Proof that G is abelian if G modulo its centre is cyclic

Proof

We write each element of $G$ as a power of one element times a central element.

1. Let $G$ be a group with identity $e$ and centre $Z = Z(G)$. Assume that the quotient group $G/Z$ is cyclic, with generator $tZ$ for some $t \in G$.
2. The projection $\pi \colon G \to G/Z$, $\pi(x) = xZ$, satisfies $\pi(xy) = (xy)Z = (xZ)(yZ) = \pi(x) \pi(y)$. So $\pi$ is a group homomorphism.
3. For $k \ge 0$, $\pi(t^k) = \pi(t)^k$ by induction on $k$. The case $k = 0$ holds because homomorphisms preserve the identity, and $\pi(t^{k+1}) = \pi(t^k) \pi(t)$.
4. For $k > 0$, $\pi(t^{-k}) = \pi((t^{-1})^k) = \pi(t^{-1})^k = (\pi(t)^{-1})^k = \pi(t)^{-k}$, since homomorphisms preserve inverses.
5. By steps 3 and 4, $(tZ)^k = t^k Z$ for each integer $k$.
6. Let $x \in G$. By the definition of $\langle tZ \rangle$ and step 5, $xZ = t^m Z$ for some integer $m$.
7. Since $e \in Z$, $x = xe$ lies in the left coset $xZ = t^m Z$. So $x = t^m z_1$ for some $z_1 \in Z$.
8. Let $x, y \in G$. By step 7, $x = t^m z_1$ and $y = t^n z_2$ with integers $m, n$ and $z_1, z_2 \in Z$.
9. Since $z_1$ commutes with every element, $xy = t^m z_1 t^n z_2 = t^m t^n z_1 z_2$. By the laws of exponents, $xy = t^{m+n} z_1 z_2$.
10. In the same way, $yx = t^n z_2 t^m z_1 = t^n t^m z_2 z_1 = t^{m+n} z_2 z_1$.
11. Since $z_1 \in Z$, $z_1 z_2 = z_2 z_1$. By steps 9 and 10, $xy = yx$.
12. Step 11 holds for all $x, y \in G$. So $G$ is abelian.