# For every positive integer n there is a unique cyclic group of order n

1. May 3, 2014

### BomboshMan

Hi,

I can't understand why the statement in the title is true. This is what I know so far that is relevant:

- A subgroup of a cyclic group G = <g> is cyclic and is <g^k> for some nonnegative integer k. If G is finite (say |G|=n) then k can be chosen so that k divides n, and so order of g^k is n/k.

- Let G = <g> be a cyclic group. If g has infinite order, then G is isomorphic to the integers with addition. If o(g) = n, then G is isomorphic to Z/nZ with addition.

And then apparently the last point implies the statement in my title and I have no idea how!

Thanks,

Matt

2. May 3, 2014

### Fredrik

Staff Emeritus
I'm not entirely familiar with the standard terminology (or this theorem), so you will have to explain a few things. The order of a group is the cardinality of its underlying set, right? But what do you mean by order of g? Isn't g an element of the group? Was that supposed to be "order of G"? What do you mean by <g>? I would guess that it's the smallest subgroup that has {g} as a subset. What does o(g) mean? Is it "order of g", where "g" actually means G? In that case, I'd say that the "last point" (the sentence that contains o(g)) is just a slightly stronger statement than the one in the thread title, and there's nothing to prove.

I assume that "unique" means "unique up to isomorphism".

3. May 3, 2014

### jbunniii

I agree with Fredrik that there seems to be nothing to prove, if you already have the quoted fact:
In other words, up to isomorphism, there is exactly one cyclic group of order $n$, namely $\mathbb{Z}/n\mathbb{Z}$.

@Fredrik - you are correct about the meanings of all of the terms, which are all standard usage in group theory:

Order of $G$ = cardinality of the underlying set

$\langle g \rangle$ = subgroup generated by $g$ = smallest subgroup of $G$ which contains $g$. This is always a cyclic group consisting of the elements $g^n$ for all integers $n$. If the powers are all distinct, then we have an infinite cyclic group, otherwise there is some smallest positive $n$ such that $g^n$ is the identity. In this case, there are $n$ distinct elements and the group is isomorphic to $\mathbb{Z}/n\mathbb{Z}$.

$o(g)$ = order of $g$ = order of $\langle g \rangle$ ("order" is overloaded here, but the meaning is always clear from the context)

Last edited: May 3, 2014
4. May 3, 2014

### BomboshMan

What exactly does 'up to isomorhism' mean, and why does this mean there isn't more than one?

5. May 3, 2014

### Fredrik

Staff Emeritus
It means that there may be infinitely many, but they're all isomorphic to each other.

6. May 3, 2014

### Erland

It means that all cyclic groups of order n are isomorpic to each other.

7. May 3, 2014

### BomboshMan

Ah! That makes sense now. Thanks very much!