LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-11

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

Statement

If the quotient group $G / Z (G)$ is cyclic, then $G$ is abelian.

Facts & Assumptions

Given: A group $G$ such that $G / Z (G)$ is cyclic.

[L1]

The center $Z (G)$ consists of the elements commuting with every element of $G$ (The center $Z (G)$ of a group).

[L2]

The center is a normal subgroup of $G$ (The center of a group is a normal subgroup).

[L3]

Multiplication in $G / Z (G)$ is $(g Z (G)) (h Z (G)) = g h Z (G)$ (The quotient group $G / N$ and coset product $(g N) (h N) = g h N$ ).

[L4]

A cyclic group is generated by one element (The subgroup $⟨ S ⟩$ generated by a subset, the cyclic subgroup $⟨ g ⟩$ , and cyclic groups).

[L5]

The subgroup generated by an element is the set of its integer powers ( $⟨ g ⟩ = {g^{n} : n \in Z}$ , and every cyclic group is abelian).

[L6]

Powers of one group element commute with one another (Exponent laws in a group: $g^{m + n} = g^{m} g^{n}$ and $(g^{m})^{n} = g^{m n}$ for all $m, n \in Z$ , and $(g h)^{n} = g^{n} h^{n}$ when $g$ and $h$ commute).

[L7]

Equality of left cosets $a H = b H$ is equivalent to $a^{- 1} b \in H$ ( $x \in a H$ iff $a^{- 1} x \in H$ , and $a H = b H$ iff $a^{- 1} b \in H$ ).

Proof

technique · direct

1.1

Choose $a Z (G)$ generating $G / Z (G)$ . By [L3]--[L6], arbitrary $x, y \in G$ satisfy $x Z (G) = a^{m} Z (G)$ and $y Z (G) = a^{n} Z (G)$ for some integers $m, n$ . By [L7], $a^{- m} x$ and $a^{- n} y$ lie in $Z (G)$ ; setting $z = a^{- m} x$ and $w = a^{- n} y$ gives $x = a^{m} z$ and $y = a^{n} w$ .

L1L2L3L4L5L6L7choose

1.2

The elements $z, w$ commute with every element by [L1], and $a^{m}$ commutes with $a^{n}$ by [L6].

L1L6

2.1

Therefore $x y = a^{m} z a^{n} w = a^{m + n} z w = a^{n + m} w z = a^{n} w a^{m} z = y x$ . Since $x, y$ were arbitrary, $G$ is abelian.

step 1.1step 1.2L6algebra∎

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Dependency tree · two levels

31 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

T. W. Judson, Abstract Algebra: Theory and Applications, Corollary 14.16, proof (standard reference, not scraped)

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