I have a follow up to this question: Rank of the quotient of an Abelian group by its torsion part?.
So my understanding is given a finitely generated abelian group $G$ and its torsion subgroup $T_G$ then $T_G$ is fully characteristic (preserved under all endomorphisms of $G$) and $G/T_G$ is free abelian with rank equal to $G$.
Does this mean $G/T_G$ is a characteristic subgroup of $G$ which is of same rank as $G$ (more precisely the coset representatives of $T_G$ in $G/T_G$ is the subgroup $H \cong Z^n$)? Intuitively I'd say something like for any $g \in G$ that was not finite order, then any automorphism of $G$ must map $g$ to another element of infinite order as the torsion subgroup is fully characteristic and if $g$ was mapped to a finite order element it was contradict that.
I'm not sure how to go about it from here. I feel like from the perspective of the fundamental theorem of finitely generated abelian groups where $G \cong \mathbb{Z}^n \times \mathbb{Z}_{k_1} \times \dots$ then $G/T_G$ would be the free abelian portion, and it would be characteristic. I'm struggling to formalise it however, I also wasn't able to find much on this so I feel like I may be missing something completely.
It's true that in a finitely generated abelian group $G$, the elements of finite order form a characteristic subgroup, called the torsion subgroup $T_G$.
But $G / T_G$ is not a subgroup of $G$. How could it be? Each element of a subgroup of $G$ is also an element of $G$ itself, but each element of $G / T_G$ is a coset of $T_G$ which is a certain subset of $G$. An element of $G$ and a subset of $G$ are two different set theoretical objects and cannot be equal to each other.
In the comments you also asked whether the coset representatives of $T_G$ form a subgroup. That question does not quite make sense because coset representatives are not unique. Nonetheless it is true is that coset representatives of the torsion subgroup can be chosen so as to form a subgroup. For example, assuming that $$G = \mathbb Z^n \times (\mathbb Z_{i_1} \times \cdots \times \mathbb Z_{i_k}) $$ it follows that the subgroup $Z^n \times \{0\}$ consists of coset representatives of the torsion subgroup (where $0$ denotes the identity element of the torsion subgroup $\mathbb Z_{i_1} \times \cdots \times \mathbb Z_{i_k}$).
However, that subgroup $\mathbb Z^n \times \{0\}$ is not always characteristic.
One obtains a counterexample from the very simplest case of an infinite, finitely generated abelian group with nontrivial torsion, namely the group $$G = \mathbb Z \times \mathbb Z_2 $$ For elements of $\mathbb Z_2$ let me use the notation $[i] = i + 2 \mathbb Z \in \mathbb Z / 2 \mathbb Z = \mathbb Z_2$. Then the subgroup $\mathbb Z \times \{[0]\}$ is not characteristic because that subgroup is not taken to itself by the automorphism $\phi : \mathbb Z \times \mathbb Z_2 \to \mathbb Z \times \mathbb Z_2$ that is defined by the formula $$\phi(m,[n]) = (m,[n+m]) $$ Notice in particular: $\phi(1,[0])=(1,[1])$ has infinite order but is not contained in $\mathbb Z \times \{0\}$.