Let $a, b, c, d, e, f, g$ denote integers, with $a$ and $c$ both non-zero, and consider the matrix

$$ M=\left(\begin{matrix} a&b&d&f \\ 0&c&e&g \end{matrix}\right) $$

Let $G$ be the quotient of the group $\Bbb Z^2$ by the column space of $M$. That is, $G$ is presented by two generators, say $x,y$, which commute, and which satisfy the four relations $ax=0$, $bx+cy=0$, $dx+ey=0$, $fx+gy=0$.

What is the order of $x$ in $G$, in terms of $a, b, c, d, e, f, g$?

This question is a challenge because I have an answer (in fact, I have two approaches that yield the same answer). The point of asking is that this problem has a huge teaching/learning potential, and to see whether there could be yet another approach.

Hint: the problem is elementary if you remove the last two columns. There is a twist if you add $d\choose e$, and yet quite another twist if you add the last column. Then, adding more columns would not make the problem more interesting. In fact, the problem given with a $2\times 4$ matrix as above is essentially as difficult as the same problem given with an arbitrary matrix.

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    $\begingroup$ If you are looking for a third approach, it would help us help you if we had any idea what the first two are. $\endgroup$
    – Arthur
    Jul 31, 2019 at 14:02
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    $\begingroup$ @Arthur The main point of my question is to let you have fun, because it was a lot of fun to me. Whether there is another approach is of course interesting to me, but I'm in no rush to find out. $\endgroup$ Jul 31, 2019 at 14:05
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    $\begingroup$ So it's posed as a challenge / puzzle. Fair enough. I just read it as you mainly looking for the third approach, hence my comment. $\endgroup$
    – Arthur
    Jul 31, 2019 at 14:08
  • $\begingroup$ @Arthur No worries. I'll add a spoiler box with my solutions later on if noone finds anything. $\endgroup$ Jul 31, 2019 at 14:15


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