# Does this set along with $\text{mod } n$ form group?

Does $$\left\{0, 1, 2\right\}$$ along with the operation of addition $$\text{mod } 6$$ form a group?

I have many practice questions like this, and I know I have to check closure, associativity, identity, and inverse to see if this forms a group.

I'm not completely clear on how to do this in the given context. Am I supposed to see if $$0,1,2$$ integers can be closed under addition in $$\text{mod } 6$$, are associative, have additive identity and inverses?

If so, how would I go about that for this? Is there any methods that would help a beginner understand this problem better?

• Well, what is $2+2\pmod 6$?
– lulu
Jan 25, 2022 at 1:17
• $1 + 2 = 3 \mod 6$ which is not in the set Jan 25, 2022 at 1:17
• So did I interpret the question correctly? Jan 25, 2022 at 1:18
• if you posed the question correctly then, yes. But it should be instantly clear to you that this set is not closed under addition $\pmod 6$.
– lulu
Jan 25, 2022 at 1:22
• @ayeayemaung $1+2 = mod 6 = 3$ and $3$ not element of set, thus it's not a group since its not closed under addition? Jan 25, 2022 at 1:23

What you are basically asking if the set $$\{\overline{0}, \overline{1}, \overline{2}\}$$ is a subgroup of $$\mathbb{Z}/6\mathbb{Z}$$, the integers mod $$6$$. Well, the set is not closed since $$\overline{2} + \overline{2} = \overline{4}$$ is not an element of it.
• so $2+2=4mod6 = 4$, which is not in the set {$0,1,2$}, so it's not closed under addition? Jan 25, 2022 at 15:43
• suppose the set was {$0,2,4$}, this would be closed under addition in $mod6$ right? Jan 25, 2022 at 15:45
• So for {$0,2,4$}, I would still need to make sure it has additive inverses for $0,2,4$ in the set, and check for the identity, which is just $0$? And once I do that, then I can say it's a subgroup? Jan 25, 2022 at 15:53