In characteristic three with $n=5$, a constant <vector> $(c,c,c,c,c)$ belongs to the augmentation <submodule> only if $5c=0$, which forces $c=0$. Thus $S^{(4,1)}$ has no trivial <submodule>. Part (b) proves that it is \b[simple], of dimension four, and therefore equals $D^{(4,1)}$.

Solution

Past exam of the Lie algebras and representation theory of symmetric groups course of the University of Oxford / 2007 / B / 8 / d / Solution

 Solution (source code)