Why do generating sets need not contain the inverses of their elements?

Question

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?

Answer 1

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.

Answer 2

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 $(\mathbb Z, +)$.

Answer 3

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.