I've copied over this question from what I asked on Mathematics Stack Exchange, in the hope that some experts can provide some relevant insight.

Suppose we have row vectors $x_1$, $x_2$ , $y_1$ , $y_2 \in \mathbb{Z}^2$, consider the $4$ by $4$ matrix $\begin{bmatrix} x_1 & y_1 \\ -y_1 & x_1 \\ x_2 & y_2 \\ -y_2 & x_2 \end{bmatrix}$.

Let $M_1 = \begin{bmatrix} x_1 & y_1 & (0, 0) \\ -y_1 & x_1 & y_1 \\ x_2 & y_2 &(0, 0) \\ -y_2 & x_2 & y_2 \end{bmatrix}$ and $M_2 =\begin{bmatrix} x_1 & y_1 \\ -y_1 & x_1 \\ x_2 & y_2 \\ -y_2 & x_2 \\ (0, 0) &y_1 \\ (0, 0) &y_2 \end{bmatrix}$.

The last invariant factor of $M_i$ is defined to be the $\gcd$ of the determinants of all $4$ by $4$ minors of $M_i$.

**Question**:If the last invariant factor of $M_2$ is $1$, does this implies that the last invariant factor of $M_1$ is $1$?

My idea was to apply some elementary row and column operations on $M_1$ and $M_2$, call the resulting matrix $M_1'$ and $M_2'$, and show that all of the $4$ by $4$ minors in $M_2'$ can be found in $M_1'$. However, it is not obvious to me how to find $M_1'$ and $M_2'$.

There is also another potential idea, let $r_1, \ldots, r_4$ denotes the rows of $M_1$, and $r'_1, \ldots r'_6$ denote the rows of $M_2$, then an eqvaiblent statement to the question would be: $$ \text{ $\frac{\mathbb{Z}^4 }{ \langle r'_1, \ldots, r'_6 \rangle}$ is torsion free implies that $\frac{\mathbb{Z}^6 }{ \langle r_1, \ldots, r_4 \rangle}$ is torsion free. } $$

Thank you for reading, any idea for this would be really appreciated.