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 .)
Comments
Leave a comment