Why is the Integer Group Z Considered Cyclic?

[Zorba]

This might sound like a silly question, but based on

Definition: A group G is called cyclic if there is g\in G such that \langle g \rangle = G

And if we take (\mathbb{Z},+) the set of integers with addition as the operation, then why is it considered cyclic? Because the problem I am having is that if you say 1 is the generator, well you can get the positive integers but not the negative, and vice versa with -1...

So you need two elements to generate the group rather than one, so it's not cyclic?

 

[Deveno]

in (Z,+), -1 = (1)^-1.

the subgroup <g> is not "all positive powers of g" but rather ALL powers of g, it is the smallest group containing g.

since every group must contain inverses, g^-1 is considered as: generated by g.

it is a happy accident that for elements g of finite order, g^-1 turns out to be a positive power of g. this does not happen in free groups, for example.

both 1 and -1 are considered to be generators of Z.

if you imagine a cyclic group to be a circle that can only rotate 1/n-th of a revolution, than an "infinite circle" is just a line. whereas with a finite circle going backwards is the same as going forwards some other amount, on a line, you have two essentially different directions.

<g> = {g^k : k in Z}, NOT (g^k: k in N}. it's just that for finite order g's, you don't need the negative powers.