# Thread: Modules over group algebra

1. ## Modules over group algebra

I'm having a little trouble understanding (intuitively or otherwise) what is meant by a module over a group algebra FG (F a field, G a finite group) and how it relates to transformations.

Identify $S_3$ with the subgroup of $S_4$ which fixes 4 ( $S_n$ is of course the symmetric group on $n$ letters). Then $S_3$ acts on $S_4$ by conjugation (???). Find the character of the linearization $\mathbb{C} S_4$ as a $\mathbb{C} S_3$-module, and determine the multiplicities of each simple $\mathbb{C} S_3$-module in $\mathbb{C} S_4$.

I would be very very grateful for any light anyone could shed on this question; especially any general comments about FG-modules and representations.

2. Originally Posted by AtticusRyan
I'm having a little trouble understanding (intuitively or otherwise) what is meant by a module over a group algebra FG (F a field, G a finite group) and how it relates to transformations.

Identify $S_3$ with the subgroup of $S_4$ which fixes 4 ( $S_n$ is of course the symmetric group on $n$ letters). Then $S_3$ acts on $S_4$ by conjugation (???). Find the character of the linearization $\mathbb{C} S_4$ as a $\mathbb{C} S_3$-module, and determine the multiplicities of each simple $\mathbb{C} S_3$-module in $\mathbb{C} S_4$.

I would be very very grateful for any light anyone could shed on this question; especially any general comments about FG-modules and representations.

to get you started:

well, a (finite dimensional) vector space $V$ over a field $F$ is called an $FG$ module ( $G$ is a group usually finite) or a $G$ module if there exists a homomorphism $\rho: G \longrightarrow \text{GL}(V).$

in your problem: $F=\mathbb{C}, \ G=S_3, \ V=\mathbb{C}S_4.$ the homomorphism $\rho$ is defined by $\rho(\sigma)(\alpha)=\sigma \alpha \sigma^{-1},$ for all $\sigma \in S_3, \ \alpha \in \mathbb{C}S_4.$ since the value of $\chi,$ the character corresponding

to the representation $\rho,$ over a conjugacy class is constant, you only need to find $\chi(\sigma_j)=\text{Tr}(\rho(\sigma_j)), \ j=1,2,3,$ where $\sigma_1=1_{S_3}, \ \sigma_2=(1 \ 2), \ \sigma_3=(1 \ 2 \ 3).$

3. I think I understand; could you perhaps help me find the matrix $\rho(1_{S_3})$? In particular what is our basis of $V$? This is the kind of stuff I don't fully understand how to do.

Obviously $\chi(1_{S_3})=\dim(V)$ but what is the dimension of $V$ as a $\mathbb{C}S_3$-module?

4. Originally Posted by AtticusRyan
Identify $S_3$ with the subgroup of $S_4$ which fixes 4 ( $S_n$ is of course the symmetric group on $n$ letters). Then $S_3$ acts on $S_4$ by conjugation (???). Find the character of the linearization $\mathbb{C} S_4$ as a $\mathbb{C} S_3$-module, and determine the multiplicities of each simple $\mathbb{C} S_3$-module in $\mathbb{C} S_4$.
This is my attempt. Hope it helps.

Lemma 1. Let N be a normal subgroup of G and let U be a $\mathbb{C}(G/N)$-module. Then U admits a canonical $\mathbb{C}G$-module structure, with a subspace of U being a $\mathbb{C}G$-submodule iff it is a $\mathbb{C}(G/N)$-submodule. If $\psi$ is the character of the $\mathbb{C}(G/N)$-module U, then the character of the $\mathbb{C}G$-module U is $\psi \cdot \mu$, where $\mu:G \rightarrow G/N$ is the natural map.

Lemma 1 says that given $g \in G$, $u \in U$ and $N \triangleleft G$, we can define gu = (gN)u such that the linear transformation of U induced by g under the action of G on U is the same as that induced by gN under the action of G/N on U. This gives U a $\mathbb{C}G$-module structure where $\mathbb{C}G$-submodule of U are exactly the $\mathbb{C}(G/N)$-submodule of U.

Let N={1, (1 2)(3 4), (13)(24), (14)(23)}. Then $S_4/N \cong S_3$, where $N \triangleleft S_4$. Now we lift the characters of $S_3$ to the characters of $S_4$.

The character table of $S_3 \cong S_4/N$ is

$\begin{matrix}
& N & (12)N & (123)N\\
\phi^1 (trivial) & 1 & 1 & 1\\
\phi^2 (sign) & 1 & -1 & 1\\
\phi^3 (permutation) & 2 &0 & -1
\end{matrix}$

Since $S_3$ have three conjugacy classes, there are three irreducible characters in $S_3$. Note that characters are constant in conjugacy classes.

Now we use the lifting property and find the character table of $S_4$. The character table of $S_4$ is

$\begin{matrix}
& 1 & (12) & (123) & (1234) & (12)(34)\\
\chi^1 (trivial) & 1 & 1 & 1 & 1 & 1\\
\chi^2 (sign) & 1 & -1 & 1 & -1 & 1\\
\chi^3 (permutation) & 2 &0 & -1 & 0 & 2\\
\cdots & \cdots& \cdots& \cdots& \cdots& \cdots\\
\end{matrix}$

The fourth and fifth column is simply acquired by using the lifting property that $\chi^i(1234) = \phi^i((12)N)$ and $\chi^i((12)(34)) = \phi^i(N)$.

While $S_3$ has three conjugacy classes, $S_4$ has five conjugacy classes. This implies that $S_4$ has five irreducible characters and two of them cannot be lifted from $S_3$. To get $\chi^4$, suppose $S_4$ act transitively on the set X ={1, 2, 3, 4}; let $\pi$ be the character of the $\mathbb{C}S_4$-module $\mathbb{C}X$. $\pi(g)$ equals the number of fixed points under the action of g on X, and we have $\chi^4 = \pi - \chi^1$ (I'll leave it to you to verify that $\chi^4$ is irreducible). Using the fact that $\chi^2\chi^4$ is irreducible, the last irreducible character is $\chi^5 = \chi^2\chi^4$.

Anyhow, I was not able to figure out the multiplicity of $\mathbb{C}S_3$-module in $\mathbb{C}S_4$.