# Math Help - FG-modules

1. ## FG-modules

Hello, this is my first post.

I was wondering how, if we have a finite group $G$ and a field $F$ and an $FG$-module $FG$, how we would show there is no $FG$-submodule complementary to $V=\left\{\sum_{h\in G}x_h h\mid \sum_{h\in G} x_h=0\right\}$? Obviously a proof by contradiction but how would I start?

Thanks,

John

2. Originally Posted by JohnnyFrankston
Hello, this is my first post.

I was wondering how, if we have a finite group $G$ and a field $F$ and an $FG$-module $FG$, how we would show there is no $FG$-submodule complementary to $V=\left\{\sum_{h\in G}x_h h\mid \sum_{h\in G} x_g=0\right\}$? Obviously a proof by contradiction but how would I start?

Thanks,

John
this is certainly not true! for example if F is a field of characteristic 0, then FG becomes a semisimple algebra (Maschke's theorem) and thus every ideal (FG submodule) of FG has a complement.

3. Originally Posted by NonCommAlg
this is certainly not true! for example if F is a field of characteristic 0, then FG becomes a semisimple algebra (Maschke's theorem) and thus every ideal (FG submodule) of FG has a complement.
Oh of course, I meant to say that $|G|= 0\in F$. Otherwise yes maschke comes to the rescue haha

4. Originally Posted by JohnnyFrankston
Oh of course, I meant to say that $|G|= 0\in F$. Otherwise yes maschke comes to the rescue haha
suppose $FG=V \oplus W$ and let $e=\sum_{g \in G} g.$ see that $ge=e$ for all $g \in G$ and thus $ve=0,$ for all $v \in V.$ we have $1=v+w,$ for some $v \in V, \ w \in W$ and so $e=we.$ now suppose that $e \in V.$

then since $V,W$ are $FG$ submodules, we'll have $e=we \in V \cap W = \{0 \}, \color{red}^*$ which is nonsense. hence $e \notin V,$ which means that $|G|,$ as an element of $F,$ is non-zero. Q.E.D.

$\color{red}*$: note that $V$ is an ideal of $FG,$ i.e. it's both left and right $FG$ sumodule. so if we assume that $W$ is a right $FG$ submodule and $e \in V,$ then $we \in V \cap W = \{ 0 \}.$

5. Originally Posted by NonCommAlg
suppose $FG=V \oplus W$ and let $e=\sum_{g \in G} g.$ see that $ge=e$ for all $g \in G$ and thus $ve=0,$ for all $v \in V.$ we have $1=v+w,$ for some $v \in V, \ w \in W$ and so $e=we.$ now suppose that $e \in V.$

then since $V,W$ are $FG$ submodules, we'll have $e=we \in V \cap W = \{0 \}, \color{red}^*$ which is nonsense. hence $e \notin V,$ which means that $|G|,$ as an element of $F,$ is non-zero. Q.E.D.

$\color{red}*$: note that $V$ is an ideal of $FG,$ i.e. it's both left and right $FG$ sumodule. so if we assume that $W$ is a right $FG$ submodule and $e \in V,$ then $we \in V \cap W = \{ 0 \}.$
Thanks a lot for that - the two bits I don't quite understand I've highlighted in red in the quote above.

Essentially I don't understand why $ge=e$ for all $g \in G$ implies $ve=0$ for all $v\in V$, nor do I understand why $e\not\in V$ implies anything about the order of the group.

6. Originally Posted by JohnnyFrankston
Thanks a lot for that - the two bits I don't quite understand I've highlighted in red in the quote above.

Essentially I don't understand why $ge=e$ for all $g \in G$ implies $ve=0$ for all $v\in V$, nor do I understand why $e\not\in V$ implies anything about the order of the group.
if $v=\sum_{g \in G}x_gg \in V,$ then $\sum_{g \in G}x_g = 0$ and thus $ve=\sum_{g \in G} x_g ge=\sum_{g \in G}x_ge = \left(\sum_{g \in G}x_g \right)e = 0.$

we have $e=\sum_{g \in G}x_g g,$ where $x_g = 1,$ for all $g \in G.$ now $e \notin V$ means that $|G|=\sum_{g \in G}1 =\sum_{g \in G} x_g \neq 0.$ (see the definition of $V$ again!)