finding a four elements group and binary operation upon it that has closure, associativity, an identity element and an inverse to each element.

Question

the question is in the heading. I tried using the group {0,1,2,3} and the binary operation $*$ defined as $a*b=|a-b|$ which I proved to have closure, an identity and an inverse, but am unable to prove associativity; $(a*b)*c=|a-b|*c=||a-b|-c|$ and $a*(b*c)=a*|b-c|=|a-|b-c||$. I cant seem to show they are equal (maybe I could do it by separating into cases, $x>y$, $x<y$, $x=y$ so that $x,y\in \{0,1,2,3\}$, but there seems to be a lot of cases to check, and I'm sure there's an easier solution).

Answer 1

If you have a $2$-element group, you can take the Cartesian product of it with itself to get a four-element group. This gives rise to a group called the Klein four-group, which is (up to isomorphism) the only group with four elements other than Hank Igoe’s sensible and obvious suggestion in the comments, $\Bbb{Z}/4\Bbb{Z}$. Hank’s is probably better if you insist on labeling the elements of your group as “$0$”, “$1$”, “$2$”, “$3$”.

The number of elements is small enough that you could verify by hand if your operation gave rise to a group. I think that your operation does not, because it lacks associativity. If it had associativity, I should be able to find an element $b$ so that $1 \star b = 3$ by left-starring by the inverse of $1$. But clearly there’s no element in $\{ 0, 1, 2, 3 \}$ at a distance of $3$ from $1$.