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_{k} ker(*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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