Is it true that "There is a part of $\mathrm{GL}_n(F)$ which is isomorphic to $\mathrm{PSL}_n(F)$, $n \geq 4$"? So since $\mathrm{PSL}_n(F)$ is not solvable (it is simple non-abelian group), the former linear group is simply not solvable.
Linear groups $\mathrm{PSL}_n(F)$ and $\mathrm{GL}_n(F)$
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abstract-algebra
1 Answers
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If $\operatorname{GL}_n$ were solvable, then its subgroup $\operatorname{SL}_n$ would be solvable, and in turn the quotient $\operatorname{PSL}_n$ of the latter would also be solvable.
I doubt $\operatorname{GL}_n$ contains a subgroup isomorphic to $\operatorname{PSL}_n$.
Later: In that book, in exercise 324 on page 123 sections are defined as quotients of subgroups (I would say subquotients) so the translation done in the body of this question is wrong.
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1$A$ section is a quotient of a subgroup, as in a quotient of SL_n(F). – 2011-03-25