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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2$\begingroup$ 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. $C_n(q)$. $\endgroup$– S. Carnahan ♦Commented Mar 4, 2011 at 18:00
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1$\begingroup$ 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. $\endgroup$– Frieder LadischCommented Mar 5, 2011 at 11:53
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3 Answers
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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 in the class of finite simple groups by its character degrees with multiplicities, 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. As mentioned in comments, Tong-Viet (MR2905242) has shown that finite simple groups are determined by their character degrees with multiplicities among all finite groups. Needless to say that all this depends heavily on the classification.
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1$\begingroup$ Sorry, but I think something is not true in your answer. By the time before 2000s, the problem of characterizing finite simple groups with their irreducible character degrees (with multiplicities) was open. So It could not be deduced from the result of Lanadzuri et al. $\endgroup$ Commented Dec 29, 2015 at 16:29
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1$\begingroup$ More evidence: Let $S$ be a finite simple group and $G$ a finite group such that $S$ and $G$ have the same irreducible character degrees (counting multiplicities). Then we have the isomorphism $\mathbb{C}S \cong \mathbb{C}G$ of their complex group algebras, and it was not until 2012 that Tong-Viet in a serie of papers proved that in this situation $S \cong G$. (See Theorem 1.1 of "Simple classical groups of Lie type are determined by their character degrees. Journal of Algebra (357) (2012)" $\endgroup$ Commented Dec 29, 2015 at 16:36
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1$\begingroup$ In fact it would be better to say that " By the result of Landazuri et al, a finite simple group is determined by its irreducible character degrees (counting multiplicity) in the class of finite simple groups..." $\endgroup$ Commented Jan 2, 2016 at 20:18
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1$\begingroup$ @FarrokhShirjian: Thank you very much for your comment, I have edited to clarify and to incorporate your useful information. I silently assumed "in the class of finite simple groups" since the question asked for two simple groups with the same character table. $\endgroup$ Commented Jan 11, 2016 at 10:14
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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:
a finite simple group is determined up to isomorphism by its order, with just a couple of exceptions.
answered Mar 4, 2011 at 18:02
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3$\begingroup$ Well, an infinite number of exceptions. $\endgroup$– Steve DCommented Mar 4, 2011 at 20:30
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1$\begingroup$ @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. $\endgroup$ Commented Mar 5, 2011 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.
answered Mar 4, 2011 at 17:54