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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.

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  • $\begingroup$ 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. $\endgroup$ – JMoravitz Feb 19 '20 at 17:36
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    $\begingroup$ Welcome to math SE. Have a look at mathjax for your mathematical expressions. $\endgroup$ – Alain Remillard Feb 19 '20 at 17:37
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    $\begingroup$ 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... $\endgroup$ – JMoravitz Feb 19 '20 at 17:37
  • $\begingroup$ 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. $\endgroup$ – Shatabdi Sinha Feb 19 '20 at 17:40
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    $\begingroup$ After you ask a question here, if you get an acceptable answer, you should "accept" the answer by clicking the check mark $\checkmark$ next to it. This scores points for you and for the person who answered your question. You can find out more about accepting answers here: How do I accept an answer?, Why should we accept answers?, What should I do if someone answers my question?. $\endgroup$ – Shaun Feb 19 '20 at 18:46
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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, $\langle 2 \rangle = \{0, 2\} \leq \{0, 1, 2, 3\} = \mathbb Z_4$.

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  • $\begingroup$ Good approach to teach! $\endgroup$ – Mikasa Mar 20 '20 at 8:36
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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 $[-2]_4=[2]_4$.

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

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