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The first sentence in the Wikipedia article entitled "Cyclic Groups" states that "In algebra, a cyclic group is a group that is generated by a single element".

How is this consistent with addition on the set of integers being considered a cyclic group. What would be the single element that generates all the integers.?

Please don't tell me it is the element 1 :)

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    $\begingroup$ Why don't you want to be told it is the element $1$? Surely, every integer is either of the form $1+1+\ldots + 1+1$ or the negative thereof. $\endgroup$ – Milo Brandt Apr 19 '15 at 18:18
  • $\begingroup$ Some texts even take it is a definition: A group $G$ is cyclic if it is isomorphic to $\mathbb Z$ or to $\mathbb Z / n \mathbb Z$ for some $n$. Aluffi's Algebra: Chapter 0 on page 67. $\endgroup$ – Andrey Kaipov Apr 19 '15 at 18:32
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It is the element $-1$.

On a more serious note, the definition of "generates" includes allowing the inverse of the generating elements. For any group $G$, and element $g\in G$, the subgroup generated by $g$ is $$\{g^n:n\in\mathbb{Z}\}$$ not $$\{g^n:n\in\mathbb{N}\}$$ (the latter is not a subgroup unless $g$ has finite order).

Observe that $g^{-1}$ is always in $\{g^n:n\in\mathbb{Z}\}$.

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  • $\begingroup$ OK, I meant please don't tell me it is the element 1 or -1 without explaining how I can generate the negative integers with 1 or how I can generate the positive integers with -1. $\endgroup$ – Geoffrey Critzer Apr 19 '15 at 18:21
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    $\begingroup$ @GeoffreyCritzer: It seems you're not working from a correct definition of "generate". I've updated my answer. $\endgroup$ – Zev Chonoles Apr 19 '15 at 18:24
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    $\begingroup$ I find it a bit confusing to use the multiplicative definition, without a word about this, when the situation in question uses additive notation and multiplicative notation would make sense there too yet give something else. $\endgroup$ – quid Apr 19 '15 at 18:28
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    $\begingroup$ @quid: When talking about general groups (notably including non commutative ones) one always uses multiplicative notation. The definition of what a single element generates applies to any group, so it should use multiplicative notation. More likely though "generates" will be actually defined for any subset of elements, and the definition will look different still. But it is reasonable to specialise for the singleton case here. $\endgroup$ – Marc van Leeuwen Apr 19 '15 at 23:33
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    $\begingroup$ @MarcvanLeeuwen yes, I still think it would be clearer and nothing lost if it was said that 'multiplicative notation is used because...' as opposed to just doing it without a word about it. $\endgroup$ – quid Apr 19 '15 at 23:46
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It is the element $1$ (or the element $-1$).

Why? The subgroup $\langle g\rangle\subset G$ generated by an element $G$ is defined as the smallest subgroup to contain $g$. Since $1$ is in $\langle 1\rangle$ (in $\mathbb Z$) and any subgroup is closed under inverses, $-1$ is also in $\langle 1\rangle$ (since it is the inverse of $1$). Clearly all positive integers are there, and so are their inverses. You get $0$ by the identity group axiom (that the additive identity must be an element of your additive subgroup).

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    $\begingroup$ This is the real answer. The fact that $\langle g\rangle$ contains all $g^n$ with $n\in\mathbb Z$ and not necessarily just with $n\in\mathbb N$ is a consequence of this. $\endgroup$ – Hagen von Eitzen Apr 19 '15 at 21:21

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