# Character tables and simple groups.

Is it known if there are two finite non-isomorpic non-abelian simple groups with the same character table? Does this answer change if the subsidiary information (like the orders and sizes of the conjugacy classes) is included? (There are no examples of this in the "ATLAS")

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If you look at the characters of identity, you can reconstruct the order of the group by summing squares. Then you just have to check the following order coincidences: $A_3(2)$ vs. $A_2(4)$, and $B_n(q)$ vs. $C_n(q)$. –  S. Carnahan Mar 4 '11 at 18:00
You can also reconstruct the sizes of conjugacy classes, since the sum of squares of absolute values of a column of the character table gives the centralizer size of the corresponding group element. –  Frieder Ladisch Mar 5 '11 at 11:53
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## 3 Answers

There are no such two non-isomorphic groups. As is pointed out in Neil Strickland's answer, up to some exceptions, a finite simple group is determined by its order. For these exceptions it is known that the smallest character degree larger than 1 is different (For the infinite series $B_n(q)=O_{2n+1}(q)$ vs $C_n(q)=PSp_{2n}(q)$ for $q$ odd this follows from the results in Landazuri and Seitz (J. Algebra 32 (1974), 418–443, MR0360852 (50 #13299)). This means that a finite simple group is determined by its character degrees (with mulitplicities), and led Huppert (Illinois J. Math. 44, 4 (2000), 828-842, MR1804317 (2001k:20009)) to the conjecture:

Conjecture: If two finite groups $G$ and $H$ have the same set of character degrees (without counting multiplicities) and $G$ is nonabelian simple, then $H\cong G\times A$ for some abelian $A$.

This has been verified for some simple groups, but is still open to the best of my knowledge. Needless to say that all this depends heavily on the classification.

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The order of $G$ is the sum of the squares of the character degrees, so is determined by the character table. According to this question:

Number of finite simple groups of given order is at most 2 - is a classification-free proof possible?

a finite simple group is determined up to isomorphism by its order, with just a couple of exceptions.

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Well, an infinite number of exceptions. –  Steve D Mar 4 '11 at 20:30
@Steve D: This type of question does have a long history, going back at least to a pair of papers by Emil Artin in Communications in Pure & Applied Mathematics (1955) partly inspired by Chevalley's just announced discovery of new finite simple groups. –  Jim Humphreys Mar 5 '11 at 14:27
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I believe the answer to the basic question here is that no one expects to find two such non-isomorphic groups. However, even with the classification of finite simple groups (probably) in hand, the question is not easy to resolve. There are however a number of papers showing that specific types of finite simple groups are characterized uniquely by their character tables. Here are random samples of the older literature from MathSciNet:

MR0316551 (47 #5098), Lambert, P. J., Characterizing groups by their character tables. I. Quart. J. Math. Oxford Ser. (2) 23 (1972), 427–433.

MR0404415 (53 #8217), Pahlings, H., Characterization of groups by their character tables. I, II. Comm. Algebra 4 (1976), no. 2, 111–153; ibid. 4 (1976), no. 2, 155–178.

I'm not at all an expert on these developments in finite group theory, but it's important to recognize the difficulty of the underlying problem.

ADDED: Neil Strickland has pointed to a very useful earlier discussion on MO, which indicates that the question is indeed resolvable if the classification is known. The earlier work I sampled tried to deal with characterizations by character tables without using the classification, in some cases successfully. The groups of Ree-type were certainly a big obstacle for some years.

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