How many $\alpha \in S_n$ are such that $\alpha^2 = 1$?

Question

This is not for homework, but I am not great at counting arguments and would like some feedback. The question asks

Let $n \in \mathbb{N}$. How many $\alpha \in S_n$ are there such that $\alpha^2 = 1$?

I know that, if $\alpha^2 = 1$, then either $\alpha = 1$ or $\alpha$ is the product of disjoint transpositions.

If $\alpha = (i, j)$ is a single transposition, then there are $\frac{1}{2^1 \cdot 1!} \binom{n}{2}$ such $\alpha$ (the $2^1$ and $1!$ are put in the denominator to help in noticing the pattern later).

If $\alpha = (i, j)(k, l)$ is the product of $2$ disjoint transpositions, then there are $\frac{1}{2^2 \cdot 2!} \binom{n}{2} \binom{n-2}{2}$ such $\alpha$, where the $2^2$ appears in the denominator to account for the cyclic permutations of each transposition, and the $2!$ appears to account for the permutation of the transpositions themselves.

If $\alpha$ is the product of $3$ disjoint transpositions, then there are $\frac{1}{2^3 \cdot 3!} \binom{n}{2} \binom{n-2}{2} \binom{n-4}{2}$ such $\alpha$.

Extrapolating from this, I find that the total number of $\alpha \in S_n$ such that $\alpha^2 = 1$ is $$ 1 + \sum_{i=1}^{\lfloor \frac{n}{2} \rfloor} \frac{1}{2^i \cdot i!} \prod_{k=0}^{i-1} \binom{n-2k}{2}. $$

Does this look OK? It looks like a rather ugly answer to me, so I have my doubts. Any input would be welcomed.

For $n = 3$ your formula for the number of 2-cycles yields yields a non-integer. I think you are overdividing. — Andreas Caranti, Nov 27 '13 at 18:03
But then how many products of two disjoint 2-cycles do you get when $n=4$? I think you get $24/16$, which can't be right. — Andreas Caranti, Nov 27 '13 at 18:20
@AndreasCaranti Perhaps I should take away the $2^i$ from the sum altogether? — tylerc0816, Nov 27 '13 at 18:20
See also https://math.stackexchange.com/questions/795541/ for the same question. — darij grinberg, Oct 02 '19 at 18:10

score 5 · Accepted Answer · answered Nov 27 '13 at 18:01

5

The way I was taught it, there are $$ \frac{n(n-1)}{2} $$ 2-cycles, $$ \frac{n(n-1)(n-2)(n-3)}{2^2 \cdot 2} $$ products of two disjoint 2-cycles, and in general $$ \frac{n(n-1) \cdot \dots \cdot (n-2k+2)(n - 2 k +1)}{2^k \cdot k!} $$ products of $k$ disjoint 2-cycles, provided $2 k \le n$.

answered Nov 27 '13 at 18:01

Andreas Caranti

70,862

It looks like these correspond to $\binom{n}{2}$, $\frac{1}{2!} \binom{n}{2} \binom{n-2}{2}$, ..., and $\frac{1}{k!} \prod_{i=0}^{k-1} \binom{n-2k}{2}$, so I was overdividing, like you said, by a factor of $2^k$ each time. Thank you for your help! – tylerc0816 Nov 27 '13 at 18:24
@tylerc0816, you're welcome! – Andreas Caranti Nov 27 '13 at 18:33

score 2 · Answer 2 · answered Nov 27 '13 at 18:42

These permutations are called involutions. The counting function for involutions on $n$ elements is documented at OEIS here. You'll be able to find explicit formulas, recurrence relations, asymptotics, and generating functions there, along with some references. OEIS (Online Encyclopedia of Integer Sequences) is a pretty nice resource in general for finding what's known about various integers, especially if you can compute the first several terms to search for.

How many $\alpha \in S_n$ are such that $\alpha^2 = 1$?

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