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Let $q$ be a prime power, $G=GL_n(q)$ and $P=q^{km}{:}(GL_k(q) \times GL_m(q))$ be a parabolic subgroup of $G$, where $k+m=n$. What is the commutator group $P'$ of $P$?

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The module defined by the action of ${\rm GL}_k(q) \times {\rm GL}_m(q)$ on the elementary abelian subgroup $q^{km}$ is the tensor product of the natural modules for the two direct factors, and hence it is irreducible. So the subgroup $q^{km}$ is in the commutator subgroup, which is therefore equal to $q^{km}:({\rm SL}_k(q) \times {\rm SL}_m(q))$ (except possibly in very small cases like $G = {\rm GL}_2(2)$).

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