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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.

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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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    The first statement is exactly true. But there is a problem in " A Course on Group Theory by J.S.Rose" telling:2011-03-23
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    p 148: Let n is an integer, n>1, and F a field. since the group GL_n(F) has a section isomorphic to PSL_n(F), when n>2 then the former group is not solvable. I am thinking about the section mentioned there.2011-03-23
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    A section is a quotient of a subgroup, as in a quotient of SL_n(F).2011-03-25