# Modules of a group over different fields.

Let $G$ be the cyclic group of order 2 and let $V$ be the regular $\mathbb{F}G$-module of $G$. Determine the $\mathbb{F}G$-submodules of $V$ where $\mathbb{F}$ has each of zero or positive characteristic.

Remark: The group algebra $\mathbb{F}G$ and the trivial subspace $\{0\}$ are clearly two submodules of $\mathbb{F}G$. Thus one needs to find (if there's any) all 1-dimensional submodules.

Let $\lambda,\mu \in \mathbb{F}$ not both zero such that $W:=\langle \lambda a+ \mu 1 \rangle$ is a submodule of $\mathbb{F}G$. By $(\lambda a+\mu 1)a=\mu a+\lambda \in W$ to get the exisistence of a $k \in \mathbb{F}$ such that $\mu a+\lambda 1=k(\lambda a+\mu 1)$. Obseve that $\mu, \lambda,k$ are all non-zero (as one is zero would then imply the other two are zero), thus $k=\frac{\lambda}{\mu}=\frac{\mu}{\lambda}$ imples $k^2=\mu^2$, so either $\lambda=\mu$ or $\lambda=-\mu$

Now, if $char(\mathbb{F})=0$ then $W$ can be $\langle a+1 \rangle$ or $\langle a-1 \rangle$.

On the other hand, if $char(\mathbb{F})=p$ for some prime $p$, then $\lambda=tp-\mu$ or $\lambda=tp+\mu$ for any $t \in \mathbb{Z}$, the correspoding $W$ is thus $\langle \lambda a+(tp-\lambda) 1\rangle$ or $\langle a-(tp-\lambda) 1 \rangle$.

Does above imply the $\mathbb{F}G$ submodules are independent of $char(\mathbb{F})$? Since in the $char(\mathbb{F})>0$ case one has $tp=0,\forall k \in \mathbb{Z}$ ?

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Well, no, it doesn't. In your example, when $V$ is the regular $FG$-module for $G$ cyclic of order $2,$ the number of proper non-zero submodules is different. If $G = \langle g \rangle$ , then when $F$ has characteristic zero ( or any characteristic other than $2$ ), there are two proper non-zero submodules, the $1$-eigenspace of $g$ and the $-1$-eigenspace of $g$. When $F$ has characteristic $2$, any proper non-zero submodule of $V$ must be $1$-dimensional, so must be contained in an eigenspace of $g$, and must in fact be the $1$-eigenspace of $g$, since $(g-I)^{2} = 0$ and the $1$-eigenspace of $g$ is just $1$-dimensional. Hence in this case, $V$ has a unique proper non-zero submodule.
Thanks! I've not been introduced to eigenspaces of elements of $G$ yet ;) My understading of your comment on the case $char(F)=2$ is as follows: since $1+1=0$ so $-1=1$, thus $<a+1>=<a-1>$, so there is only one submodule, namely $<a+1>$ in this case. However in the rest cases $a+1 \neq a-1$ and so the two 1-dim submodules are distinct. Is my interpretation valid? –  user31899 Sep 29 '12 at 12:50