If a group has prime order it is isomorphic to $\mathbb{Z}_{p}$?

Question

If a group $G$ has prime order it is isomorphic to $\mathbb{Z}_{p}$?

What is the intuition behind this? I guess the bijectivity is easy to prove, but how can we prove that the group operations are preserved?

I saw a proof that talked about cyclic groups with finite order (say $n$) are isomorphic to $\mathbb{Z}_{n}$? But a proof of the above will be very helpful, especially if someone could also provide some intuition about it.

Thanks!

What can be the order of a non-zero element of $G$ if $|G|=p$ is prime? — Thomas Andrews, Oct 25 '21 at 17:17
$0$ is not an order. Can an element of a finite group have infinite order? — Thomas Andrews, Oct 25 '21 at 17:18
Since $G$ is finite, $g,g^2,g^3,\dots,$ must have some $g^i=g^j$ for $i<j.$ Multiply both sides by $g^{-i}=(g^{-1})^i$ and you get $g^{j-i}=e.$ So $g$ must have finite order. — Thomas Andrews, Oct 25 '21 at 17:20

score 1 · Answer 1 · answered Oct 25 '21 at 17:32

Let $G$ be a group with order $p$ where p is a prime and let $g \in G$. By Lagrange's theorem we have that $|g|$ is a factor of $|G|$. Since $|G|=p$, the only factors of $|G|$ are $1$ and $p$.

Therefore $|g| = 1$ or $|g| = p$. Recall that the only element of a group whose order is $1$ is the identity element. Therefore, since $|G|$ is a prime number, $G$ must have at least one non-trivial (non-identity) element. Thus $G$ has an element of order $p$ and thus $G$ is a cyclic group (and so isomorphic to $\mathbb{Z}_p$)

If a group has prime order it is isomorphic to $\mathbb{Z}_{p}$?

1 Answers1