Is the set $\{0,1,2,3,4,5\}$, with the binary operation of "addition, then modulo $3$", a group?

Question

Consider a binary operation on the finite set $S=\{0,1,2,3,4,5\}$ where the operation is "addition, then modulo 3" ($+_3$). Does $S$ form a group under the binary operation $+_3$?

Now this binary operation follows closure property . It's associative. There exists identity element which is $0$. And for each element of the set $S$ there exists its inverse such that $aa^{-1}=0$.

It follows all requirements to be a group.

But it violates some other properties of group like left and right cancellation laws:

If $ax=ay\implies x=y$ $0+_32=0+_35\implies 2=5$

And equations $a+_3x=b$ and $y+_3a=b$ don't have unique solutions.

Answer 1

This is in fact, not a group. We can just check via brute force that it has no identity element:

$0$ is not the identity, since $3+_30=0\not=3$.

$1$ is not the identity, since $3+_31=1\not=3$.

$2$ is not the identity, since $3+_32=2\not=3$.

$3$ is not the identity, since $3+_33=0\not=3$.

$4$ is not the identity, since $3+_34=1\not=3$.

$5$ is not the identity, since $3+_35=2\not=3$.

Remember, for an element $e$ to be the identity, we would need that $x+_3e=x$ for every element $x$. But the above computations show that no such $e$ exists here.

Answer 2

If your operation is addition modulo $3$, then if elements too are from the modulo $3$ set; then is a group with $0\equiv \bmod 3$.

Then identity $=0$ and element $3$ doesn't exist in the set, and satisfies the requirements shown for $0$ in the selected answer.

But, here $0,3; 1,4; 2,5$ exist in the set.

So, the identity $\neq 0$.