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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?

darij grinberg
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Leox
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  • I don't think your usage of cycle type matches your definition, as a permutation with two $1$-cycles and two $2$-cycles can't be in $S_4$, nor can a permutation in $S_6$ have a $3$-cycle and a $4$-cycle. Perhaps you mean "$\lambda_i$ is the length of the $i$-th cycle"? – universalset Dec 17 '13 at 20:03
  • yes, corrected.. – Leox Dec 17 '13 at 20:15
  • Also you aren't going to have $n$ cycles, usually (you use $n$ for two different things in the above -- maybe use $k$ for the number of cycles?) – universalset Dec 17 '13 at 20:25
  • well.. the idenity permutation is considered as $n$ cycles $(1)(2)\cdots(n).$ – Leox Dec 17 '13 at 20:29
  • True, but all other permutations will have fewer cycles. Using $n$ for the number of cycles implies that the number of cycles will always be $n$, which is not the case. Better to use a different variable for potentially different numbers. – universalset Dec 17 '13 at 20:30
  • I have edited your question to clarify it based on the comments. Let me know if it is not correct. – universalset Dec 17 '13 at 21:08
  • @ universalset..Yes, that is correct.. Thank you! – Leox Dec 17 '13 at 21:18

1 Answers1

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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 $\pi$.

darij grinberg
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universalset
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  • @universalset..I need some time to undestand and check it. But thank you anyway. – Leox Dec 17 '13 at 22:03
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    By the way, there is another way to phrase the question (and answer) in terms of group actions: We're looking at the action of the group $\langle \pi\rangle$ on the set ${{i,j}\ |\ i, j\in{1,\ldots,n}\text{ and } i\neq j}$. The results I mention can then be derived by careful application of the orbit-stabilizer theorem. – universalset Dec 17 '13 at 22:13
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    What you are describing is the cycle structure of the edge permutation group of the complete graph on $n$ nodes induced by vertex permutations from the symmetric group. There is a discussion of how the cycle structure of the induced permutation can be computed efficiently and some relevant Maple code at this MSE link. – Marko Riedel Dec 18 '13 at 01:09
  • I have found the answer on my question in Harary,Palmer book, page 84 – Leox Dec 18 '13 at 19:48