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.

Definition

The abelianisation of a group \(G\) is defined as \(G/[G, G]\) where \([G, G]\) is the commutator subgroup.

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.

Theorem

For a group \(G\) with subgroup \(H\), \(G/H\) is abelian if and only if \(H\) contains that commutator subgroup, that is \([G, G] \subseteq H\).

Proof

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

\[\begin{align*} (aH)(bH) = (bH)(aH) &\iff abH = baH \\ &\iff ab(ba)^{-1} \in H \\ &\iff aba^{-1}b^{-1} \in H. \\ \end{align*}\]

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 \(aba^{-1}b^{-1} \in H\) for all \(a, b \in G\), that is

\[ \{aba^{-1}b^{-1} : a, b \in G\} \subseteq H.\]

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

Corollary

For any group \(G\), \([G, G]\) is the smallest (in terms of inclusion) subgroup of \(G\) such that \(G/[G, G]\) is abelian.

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.

Corollary

A group \(G\) is abelian if and only if \([G, G] = \mathrm{id}\).

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 \([G, G] = \mathrm{id}\).