Does (Z, +) have two generators but infinitely many generating sets?

We say the group of integers under addition Z has only two generators, namely 1 and -1.

However, Z can also be generated by any set of 'relatively prime' integers. (Integers having gcd 1).

I have two questions here. Couldn't find a satisfactory answer anywhere.

1. If a group is generated by a set consisting of a single element, only then is it cyclic?

2. Does 'generator' mean a single generating element?

3. Is it correct to say '(Z, +) has two generators but infinitely many generating sets'?

Thank you for your help!

• Note also that $\{6,10,15\}$ is a generating set in the same sense. It is also a minimal generating set, in that any proper subset fails to generate $\mathbb Z$. There are minimal generating sets of arbitrary finite size. – Mark Bennet Nov 26 '14 at 14:06
• @MarkBennet Okay. So if we had {p,q} where gcd(p,q) = 1, then {p,q} would be an example of a minimal generating set for ℤ. By the way, only if the generating set has 1 element, we say the group is cyclic. Right? – Deepabali Roy Nov 26 '14 at 14:16
• Indeed. If a group has a generating set containing one element it is a cyclic group. – Mark Bennet Nov 26 '14 at 15:35

The answer to all your questions is yes. By definition a cyclic group is a group which is generated by a single element (or equivalently, by a subset containing only one element). Such an element is called a generator.

$(\mathbf{Z},+)$ of course has infinitely many generating subsets, be it only because any subset containing $1$ or $-1$ is generating, and there are of course infinitely many such subsets. There are more interesting generating subsets however, such as those containing two relatively prime integers.

• Thanks a lot! Just one thing, like you said any subset containing 1 or -1 will generate Z but it is important to note that a subset containing both 1 and -1 will not generate Z as that would mean the elements aren't relatively prime. – Deepabali Roy Nov 26 '14 at 15:03
• "a subset containing both 1 and -1 will not generate Z" It certainly will. I'm not sure you understand what "generating subset" means exactly... – fkraiem Nov 26 '14 at 15:06
• I understand what a "generating subset" means. I got confused regarding the definition of 'relatively prime'. I rechecked the definition. 1 and -1 are relatively prime and any subset containing both will be a subset of relatively prime integers and hence generate Z. I hope I am finally right! Thanks a lot :) – Deepabali Roy Nov 26 '14 at 15:25

You are correct. $(\mathbb{Z},+)$ is a cyclic group with generator $(\{+1\})$ or $(\{-1\})$ And it has many generating subgroup. May be it is important to just notice that $(\mathbb{Z},+)$ is cyclic group because it can be generated by a single element.

• The OP is completely correct, not "kind of correct". What you claim is "the correct thing" doesn't even address the question. – Mirko Nov 26 '14 at 14:09
• @Susobhan Just wanted to confirm, if the generating set has only 1 element, we say that the concerned group is cyclic. Right? – Deepabali Roy Nov 26 '14 at 14:17
• yes, correct. $(\mathbb{Z},+)$ is a cyclic subgroup. – Babai Nov 26 '14 at 14:19
• What's a "generating subgroup"? – fkraiem Nov 26 '14 at 14:39
• @fkraiem Eg: if 'a' is an element in a group G. Then <a> is a cyclic subgroup of G generated by a. In this case, <a> = G we say <a> is a generating subgroup of G. – Deepabali Roy Nov 26 '14 at 14:53