# $\mathbb{Z}^{3}/\langle(b,6,0)\rangle$ 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.

In $\def\Z{\Bbb Z}\Z^n$, every saturated subgroup (intersection with a subspace of $\Bbb Q^n$) is a direct factor (there is a complementary subgroup, forming a direct sum with it). This means that for the quotient you get a free factor $\Z^2$, and possibly a torsion factor from the minimal saturated subgroup containing $(b,6,0)$. If $d=\gcd(b,6)$ then $(b/d,6/d,0)$ spans that saturated subgroup, and one gets a torsion factor isomorphic to $\Z/d\Z$ (so the torsion is trivial if $d=1$). All in all $\Z^3/\langle(b,6,0)\rangle\cong \Z^2\oplus(\Z/d\Z)$, as you guessed.
You need the Smith Normal Form of $[b \ 6 \ 0]$, which is $[\gcd(b,6) \ 0 \ 0 ]$. Thus the answer is $G \cong \mathbb{Z}^3/ \langle \gcd(b,6),0,0 \rangle=\mathbb{Z}_d \oplus \mathbb{Z}^2$.