In Sehgal's book, (Topics in group rings-page 61) while finding units of $\Bbb{Z}S3$ , I encountered these congruence relations given in the pic below.
My question is how did he reduce these six congruence relations into those three (6.3). By making matrix of coefficients and row reducing gives Identity matrix. Can some one please explain how to simplify these six relations as in the pic.
Thanks!
When you did rowops, I am sure you divided by $2$ and $3$ along the way. This is not valid in $\bmod6$ because they are factors of $6$. For example, $4=10\bmod6$; but divide by $2$ and you get $2=5\bmod6$, which is wrong.
If the six equations are correct $\bmod2$ and $\bmod3$, then they are correct $\bmod6$.
Simply look at all six equations $\bmod2$, you find they are $x1+x2\equiv0$ or $x1-x2\equiv0$, because all other terms are even. But $x1+x2$ is even whenever $x1-x2$ is even, so these are the same equation $\bmod2$.
Add equations D and E, you get $2x1-2x3+4x5$ is a multiple of 6. So $x1-x3+2x5$ is a multiple of $3$ (not 6), so $x1-x3-x5$ is a multiple of 3, so $x1\equiv x3+x5\bmod 3$.
Try to do similar arithmetic to prove the other equations.