# Applying the fundamental theorem of finitely generated abelian groups to the group $\Bbb Z^3/((1,0,3),(-1,2,1))$

Consider two vectors $$v_1=(1,0,3), v_2=(-1,2,1)$$ in $$\Bbb Z^3$$. Let $$A$$ be the subgroup generated by $$v_1,v_2.$$ Then $$\Bbb Z^3/A$$ would be a finitely generated abelian group, so by the fundamental theorem of finitely generated abelian groups, $$\Bbb Z^3/A$$ can be expressed as a finite direct sum of $$\Bbb Z$$'s and $$\Bbb Z_m$$'s. Since $$v_1,v_2$$ is linearly independent, $$A$$ should be free of rank $$2$$. So $$\Bbb Z^3/A$$ should have rank $$1$$. But how should I find the torsion part?

Lets generalize a little. Let $$\vec{v_1}, \ldots, \vec{v_m}$$ be vectors in $$\mathbb{Z}^n$$. To compute the quotient group $$\mathbb{Z}^n / \langle \vec{v_1}, \ldots, \vec{v_m}\rangle$$, make the $$m \times n$$ matrix whose rows are the $$\vec{v_i}$$.
Now swapping two rows doesn't change the isomorphism class of the quotient; that just corresponds to relabeling the vectors $$\vec{v_i}$$. Similarly, you can multiply any row by $$\pm 1$$ without changing the isomorphism class of the quotient, and (most usefully) you can replace any row by itself plus an arbitrary scalar multiple of another; that is, the isomorphism class of the quotient is preserved by $$\mathbb{Z}$$-linear elementary row operations on the matrix.
Similarly, we can do change of basis operations in the ambient space $$\mathbb{Z}^n$$, which correspond to column operations on your matrix.
In your example, we have \begin{align*} \begin{pmatrix}1 & 0 & 3 \\ -1 & 2 & 1 \end{pmatrix} & \sim \begin{pmatrix}1 & 0 & 3 \\ 0 & 2 & 4 \end{pmatrix} \\ &\sim \begin{pmatrix}1 & 0 & 0 \\ 0 & 2 & 0 \end{pmatrix}, \end{align*} using a row operation at step one and two column operations at step two. We see the quotient is isomorphic to $$\frac{\mathbb{Z}}{\mathbb{Z}} \oplus \frac{\mathbb{Z}}{2\mathbb{Z}} \oplus \frac{\mathbb{Z}}{0\mathbb{Z}} = \mathbb{Z} \oplus \frac{\mathbb{Z}}{2\mathbb{Z}}$$
• (for an explicit example of a two-torsion element in the quotient, note that $2*(0, 1, 2) \in A$ but $(0, 1, 2) \not \in A$). – hunter Apr 30 '20 at 4:18