# Why is $\mathbb{Z}$ equal to $\langle 1 \rangle$ or $\langle -1 \rangle$?

As a homework assignment for my math lecture I had to prove something, and I used the fact that $\mathbb{Z} = \langle 1 \rangle$. The professor asked how do I know that and I answered him that it's natural, but he wasn't satisfied with that.

What other kind of explanation should I give? Why is $\mathbb{Z} = \langle 1 \rangle$ (or $\langle -1 \rangle$)?

• I'm the \langle \rangle fairy, here to let you know that $\langle, \rangle$ plays nicer with TeX than <, > does :) – Patrick Stevens Apr 26 '17 at 21:58

If we're being extremely pedantic, I'd say to start by noting that the map $a\mapsto 1$ extends to a unique map (a la the universal property of free groups) from $\langle a \rangle \to \mathbb{Z}$, specifically, it sends $a^n \to n$ for $n\in\mathbb{Z}$. It is immediate that this is a bijection, and hence is a group isomorphism.

Likewise, we could have taken our map $\langle a \rangle \to \mathbb{Z}$ to be the unique extension of the map $a\mapsto -1$, and it would send $a^n$ to $-n$, which is also a bijection and so a group isomorphism.

• This works if we define $\Bbb Z$ as a free group over one generator, but if we constructed $\Bbb Z$ from $\Bbb N$, then I think we still need to use induction from Peano's axioms. – Ennar Apr 26 '17 at 21:59
• This works regardless of the definition you use for $\mathbb{Z}$. What does depend on the definition you use for $\mathbb{Z}$, is how you justify that the induced maps give $a^n \mapsto n$ and $a^n \mapsto -n$, respectively. I'd reckon in most cases (unless you're already a free group on one generator, but then there'd be nothing to show) you can just do a simple induction proof on positive integers and non-positive integers. – Hayden Apr 26 '17 at 22:04
• Where by $<a>$ we understand class of a? – theSongbird Apr 26 '17 at 22:10
• Well, yes, that's what I meant, you do need that $n=1+1+\ldots+1$ and that these are all positive integers, which is what induction as an axiom gives. For negative integers you just use inverses in $\Bbb Z$. – Ennar Apr 26 '17 at 22:11
• @theSongbird By $\langle a \rangle$ I meant the free group generated by $\{a\}$. I used $a$ rather than $1$ to make sure it was clear that we were considering $\langle a \rangle$ as a free group on a single generator, rather than a subgroup of $\mathbb{Z}$. – Hayden Apr 26 '17 at 22:13

This might be naive depending on the level of pedantness :)

Considering the group $(\mathbb{Z}, +)$ and $x \in \mathbb{Z}$, $\langle x \rangle$ represents the subgroup comprising all elements that can be expressed as the finite combination of $x$ and its inverse $-x$ under the $+$ operation over $\mathbb{Z}$ (taking this for definitions). This means that $\langle 1 \rangle$ and $\langle -1 \rangle$ are the same, so we can concentrate only on one of them.

$\forall x \in \mathbb{Z}:\langle x \rangle \subseteq \mathbb{Z}$ is a consequence of the closeness of operation $+$ in $\mathbb{Z}$ (descends from $\mathbb{Z}$ being a group). This is true for $\langle 1 \rangle$ too.

On the other hand, $x \in \mathbb{Z} \Rightarrow x \in \langle 1 \rangle$ that is $\mathbb{Z} \subseteq \langle 1 \rangle$. In fact, every $x \in \mathbb{Z}$ can be expressed as:

• the sum of $1$ for $x$ times if $x > 0$, so it's an element of $\langle 1 \rangle$;
• $(1 + (-1)) = 0$ if $x = 0$, so it's an element of $\langle 1 \rangle$;
• the sum of $-1$ for $-x$ times if $x < 0$, so again it's an element of $\langle 1 \rangle$.

The first and third bullet might be objected to and at that point you would probably have to ask Peano for help.

Now we have $\langle 1 \rangle \subseteq \mathbb{Z}$ and $\mathbb{Z} \subseteq \langle 1 \rangle$, so $\mathbb{Z} = \langle 1 \rangle$.