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Cycle index

Cycle indices of some permutation groups

Identity group EnE_n

This group contains one permutation that fixes every element (this must be a natural action).

Z(En)=a1n.Z(E_n) = a_1^n.

Cyclic group CnC_n

A Cyclic group, CnC_n is the group of rotations of a regular nn-gon, that is, nn elements equally spaced around a circle. This group has φ(d)\varphi(d) elements of order dd for each divisor dd of nn, where φ(d)\varphi(d) is the Euler φ-function, giving the number of natural numbers less than dd which are relatively prime to dd. In the regular representation of CnC_n, a permutation of order dd has n/dn/d cycles of length dd, thus

Z(Cn)=1ndnφ(d)adn/d.Z(C_n) = \frac{1}{n} \sum_{d|n} \varphi(d) a_d^{n/d}.

Dihedral group DnD_n

The Dihedral group is like the cyclic group, but also includes reflections. In its natural action,

Z(Dn)=12Z(Cn)+{12a1a2(n1)/2,n\mboxodd,14(a12a2(n2)/2+a2n/2),n\mboxeven.Z(D_n) = \frac{1}{2} Z(C_n) + \begin{cases} \frac{1}{2} a_1 a_2^{(n-1)/2}, & n \mbox{ odd, } \\ \frac{1}{4} \left( a_1^2 a_2^{(n-2)/2} + a_2^{n/2} \right), & n \mbox{ even.} \end{cases}

The alternating group AnA_n

The cycle index of the alternating group in its natural action as a permutation group is

Z(An)=j1+2j2+3j3++njn=n1+(1)j2+j4+k=1nkjkjk!k=1nakjk.Z(A_n) = \sum_{j_1+2 j_2 + 3 j_3 + \cdots + n j_n = n} \frac{1 + (-1)^{j_2+j_4+\cdots}}{\prod_{k=1}^n k^{j_k} j_k!} \prod_{k=1}^n a_k^{j_k}.

The numerator is 2 for the even permutations, and 0 for the odd permutations. The 2 is needed because

1An=2n!.\frac{1}{|A_n|}=\frac{2}{n!}.

The symmetric group SnS_n

The cycle index of the symmetric groupSnS_n in its natural action is given by the formula

Z(Sn)=j1+2j2+3j3++njn=n1k=1nkjkjk!k=1nakjk.Z(S_n) = \sum_{j_1+2 j_2 + 3 j_3 + \cdots + n j_n = n} \frac{1}{\prod_{k=1}^n k^{j_k} j_k!} \prod_{k=1}^n a_k^{j_k}.

This formula is obtained by counting how many times a given permutation shape can occur. There are three steps: first partition the set of nn labels into subsets, where there are jkj_k subsets of size kk. Every such subset generates k!/kk!/k cycles of length kk. But we do not distinguish between cycles of the same size, i.e. they are permuted by SjkS_{j_k}. This yields

n!k=1n(k!)jkk=1n(k!k)jkk=1n1jk!=n!k=1nkjkjk!.\frac{n!}{\prod_{k=1}^n (k!)^{j_k}} \prod_{k=1}^n \left( \frac{k!}{k} \right)^{j_k} \prod_{k=1}^n \frac{1}{j_k!} = \frac{n!}{\prod_{k=1}^n k^{j_k} j_k!}.

There is a useful recursive formula for the cycle index of the symmetric group. Set Z(S0)=1Z(S_0) = 1 and consider the size ll of the cycle that contains nn, where 1ln1 \le l \le n. There are (n1l1)\binom{n-1}{l-1} ways to choose the remaining l1l-1 elements of the cycle and every such choice generates l!l\frac{l!}{l} different cycles.

This yields the recurrence

Z(Sn)=1n!gSnk=1nakjk(g)=1n!l=1n(n1l1)  l!l  al  (nl)!  Z(Snl)Z(S_n) = \frac{1}{n!} \sum_{g\in S_n} \prod_{k=1}^n a_k^{j_k(g)} = \frac{1}{n!} \sum_{l=1}^n \binom{n-1}{l-1} \; \frac{l!}{l} \; a_l \; (n-l)! \; Z(S_{n-l})

or

Z(Sn)=1nl=1nal  Z(Snl).Z(S_n) = \frac{1}{n} \sum_{l=1}^n a_l \; Z(S_{n-l}).

See also