the number of $m$-cycles is $\frac{n!}{m(n-m)!}$:

There are $\frac{n!}{(n-m)!}$ possible permutations of $m$ elements from possible $n$ elements ($n$ choices for the first one, $n-1$ for the next one etc.). Since every $m$-cycle can be permuted $m$ times cyclically to get the same cycle, we will get $\frac{n!}{m(n-m)!}$ $m$-cycles.

Now let $m = n-1$. This yields $\frac{n!}{n-1} = n \cdot (n-2)!$ many $(n-1)$-cycles.

Answer 5

There are ${}^{n-1}P_n=\dfrac {n!}1=n!$ $(n-1)$-cycles that can be formed.

However, we divide by $n-1$, because you can cyclically permute a $k$ cycle $k$ ways and get the same cycle.

So, $\dfrac {n!}{n-1}=n(n-2)!$.