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So, there are several standard ways of proving irreducibility/reducibility for representations over fields where the characteristic doesn't divide $|G|$ such as Maschke's theorem, jordan normal form, using eigenvectors, etc. However, I was wondering, how does this change over a field $F$ of characteristic that does divide $|G|$? Is there any simple example of how one would approach this? As an example, how would one show that the permutation representation of the symmetry group $S_{3}$ is not completely reducible over a field $F$ of characteristic 3?

user211685
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  • For your example, there is a fact that every irreducible representation of a $p$-group over a field of characteristic $p$ must be $1$-dimensional and trivial. (See http://math.stackexchange.com/questions/572564/any-irreducible-representation-of-a-p-group-over-a-field-of-characteristic-p .) – darij grinberg Jan 30 '15 at 02:56
  • @darijgrinberg is $S_3$ a 3-group? In this case I'd just play around and find an explicit counterexample-this is a small enough group and dimension that that's plausible. – Kevin Carlson Jan 30 '15 at 03:07
  • Oops! Sorry, I didn't read this properly. – darij grinberg Jan 30 '15 at 03:09
  • Is there a direct way to show that the representation would not be completely reducible over char 3? How would you do this using a counterexample? – user211685 Jan 30 '15 at 03:32
  • Compute its endomorphism ring, for example. – Mariano Suárez-Álvarez Jan 30 '15 at 04:30

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Let $V = k^3$ be the permutation representation of $S_3$ and consider $M = k\left[\begin{smallmatrix} 1 \\ 1 \\ 1 \end{smallmatrix}\right]$. This is a one dimensional invariant subspace. If $V$ is completely reducible then there is some two dimensional invariant subspace $N$ such that $V = N \oplus M$.

Assume such an $N$ exists and assume $\left[\begin{smallmatrix} a \\ b \\ c \end{smallmatrix}\right]$ is a vector in $N$. As $N$ is invariant I can apply the permutation $(1 \ 2 \ 3)$ to this and get that $\left[\begin{smallmatrix} c \\ a \\ b \end{smallmatrix}\right]$ is in $N$. Similarly $\left[\begin{smallmatrix} b \\ c \\ a \end{smallmatrix}\right]$ is in $N$. Add them all together and I get that $(a + b + c)\left[\begin{smallmatrix} 1 \\ 1 \\ 1 \end{smallmatrix}\right]$ is in $N$. But $N \cap M = 0$ so it must be that $a + b + c = 0$.

The set of vectors satisfying $a + b + c = 0$ is a two dimensional subspace of $V$ and it contains $N$, so it must equal $N$. But it also contains $M$, and this contradicts the assumption that $N \cap M = 0$.

Jim
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