Isomorphic Groups of prime/not prime orders

Question

I believe the following 2 facts are true...

1) Any group of prime order is a cyclic group

2) Any cyclic group of order n is isomorphic to $(\mathbb{Z}_n, +)$

So for example, every group of order $101$ is isomorphic to $\mathbb{Z}_{101}$, correct?

But what if the group order is not prime, for instance, is every group of order 24 isomorphic to $\mathbb{Z}_{24}$? Can this fail?

Answer 1

If you do not have a prime n, it does not have to be isomorphic to to $\mathbb{Z}_n$. For example, there are two groups of order 4. $\mathbb{Z}_4$ and $\mathbb{Z}_2\times\mathbb{Z}_2$. And $\mathbb{Z}_2\times\mathbb{Z}_2$ is not isomorphic to $\mathbb{Z}_4$, since $\mathbb{Z}_2\times\mathbb{Z}_2$ is not cyclic.

Yes, every group of order $p$ is cyclic. And every group of primeorder is isomorphic to $(\mathbb{Z}_p, +)$

Answer 2

Your first and second statements are both correct. If a group has prime order then it will be cyclic. Moreover any cyclic group of order $n$ will be isomorphic to $(\mathbb{Z}_n,+)$.

There's a nice theorem that gives a partial answer to your later question. Let $G$ be a finite abelian group of order $n$. Then there exist positive integers $a_1,a_2,\ldots,a_n$, where $a_i|a_{i+1}$, such that $a_1a_2\cdots a_d=n$ and $$G\cong \mathbb{Z}_{a_1}\times\mathbb{Z}_{a_2}\cdots\times \mathbb{Z}_{a_d}$$

So in the case of abelian groups of order 24 our options for abelian groups are $$\mathbb{Z}_{24},\quad\mathbb{Z}_{12}\times\mathbb{Z}_{2}$$

This is a specific case of the Structure Theorem for Finitely Generated Abelian Groups which generalizes the above for any finitely generated abelian group. Unfortunately groups that are not abelian aren't well behaved enough to have such a nice classification. In total there are actually 15 different groups of order 24.