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Does $0$ have an additive inverse in $Z_6$ of itself?

I have the set {$0,2,4$}, and I want to check if it satisfies the property of having inverses under the operation of addition in mod $6$.

So, I need to make sure $0,2,4$ all have an inverse within the set.

$2+4 = 0$ mod$6$, so the additive inverse of $2$ is $4$.

$4+2 = 0$ mod$6$, so additive inverse of $4$ is $2$.

Now for $0$, I'm wondering if this is true and works to satisfy the property:

$0+0 = 0 = 0$ mod$6$, so additive inverse of $0$ is $0$. Is this correct, or does $0$ not have an inverse element within the set here?

Also how would I go about showing that this set is associative under addition mod$6$?

Would it be $((a+b)+c)$ mod$6$ = $(a+(b+c))$ mod$6$, because addition of integers is associative? So, since $(a+b)+c = a+(b+c) = m$, some $m$.

$((a+b)+c)$ mod$6$ = $m$ mod$6$ = $(a+(b+c))$ mod$6$.

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  • $\begingroup$ No, here, the neutral element is $0$, and it must be its own inverse. $\endgroup$
    – Lubin
    Jan 28, 2022 at 3:18
  • $\begingroup$ @Lubin so is my argument for $0$ having an additive inverse of $0$ correct? $\endgroup$
    – eddie
    Jan 28, 2022 at 3:19
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    $\begingroup$ In any group, the additive identify is its own inverse. $\endgroup$ Jan 28, 2022 at 3:29
  • $\begingroup$ And to prove it is unique, let $x$ be another element such that $x+0=0$. Then, since $0$ is the additive identity, $x+0=x$ so $x=0$. $\endgroup$
    – John Douma
    Jan 28, 2022 at 3:32

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The additive identity is always its own additive inverse, so you are correct. Moreover, there's no reason to think that an element cannot be its own additive inverse. For example, $4\in\Bbb Z_8$ is its own additive inverse, since $4+4=0$.

Now, for some $k,m\in\Bbb Z$ with $0\leq m<6$, we can write $$(a+b)+c=6k+m=a+(b+c).$$

Therefore,

$$((a+b)+c)\bmod 6=m=(a+(b+c))\bmod 6.$$

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  • $\begingroup$ In fact, $x$ being its own additive inverse is equivalent to $2x=0$. $\endgroup$
    – Kenta S
    Jan 28, 2022 at 15:33
  • $\begingroup$ why is it equal to $6k+m$? and how does the next step follow from there where $m$ is isolated? $\endgroup$
    – eddie
    Jan 28, 2022 at 16:13
  • $\begingroup$ We can always write a number as a multiple of $6$ plus some remainder. For example, $17=6(2)+5$ and $21=6(3)+3$. This is known as the quotient-remainder theorem. $\endgroup$
    – Bonnaduck
    Jan 28, 2022 at 16:16
  • $\begingroup$ In general, $x=yk+m\iff x\bmod y=m$ for some $k\in\Bbb Z$ and $0\leq m<y$. $\endgroup$
    – Bonnaduck
    Jan 28, 2022 at 16:16

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