How to prove that End(R) is isomorphic to R? I tried to prove this: 
$R$ as a ring is isomorphic to $End(R)$, where $R$ is considered as a left $R$-module. The map of isomorphism is
$$F:R\to End(R), \quad   F(r)=fr,$$ 
where $fr(a)=ar$. And I define the multiplication in $End(R)$ by $(.)$, where $h.g=g\circ h$ for $g,h$ in $End(R)$. 
Is this true?

It's true that: $a\mapsto ar$ is a left module homomorphism.
If we call this map $a\mapsto ar$ by $\theta_r$, then indeed $\theta:R\to End(_RR)$.


*

*Check that it's additive.

*Check that it's multiplicative. (You will absolutely need your rule that $f\circ g=g\cdot f$. The $\cdot$ operation you have given is the multiplication in $(End(R))^{op}$)

If you instead try to show that $R\cong End(R_R)$ in the same way (with $\theta_r$ denoting $a\mapsto ra$), you will have better luck. Can you spot where the two cases are different?
