The functor $T:FAb\to FAb$ from finitely generated abelian groups to finitely generated abelian groups taking a group to its torsion subgroup is left exact.
Suppose there is a functor $G:FAb\to FAb$ which is a left adjoint to $T$, so that for all f.g. abelian groups $M$ and $N$ we have $\hom(GM,N)=\hom(M,TN)$. In particular taking $M=\mathbb Z$, we see that if we set $X=G\mathbb Z$, there is a natural isomorphism $TN=\hom(X,N)$.
There is no f.g. abelian group $X$ with this property.
Indeed, suppose there is such an $X$ and that $X=\mathbb Z^l\oplus F$ with $F$ a finite abelian group. If $p$ is a prime number which does not divide the order of $F$, then $\hom(X,\mathbb Z/p\mathbb Z)=\hom(\mathbb Z^l,\mathbb Z/p\mathbb Z)$ has $p^l$ elements, and has to be isomorphic to $\mathbb Z/p\mathbb Z$: it follows that $l=1$. Now, if $N$ is an infinite f.g. abelian group, we have now that $\hom(X,N)$ is infinite, while the torsion subgroup $TN$ is necessarily finite. This is absurd.
The functor $T:FAb\to FAb$ from finitely generated abelian groups to finitely generated abelian groups taking a group to its torsion subgroup is left exact.
Suppose there is a functor $G:FAb\to FAb$ which is a left adjoint to $T$, so that for all f.g. abelian groups $M$ and $N$ we have $\hom(GM,N)=\hom(M,TN)$. In particular taking $M=\mathbb Z$, we see that if we set $X=G\mathbb Z$, there is a natural isomorphism $TN=\hom(X,N)$.
There is no f.g. abelian group $X$ with this property.
Indeed, suppose there is such an $X$ and that $X=\mathbb Z^l\oplus F$ with $F$ a finite abelian group. If $p$ is a prime number which does not divide the order of $F$, then $\hom(X,\mathbb Z/p\mathbb Z)=\hom(\mathbb Z^l,\mathbb Z/p\mathbb Z)$ has $p^l$ elements, and has to be isomorphic to $\mathbb Z/p\mathbb Z$: it follows that $l=1$. Now, if $N$ is an infinite f.g. abelian group, we have now that $\hom(X,N)$ is infinite, while the torsion subgroup $TN$ is necessarily finite. This is absurd.