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I recently learned about generating sets, and a common elementary example that is provided is the sets $\{1\}$ and $\{-1\}$, both of which independently generate $(\mathbb{Z},+)$. I understand why each of them in their own right is a generating set of $(\mathbb{Z},+)$ given the definition of a generating set. My question is why generating sets are defined such that they do not need to contain the inverses of their elements. That is, why aren't generating sets defined such that $\{-1,1\}$ is a generating set of $(\mathbb{Z},+)$ and $\{-1\}$ and $\{1\}$ are not?

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    $\begingroup$ The reason the definition of "generating set" is as it is, is precisely so that $\{1\}$ alone generates that group. $\endgroup$ – vadim123 Dec 18 '18 at 18:47
  • $\begingroup$ If $\{1\}$ doesn't generate $\Bbb Z$, then what does it generate? $\endgroup$ – Lord Shark the Unknown Dec 18 '18 at 18:48
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    $\begingroup$ I think what you're looking for is a generating set for the semigroup $(\mathbb Z, +)$ rather than for the group $(\mathbb Z, +)$. $\endgroup$ – Robert Israel Dec 18 '18 at 18:50
  • $\begingroup$ An alternate definition of "generate" is that $\langle g\rangle$, for some $g$ in some group $G$ isn't directly dependent on what you can reach with arithmetic using $g$, but rather as the smallest subgroup of $G$ which contains $G$. This way there is no issue with $\langle 1\rangle =\Bbb Z$. The use of the word "generate" does get slightly more cryptic with this definition, though. $\endgroup$ – Arthur Dec 18 '18 at 18:51
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When generating an algebraic structure from a subset, you are allowed to use all operations and axioms available to generate new elements.

So of course when you talk about $\mathbb Z$ as a group, you have $1+1$ and $1+1+1$ via the operation, but you also have the existence of $-1$ from the axioms, so including it with $1$ in a generating set would be redundant.

If instead you were merely talking about a semigroup, in $\mathbb Z$ generated by $1$ and you did not have the inverses axiom, then you could only say that $1$ can generate $2, 3,\ldots$ and not $0$ or $-1$ or anything else. You could only use the operation and elements already known to be in the semigroup.

Suppose more generally you're in a larger ring like $\mathbb Z[x]$, and you examine how it is generated as a $\mathbb Z[x]$ module by $1$. Not only can you generate $\mathbb Z$ with $1$ (because it is an additive group) but additionally you can use the module action to generate $x=1\cdot x$, and after that, you get everything in $\mathbb Z[x]$. There is no need to add anything else to the generating set.

So, you can see how "generating set" heavily depends on the conrtext of the algebraic object in question.

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  • $\begingroup$ This is a very thorough explanation, thank you! $\endgroup$ – MelonBaller09 Dec 19 '18 at 20:52
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Yes, $\{1\}$ is a generating set of the group $(\mathbb Z, +, 0, -)$. You can reach all of $\mathbb Z$ from $1$ using the operations $+, 0, -$.

No, $\{1\}$ is not a generating set of the monoid $(\mathbb Z, +, 0)$ or the semigroup $(\mathbb Z, +)$.

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  • $\begingroup$ Now that is interesting. From what I've read so far, groups are presented as a double consisting of the set and the operation (i.e. (Z,+) is presented as a group instead of a semigroup). I would think that the notation you used for a group would be more common to avoid confusion with monoids and semigroups. $\endgroup$ – MelonBaller09 Dec 18 '18 at 22:01
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    $\begingroup$ Well, the notation that minimizes confusion will depend on the context. It's more important to be precise about the signature of the structure in universal algebra, which is the natural setting for your question. $\endgroup$ – Chris Culter Dec 18 '18 at 22:23
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Just a guess: $\left< S \right>$ is always a subgroup for any subset $S$. If you change the definition of $\left< \cdot \right>$ as you would, then this would not be the case anymore.

Edit: Another definition of $\left< S \right>$ is that it is the smallest subgroup containing $S$, i.e. $\left< S \right> = \bigcap \{H \subseteq G \space | \space S \subseteq H \wedge H \text{ subgroup}\}$. Maybe this definition is more intuitive.

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  • $\begingroup$ I think this alternative definition is part of what is adding to my confusion. If <S> is the smallest subgroup containing S, and S is the generating set {1}, then it cannot possibly be the case that <S>=({1},+), since ({1},+) does not meet the requirements of a group (closure under +, no inverses, no identity elements). $\endgroup$ – MelonBaller09 Dec 18 '18 at 22:14
  • $\begingroup$ I don't understand where your confusion is :) as you say, $\left< \{ 1 \} \right>$ cannot be $\{1\}$, since it has to be closed under $+$, have inverses and an identity element. In fact, the only subgroup $H \subseteq \mathbb{Z}$ that contains $\{ 1 \}$ is $\mathbb{Z}$ and thus $\left< \{ 1 \} \right> = \mathbb{Z}$ $\endgroup$ – Jakob B. Dec 19 '18 at 16:42
  • $\begingroup$ Okay, I think we're on the same page. I think what was confusing me was the notation. I didn't understand that <S> refers to the subgroup generated by S, and doesn't refer to the group formed by the set S and some operation (e.g. (S,+)). $\endgroup$ – MelonBaller09 Dec 19 '18 at 20:51

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