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Denote $S_n$ symmetric group on $n$ letters.

Denote $C_n$ to be set of cycles in $S_n$ of length $n$.

Denote $I_n$ to be set of involutions in $S_n$.

What are the cardinalities of $C_n$, $I_n$ and $I_n\cap C_n$?

What is the probability that a random member of $S_n$ is an involution?

  • For $I_n \cap C_n$, can you think of any cycles $\tau$ that satisfy $\tau^2 = 1$? Do you notice anything special about their lengths? – pjs36 Mar 05 '16 at 21:54
  • For $I_n$ choose the fixed points and partition the rest into two-cycles, counted with a multinomial coefficient. – Marko Riedel Mar 05 '16 at 22:03

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Each cycle of length $n$ corresponds to $n$ permutations of the $n$ letters, one for each of the $n$ possible points at which to break open the cycle. There are $n!$ permutations of the $n$ letters, so there are $\frac{n!}n=(n-1)!$ cycles of length $n$.

Suppose that $\sigma=(p_1p_2\ldots p_n)$ is an involution. Clearly

$$\sigma^2=(p_1p_3\ldots p_np_2p_4\ldots p_{n-1})$$

if $n$ is odd, and

$$\sigma^2=(p_1p_3\ldots p_{n-1})(p_2p_4\ldots p_n)$$

if $n$ is even. Suppose that $n\ge 3$; in both cases we have $p_1=\sigma^2(p_1)=p_3$, contradicting the assumption that $n\ge 3$, so $n\le 2$. I’ll leave the details of the cases $n=1$ and $n=2$ to you.

For $|I_n|$, try to show that the involutions are precisely the permutations that can be expressed as products of pairwise disjoint transpositions. Thus, you get one involution for every partition of $[n]=\{1,2,\ldots,n\}$ into sets of cardinality $1$ and $2$. To form such a partition, first choose a subset $S$ of $[n]$ whose members will be the fixed points of the permutation, and then count the partitions of $[n]\setminus S$ into pairs; this question and its answers may be helpful. You may want to consider separately the cases of odd and even $n$. You may also find that you have to leave the answer in the form of a summation.

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Brian M. Scott
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    Helpful answer. (+1). The involutions can also be extracted from the exponential generating function $$\mathfrak{P}(\mathfrak{C}_{\le 2}(\mathcal{Z}))\quad\text{which is}\quad\exp\left(z+\frac{z^2}{2}\right).$$ – Marko Riedel Mar 05 '16 at 22:19
  • @BrianM.Scott are you saying $I_n\cap C_n=\emptyset$? –  Mar 05 '16 at 22:42
  • @Arul: Yes, if $n>2$. – Brian M. Scott Mar 05 '16 at 22:44
  • @BrianM.Scott Also what is the asymptotic of $|I_n|$? –  Mar 05 '16 at 23:17
  • The asymptotics are explained in the slides part 8 from the book Analytic Combinatorics by Flajolet and Sedgewick which are available for download at the book's website. – Marko Riedel Mar 05 '16 at 23:22
  • @Arul: The asymptotics are messy, and I’ve never looked into them, but as Marko says above, they can be found in Flajolet & Sedgewick, Analytic Combinatorics, which is freely available here; the discussion starts on p. $558$. – Brian M. Scott Mar 05 '16 at 23:32
  • @BrianM.Scott So probability of random permutation being an involution is roughly $\frac{n^{n/2}}{n^n}$? –  Mar 05 '16 at 23:42
  • @Arul: If I’m reading $(38)$ correctly, it’s roughly $$\frac{e^{n/2+\sqrt{n}}}{2e^{1/4}\sqrt{\pi n}n^{n/2}};.$$ – Brian M. Scott Mar 05 '16 at 23:46