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A new characterization for the simple group \(\mathrm{PSL}(2,p^2)\) by order and some character degrees. (English) Zbl 1363.20031
For a finite group \(G\), let \(X_1(G)\) denote the set of irreducible complex character degrees of \(G\), counting multiplicitites, i.e. the first column of the character table of \(G\). It is a well-known fact from basic representation theory that \(X_1(G)\) completely determines the group algebra \(\mathbb C G\). This means that \(X_1(G)=X_1(H)\) for two finite groups \(G\) and \(H\) if and only if \(\mathbb C G\simeq \mathbb C H\).
There are examples, when the isomorphism \(\mathbb C G\simeq \mathbb C H\) does not imply that \(G\simeq H\). In contrast, H. P. Tong-Viet [Monatsh. Math. 166, No. 3–4, 559–577 (2012; Zbl 1255.20006); Algebr. Represent. Theory 15, No. 2, 379–389 (2012; Zbl 1252.20005); J. Algebra 357, 61–68 (2012; Zbl 1259.20008)] proved recently that if \(S\) is a finite simple group and \(G\) is any finite group such that \(X_1(G)=X_1(S)\) (or, equivalently, \(\mathbb C G\simeq \mathbb C S\)), then \(G\simeq S\) must hold.
In this paper, the authors strengthen this result for the case \(S\simeq \mathrm{PSL}(2,p^2)\), where \(p\) is an odd prime. Their main theorem says that if \(G\) is a finite group such that \(|G|=|\mathrm{PSL}(2,p^2)|,\;p^2\in X_1(G)\) but no element of \(X_1(G)\) is divisible by \(2p\), then \(G\simeq \mathrm{PSL}(2,p^2)\).

20D60 Arithmetic and combinatorial problems involving abstract finite groups
20D06 Simple groups: alternating groups and groups of Lie type
20G40 Linear algebraic groups over finite fields
20C15 Ordinary representations and characters
20D05 Finite simple groups and their classification
Full Text: DOI Link
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