# Why is $\{0, 2\}$ a subgroup of $\mathbb Z_4$?

I feel like this should be obvious but why is $$\{0, 2\}$$ a subgroup of $$\mathbb Z_4$$? So, $$\langle 2\rangle=\{0,2\}$$. Shouldn't this set contain the inverse ($$-2$$)? Or does it have to do with the fact that $$(-2)(-2)=4=0$$? Please advise.

• What is the difference between $2$ and "$-2$"? Recall, the minus sign merely means "the additive inverse of" which does not necessarily need to appear in other representations of the element. – JMoravitz Feb 19 '20 at 17:36
• Welcome to math SE. Have a look at mathjax for your mathematical expressions. – Alain Remillard Feb 19 '20 at 17:37
• As for "does it have to do with the fact that $(-2)(-2)=4=0$" No, it doesn't. It has to do with the fact that $2 + 2 = 0$. Addition is what is important here, not multiplication. You will find that the additive inverse of $0$ is $0$, the additive inverse of $1$ (which you might when convenient decide to notate as $-1$) is equal to $3$, the additive inverse of $2$ (which you might when convenient decide to notate as $-2$) is equal to $2$, and so on... – JMoravitz Feb 19 '20 at 17:37
• There's a problem with the title of the question. 2 is not a subgroup of $\mathbb Z_4$, it's an element. I guess you mean the subgroup generated by 2. – Shatabdi Sinha Feb 19 '20 at 17:40
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$$\langle 2\rangle = \{0, 2\}$$ is a subgroup of the group $$\mathbb Z_4 = \{0, 1, 2, 3\}$$ under modular arithmetic, modulo $$4$$.

The identity of this group is $$0$$, and because $$2+2 \equiv 0 \pmod 4$$, it has order two, and hence $$2$$ generates a group (subgroup) of order 2. In fact, the additive inverse of $$2$$ is $$2$$.

That is, $$\langle 2 \rangle = \{0, 2\} \leq \{0, 1, 2, 3\} = \mathbb Z_4$$.

• Good approach to teach! – Mikasa Mar 20 '20 at 8:36

The elements of $$\Bbb Z_4$$ are not technically $$0$$, $$1$$, $$2$$ and $$3$$; rather, they are equivalence classes of integers with respect to the divisibility of their differences by $$4$$, like so: $$[a]_4:=\{b\in\Bbb Z: 4\mid a-b\}.$$ The operation of the group is defined by $$[x]_4+_4[y]_4=[x+y]_4$$.

Thus, since $$4\mid (-2)+2=0$$, we have $$[-2]_4=[2]_4$$.

• Why the downvote? – Shaun Feb 19 '20 at 19:00
• No, it isn't, @amWhy. – Shaun Mar 11 '20 at 16:58
• But $-2\equiv 2\pmod{4}$. – Shaun Mar 11 '20 at 17:02