Group of Order Prime Squared is Abelian

Theorem
A group whose order is the square of a prime is abelian.

Proof
Let $$G$$ be a group of order $$p^2$$, where $$p$$ is prime.

Let $$Z \left({G}\right)$$ be the center of $$G$$.

By Lagrange's Theorem, $$\left|{Z \left({G}\right)}\right| \backslash \left|{G}\right|$$, so $$\left|{Z \left({G}\right)}\right| = 1, p$$ or $$p^2$$.

By Center of Group of Prime Power Order is Non-Trivial, $$\left|{Z \left({G}\right)}\right| \ne 1$$.

Now suppose $$\left|{Z \left({G}\right)}\right| = p$$.

Then:

$$ $$ $$

So $$G / Z \left({G}\right)$$ is non-trivial, and of prime order and therefore cyclic.

But from Cyclic Quotient Group of Center, it can not be.

Therefore $$\left|{Z \left({G}\right)}\right| = p^2$$ and therefore $$Z \left({G}\right) = G$$.

Therefore $$G$$ is abelian.