Check whether $G$ is group or not

Question

Let $G=\{0,1,2\}$ define $*$ on $G$ such that $a*b=|a-b|$. Check if $G$ is a group.

Edit : As @egreg correctly pointed out if you draw the Cayley table then one can directly see it is not a group, I was making a mistake in drawing up Cayley table.

My solution : I can prove that $G$ is closed, has an identity ($0$), has an inverse (each element is inverse of itself).

What I am not sure is whether the group is associative under this binary operation, at the face of it $|a-|b-c||\neq||a-b|-c||$ but if we check all possibilities then both do come out to be equal if $a,b,c \in G$.

Any hint/help will be appreciated

Answer 1

The Cayley table (that also shows closure, identity and inverses) is \begin{array}{c|ccc} * & 0 & 1 & 2 \\ \hline 0 & 0 & 1 & 2 \\ 1 & 1 & 0 & 1\\ 2 & 2 & 1 & 0 \end{array} Since in the second row the element $1$ appears twice, the set is not a group.

In a group, from $xy=xz$ one deduces $y=z$; here, instead, $1*0=1*2$, but $0\ne2$.

Answer 2

The operation $*$ is not associative, since:\begin{align}\bigl||2-1|-1\bigr|&=0\\&\neq2\\&=\bigl|2-|1-1|\bigr|\end{align}and therefore, $(G,*)$ is not a group.