Prove that the order of Z(G) is either <e> or G

Let $\displaystyle G $ be a group of order $\displaystyle pq $, with $\displaystyle p $ and $\displaystyle q $ (not necessarily distinct) primes. Prove that the center $\displaystyle Z(G) = <e> $ or $\displaystyle Z(G) = G $.

I started off by doing a proof by contradiction.

Suppose that $\displaystyle Z(G) \neq <e> $, $\displaystyle Z(G) \neq G $

Then $\displaystyle |Z(G)| $ divides either $\displaystyle p $ or $\displaystyle q $, which implies that $\displaystyle |Z(G)| = p $ or $\displaystyle |Z(G)| = q $

We know that $\displaystyle Z(G) $ is a normal subgroup.

Consider the quotient group $\displaystyle G/Z(G) $.

Using Larange's Theorem, we know that $\displaystyle |G|/|Z(G)| $ (*).

I broke it into two cases: 1: $\displaystyle |Z(G)| = p $, 2: $\displaystyle |Z(G)| = q $.

If under case 1, (*) evaluates to $\displaystyle p $. Otherwise, in case 2, (*) evaluates to $\displaystyle q $.

This is as far as I've gotten and I'm pretty stumped as how else to follow up after.