Every Ring is Isomorphic to a Subring of an Endomorphism Ring of an Abelian Group Show that for every ring $(R,+,\cdot)$, there is an abelian group, $(A,+)$, such that $R$ is isomorphic to a subring of $(\operatorname{End}(A),+,\circ)$.
$(\operatorname{End}(A),+,\circ)$ is the set of homomorphisms of $A$ that form a ring under function addition and composition.
I am thinking to let $\operatorname{End}(A)$ be the group such that $A$ is the abelian group $(R,+)$ and create a ring homomorphism from $(R,+,\cdot)$ into $\operatorname{End}((R,+))$.
Thoughts?
 A: You are correct, (madame or) sir.
This is essentially the ring-theoretic analogue of Cayley's Theorem for groups.
Also, this issue (as a question) came up a while back on Math Overflow.
Added: I had missed that the explicit definition of the map was not contained in the OP's question.  A natural ring embedding from $R$ to $\operatorname{End}(R,+)$ is 
$r \mapsto \bullet r: (x \in R \mapsto xr)$. 
[Or possibly $r \mapsto r \bullet: (x \in R \mapsto rx)$, depending upon your conventions on composition.] 
A: I think instead of using $(R,+)$ as the candidate abelian group, you should use
$$
A = \oplus_{a\in R}A_a
$$
where $A_a$ is the cyclic group generated by the element $a$.
Let $f \in End(A)$. Then we can think of $f$ as
$$
\oplus_{a\in R}f_a
$$
where $f_a \in End(A_a)$. 
Finally, we can define a map $\phi: R \rightarrow End(A)$ such that for $b \in R$
$$
\phi(b) = \oplus_{a\in R}f_a
$$
where $f_a$ is the trivial map if $a \neq b$;
and $f_a$ is the identity map if $a = b$.
A: You can consider $(R,+)$, the underlying abelian group. Then, an injective morphism is
\begin{align*}
\lambda:R&\to \text{End}_{\textsf{Ab}}(R)\\
r&\mapsto \lambda_r
\end{align*}
where $\lambda_r$ is left-multiplication by $r$.
