11

Is $\mathbb{Z}^2$ cyclic? What does it mean for a group to be cyclic? Is it just that it has one generator?

Thanks

azimut
  • 22,696
user51327
  • 451
  • 3
  • 6
  • 12
  • Sorry for the misfire: yes, a cyclic group has exactly one generator. No $\mathbb Z^2$ is not cyclic. – amWhy Mar 03 '13 at 21:01
  • 5
    A cyclic group need not have exactly one generator. Any group of prime order is cyclic and every element, other than the identity, is a generator. Take $\mathbb{Z}_p$ under addition. – J.H. Mar 03 '13 at 21:09
  • 3
    If $G$ is a cyclic group of order $n$ then $G$ has $\phi(n)$ generators. In particular $\mathbb Z_2$ and $\mathbb Z_1$ are the only groups with exactly one generator. – JSchlather Mar 03 '13 at 21:13
  • 4
    A cyclic group is a group which can be geenrated by a single element. In general, there are more than one such generator. If $G$ is infinite cyclic, then $G\simeq \mathbb{Z}$ is generated by $1$ and $-1$. And if $G$ is finite cyclic, see JSchlater's comment. – Julien Mar 03 '13 at 21:32
  • 1
    What you mean to say is that cyclic groups have a generator, not one generator. To see this easily note that if $x$ generates a group, so does $x^{-1}$. – Alexander Gruber Mar 03 '13 at 21:59
  • 6
    I think this might be a linguistic problem rather than a mathematical one! Perhaps the people who said that a cyclic groups has exactly one generator intended this to mean that it can be generated by exactly one element. – Derek Holt Mar 03 '13 at 22:21
  • 2
    @Derek Yes I agree. – Alexander Gruber Mar 04 '13 at 18:39

6 Answers6

12

No, it isn't.

For all $a,b\in\mathbb{Z}$ it holds that $\langle (a,b)\rangle = \{(ka,kb) \mid k\in\mathbb{Z}\} \neq \mathbb{Z}^2$. So $\mathbb{Z}^2$ is not generated by a single generator and hence not cyclic.

EDIT

To address the comments "There is no argument, you just rephrase the claim." and "Nothing is wrong, but in fact this is because nothing is there." by Martin Brandenburg, I add details. By the way, I don't agree with these comments, since point 1. below is basic knowledge and 2. is quite obvious.

1. If $(G,\cdot)$ is a group and $g\in G$, then $\langle g\rangle = \{g^k \mid k\in \mathbb Z\}$.

By definition, $\langle g\rangle$ is the intersection of all subgroups of $G$ containing $g$. It is straightforward to check that $\{g^k \mid k\in \mathbb Z\}$ is a subgroup of $G$, showing "$\subseteq$". As the intersection of subgroups, $\langle g\rangle$ is a subgroup. Since any subgroup containing $g$ must contain all the $g^k$, too, we get "$\supseteq$".

Application to the additively written group $\mathbb Z^2$ yields $\langle (a,b)\rangle = \{k\cdot (a,b) \mid k\in\mathbb Z\} = \{(ka,kb) \mid k\in\mathbb Z\}$.

2. $\{(ka,kb) \mid k\in\mathbb Z\} \neq\mathbb Z^2$

If $a = 0$, then $(1,0)\notin \{(ka,kb) \mid k\in\mathbb Z\}$ and otherwise $(a,b+1)\notin \{(ka,kb) \mid k\in\mathbb Z\}$.

azimut
  • 22,696
  • 3
    Your "argument" is that it is not cyclic because no element generates the group ... – Martin Brandenburg Apr 22 '13 at 19:49
  • 4
    @MartinBrandenbug: What exactly is the point of your comment? Argument in quotation marks and ellipsis point out that something should be wrong here. What are you objecting to and what are you adding? – Martin Apr 22 '13 at 20:15
  • 1
    @MartinBrandenburg: Exactly. What's the problem? – azimut Apr 22 '13 at 21:04
  • 3
    There is no argument, you just rephrase the claim. The only argument so far is contained Alexander's answer. – Martin Brandenburg Apr 23 '13 at 09:42
  • 2
    @MartinBrandenburg: I just don't get it. The question was if the group $\mathbb Z^2$ is cyclic. By definition $\mathbb Z^2$ is cyclic iff it is generated by a single element. However for any element $(a,b)\in\mathbb Z^2$, we see that $\langle(a,b)\rangle \neq \mathbb Z^2$. What on earth is wrong with this argument? – azimut Apr 23 '13 at 11:37
  • 4
    Nothing is wrong, but in fact this is because nothing is there. – Martin Brandenburg Apr 23 '13 at 12:45
  • @MartinBrandenburg: Alexander Gruber was kind enough to help me to understand what is going on. So please comment on my modified answer. – azimut Apr 25 '13 at 08:31
  • 3
    @azimut: If it makes you feel better, I'm boggled by Martin's response too, as your original answer sets a waypoint that makes a proof strategy clear, and arguably the path to the waypoint, then from the waypoint to the conclusion are both sufficiently obvious as to not require proof. (although also arguably, some students who have trouble with proofs should be made explicitly aware that there are still things to prove) –  Apr 30 '13 at 21:46
  • @Hurkyl: Thank you! When Martin wrote his comments, the comments started getting upvoted and my answer downvoted, which really embarrassed me. – azimut Apr 30 '13 at 22:14
  • 1
    @Martin My answer is not "pretending" to be anything. I briefly exposit the nature of cyclic groups in general, then allude to the content of the proof in the spoilertext. If OP or another reader desires an absolutely formal proof, they can feel free to formalize it, but that isn't how I write my answers. – Alexander Gruber May 05 '13 at 00:08
  • @Martin Alright. I hope my response to azimut's question was not one of those comments - for the record, I like (and upvoted) his answer. – Alexander Gruber May 05 '13 at 00:24
  • @Alexander: No, not at all, I liked how you helped azimut out to understand the obscure objections. I should have remained silent and removed my previous comments because they only contribute noise. Sorry to you and azimut. – Martin May 05 '13 at 00:36
  • Martin, there is no reason at all to apologize for your comments. Your and Alexander's support was really helpful for me. However, since @MartinBrandenburg was the one who started this whole debate, I think an additional comment of him would be more than appropriate. – azimut May 05 '13 at 09:47
11

There are only two kinds of cyclic groups: $\mathbb{Z}$ and $\mathbb{Z}/\left(n\mathbb{Z}\right)$. This is easy to see. If $G$ is an infinite cyclic group generated by $x$, then $G=\{x^m:m \in \mathbb{Z}\}$, which suggests the isomorphism $x^m\mapsto m$. The same argument works for $\mathbb{Z}/\left(n\mathbb{Z}\right)$.

Since $\mathbb{Z}^2$ is infinite, it would have to be isomorphic to $\mathbb{Z}$, which is easily shown to be impossible.

Letting $\phi:\mathbb{Z}\rightarrow \mathbb{Z}^2$ and $\phi(1)=(x,y)$, we have by the homomorphic property that $\phi(m)=(mx,my)$ for any $m\in \mathbb{Z}$. However, then there's no $z\in \mathbb{Z}$ for which $\phi(z)=(x,y+1)$.

Alexander Gruber
  • 26,963
  • Alexander, I need your help. May I ask you to please read the comments to my answer and help me to understand what's going on there? – azimut Apr 23 '13 at 20:47
  • 1
    @azimut MartinBrandenburg wants you to justify the step $\langle (a,b) \rangle \not= \mathbb{Z}^2$. (In my answer, this is proved in the spoilertext.) His point is that your answer seems to assume the conclusion - if we are allowed to assume that no element generates the group, we have already shown that the group is not cyclic. So your answer skips over the content of the proof. – Alexander Gruber Apr 24 '13 at 16:56
  • Thank you so much! I've added more arguments now. Being accused of "just rephrasing the claim" is certainly not a good feeling. – azimut Apr 25 '13 at 08:35
8

In general one has the following for finite groups. Let $G$ and $H$ be cyclic groups then $G \times H$ is cyclic if and only if $\gcd(|G|,|H|)=1$. If $G$ is an infinite group and $H$ is any non-trivial group then $G \times H$ is never cyclic.

We call a group $G$ (written multiplicatively) cyclic if there exists $g \in G$ such that $\{g^n : n \in \mathbb Z\}=G$. Or rather that $G$ is generated by a single element. As I've noted in my comment, this almost never means there is only one generator.

JSchlather
  • 15,427
5

$\mathrm{End}(\mathbb{Z}^2) = M_2(\mathbb{Z})$ is not commutative, hence $\mathbb{Z}^2$ is not cyclic.

Martin Brandenburg
  • 163,620
2

Notice that, if $G$ is a cyclic group, for any $g,h \in G$, there exist $m,n \in \mathbb{Z}_{\neq 0}$ such that $g^m=h^n$. But in $\mathbb{Z}^2$, $h \cdot (1,0)= (h,0) \neq (0,k)=k \cdot (0,1)$ for all $k,h \neq 0$.

Seirios
  • 33,157
0

Suppose $m,n\in\mathbb{Z}$ are such that $\mathbb{Z}\times\mathbb{Z}=⟨(m,n)⟩$, then there exists $x\in\mathbb{Z}$ s.t. $x(m,n)=(1,1)\implies xm=1=xn\implies x=\frac{1}{m}=\frac{1}{n} \implies m=n=1$ Well, now it must also be true that there exists $y \in \mathbb{Z}$ s.t.∶ $y(m,n)=(0,1)\implies ym=0,yn=1 \implies m=\frac{0}{y}=0 ,n=\frac{1}{y}\implies y=1, m=0 ,n=1$ Contradiction...

Martin Sleziak
  • 53,687
narges
  • 11