Proving that $\{f \in End(A): \forall a \in A:|a|<\infty \implies f(a)=0\}$ is an ideal Let $A$ be an abelian group. I need to prove that 
$I = \{f \in End(A): f(a)= 0 \ \text{for all $a$ of finite order}\}$,
is an ideal of $\text{End}(A)$. It isn't hard to prove that $I$ is a subgroup of $End(A)$, but it is quite hard to prove that:
$g(x) \cdot f(x)$ and $f(x) \cdot g(x)$ are in $I$ with $f(x)\in I$ and $g(x) \in End(A)$.
I thought that if you multiply $g(x)$ and $f(x)$ and you take an element $a$ that $g(a)*f(a) = g(a)*0 = 0$, so $g(x)f(x)=0$ for $x$ with finite order so $g(x)f(x) \in I$.
Is this proof correct?
 A: If $A$ is an abelian group, the multiplication on $\operatorname{End}(A)$ is the composition. It's customary to write $A$ additively and I'll use this convention.
It's clear that $0\in I$. If $f,g\in I$, then $(f+g)(a)=f(a)+g(a)=0+0=0$ for all $a$ of finite order.
Now, let $f\in I$ and $g$ be an arbitrary endomorphism. Saying $a$ has finite order means that $na=0$ for some integer $n>0$.
Since $ng(a)=g(na)$, we see that if $a$ has finite order, then also $g(a)$ has finite order. Therefore
$$
(fg)(a)=f(g(a))=0
$$
by hypothesis. Proving that $gf\in I$ is even easier: if $a$ has finite order, then
$$
(gf)(a)=g(f(a))=g(0)=0.
$$
A: You can also prove this in the following way: let $T = \{ a \in A \hspace{1mm} | \hspace{1mm} a \text{ of finite order}\}$ be the so-called torsion subgroup of $A$. Observe that the set $I$ that you describe is precisely the kernel of the following ring homomorphism
$\phi: \text{End}(A) \rightarrow \text{End}(T), \hspace{1mm} f \mapsto f|_{T}$,
where $f|_{T}$ denotes the restriction of $f$ tot $T$. Now use the fact that the kernel of every ring homomorphism is an ideal (every ideal even arises in this way), to conclude the proof. 
An interesting question, to which I don't know the answer myself, is whether the map $\phi$ is surjective or not.
