Property of G-modules involving the invariant elements under the G-action I am stuck at some basic fact I would like to prove. I tried proving it using $G-$orbits and cardinalities, but without success.
Let $p$ be some prime number, $G$ be a finite $p-$group and $A$ a $G-$module such that $pA=0$. Recall that $A^{G}:=\{a\in A \quad | \quad g.a=a \quad \text{for all }g\in G  \}$, where $.$ represents the action of $G$ on $A$. 
SHOW: If $A^{G}=0$, then $A=0$.
 A: Assume that $A \neq 0$. Since $pA=0$ we can consider it as a vector space over $F_p$, and by taking the $G$-submodule generated by some nonzero element, we can assume that it is finite dimensional. In particular an action of $G$ is just a homomorphism $\varphi:G \to GL_m(F_p)$.
Suppose first that $G$ is cyclic generated by $g$. Since $g^{p^m}=e$ where $|G|=p^m$, its image $\varphi(g)\in GL_m(F_p)$ satisfies $0=\varphi(g)^{p^m}-1=(\varphi(g)-1)^{p^m}$. Hence 1 is an eigenvalue of $\varphi(g)$, and its corresponding eigenvector is an invariant element in $A$.
For general $G$, let $H$ be a normal subgroup of index $p$. By induction $A^H \subseteq A$ is a nonzero subspace. Let $g \in G$ such that $G=<g,H>$. Since $g^{-1}Hg=H$ we get that for any $a\in A^H$ and $h\in H$ we have that $g^{-1}hga=a$ so that $h(ga)=ga$, and therefore $ga\in A^H$. It follows that $A^H$ is a $G$ module, invariant under $H$, so it has an induced $G/H$ action, which is a cyclic group. Thus $\emptyset \neq (A^H)^{G/H} \subseteq A^G$ from the previous case.
