Answer to Question #23885 in Abstract Algebra for

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 {e1, e2}on V , the G-action is described by (12)e1 = e2,(12)e2 = e1; (123)e1 = e2, (123)e2= e1 + e2. Using these, it is easy to check that the k-spanof x = e1 ⊗ e1, y =e2 ⊗ e2, and z = e1 ⊗ e2 + e2 ⊗ e1 is a 3-dimensional kG-submodule A ⊆ W (with B = k · z as a trivial kGsubmodule).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 kGmodule is trivial.But if w = a(e1 ⊗ e1) + b(e2 ⊗ e2) + c(e1⊗ e2) + d(e2 ⊗ e1), (12)w = w implies thatc = d so w ∈ A, a contradiction.This shows that W is notsemisimple. (In fact, (123)w = wimplies 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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