Proof that $n\Bbb Z \leq \Bbb Z$ and are the only subgroups of $\Bbb Z$

Question

My challenge is

Prove that if $n = 0,1,2,\ldots$ and $n\Bbb Z = \lbrace nk: k \in \Bbb Z \rbrace$, show that $n\Bbb Z$ is a subgroup of $\Bbb Z$ and are the only subgroups.

I handled the first point by saying that

i) Identity is 0, and $0\in n\Bbb Z$ and $0 \in \Bbb Z$

ii) For any $a,b \in n\Bbb Z$ $a+b = na_2 + nb_2 = n(a_2+b_2) \in n \Bbb Z$

iii) For any $na \in n\Bbb Z$, its inverse $n(-a) \in n\Bbb Z$.

For the second point, I said that all of the subgroups should be infinite, because for any finite group the largest element $a+a \notin H$, where $H$ is the potential subgroup, which is a contradiction. So $H$ must be of infinite order. For any $a \in H, -a$ should also be in $H$. The difference between any $a,b \in H$ should also be in $H$, i.e. $(a-b) \in H$. This leaves only the whole group or any group of the form $n\Bbb Z$.

Is my proof complete?

How do you know it leaves only groups of the form $n\Bbb Z$ (of which $\Bbb Z$ is an example I might add)? — anon, May 27 '15 at 19:24
@quid case n = 0 is the identity. That's a trivial subgroup. — SalmonKiller, May 27 '15 at 19:26
Yes, what I meant is that "the largest element" type argument only works if there is an element other than $0$. And the group nZ for n=0 is also not infinite. So what you say "all subgroups are infinite" is false. It is basically right, but you need to treat that special case n=0. — quid, May 27 '15 at 19:27
Your usage of "proper" is unusual (as normally it will mean not the full group) But I get what you mean. As said it was only a tangential remark. But still you should say this somewhere. — quid, May 27 '15 at 19:31
@quid Proper means not the identity and not the full subgroups. — SalmonKiller, May 27 '15 at 19:31
At least MathWorld http://mathworld.wolfram.com/ProperSubgroup.html and Wikipedia http://en.wikipedia.org/wiki/Subgroup agree with me. — quid, May 27 '15 at 19:33

score 7 · Answer 1 · answered May 27 '15 at 19:27

7

Hint: A simple way to tackle this is by considering $$n = \min \{x \in H ; x> 0\}$$

where $H \leq \mathbb Z$.

The inclusion $H \supseteq n\mathbb Z$ is clear.

For the other one take any $h \in H$ and use the division algorithm and the fact that $n$ is minimum.

answered May 27 '15 at 19:27

Aaron Maroja

17,571

1

Using a similar idea, could you characterize all subgroups of $\mathbb R$? – Aaron Maroja May 27 '15 at 19:33
2

No, but one can prove that a subgroup of $\mathbb{R}$ is either of the form $a\mathbb{R}$, with $a\ge0$ or it is dense. The set of subgroups of $\mathbb{R}$ is very big. – egreg May 27 '15 at 20:14
1

I mentioned this in the comments of the answer above. Unless my english is incorrect, that is what I meant by characterize. – Aaron Maroja May 27 '15 at 20:17
In case someone needs all the details https://proofwiki.org/wiki/Subgroups_of_Additive_Group_of_Integers :-) – the gods from engineering Mar 30 '21 at 15:12

Cameron Williams · Accepted Answer · 2015-05-27T19:26:22.337

5

You're almost there but not quite. You've shown that any subgroup of $\Bbb Z$ has a copy of $n\Bbb Z$. The way you want to go about it is to consider the least positive element in the subgroup $H$ and show that element generates $H$. (You'll end up doing a proof by contradiction.)

edited May 27 '15 at 19:26

answered May 27 '15 at 19:24

Cameron Williams

29,432

What do you mean by generates $H$? – SalmonKiller May 27 '15 at 19:25
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@SalmonKiller Generates means that every element is a multiple (or power depending on how you want to view it) of that specific element. For instance, $n\Bbb Z$ is generated by $n$ or $-n$. – Cameron Williams May 27 '15 at 19:26
So the argument will go like if $n$ does not divide an element $v \in H$, then the difference between them is not in $H$? – SalmonKiller May 27 '15 at 19:32
@SalmonKiller You're on the right track. If you use the division algorithm, then what can you say about the remainder? How does this give a contradiction? – Cameron Williams May 27 '15 at 19:33
Can you please edit your question with the division algorithm argument, cause I am not quite catching it. – SalmonKiller May 27 '15 at 19:36
You can write $v = kn + r$. If $v\in H$ but $v$ isn't a multiple of $r$, note that $v - kn = r\in H$, right? Can you take it from there? – Cameron Williams May 27 '15 at 19:38
Ah. I see it know. So because $r < n$, it's a contradiction. – SalmonKiller May 27 '15 at 19:39
@SalmonKiller And $r \in H$. – Aaron Maroja May 27 '15 at 19:42
@SalmonKiller You're exactly right. – Cameron Williams May 27 '15 at 19:42
@AaronMaroja so $r$ can't be in $H$. – SalmonKiller May 27 '15 at 19:42
@SalmonKiller So $r = 0$. – Aaron Maroja May 27 '15 at 19:43
@AaronMaroja Yeah. – SalmonKiller May 27 '15 at 19:46
@SalmonKiller an excelent exercise is to show that all subgroups of $\mathbb R$ are either $n \mathbb Z$ or dense in $\mathbb R$. – Aaron Maroja May 27 '15 at 19:47
@AaronMaroja I haven't studied topology enough yet, so I have no idea what dense sets are. :) – SalmonKiller May 27 '15 at 19:48
@SalmonKiller I see. Just put a pin in it then. It's rewarding. – Aaron Maroja May 27 '15 at 19:52
1

@AaronMaroja Certainly. – SalmonKiller May 27 '15 at 19:53

score 0 · Answer 3 · answered Sep 29 '23 at 17:02

To write out your question completely: Prove that the subgroups of $\mathbb{Z}$ under the addition binary operation are precisely the groups $n\mathbb{Z}$ under addition.

Let H be a subgroup of $\mathbb{Z}$

(Note: because the binary operation is addition (and $H$ is a group), this means that if $a\in H$ then $-a \in H$. This means that if there is a non zero element in $H$, then there will always be a positive element in $H$)

Let $n$ be the least positive integer (i.e. an integer $ n > 0$) such that $n \in H$, the due to the addition binary operation on the group $H$, $n+n = 2n\in H$ and $n+n+n = 3n \in H$ and generally $n +n+...+n =tn$.
(Note: same applies for $-n+-n = -2n\in H$ and $-n+-n+-n = -3n \in H$ etc...)

This shows us that $H$ contains all the multiples of $n$.

Now let $p \in n\mathbb{Z}$, thus we can find some integer $q$ such that $p = qn$. Because $H$ contains all the multiples of $n$, this means that $p = qn \in H$.

Hence this shows that $n\mathbb{Z} \subseteq H$

Now for the reverse direction.

Let $k \in H$. Then by division algorithm, $k = nq +r$ where $n$ is the least positive integer in $H$ and $0\leq r < n$. As shown above, $nq \in H$. This means $-nq \in H$ due to the inverse property of groups. Thus $k - nq = r \in H$. But $n$ is the least positive integer in $H$, so if $0<r < n$, the $r$ would be the least positive integer in $H$ and hence a contradiction to "$n$ is the least positive integer in $H$". Thus the only other option is that $r = 0$. Thus $k = nq +r = nq \in n\mathbb{Z}$. Thus for all $k \in H$ we get $k \in n\mathbb{Z}$. Thus $H \subseteq n\mathbb{Z}$

Thus because $n\mathbb{Z} \subseteq H$ and $H \subseteq n\mathbb{Z}$, we can finally conclude that $H = n\mathbb{Z}$ where $H$ is an arbitary subset of $\mathbb{Z}$

Proof that $n\Bbb Z \leq \Bbb Z$ and are the only subgroups of $\Bbb Z$

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