Cycle type of induced permutation Let $m = \binom{n}{2}$ and $S_n, S_m$  be  the symmetric groups,  $S_n  \subset S_m$. Let  $\pi \in S_n$ and let $\pi$ have the the cycle type  $[λ_1,λ_2,\dots,λ_k]$, $\lambda_1+\lambda_2+ \cdots+\lambda_k=n$ where  $λ_i$ is the  length of $i$-th cycle. $\pi$ induces a permutation $\pi^\prime$ in $S_m$ by permuting the set of pairs $\{ \{i,j\}\ |\ i,j \in \{ 1, \ldots, n\}\text{ and } i \neq j \}$. We may ask what is the cycle type of $\pi^\prime$ considered as element of $S_m$?
For arbitrary permutation we may calculate the cycle type of the corresponding induced permutation  by hand. For example, let $n=4,m=6$, and a permutation $\pi \in S_4$  have the cycle type $[2,2]$. Then after some direct  calculation we get that the induced permutation in $S_6$  has  the cycle type $[2,2,1,1].$
Question. Is there any effective algorithm for calculating the cycle type of the induced permutation?
 A: For every unordered pair of distinct cycles in $\pi$ of lengths $\lambda$ and $\mu$, we get $\gcd(\lambda, \mu)$ cycles of length $\text{lcm}(\lambda, \mu)$ in $\pi^\prime$.  These correspond to the orbits of $\{i,j\}$, where $i$ is in the first cycle and $j$ in the second, under repeated application of $\pi^\prime$.
For every cycle in $\pi$ of length $\lambda > 1$, we also get the permutations of the pairs of elements from that cycle.  All cycles in $\pi^\prime$ thus obtained have length $\lambda$, except when $\lambda$ is even, and in that case exactly one cycle (the orbit of $\{1, 1+\frac{\lambda}{2}\}$ if we assume that the cycle is $(1, \ldots, \lambda)$) has length $\displaystyle \frac{\lambda}{2}$.  So from this cycle of length $\lambda$ we get $\displaystyle \left\lfloor \frac{\lambda-1}{2}\right \rfloor$ cycles of length $\lambda$, and an additional single cycle of length $\displaystyle \frac{\lambda}{2}$ if $\lambda$ is even.
Combining these should give you an effective way of computing the cycle type of $\pi^\prime$ given the cycle type of $\pi$.
