Let $G$ be a group , $F$ be a field , $n$ be a positive integer , a map $h:G \times F^n \to F^n$ is called a linear action if there is a group homomorphism $f:G \to GL(n,F)$ such that $h(g,v)=f(g)v,\forall g\in G , v \in F^n $ ; then is it true that for any finite group $G$ and field $F$ , there is a positive integer $n$ , a linear action $h$ and a non-zero vector $v \in F^n$ such that $h(g,v)=v , \forall g \in G$ ?
I tried as follows : Let $|G|=n$ , then we can embed $G$ in $S_n$ which we can embed in $GL(n,F)$ by
$\sigma \in S_n \to (e_{\sigma(1)} .... e_{\sigma(n)})\in GL(n,F)$ ; so we have an injective group homomorphism $f:G \to GL(n,F)$ , now is this the required homomorphism which induces the action ? If so then what is my non-zero vector $v$ ?
Please help . Thanks in advance
EDIT : From leibnewtz 's comment below , we can always have the trivial action ; but I am looking for a non-trivial action ..
Your example works since you can take the non-zero vector $v := e_1 + e_2 + \dots + e_n$ which is fixed by all $g \in G \leq S_n$.
It works also (probably a little bit easier) if you simply take $G \hookrightarrow S_n \hookrightarrow S_{n+1}$, where $S_n$ is the subgroup of permutations which fix $n+1$. In the picture of your linear action (but now) onto $F^{n+1}$ this means fixing the (basis) vector $(0, \dots, 0, 1)$.