If G/Z(G) is Cyclic then G is Abelian

# If G/Z(G) is Cyclic then G is Abelian

 Proposition 1: Let $G$ be a group. If $G/Z(G)$ is cyclic then $G$ is abelian.

Note that $Z(G)$ is always a normal subgroup of $G$, and so the quotient $G/Z(G)$ is well-defined.

• Proof: Suppose that $G/Z(G)$ is a cyclic group. Then there exists an $gZ(G) \in G/Z(G)$ such that:
(1)
\begin{align} \quad G/Z(G) = \langle gZ(G) \rangle \end{align}
• Let $h \in G$. Then $hZ(G) \in G/Z(G)$. Then there exists an $n \in \mathbb{Z}$ such that:
(2)
\begin{align} \quad hZ(G) &= (gZ(G))^n \\ &= \underbrace{(gZ(G))(gZ(G))...(gZ(G))}_{n \: \mathrm{times}} \\ &= g^nZ(G) \end{align}
• Therefore $(g^n)^{-1}h \in Z(G)$. So there exists an $i \in Z(G)$ such that:
(3)
\begin{align} \quad i = (g^n)^{-1}h \end{align}
• So $h = g^ni$. So for every $h \in G$ there exists an $n \in \mathbb{Z}$ and an $i \in Z(G)$ such that $h = g^ni$. Let $h_1, h_2 \in G$ and write $h_1 = g^{n_1}i_1$, $h_2 = g^{n_2}i_2$ where $n_1, n_2 \in \mathbb{Z}$ and $i_1, i_2 \in Z(G)$. Then:
(4)
\begin{align} \quad h_1h_2 = (g^{n_1}i_1)(g^{n_2}i_2) \end{align}
• Since $i_1 \in Z(G)$ we have that $i_1g^{n_2} = g^{n_2}i_1$, so:
(5)
\begin{align} \quad h_1h_2 &= g^{n_1}g^{n_2}i_1i_2 \\ &= g^{n_1 + n_2} i_1i_2 \\ &= g^{n_2 + n_1} i_1i_2 \\ &= g^{n_2} g^{n_1} i_1i_2 \end{align}
• Since $i_1, i_2 \in Z(G)$ we have that $i_1i_2 = i_2i_1$, so:
(6)
\begin{align} \quad h_1h_2 &= g^{n_2} g^{n_1} i_2i_1 \end{align}
• And lastly, since $i_2 \in Z(G)$ we have that $g^{n_1}i_2 = i_2g^{n_1}$, so:
(7)
\begin{align} \quad h_1h_2 &= g^{n_2} i_2g^{n_1}i_1 \\ &= h_2h_1 \end{align}
• Since this holds for all $h_1, h_2 \in G$ we have that $G$ is abelian. $\blacksquare$