$R$ is a commutative ring and $M$ is a left $R$ module.
Define
$\cdot : M*R \to M$ as $(x,a) \to x\cdot a$ by,
$$x\cdot a=ax$$
To prove : $M$ is a right $R$ module.
Proof:
Let $x,y \in M$ and $a,b\in R$,
Then the first two properties i.e.
$(x+y)\cdot a = x\cdot a+ y\cdot a$ and $x\cdot (a+b) = x\cdot a+x\cdot b$ are very easy to verify.
I am a little confused about the third one i.e.
$$x\cdot (ab) = (ab)x= a(bx)= a(xb)= (ax)b= (x\cdot a)b$$
Is it correct to use the commutativity of $R$ to switch $bx$ to $xb$, because $x\in M$ but $M$ is a module and hence a vector space over a ring and hence can i say that elements of $M$ are also in $R$ ?