If $G=A\times B$, where $A$- torsion group, $B$-free abelian group, how to show that $A=G_{tor}$

Question

$G$ is a finitely generated Abelian group with $G=A\times B$ where $A$ is a torsion group and $B$ a free Abelian group.

I know that every finitely generated Abelian group can be factored into cyclic groups as follows $G=\mathbb{Z}_{a1}\times \mathbb{Z}_{a2}\times ... \times \mathbb{Z}^d$ where $a_1$ divides $a_2$ and so on.

Am I correct in saying that each $\mathbb{Z}_{aj}$ is a torsion group , $\mathbb{Z}^{d}$ is a free Abelian group and since the product of torsion groups is a torsion group, the group $A$ in the product corresponds to $\mathbb{Z_{a1}}\times \mathbb{Z_{a2}}\times..$? So to show that $A=G_{tor}$ where $G_{tor}$ is the torsion subgroup of G, can I say that $G_{tor}$ has two components, the first of which are elements of $A$ and the second is $0$ and hence $G_{tor} = A$? Any hints are appreciated to make me understand more concretely. Thanks.