Let group $G$ acts on the group $V$. Then there exists a homomorphism $\phi: G \longrightarrow {\rm Aut}(V)$. I want to know why we can consider $V$ as a module?
I saw this book: Gorenstein(Finite groups). He said that if there exists a homomorphism $\rho$ of $G$ to ${\rm GL}(V)$, then by definition
$$vx=v(\rho(x)) \ \forall v \in V,$$
$V$ is a $G$-module. Now i cant understand if there exists a homomorphism $\phi: G \longrightarrow {\rm Aut}(V)$, then why $V$ is a module?
I think$V$ needs to be abelian for this to make sense. An arbitrary abelian group $V$, can be regarded as a ${\mathbb Z}$-module, so $V$ becomes a ${\mathbb Z}G$-module under the action.
But if $V$ is an elementary abelian $p$-group for some prime $p$ (i.e. abelian of exponent $p$), then we can consider $V$ as a module over the finite field ${\mathbb F}_p$ of order $p$, and so $V$ becomes a ${\mathbb F}_pG$-module under the action. This is useful, because it enables us to use results from representation theory, such as Maschke's Theorem, to deduce properties of the action.