# Answer to Question #23885 in Abstract Algebra for john.george.milnor

Question #23885

For G = S3 and any field k of characteristic 2, view V = ke1 ⊕ ke2 ⊕ ke3/k(e1 + e2 + e3) as a (simple) kG-module with the permutation action. Show that W = V ⊗k V with the diagonal G-action is not a semisimple kG-module.

Expert's answer

With the basis

*{e*1*, e*2*}*on*V*, the*G*-action is described by (12)*e*1 =*e*2*,*(12)*e*2 =*e*1; (123)*e*1 =*e*2*,*(123)*e*2=*e*1 +*e*2*.*Using these, it is easy to check that the*k*-spanof*x*=*e*1*⊗**e*1*, y*=*e*2*⊗**e*2*,*and*z*=*e*1*⊗**e*2 +*e*2*⊗**e*1 is a 3-dimensional*kG*-submodule*A**⊆**W*(with*B*=*k · z*as a trivial*kG*submodule).We claim that A is not a kG-direct summand of W. Indeed, if*W*=*A**⊕**k · w*is a*kG*-decomposition, we must have*w**∈**WG*(*G*-fixed points in*W*) since*G/*[*G,G*]*∼**{±*1*}*implies that any 1-dimensional*kG*module is trivial.But if*w*=*a*(*e*1*⊗**e*1) +*b*(*e*2*⊗**e*2) +*c*(*e*1*⊗**e*2) +*d*(*e*2*⊗**e*1)*,*(12)*w*=*w*implies that*c*=*d*so*w**∈**A*, a contradiction.This shows that*W*is notsemisimple. (In fact, (123)*w*=*w*implies further that*a*= 0, so we have*WG*=*k · z*=*B*.The*kG*-composition factors of*W*are the trivial*G*-modules*B*,*W/A*, together with*A/B**∼**V*.)
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