# Is cycle index unique for every distinct group up to isomorphism?

Are two groups isomorphic iff their cycle index is the same? Note that for every group there exists a permutation group to which it is isomorphic.

-
No...because $\{1, (123), (132)\}$ has cycle index $\frac{1}{3}(a_1^3+2a_3)$ while $\{1, (123)(456), (132)(465)\}$ has cycle index $\frac{1}{3}(a_1^3+2a_6)$. These groups are isomorphic as they are both cyclic of order $3$. – user1729 May 21 '12 at 10:01
Thanks for the contradiction example. But are there any normalized forms of the groups? A systematic way we could rewrite the later group into the first one? Perhaps my 2nd part of the original question is, are there any 2 groups that are not isomorphic and have the same cycle index? – Dávid Tóth May 21 '12 at 12:31
I'm not sure about the second part, which is why I didn't post that as an answer! – user1729 May 21 '12 at 13:06
Sorry... what is the definition of cycle index? – Arturo Magidin May 21 '12 at 16:32

[two groups] having the same cycle index need not be identical. In fact they need not even be isomorphic. Namely, let p be an odd prime and $m \geq 3$ be an integer ($p=m=3$ is the simplest example). It is well known that there is a nonabelian group of order P^m in which every element except the identity has order p. Let B be the regular representation of this group. Let A be the regular representation of the abelian group of order p^m and type (p,p,...,p). Then A and B are permutation groups of order and degree p^m = d with the same cyclic index $$d^{-1}(d_1^d + (d - 1)s^{d/p}_p)$$ for each permutation of A and B other than the identity contains p^{m-1} cycles of length p.