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Superstable subgroups of \(\text{SL}_2(K)\) and \(\text{PSL}_2(K)\). (Sous-groupes superstables de \(\text{SL}_2(K)\) et de \(\text{PSL}_2(K)\)). (French) Zbl 1095.03021

The first author has shown [“Sous-groupes de SL\(_2(K)\) sur un corps superstable”, in: L. Bélair (ed.) et al., Model theory and applications. Quaderni di Matematica 11, 297–306 (2002; Zbl 1084.03033)] that a definable subgroup of \(\text{SL}_2(K)\), where \(K\) is a superstable field, is either soluble or a conjugate of SL\(_2(k)\) for some definable subfield \(k\); this generalized a result by the second author [“Quelques modestes remarques a propos d’une conséquence inattendue d’un résultat supprenant de Monsieur Frank Olaf Wagner”, J. Symb. Log. 66, 1637–1646 (2001; Zbl 1005.03039)], where \(K\) was assumed to be of finite Morley rank. In this paper the result is generalized to groups superstable in the group language, i.e.the underlying field and the linear action need not be definable, with the proviso that the conjugation may take place in \(K(\alpha)\), where \(\alpha^2\in K\).
The main new ingredient is the definition of a generic copy of the small field \(k\) on the set of traces of \(G\) via the trace formula established by the second author [loc.cit.]. Consequently, the methods seem restricted to dimension \(2\).
With slightly more effort, the analogous result is shown for a superstable subgroup of PSL\(_2(K)\). (This would be immediate if \(K\) were known to be superstable, but for \(G\leq \text{SL}_2(K)\), superstability of \(G/Z(\text{SL}_2(K))\) need not a priori imply superstability of \(G\).)
As corollaries, the authors obtain that a superstable subgroup of a free group is commutative, and that there is no linear bad group of rank \(3\).

MSC:

03C45 Classification theory, stability, and related concepts in model theory
20G15 Linear algebraic groups over arbitrary fields
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References:

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