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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.

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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

1 Answer 1

No, as per Graphical Enumeration by Hararay and Palmer, p.37:

[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.

This is a quote from a translation of Polya, I can give the full citation if you want (it's in German).

Note that the term "identical," in this context is isomporphism and isomorphism is homomorphism (the reference text is very old).

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