What is the generator of the cyclic group (Z,+)?

[LagrangeEuler]

I do not understand why $(Z.+)$ is the cyclic group? What is a generator of $(Z,+)$?
If I take $<1>$ I will get all positive integers. If I take $<-1>$ I will get all negative integers. I should have one element which generates the whole group. What element is this?

 

[PeroK]

I do not understand why $(Z.+)$ is the cyclic group? What is a generator of $(Z,+)$?
If I take $<1>$ I will get all positive integers. If I take $<-1>$ I will get all negative integers. I should have one element which generates the whole group. What element is this?

The positive integers do not form a subgroup, so they cannot be the subgroup generated by the element $1$. Something is wrong with the definitions you are using.

 

[Euge]

If $(G, \cdot)$ is a group and $g\in G$, the cyclic subgroup generated by $g$ is $\langle g\rangle = \{g^n : n\in \mathbb{Z}\}$, where $g^0 = e$, $g^n= g\cdot g\cdots g$ ($n$ times) if $n$ is a positive integer and $g^n = g^{-1}\cdot g^{-1}\cdots g^{-1}$ ($-n$ times) is $n$ is a negative integer.

If $G$ is additive, $ng$ is written in place of $g^n$. So when $G$ is additive $\langle g\rangle = \{ng : n\in \mathbb{Z}\}$. In the case $G = \mathbb{Z}$, the cyclic subgroup $\langle 1\rangle = \{n\cdot 1 : n\in \mathbb{Z}\} = \{n : n\in \mathbb{Z}\} = \mathbb{Z}$. Similarly $\langle -1\rangle = \mathbb{Z}$. So indeed, $\mathbb{Z}$ is cyclic, and both $-1$ and $1$ are generators of the group.

 

[PeroK]

If $G$ is abelian,

I think you mean additive.

 

[Euge]

I think you mean additive.

Corrected, thanks!

 

[Steve4Physics]

Can I add, as a non-mathematician….

If a finite group is cyclic, the group can be generated by a single element.

But if an infinite group (such as integers under addition) is cyclic, the group can be generated by a single element or its inverse. So in the present case we can use 1 or -1.

 

[Euge]

Can I add, as a non-mathematician….

If a finite group is cyclic, the group can be generated by a single element.

But if an infinite group (such as integers under addition) is cyclic, the group can be generated by a single element or its inverse. So in the present case we can use 1 or -1.

A group element has the same order as its inverse, so for any group $G$ and $g\in G$, the cyclic subgroup $\langle g\rangle = \langle g^{-1}\rangle$.

 

[Office_Shredder]

A fun exercise is proving in an infinite cyclic group that $g$ and $-g$ are the only generators, which is not true for finite groups - e.g. the cyclic group of order $p$ for $p$ prime has every element as a generator except for the identity.