# Rings act on modules

$R$ is a ring, then a left $R$ module is an abelian group $V$ together with a multiplication map $$R \times V \to V, (r,v) \to rv$$ satisfies some natural axioms. I learned that rings act on modules the way groups act on sets, so how can you explain that based on the definition? and from the definition,is that means $rv \in V$ ? But since $R$ is a ring and $V$is an abelian group, so the product of an element of ring and an element of group is still in that group ??
An action of a group $G$ on a set $S$ is usually defined as a group homomorphism $\eta:G\to{\rm Sym}\; S$, and we define the action of $G$ on $S$ by declaring that $g\cdot s=\eta(g)(s)$, i.e. for each $g\in G$, $\eta(g)$ is a bijection of $S$, and $g\cdot s$ means "apply $\eta(g)$ to $s$".
Similarly, a module $M$ over a ring $A$ is an abelian group with a ring homomorphism $\nu:A\to{\rm End}\; M$, and the multiplication $a\in A$, $m\in M$ is defined by declaring that $a\cdot m=\nu(a)(m)$, that is, for every $a\in A$ we have an endomorphism of the abelian group $M$, $\nu(a):M\to M$; and $a\cdot m$ means "apply $\nu(a)$ to $M$."