If G/Z(G) is cyclic, then G is abelian

If G/Z(G) is cyclic, then G is abelian.

Proof. Recall that that center of G is defined as
Z(G) = \{z \in G \ | \ \forall g \in G, gz = zg\}.
We have that G/Z(G) is cyclic, so there exists an x \in G such that
G/Z(G) = \langle xZ(G) \rangle
where x is the representative of the coset xZ(G). Let g \in G. Then gZ(G) = (xZ(G))^m = x^mZ(G) for some m \in \mathbb{Z}, where the second equality is proven here because Z(G) is a normal subgroup of G. Further, we know that aH = bH if and only if b^{-1}a \in H, where a,b \in G and H is a subgroup of G. So since gZ(G) = x^mZ(G), we have that (x^m)^{-1}g \in Z(G). Therefore, there exists an z \in Z(G) such that (x^m)^{-1}g = z if and only if g = zx^m. After seeing the above, we can finally prove that G is abelian. Take g_1,g_2 \in G such that g_1 = x^kz_1 and g_2 = x^lz_2 where k,l \in \mathbb{Z} and z_1,z_2 \in Z(G). So we get
g_1g_2 &= x^kz_1x^lz_2 \\
&= x^kx^lz_1z_2 \quad \text{because } z_1x^l = x^lz_1 \text{ as } z_1 \in Z(G)\\
&= x^{k + l}z_1z_2 \\
&= x^lx^kz_1z_2 \\
&= x^lx^kz_2z_1 \quad \text{because } z_1z_2 = z_2z_1 \text{ as } z_1,z_2 \in Z(G) \\
&= x^lz_2x^kz_1 \quad \text{because } x^kz_2 = z_2x^k \text{ as } z_2 \in Z(G) \\
&= g_2g_1,
where we have proven that G is abelian.

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