Additive Inverse in $\mathbb{Z}$ mod $6$ for $0$

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$$6$$.

• No, here, the neutral element is $0$, and it must be its own inverse. Jan 28, 2022 at 3:18
• @Lubin so is my argument for $0$ having an additive inverse of $0$ correct? Jan 28, 2022 at 3:19
• In any group, the additive identify is its own inverse. Jan 28, 2022 at 3:29
• 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 $x=0$. Jan 28, 2022 at 3:32

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,

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

• In fact, $x$ being its own additive inverse is equivalent to $2x=0$. Jan 28, 2022 at 15:33
• why is it equal to $6k+m$? and how does the next step follow from there where $m$ is isolated? Jan 28, 2022 at 16:13
• 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. Jan 28, 2022 at 16:16
• In general, $x=yk+m\iff x\bmod y=m$ for some $k\in\Bbb Z$ and $0\leq m<y$. Jan 28, 2022 at 16:16