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I am trying an exercises on determining all isomorphism types of simple $KG$-modules with $G=S_3$, the symmetric group of degree $3$.

If $K$ is algebraically closed then we can use the following result :

Let $K$ be an algebraically closed field, $V$ a simple $KG$-module and $A$ an abelian subgroup of $G$, then

$$\dim_kV\leq |G:A|$$

Hence, if $K$ is algebraically closed, we can apply this result for $A=<(12)>$ and obtain that $dim_k(V)\leq 3$, which implies that there are $3$ different isomorphism types of simple $KG$-modules.

In case, $K$ is not algebraically closed, then we can not apply the result above. Could you give me some hints? Thanks in advance.

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Are you interested in the case of fields with positive characteristic or not? The characteristic 0 case is typically easier. – rschwieb Jan 21 '13 at 12:57
I think I should also consider the case when char K is positive. – 9999 Jan 21 '13 at 13:08
I don't see why you think $\dim_K(V)\leq 3$ would mean there are only three isoclasses of $K[G]$ modules. If $G$ were a direct sum of $5$ groups of order 2, and $N$ were a subgroup of index 2, then over $\Bbb C$ your result says that $\dim_{\Bbb C}V\leq 2$, but $K[G]\cong \oplus_{i=1}^5\Bbb C$ has five isoclasses of simple modules. – rschwieb Jan 21 '13 at 15:34
$G$ also has an abelian subgroup $A=\langle (123)\rangle$ of index two implying the inequality $\dim_k V\le 2$. The number of non-isomorphic simple modules is still three, because there are two non-isomorphic one-dimensional modules. – Jyrki Lahtonen Jan 21 '13 at 15:37
up vote 4 down vote accepted

Here's some notes on what you could do. I might write $R$ for $K[S_3]$ at times.

There should be three cases: $char(K)=2$, $char(K)=3$ and "otherwise".

Overarching everything here is the basic fact that the isoclasses of simple modules for a semisimple ring are known once you have factored it into simple rings. If $R=\oplus R_i$ is a factorization of a semisimple ring into simple rings, then all simple left $R$ modules look like $\oplus I_i$ where each $I_i$ is a left ideal of $R_i$, exactly one of them is nonzero, and the one that is nonzero is a minimal left ideal of its ring.

In the "otherwise" case, the group ring will be semisimple by Maschke's theorem, and you can go about trying to deduce what it looks like. Let $N$ be the rotations in $S_3$. Since $K[S_3]$ is semisimple, we know that the kernel of the projection to $K[S_3/N]$ is a summand of $R$ (two dimensional) and that its kernel $\omega(N)$ is another summand ring of dimension four over $K$. The factor rings are also semisimple.

Now considering dimensions and the Artin-Wedderburn structure of $R$, the commutative ring $K[S_3/N]$ could in theory be a two dimensional field extension of $K$, or else it is just $K\oplus K$. The former case is ruled out when you notice that $\frac{1}{2}(1+\tau)$ is a nontrivial idempotent in this ring, and so it cannot be a field. ($\tau$ is the reflection in $S_3$.) Already here you can see there are at least three nonisomorphic simple modules for $R$.

The kernel $\omega(N)$ is four dimensional over $K$, and so in principal it could be a $2\times 2$ matrix ring over $K$, or else some mixture of $K$ and two degree extensions over $K$. It cannot just be $K\oplus K\oplus K\oplus K$ because that would make $R$ commutative.

If $K$ happened to be algebraically closed, then the extensions are ruled out, and it would have to be the matrix ring. That would mean there are exactly three isoclasses of simple $R$ modules.

In the other two cases, things would be similar except $rad(R)$ is nonzero. We would make use of the fact that $R$ and $R/rad(R)$ have the same simple modules, and that $R/rad(R)$ is again a semisimple ring. Things would be a bit easier to deduce because of the lower dimension.

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