Abelianisation of a Group

The abelianisation of a group takes an arbitrary group and generates from it, in some sense, a maximal abelian quotient group.

The followings results build the necessary theory to deduce that the commutator subgroup is the minimal choice for $[G,G]$, thus leading to the maximal $G/[G,G]$ such that $G/[G,G]$ is abelian.

Proof

Suppose $G$ is a group and $H$ is a subgroup. If $aH,bH\in G/H$, then we have the equivalence

If $[G,G]\subseteq H$, then the final condition above is trivially true for all $a,b\in G$, and thus $G/H$ is abelian.

If $G/H$ is abelian, then $ab{a}^{-1}{b}^{-1}\in H$ for all $a,b\in G$, that is

The smallest subgroup which contains this set on the left is exactly the commutator subgroup $[G,G]$.

Proof

By the first result, $G/H$ is abelian if and only if $[G,G]\subseteq H$, so the minimal choice for $H$ is $[G,G]$ itself.

Proof

We note that $G\cong G/\{\mathrm{id}\}$ and therefore $G$ is abelian if and only if $G/\{\mathrm{id}\}$ is. However by the first result, $G/\{\mathrm{id}\}$ is abelian if and only if $[G,G]\subseteq \{\mathrm{id}\}$ which is if and only if $[G,G]=\mathrm{id}$.