Why is the Integer Group Z Considered Cyclic?

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SUMMARY

The Integer Group Z, defined as the set of integers under addition, is considered cyclic because it can be generated by a single element, either 1 or -1. This means that every integer can be expressed as a power of these generators, including both positive and negative integers. The subgroup generated by g includes all integer powers of g, demonstrating that Z is indeed cyclic despite initial misconceptions about needing two elements for generation. The distinction lies in the understanding that both positive and negative powers are included in the definition of cyclic groups.

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Zorba
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This might sound like a silly question, but based on

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

And if we take [tex](\mathbb{Z},+)[/tex] 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?
 
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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.
 

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