How commutative?

There are a couple standard proofs I know of, both of which follow by first assuming the index is a prime $p$ :

• Let $a \in G \setminus Z$ . Then the centralizer of $a$ inside $G,$ $C_G(a) = \{g \in G : ag = ga\}$ is a subgroup such that $Z \leq C_G(a) \leq G.$ In terms of the index, this means $\begin{aligned} p = |G : Z| = |G : C_G(a)| \times |C_G(a) : Z| &\implies |G : C_G(a)| = 1 \text{ or } |C_G(a) : Z| = 1 \\ &\implies C_G(a) = G \text{ or } C_G(a) = Z \end{aligned}$ However, by definition, we can see that $C_G(a) = G \implies a \in Z$ and $a \in C_G(a) = Z \implies a \in Z$ , so both cases end in contradiction.
• Since the center is a normal subgroup of $G,$ the quotient $G/Z$ is a group of prime order $|G : Z| = p$ and therefore it is cyclic. From this, we can see that there is some $a \in G$ such that every element $g \in G$ can be written as $g = a^nz$ where $0 \leq n < p$ and $z \in Z$ . It's now an easy exercise to show that $G$ is abelian (i.e. $Z = G \implies |G:Z| = 1$ ) which is a contradiction