Prologue~
“What is it indeed that gives us the feeling of elegance in a solution, in a demonstration? It is the harmony of the diverse parts, their symmetry, their happy balance; in a word it is all that introduces order, all that gives unity, that permits us to see clearly and to comprehend at once both the ensemble and the details.”
—From Science et Méthode (1908), Livre Premier, Chapitre 2, p. 25; Translated in The Foundations of Science: Science and Hypothesis, The Value of Science, Science and Method (1913), p. 372. by Henri Poincaré
Verse I: What is a 'nicer' G?
So I have a group $G$- (say, a non-trivial one) which can either have a commutative or non-commutative structure, that is, be abelian or non-abelian. Now, abelian groups are very 'nice' to deal with- they make things easier for us, but non-abelian groups, not so much...
So how can one make things easier in a non-abelian group, or rather- how can we work in or deal with a 'nice' part of this non-abelian group which can make things (again, say-) 'nicer'? We explore this question in this post.
Verse II: The canvas~
Let $G$ be any non-trivial group, then if $a,b \in G$, then the commutator of $a$ and $b$ is the element $ab a^{-1} b^{-1}$.
A good observation would be to see that if our $G$ is abelian, then our commutator is simply the identity element $e \in G$. Now, let us define $C$ to be the set\[ C =\{ x_1 x_2 x_3 \cdots x_n \ | \ n \geq 1, \text{each $x_i$ is a commutator in $G$} \} \] Basically, our $C$ here is the collection of all finite products of commutators in $G$. One can intuitively see that $C$ then will be a normal subgroup of $G$. \[ C \lhd G\] If not, maybe we can try thinking about it- as mentioned above, at least the identity element is a commutator and hence our $C$ is non-empty. And given any two elements in $C$, they are of the form $x_1 x_2 \cdots x_n$ and $y_1 y_2 \cdots y_n$ and since this is just a finite product of commutators, we also have their inverses existing, which then again belong to $C$. Now we can take an element $g \in G$ and apply the left conjugation action by $g$ on an element of $C$. As depicted below, we can play around!\[ g c g^{-1} = g x_1 x_2 \cdots x_n g^{-1} = (g x_1 g^{-1}) (g x_2 g^{-1}) \cdots g(x_n g^{-1}) \] Also notice that given any arbitrary element $c \in C$, we have, \[ (g c g^{-1})^{-1} = g c^{-1} g^{-1}\] and we obtain our equation back. Moreover, $g c g^{-1}$ was a product of commutators and hence it belongs in $C$.
$\therefore$ $\forall \ g \in G$ and $c \in C$, $gcg^{-1} \in C$, and hence, $C \lhd G$.
The normal subgroup $C$ of $G$ is called the commutator subgroup or the derived subgroup of $G$ and is usually denoted by $C= G' = [G,G]$.
Also, if our $G$ is abelian (to begin with,) we then obtain $C=\{e\}$ and hence one can see the commutator subgroup as one measure of how far away a group is from being abelian.
To answer the question we began with, what does it mean to abelianize a group?
Verse III: Towards Abelianization~
What it intuitively means is that whenever $g,h \in G$, we must have $hg=gh$ with respect to the group operation (which has been suppressed here). But from the sketch of the proof done above, we realise that an abelian group has $C=\{e\}$ and therefore, in particular, we want every element of the form $ghg^{-1} h^{-1}$ to be the identity. And indeed, this is all that we need to do!
That is to say, if $gh g^{-1} h^{-1} = e, \ \forall g,h \in G$, then our $G$ is abelian!
We can see this as, \[ ghg^{-1}h^{-1}= e \Rightarrow ghg^{-1}= h \Rightarrow gh =hg \]
What does this tell us though?
If we make every commutator of our group $G$ trivial, then we will abelianize $G$, and conversely, to make $G$ abelian, we have to make every commutator of our group $G$ trivial.
But, how do we do that?
We have a commutator subgroup $C$ which is normal in $G$. We can take the quotient of $G$ by $C$ (since it will give us a quotient group) and it makes good sense as the first step to play around!
Now, what we obtain will be a quotient group consisting of all the cosets of $C$ in $G$ and one can see that this very well could be (rather will be) the 'nicer', that is, the 'abelian version' of our group $G$ which we have been looking for. Now let's see if this works!
Consider the abelianization of $G$: \(G^{\mathrm{ab}} = G/[G,G]\)
Consider the quotient \[G^{\mathrm{ab}} \;:=\; G/[G,G]\] It is abelian because for any \(a,b \in G\), which is to say that, \[(a[G,G])\,(b[G,G]) \;=\; ab[G,G]\] while \[(b[G,G])\,(a[G,G]) \;=\; ba[G,G]\] But \(ab\) and \(ba\) differ by the commutator- what does that mean?
Well, since $ab= [a,b] (ba)$, the two cosets coincide.
We have $[a,b] = aba^{-1} b^{-1}$ and if we multiply on the right by $(ba)$, we obtain \[ [a,b] (ba)= aba^{-1}b^{-1} (ba) = ab a^{-1} (b^{-1} b) a = ab a^{-1} a = ab\] \[\therefore ab = [a,b](ba),\] so \[ab[G,G] = [G,G] (ba)\] since \([a,b]\in [G,G]\). Hence the cosets commute and \(G^{\mathrm{ab}}\) is abelian. Conversely, if every commutator is trivial in a quotient \(G/N\), then \([G,G]\subseteq N\). This gives the minimality of \([G,G]\): \([G,G]\) is the smallest normal subgroup of \(G\) with abelian quotient, i.e. for any normal \(N\trianglelefteq G\), \(G/N\) is abelian if and only if \([G,G]\subseteq N\).
Verse IV: Universality in Action~
Now, the abelianization of our group $G$ is not just an abelian quotient; it is the universal one.
If \(A\) is abelian and \(\varphi:G\to A\) is any homomorphism, then there exists a unique homomorphism \(\overline{\varphi}:G^{\mathrm{ab}}\to A\) such that \(\varphi = \overline{\varphi}\circ \pi\), where \(\pi:G\to G^{\mathrm{ab}}\) is the natural projection.
Why? Since \(A\) is abelian, \(\varphi([a,b])=e\) for all \(a,b\in G\), so \([G,G]\subseteq \ker\varphi\).
And by the First Isomorphism Theorem, \(\varphi\) factors uniquely through \(G/[G,G]\).
And guess what? This universal property is often the most efficient definition of abelianization.
Epilogue~
When we begin with a non-abelian group, we constantly worry about the turbulence of non-commutativity and what abelianization does for us is- it takes away that turbulence and offers us a 'nicer' environment to work with.
The process gives a precise answer: we take the quotient by the commutator subgroup, leaving the 'largest' abelian image of the original group. This is not just convenient—it is universal. Any map from our group to an abelian group must factor uniquely through this construction. That is why abelianization matters: it formalizes the idea of extracting commutativity in the most general and economical way, without discarding more structure than necessary.