$\mathbb{Z}^{3}/\langle(b,6,0)\rangle$ according to the structure theorem for finitely generated abelian groups Let $b\in\mathbb{Z}-\{0\}$. I have to classify $\mathbb{Z}^{3}/\langle(b,6,0)\rangle$ according to the  Structure theorem for finitely generated abelian groups.
I think $\mathbb{Z}^{3}/\langle(b,6,0)\rangle\cong \mathbb{Z}^{2}\oplus \mathbb{Z}_{d}$, where $d=$g.c.d.$(6,b)$, but I do not know how to prove it. I have tried to find a surjective homomorphism $\phi:\mathbb{Z}^{3}\rightarrow \mathbb{Z}^{2}\oplus \mathbb{Z}_{d}$ such that $\ker\phi=\langle(b,6,0)\rangle$, but I have not been able.

You need the Smith Normal Form of $[b \ 6 \ 0]$, which is $[\gcd(b,6) \ 0 \ 0 ]$.
Thus the answer is $G \cong \mathbb{Z}^3/ \langle \gcd(b,6),0,0 \rangle=\mathbb{Z}_d \oplus \mathbb{Z}^2$.
