Why are cyclic groups of order $p$ (prime) unique up to isomorphism?

Question

It says here that "By the classification of cyclic groups, there is only one group of each order[prime] (up to isomorphism)".

What is the intuition behind this and how would you go about proving this? An proof would be greatly appreciated.

Answer 1

Let $n\in \Bbb N$ and $A,B$ two finite cyclic groups, both of order $n$ (not necessarily prime). Then by definition of "cyclic" $A=\langle a\rangle$, $B=\langle b\rangle$ for some generartors $a,b$. Verify that $a^k\mapsto b^k$ defines a homomorphism $A\to B$ that is in fact an isomorphism.

Upon deeper reading, the statement made in the linked text is intended to mean something else (and relies on a different proof than the classification of cyclic groups!):

If $p$ is prime and $G$ a group of order $p$, then $G$ is cyclic.

This follows from Lagrange's theorem because for any $g\in G$ with $g\ne 1$, the group $\langle g\rangle$ must have order $>1$ and dividing $p$, hence $G=\langle g\rangle$ is cyclic.

Answer 2

If $G$ is cyclic with generator $g$, there is a unique surjective homomorphism $\mathbb{Z}\to G$ sending $1$ to $g$. By the First Isomorphism Theorem and the classification of subgroups of $\mathbb{Z}$, $G\cong \mathbb{Z}/n$ for some $n$.