For each abelian group $G$ let $T (G)$ be the set of all elements of finite order show that $T (G / T (G)) = 0.$ For each abelian group $G$ let $T (G)$ be the set of all elements of finite order show that $T (G / T (G)) = 0.$
I already show that $T(G)$ is a subgroup of $G$
But I'm stuck here and I don't know what equals to $0$ means if empty set of identity.
Any hint on how to do it?
 A: Saying that an abelian group $A = 0$ is an abuse of notation saying that it is isomorphic to the zero group, or equivalently that $A = \{e\}$ consisting of only the identity element. Thus, to show that $T(G/T(G)) = 0$ really means to show that it contains only the identity element. Every group has an identity, so it is impossible for them to be empty. As elements of $G/T(G)$ are cosets, the identity element of this group is $T(G)$.
Now, let's take some $a \in T(G/T(G))$ and pick some $g \in G$ such that $a = gT(G)$. This means that $g T(G)$ has finite order, so $(g T(G))^n = T(G)$ for some $n \geq 1$. By definition of coset multiplication, this means that $g^n T(G) = T(G)$. As these two cosets are equal, we necessarily have $g^n \in T(G)$. Hence, there is some $m \geq 1$ such that $(g^n)^m = e$, where $e \in G$ is the identity element. But then $g^{mn} = e$, so $g$ itself had finite order, i.e. $g \in T(G)$. But if $g \in T(G)$ then $g T(G) = T(G)$, so $a$ is the identity element.
We started with an arbitrary $a \in T(G/T(G))$ and proved that $a = T(G)$ the identity element. Thus, the only element of $T(G/T(G))$ is the identity, so as defined above, $T(G/T(G)) = 0$.
