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 *.)

## Comments

## Leave a comment