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

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.

• How is $0$ the identity element? If I add $3$ with $0$ then mod by $3$ dont I get $0\not=3$? Commented Apr 26, 2021 at 4:00
• If it violates the cancellation laws then it must violate at least one of the group axioms. It can't be a group. You need to look at the proof of the cancellation laws and find where that happens. You are very close with the last line. What is $b-a=b+a^{-1}$? Commented Jun 20, 2022 at 3:10

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.

• In my book there is a question that give an example of a semi group with identity element in which cancellation law doesnt hold and answer is $(\{0,1,2,3,4,5\},+_6)$. Is it correct ? I thing it's wrong because its group not semigroup.
– RKK
Commented Apr 26, 2021 at 16:10

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 $$\neq 0$$.