When ring is commutative, prove that left and right modules coincide I have the following definition of a left $R$-module for a ring $R$ and an abelian group $X$:
these conditions must be satisfied:
$$(\alpha+\beta)x = \alpha x + \beta x$$
$$\alpha(x+y) = \alpha x + \alpha y$$
$$\alpha(\beta x) = (\alpha\beta)x$$
$$1x = x$$
The book then asks me to define a right $R$-module and prove that, when the ring $R$ is commutative, both definitions coincide. That's what I did:
Definition for the right $R$-module:
$$x(\alpha+\beta) = x\alpha + x\beta$$
$$(x+y)\alpha = x\alpha + x\beta$$
$$(x\beta)\alpha = x(\beta\alpha)$$
$$x1 = x$$
I think now I need to prove that
$$(\alpha+\beta)x = x(\alpha + \beta)$$
$$\alpha(x+y) = (x+y)\alpha$$
$$\alpha(\beta x) = (x\alpha)\beta$$
$$1x = x1$$
at least for the last two:
$$\alpha(\beta x) = (\alpha\beta)x = (\beta\alpha)x = \beta(\alpha x)$$
but how to prove that $\alpha x = x\alpha$?
And for the unit it's easy:
$$1x = x = x1$$
but what about the first two? 
 A: The result you want to prove follows immediately from this proposition:
Proposition: Let $(X,+)$ be an abelian group. Then $X$ is a right $R$-module if only if $X$ is a left $R^{\text{op}}$-module. Here $R^{\text{op}}$ is the opposite ring of $R$, i.e., in this ring the multiplication is defined in the following way: $a\cdot_{\text{op}}b=b\cdot a$.
Proof: $\Longrightarrow$) Let $\sigma$ be the right action. Since $X$ is a right $R$-module we have $\sigma(x,\alpha)=x\cdot \alpha$, and then the axioms that define $X$ as a right $R$-module can be written in this form: $$\sigma(x+y,\alpha)=\sigma(x,\alpha)+\sigma(y,\alpha),$$ $$\sigma(x,\alpha+\beta)=\sigma(x,\alpha)+\sigma(x,\beta),$$ $$\sigma(x,(\beta\cdot\alpha))=\sigma(\sigma(x,\beta),\alpha).$$
Now we define the application $\sigma'\colon R^{\text{op}}\times X\rightarrow X$ given by $(\alpha,x)\mapsto \sigma'(\alpha,x)=\sigma(x,\alpha)$. So we have to prove that $\sigma'$ is a left action.  Let $\alpha, \beta\in R^{\text{op}}$ and $x,y\in X$. Then $$\sigma'(\alpha,x+y)=\sigma(x+y,\alpha)=\sigma(x,\alpha)+\sigma(y,\alpha)=\sigma'(\alpha,x)+\sigma'(\alpha,y),$$ $$\sigma'(\alpha+\beta,x)=\sigma(x,\alpha+\beta)=\sigma(x,\alpha)+\sigma(x,\beta)=\sigma'(\alpha,x)+\sigma'(\beta,x),$$ $$\sigma'((\alpha\cdot_{\text{op}}\beta),x)=\sigma'((\beta\cdot\alpha),x)=\sigma(x, (\beta\cdot\alpha))=\sigma(\sigma(x,\beta),\alpha)=$$
$$=\sigma(\sigma'(\beta,x),\alpha)=\sigma'(\alpha,\sigma'(\beta,x)).$$
Finally, as $\sigma'(1_R,x)=\sigma(x,1_R)=x$, we deduce that $X$ is a left $R^{\text{op}}$-module. 
$\Longleftarrow$) The proof is completely analogous. 
