Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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


share|cite|improve this question
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 3 '13 at 21:09
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 3 '13 at 21:13
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. – 1015 Mar 3 '13 at 21:32
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 3 '13 at 22:21
@Derek Yes I agree. – Alexander Gruber Mar 4 '13 at 18:39

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.


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 $(a,b+1)\notin \{(ka,kb) \mid k\in\mathbb Z\}$.

share|cite|improve this answer
Your "argument" is that it is not cyclic because no element generates the group ... – Martin Brandenburg Apr 22 '13 at 19:49
@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
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 9:42
Nothing is wrong, but in fact this is because nothing is there. – Martin Brandenburg Apr 23 '13 at 12:45
@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) – Hurkyl Apr 30 '13 at 21:46

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 $\phi(z)=(x,y+1)$.

share|cite|improve this answer
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
@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 8:35

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.

share|cite|improve this answer

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

share|cite|improve this answer
Could you please state for me the general theorem on the relationship between cyclicity and endomorphism ring being commutative? – Metin Y. Apr 23 '13 at 20:45
@MetinY., cyclic groups have commutative endomorphism rings, in particular $\operatorname{End}(\Bbb{Z}) \cong \Bbb{Z}$. – Andreas Caranti Apr 23 '13 at 20:52
@Andreas Caranti Appreciated... – Metin Y. Apr 23 '13 at 21:00
What is the benefit of this approach over just showing no generator exists? – Potato May 5 '13 at 0:40
@Potato: I don't have to do calculations. – Martin Brandenburg Jun 30 '15 at 13:53

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 $k,h \neq 0$.

share|cite|improve this answer

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...

share|cite|improve this answer
please format your answer with Mathjax. – Surb Jun 30 '15 at 9:51

protected by Zev Chonoles Jun 30 '15 at 9:41

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?

Not the answer you're looking for? Browse other questions tagged or ask your own question.