I was wondering if anyone would pick up on that. Kudos to you!
Here is my explanation of what’s going on here: you’re looking at the different cosets of the submodule of the free module Z3 spanned by the column vectors
A=⎡⎢⎣−1111−1111−1⎤⎥⎦
You can figure out the structure of the quotient systematically by looking at the Smith normal form of this matrix, and you can work that out by taking the greatest common divisor of the determinants of its minors. In this case we get d1d2d3=detA=(4), d1d2=(2) and d1=1, so we can deduce that the Smith normal form is diag(1,2,2) and the quotient is isomorphic to the Klein-four group V4. This explains why a map to V4 is the right invariant for this problem.
I was wondering if anyone would pick up on that. Kudos to you!
Here is my explanation of what’s going on here: you’re looking at the different cosets of the submodule of the free module Z3 spanned by the column vectors
A=⎡⎢⎣−1111−1111−1⎤⎥⎦
You can figure out the structure of the quotient systematically by looking at the Smith normal form of this matrix, and you can work that out by taking the greatest common divisor of the determinants of its minors. In this case we get d1d2d3=detA=(4), d1d2=(2) and d1=1, so we can deduce that the Smith normal form is diag(1,2,2) and the quotient is isomorphic to the Klein-four group V4. This explains why a map to V4 is the right invariant for this problem.