What are the simple $F[V_4]$ modules, where $F= \mathbb{F}_2$? $V_4 \cong C_2 \times C_2$ is the Klein group.
In a group algebra, there is always the 1.d. simple submodule $A_1:= F \langle \sum_{g \in V_4} g \rangle $. Suppose exists $A_1 \subsetneq A_2 \subsetneq A$.
My proof from here is simple brute force:
- $x \in A_2 \setminus A_1$. If $A_2$ is also a proper submodule, it cannot contain be a single element, by transitivity of $V_4$ onto itself.
- By symmetry, $x$ cannot be of form $\sum_{g \in V_4\setminus \{h \}} g$ as it contains $A_1$.
- Again, by symmetry, we consider when $x= e_1+e_a$. So $A_1+Ax=A_2$ is a proper submodule.
- $A_2/A_1 \cong A_1/Ax \cap A_1 = A_1 $ by 2nd isomorphism.
- Let $x'= e_1 + e_b$. Then $A/A_2 = (A_2+Ax')/A_2 \cong A_2/Ax' \cap A_2 = A_2/A_1 \cong A_1 $
Hence there is precisely one simple module up to isomorphism.
Question: Is there a simpler method that simplifies the calculation? (Is my proof correct?)