If $G/H$ is cyclic, where $H$ is a subgroup of $Z(G)$, then $G$ is abelian.

Question

Theorem 9.3 *$G/Z$ Theorem.

Let $G$ be a group and let $Z(G)$ be the center of $G$. If $G/Z(G)$ is cyclic, then $G$ is Abelian.

$\mathsf{PROOF}$ Since $G$ is Abelian is equivalent to $Z(G)=G$, it suffices to show that the only element of $G/Z(G)$ is the identity coset $Z(G)$. To this end let $G/Z(G)=\langle gZ(G)\rangle$ and let $a\in G$. Then there exists an integer $i$ such that $aZ(G)=(gZ(G))^i = g^iZ(G)$. Thus, $a=g^iz$ for some $z\in Z(G)$. Since both $g^i$ and $z$ belong to $C(g)$, so does $a$. Becaus $a$ is an arbitrary element of $G$ this means that every element of $G$ commutes with $g$ so $g\in Z(G)$. Thus, $gZ(G)=Z(G)$ is the only element of $G/Z(G)$. $\Box$

A few remarks about Theorem 9.3 are in order. First, our proof shows that a better result is possible. If $G/H$ is cyclic, where $H$ is a subgroup of $Z(G)$, then $G$ is Abelian. Second, in practice, it is the contrapositive of the theorem that is most often used - that is, if $G$ is non-Abelian, then $G/Z(G)$ is not cyclic. For example, it follows immediately from the statement and Lagrange’s Theorem that a non-Abelian group of order $pq$, where $p$ and $q$ are prime, must have trivial center. Third, if $G/Z(G)$ is cyclic, it must be trivial.

In the book Contemporary abstract algebra, page 194, there's a result corresponding to theorem 9.3 which is

If $G/H$ is cyclic, where $H$ is a subgroup of $Z(G)$, then $G$ is abelian.

My question is - is there any way to prove this result using the same method used to prove the Theorem 9.3 ? I've attached an image of the proof of Theorem 9.3.

My proof: [In short] $$\begin{align*} G/H &= \ \langle gH\rangle \\ &\implies \text{for any} \ a,b \in G,a\in g^iH \ \text{and} \ \ b\in g^jH. \\ \therefore a&=g^ih_1 \ \text{and} \ b=g^jh_2. \\ \therefore ab &= (g^ih_1)(g^jh_2) \\ &= g^{i+j}h_1h_2 \\ &= (g^jh_2)(g^ih_1) \\ &= ba. \end{align*}$$

Answer 1

The more general result can indeed be proved using the same method. May I present a few additional, general results that would help in organising the proof:

Lemma 1. Let $G$ be an arbitrary group, $H \trianglelefteq G$ a normal subgroup and $S, T \subseteq G$ subsets such that $H=\langle S \rangle$ ($S$ is a generating system for $H$) and $G/H=\langle \sigma(T) \rangle$ (the canonical image of $T$ generates the quotient), where $\sigma \colon G \to G/H$ denotes the canonical surjection. Then $G=\langle S \cup T \rangle$, i.e. $G$ is generated by the union $S \cup T$.

Lemma 2. Let $G$ be an arbitrary group and $S \subseteq G$ a generating system consisting of mutually commuting elements (in formal expression, $G=\langle S \rangle$ and $S \subseteq \mathrm{C}_G(S)$). Then $G$ is abelian.

These two lemata suffice to prove the following:

Proposition. Let $G$ be an arbitrary group and $H \leqslant \mathrm{Z}(G)$ a central subgroup such that $G/H$ is cyclic. Then $G$ is abelian.

Proof. As the quotient $G/H$ is cyclic, there exists $a \in G$ such that $\sigma(a)$ generates $G/H$ (where $\sigma$ denotes the canonical surjection, as above). It then follows by virtue of lemma 1 that $G$ is generated by $H \cup \{a\}$ ($H$ generates itself, of course) and since $H$ is central -- consisting of elements which commute with anything in $G$ -- $H \cup \{a\}$ is easily seen to consist of pairwise commuting elements and hence lemma 2 applies. $\Box$

Let me know if you would like to see proofs for the two lemata (they are very simple).