Question

Let k ⊆ K and G be as above. Show that a kG-module M is semisimple iff the KG-module MK = M ⊗k K is semisimple.

Answer 1

In given case we have that, rad $KG = (\mathrm{rad}~{}kG)\circ_kK$. Therefore,

$(\mathrm{rad}~{}KG)\cdot M^{K} = (\mathrm{rad}~{}kG\circ_{k}K)\cdot (M\circ_{k}K) = (\mathrm{rad}~{}kG\cdot M)\circ_{k}K.$

It follows that: $M$ is semisimple $\iff \mathrm{rad}~{}kG\cdot M = 0\iff (\mathrm{rad}~{}KG)\cdot M^{K} = 0\iff M^{K}$ is semisimple.