Is an abelian group $G$ always isomorphic to the direct sum of its torsion group $T$ with a free group?
Alternatively, is the quotient $G/T$ always free?
By definition, $T$ consists of all elements in $G$ having finite order. I can see that this is true when $G$ is finitely generated, but I believe it is false in general.
No. For example, consider $\mathbb{R}\setminus\{0\}$ under multiplication. The only non-trivial torsion element is $-1$, and quotienting this out you obtain the positive reals under multiplication, $\mathbb{R}_{>0}\setminus\{0\}$. Your task is then to prove that $\mathbb{R}_{>0}\setminus\{0\}$ is not free abelian. Which it isn't.