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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?

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

1 Answer 1

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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.

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  • $\begingroup$ so $2+2=4mod6 = 4$, which is not in the set {$0,1,2$}, so it's not closed under addition? $\endgroup$
    – eddie
    Jan 25, 2022 at 15:43
  • $\begingroup$ suppose the set was {$0,2,4$}, this would be closed under addition in $mod6$ right? $\endgroup$
    – eddie
    Jan 25, 2022 at 15:45
  • $\begingroup$ Yes this is a subgroup indeed! $\endgroup$ Jan 25, 2022 at 15:46
  • $\begingroup$ 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? $\endgroup$
    – eddie
    Jan 25, 2022 at 15:53
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    $\begingroup$ That makes everything much clearer, thanks so much!! :) $\endgroup$
    – eddie
    Jan 25, 2022 at 16:15

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