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I ask this question because, I see from the textbook the claim that "All subgroups of the additive group $\mathbb{Z}$ is in the form of $H = n \mathbb{Z}$ for $n = 0, 1, 2, ...$", however, $H = \{-1, 0, 1\}$ is a subset of $\mathbb{Z}$ and it also satisfies the axioms of groups and of subgroups but it is not in the form of $H = n \mathbb{Z}$ for $n = 0, 1, 2, ...$.

Therefore, could anyone help me explain this thing? Maybe I got something wrong and I want to know why I am wrong here. Thanks!

Shaun
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ZYX
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    $1\in H$ but $1+1\notin H$. If $H={0}$ is a subgroup. If $H$ is a subgroup that contains a non-zero element $a$, then it also contain $-a$. Let $n$ be the smallest positive in $H$. If $b\in H$, then $b=nq+r$ with $0\leq r<n$. Then $r=b-nq\in H$. Since $n$ was the smallest positive element of $H$, it follows that $r=0$. – plop Sep 20 '22 at 23:40
  • So we must have $na \in H$ $(n = 0, 1, 2, ...)$ if $a \in H$? – ZYX Sep 20 '22 at 23:44
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    Yes, adding (the operation of the group in this case) an element to itself (or any other of the subgroup) should still be an element of the subgroup. – plop Sep 20 '22 at 23:46
  • Thank you! I now understand it. – ZYX Sep 20 '22 at 23:47

2 Answers2

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Your example $H = \{-1, 0, 1\}$ is not a subgroup of $\mathbb{Z}$ as 1 $\in H$ but 1 + 1 = $2 \notin H$. Newcomers often overlook this fact. In a group, the binary operation (here given by $+$) must be a map from the direct product of the group with itself to the group itself. But the addition on $\mathbb{Z}$ restricted to your subset $H$ will not result in a map \begin{align*} +_{H \times H} : H \times H \to H \end{align*}, but rather just a map $H \times H \to \mathbb{Z}$. One would often say here that H is not closed under addition in technical terms.

For every $n \in \mathbb{N}, n\mathbb{Z}$ is indeed a subgroup of $\mathbb{Z}$, as for each $a, b \in n\mathbb{Z}$, you have that \begin{align*} a + b \in n\mathbb{Z} \\ -a \in n\mathbb{Z}\\ \text{and the neutral element} \quad 0 \in n\mathbb{Z} \end{align*}

I believe this already answers the question, but if you are interested to know why the $n\mathbb{Z}$ are the only subgroups of $\mathbb{Z}$, @user85667 gave a short explanation. If you have a subgroup $H$ of $\mathbb{Z}$, then there is an integer with smallest absolute value $n \in \mathbb{N}$ in $H$. Then $H$ must be equal to $n\mathbb{Z}$. Because if there is an element in $H$ that is not a multiple of $n$, then you could divide this element by $n$ and the remainder would be less than $n$. Since $H$ is a group, the remainder would be in $H$, contradicting the fact that $n$ was the element with smallest absolute value in $H$. A generalization of this proof shows why Euclidian Domains are Principal Ideal Domains, you will probably learn about these terms later once you get to Ring Theory.

fatboy892
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Each subgroup of a cyclic group is cyclic. (Why?) We have that $\Bbb Z$ is cyclic; therefore, $H\le \Bbb Z$ must be cyclic, so it can be written as $H=\langle n\rangle$ for some $n\in\Bbb N\cup\{0\}$. (Why?) But then $H=n\langle 1\rangle=n\Bbb Z$.

Note that $\{-1,0,1\}$ is not closed under addition: $1+1=2\notin \{-1,0,1\}$; therefore, it is not a group and so cannot be a subgroup.

Shaun
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