# Why are $-1$ and $1$ generators for the Set of integers under addition?

I'm reading my textbook and I'm confused why $-1$ and $1$ are generators for the group of integers under addition.

For example for 1: we have $1, 1+1=2, 1+1+1=3, 1+1+1+1=4,$ etc. So shouldn't $1$ be a generator for only the group of positive integers under addition?

For $-1$: we have $-1, -1+-1=-2, -1+-1+-1=-3$. So shouldn't $-1$ be a generator for only the group of negative integers under addition?

Or is it that both $1$ and $-1$ are generators for the set of integers under addition because in the definition of cyclic subgroup $\{a^n; n\in \Bbb Z \}$, $n$ can take on negative powers? Or am I just confused? Thanks.

• $1$ is a generator for the group of integers. $-1$ is also a generator for the group of integers. – Angina Seng Jan 31 '18 at 5:11
• Yes, I know that, let me edit my question slightly to be more precise. I'm just confused why. – kemb Jan 31 '18 at 5:13
• Yes, you're just confused. In principle, you got it, already: under addition, $+1$ generates the positive integers, and $-1$ generates the negative ones. So? – Professor Vector Jan 31 '18 at 5:13
• But my textbook says that +1 is a generator for the entire group of integers. Is this correct? – kemb Jan 31 '18 at 5:14
• If $g$ generates the group then $g$ generates $g^{-1}$, too. – Doug M Jan 31 '18 at 5:15

Syntactically, we have these distinctions:

• The semigroup generated by $1$ is the semigroup of positive integers. Every positive integer $n$ can be expressed as $1+1+\cdots+1$ with $n$ repetitions of $1$.
• The monoid generated by $1$ is the monoid of nonnegative integers. Every nonnegative integer $n$ can be expressed as $0+1+1+\cdots+1$ with $n$ repetitions of $1$.
• The group generated by $1$ is the group of integers. Every integer $n$ can be expressed as $0\pm1\pm 1\pm\cdots\pm1$ with $|n|$ repetitions of $1$ and either $+$ or $-$ depending on the sign of $n$.

And:

• The semigroup generated by $-1$ is the semigroup of negative integers. Every negative integer $n$ can be expressed as $(-1)+(-1)+\cdots+(-1)$ with $|n|$ repetitions of $-1$.
• The monoid generated by $-1$ is the monoid of nonpositive integers. Every nonpositive integer $n$ can be expressed as $0+(-1)+(-1)+\cdots+(-1)$ with $|n|$ repetitions of $-1$.
• The group generated by $-1$ is the group of integers. Every integer $n$ can be expressed as $0\pm(-1)\pm (-1)\pm\cdots\pm(-1)$ with $|n|$ repetitions of $-1$ and either $-$ or $+$ depending on the sign of $n$.

In conclusion, your intuitions are right, but misplaced. You're thinking of how to generate a semigroup, an object where all you can do is $+$. Generating a group allows you to apply $0$ and $-$ to elements as well, and with those operations, you can get to every integer from either $1$ or $-1$.

• Great, thanks, I understand now. – kemb Jan 31 '18 at 5:45

The short version is that groups have three operations1:

• There is the constant zero
• You can negate an element
• You can add a pair of elements

In your analysis, you only used the last of these three operations; however, you can construct yet more numbers when you include the use of the first two operations as well.

1: To match the example at hand, I've used the additive language rather than the multiplicative language

$(-1)=\{(-1)a; a\in \Bbb Z \}=\{-a; a\in \Bbb Z \}=\Bbb Z$ and $(+1)=\{(+1)a; a\in \Bbb Z \}=\{a; a\in \Bbb Z \}=\Bbb Z$, so $$(-1)=(1)=\Bbb Z$$

In group theory, the word "generates" has the following meaning: a set $S$ generates a group $G$ if every element of $G$ can be written as a string over $S\cup S^{-1}$ (here, if $S=\{a, b, \ldots\}$ then $S^{-1}=\{a^{-1}, b^{-1}, \ldots\}$). Equivalently, $G$ is the minimal subgroup of $G$ containing all the elements of $S$. In your example, $S=\{1\}$ so $S\cup S^{-1}=\{1, -1\}$ (there is a notation issue here...but I am sure that you can cope).

If $S=S^{-1}$ then we say that $S$ is symmetrised.

In finite groups we do not need to consider the set $S\cup S^{-1}$ as if $a\in S^{-1}$ then there exists $b\in S$ and $n\in\mathbb{N}$ such that $b^n=a$ (indeed, $b=a^{-1}$).