LemmaStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck 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 $(gZ(G))(hZ(G))=ghZ(G)$ (The quotient group $G/N$ and coset product $(gN)(hN)=ghN$ ).

[L4]

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

[L5]

The subgroup generated by an element is the set of its integer powers ( $\langle g \rangle = \{\, g^{n} : n \in \mathbb{Z} \,\}$ , 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 $(g^{m})^{n} = g^{mn}$ for all $m, n \in \mathbb{Z}$ , and $(gh)^{n} = g^{n}h^{n}$ when $g$ and $h$ commute).

[L7]

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

Proof

technique · direct

1.1

Choose $aZ(G)$ generating $G/Z(G)$ . By [L3]--[L6], arbitrary $x,y\in G$ satisfy $xZ(G)=a^mZ(G)$ and $yZ(G)=a^nZ(G)$ 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 $x=a^mz$ and $y=a^nw$ .

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 $xy=a^mza^nw=a^{m+n}zw=a^{n+m}wz=a^nwa^mz=yx$ . Since $x,y$ were arbitrary, $G$ is abelian.

step 1.1step 1.2L6algebra∎

Depends on

Used by

Every group of order p², for prime p, is abelian Corollary

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 63 results over 17 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click 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)G/Z(G) is cyclic, then GGG is abelian