Finite quotients of the multiplicative group of a finite dimensional division algebra are solvable

Rapinchuk, Andrei¹ ; Segev, Yoav² ; Seitz, Gary³

¹ Department of Mathematics, University of Virginia, Charlottesville, Virginia 22904
² Department of Mathematics, Ben-Gurion University, Beer-Sheva 84105, Israel
³ Department of Mathematics, University of Oregon, Eugene, Oregon 97403-1226

Journal of the American Mathematical Society, Tome 15 (2002) no. 4, pp. 929-978

Résumé

We prove that finite quotients of the multiplicative group of a finite dimensional division algebra are solvable. Let $D$ be a finite dimensional division algebra having center $K$, and let $N\subseteq D^{\times }$ be a normal subgroup of finite index. Suppose $D^{\times }/N$ is not solvable. Then we may assume that $H:=D^{\times }/N$ is a minimal nonsolvable group (MNS group for short), i.e. a nonsolvable group all of whose proper quotients are solvable. Our proof now has two main ingredients. One ingredient is to show that the commuting graph of a finite MNS group satisfies a certain property which we denote Property $(3\frac {1}{2})$. This property includes the requirement that the diameter of the commuting graph should be $\ge 3$, but is, in fact, stronger. Another ingredient is to show that if the commuting graph of $D^{\times }/N$ has Property $(3\frac {1}{2})$, then $N$ is open with respect to a nontrivial height one valuation of $D$ (assuming without loss of generality, as we may, that $K$ is finitely generated). After establishing the openness of $N$ (when $D^{\times }/N$ is an MNS group) we apply the Nonexistence Theorem whose proof uses induction on the transcendence degree of $K$ over its prime subfield to eliminate $H$ as a possible quotient of $D^{\times }$, thereby obtaining a contradiction and proving our main result.

Détail

DOI : 10.1090/S0894-0347-02-00393-4

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