Timeline for Dimension of projective cover of trivial $kG$-module
Current License: CC BY-SA 4.0
5 events
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| Aug 12, 2021 at 10:34 | comment | added | Derek Holt | @GeoffRobinson In fact that is exactly how the algorithm computes the projective indecomposable, using induction from the subgroup of order $55$. | |
| Aug 12, 2021 at 9:50 | comment | added | Master Gang | Thanks for your nice counterexample and useful comments. | |
| Aug 12, 2021 at 9:49 | vote | accept | Master Gang | ||
| Aug 12, 2021 at 9:01 | comment | added | Geoff Robinson | Another way to see this is is that $G = {\rm PSL}(2,11)$ has a subgroup $H$ of order $55$ . When char $k =3$, the module ${\rm Ind}_{H}^{G}(k)$ is projective, and has the projective cover of the trivial module as a summand . The associated (complex) permutation character is the sum of the trivial and an irreducible character of degree $11$, as the permutation action is doubly transitive. It is not possible to decompose this character as the sum of two characters which each vanish on $3$-singular elements. Hence ${\rm Ind}_{H}^{G}(k)$ is the projective cover of the trivial module. | |
| Aug 12, 2021 at 8:09 | history | answered | Derek Holt | CC BY-SA 4.0 |