Abelianization of Group is Abelian

Theorem

Let $\struct {G, \circ}$ be a group whose identity is $e$.

Let $G^{\mathrm {ab} }$ denote the abelianization of $G$.

Then $G^{\mathrm {ab} }$ is an abelian group.

Proof

Recall the definition of the abelianization of $G$:

$G^{\mathrm {ab} }$ is the quotient of $G$ by $\sqbrk {G, G}$:

$G^{\mathrm {ab} } = G / \sqbrk {G, G}$

where $\sqbrk {G, G}$ is the derived subgroup of $G$.

From Derived Subgroup is Normal, $\sqbrk {G, G}$ is a normal subgroup of $G$.

Hence the above definition is valid.

By definition of derived subgroup:

$\forall x, y \in G: \sqbrk {x, y} \in \sqbrk {G, G}$

where $\sqbrk {x, y}$ denotes the commutator of $x$ and $y$.

Hence from Quotient Group is Abelian iff All Commutators in Divisor it follows that $G^{\mathrm {ab} }$ is an abelian group.

$\blacksquare$

Sources

• 2021: Richard Earl and James Nicholson: The Concise Oxford Dictionary of Mathematics (6th ed.) ... (previous) ... (next): commutator