TITLE: How can $G$ a simple group always be isomorphic to $Z_p$ for some prime $p$?
QUESTION [6 upvotes]: Is it not a necessary condition that the order of two groups must be equal for them to be isomorphic?
Does "$G$ is a simple group of odd order" somehow imply "$|G|$ is prime"?"
If not, I don't see how it can't be the case that $|G|$ is some odd composite number, so it couldn't be isomorphic to any $Z_p$ where $p$ is prime.

REPLY [2 votes]: Is it not a necessary condition that the order of two groups must be
  equal for them to be isomorphic?

It is. Any two isomorphic groups have the same order.

Does "$G$ is a simple group of odd order" somehow imply "$\lvert G \rvert$ is prime"?"

Yes it does. This is equivalent to the Feit-Thompson theorem that every finite group of odd order is solvable, as discussed in the question Every simple group of odd order is isomorphic to $\mathbb{Z}_{p} $ iff every group of odd order is solvable.
That theorem was proved in the 255-page 1963 paper Solvability of groups of odd order.