# Answer to Question #23886 in Abstract Algebra for john.george.milnor

Question #23886

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 kG ∼ M2(k)×(k[t]/(t2)), and that kG/rad(kG) ∼ M2(k)×k.

Expert's answer

Let

*ϕ*:*kG →*End*(*_{k}*V*)*∼*M_{2}(*k*) be the*k*-algebrahomomorphism associated with the*kG*-module*V*. This map is*onto*since*V*is absolutely irreducible. Thus, dim*ker(*_{k}*ϕ*) = 2 .Now*e*= (1) + (123) +(132) is a central idempotent of*kG*, with*kG · e*=*ke**⊕**kσ**,*where*σ*=(sum over*g**∈**G)g**∈**kG.*Since*σ*_{2}= 6*σ*=0, we have*kG · e**∼**k*[*t*]*/*(*t*^{2}). By a simple computation,*ϕ*(*e*) = 0, so*kG · e*=ker(*ϕ*) (both spaces being 2-dimensional).Therefore,*kG*=*kG ·*(1*− e*)*× kG · e**∼*M2(*k*)*×*(*k*[*t*]*/*(*t*2))*.*Computing radicals, we get rad(*kG*) = rad(*kG · e*) =*k ·**σ*, so*kG/*rad(*kG*)*∼*M_{2}(*k*)*× k*.
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