# Making sense of the commutator

For a group $G$, the commutator of two elements is defined as $[a,b]=aba^{-1}b^{-1}$, and is usually said to measure the extent to which the elements $a$ and $b$ fail to commute.

I'm having some trouble making sense of the last bit: I understand that if $a$ and $b$ commute, then $[a,b]=e$. But if $a$ and $b$ don't commute, in what sense is the commutator actually capturing the extent of their failure to commute, since there is no way to talk about how "far" an element $g\in G$ is from the identity?

Am I just interpreting the word "measure" too literally here, or is there actually a way to think about commutators that makes it clear in what sense they compare the way two pairs of elements fail to commute?

• You are being too literal. – Mariano Suárez-Álvarez Jul 6 '16 at 23:43
• Thinking of a given commutator as a distance is too literal. But the size of the subgroup generated by the commutators is like a distance in that sense, because it will be contained in any normal subgroup for which the quotient is abelian. – Matt Samuel Jul 6 '16 at 23:43
• Maybe you could think it as an actual measure. Like a relation $\mu\subset G^2\times\{0,1\}$ where $\mu(a,b,1)$ says that $[a,b]=e$, or that it is truth that they commute, and $\mu(a,b,0)$ says $[a,b]\neq e$, they don't commute. (I'm not sure if this relation $\mu$ is a function, maybe you can check it). – edgar alonso Jul 6 '16 at 23:45
• Note that $[a,b] = ab(ba)^{-1}$, which you might call the "difference" between $ab$ and $ba$. Also, $[ab] = (aba^{-1})b^{-1}$, so it is also the "difference" between $b$ and its conjugate $aba^{-1}$. But you are correct that without some additional structure, we don't have a way to measure the size of this "difference" except for distinguishing between equality ($a$ and $b$ commute) and unequality ($a$ and $b$ do not commute). – Bungo Jul 6 '16 at 23:47
• For the context of an arbitrary group, you can't take this too literally. However, in a group where you have the additional structure of distance (for example, a Lie group), the notion of "extent" takes a real and important meaning. – Omnomnomnom Jul 6 '16 at 23:56

We can't really say "how non-commutative" $a$ and $b$ are, without some corresponding notion of "how much not the identity" any given element of a group $G$ is, as you point out. For some groups, we may be able to do this, but in general, there's no "universal" way.

The real value in this, is not the individual commutators $[a,b]$, but rather the commutator subgroup $[G,G]$. It should be clear $G/[G,G]$ is abelian, for:

$(x[G,G])(y[G,G])(x[G,G])^{-1}(y[G,G])^{-1} = [x,y][G,G] = e[G,G] = [G,G]$

But the story doesn't end there, if $N$ is any normal subgroup such that $G/N$ is abelian, we have $[G,G] \subseteq N$. The reason is very plain:

if for any $x,y \in G$, we have $(xN)(yN) = xyN = yxN = (yN)(xN)$, then we must have $xy(yx)^{-1} = [x,y] \in N$ for any pair $x,y \in G$. Thus $[G,G]$ is minimal among all normal subgroup $N$ that make $G/N$ abelian.

Another nice thing about this, is that the way we do it doesn't really depend on the group $G$ in the following sense: if $\phi:G \to H$ is a group homomorphism, we get a group homomorphism $\tilde{\phi}:G/[G,G] \to H/[H,H]$ of abelian groups defined by:

$\tilde{\phi}(x[G,G]) = \phi(x)[H,H]$, since a homomorphism preserves commutators:

$\phi([x,y]) = [\phi(x),\phi(y)]$.

To begin to make some sense it is convenient to see that the commutators subgroup $G'$ makes the quotient $G/G'$ be abelian. It is like killing the non-commutativity in $G$.