Group whose Derived Subgroup is Trivial is Abelian

Theorem

Let $G$ be a group with identity $e$.

Let $G'$ be the derived subgroup of $G$.

Let $G'$ be the trivial group $\set e$.

Then $G$ is abelian.

Proof

We assume by hypothesis that the derived subgroup of $G$ is $\set e$.

Aiming for a contradiction, suppose $G$ is non-abelian.

Hence by definition:

$\exists g, h \in G: g h \ne h g$

Multiplying both sides by $g^{-1} h^{-1}$:

$\exists g, h \in G: g^{-1} h^{-1} g h \ne e$

But from the definition of the derived subgroup:

$G' := \set {g^{-1} h^{-1} g h : g, h \in G}$

Hence:

$G' \ne \set e$

So by definition, $G'$ is not a trivial group.

By Proof by Contraposition it follows that $G$ is abelian.

$\blacksquare$

Also see

• Derived Subgroup of Abelian Group is Trivial