# Perfect groups of infinite commutator length: setup

At the end of my post Some trivia on derived subgroups I quickly mention that what I had been discussing up to then is essentially about the commutator length of the group. Recall that the derived subgroup is the group *generated by* all commutators, and that a group is said to be perfect if and only if it is equal to its derived subgroup. My previous post was about finite (perfect) groups and the necessity of generating the derived subgroups, seeing which groups have commutator length strictly greater than 1.

This post on the other hand concerns perfect groups with infinite commutator length: every element of the group is a product of finitely many commutators, but the commutator length of the elements can grow arbitrarily large. For this I will discuss what I've learned recently in the talk "Quasi-morphismes et longueurs de commutateurs" by Emmanuel Militon.

We will have to leave the realm of finite groups (and computational group theory) of course, which was the original motivation for the original post. But we will meet quasimorphisms (not to be confused with quasi-isomorphism or quasi-finite morphisms), apparently a rather new concept as far as I can tell. And an intriguing one too.

## Definitions

As already hinted at, we will define **commutator length** for single elements. Let $G$ be any group. The map $\textrm{cl}_G\colon G\to\mathbb{N}$ maps an element $g$ to the minimal number $n$ such that there exist $g_1,\dotsc,g_n,h_1,\dotsc,h_n\in G$ where $g=[g_1,h_1]\dotsm[g_n,h_n]$. This map can serve as a norm, because $\textrm{cl}_G(1_G)=0$ and we have a triangle inequality $\textrm{cl}_G(gh)\leq\textrm{cl}_G(g)+\textrm{cl}_G(h)$ because the minimal expressions of the two factors can be multiplied to get a not necessarily minimal expression for the product. We won't need this.

A quasimorphism on the other hand is something that is almost, but not quite, a group morphism landing in $\mathbb{R}$. More precisely a **quasimorphism** is a map (of sets) $q\colon G\to\mathbb{R}$ such that its deviation from being a group morphism is *uniformly bounded*, i.e. there exists a constant $C\geq 0$ such that for all $g,h\in G$ we have $|q(gh)-(q(g)+q(h))|\leq C$. The least such constant is called the **defect** and will be denoted $\mathrm{D}(q)$. By recurrence we can prove for any $g_1,\dotsc,g_n\in G$ that
$$
\displaystyle\left|q(g_1\cdots g_n)-\left(\sum_{i=1}^nq(g_i)\right)\right|\leq(n-1)\mathrm{D}(q)
$$

which is a relation we will need later on.

A personal note: I think *near morphism* would be a better term. Less confusion with existing concepts, and it covers the meaning better.

## Examples and homogeneity

Some less interesting examples are:

- the trivial map $q\colon G\to\mathbb{R}:g\mapsto 0$, which is (for obvious reasons) called the
*trivial quasimorphism*; - bounded maps
- actual group morphisms

In the argument about a group with infinite commutator length a far more interesting example will be given. It can be noted that the collection of quasimorphisms has the structure of a real vectorspace.

There is a subclass of quasimorphisms, which satisfies a stronger condition, and which will prove to be useful. Let's call a quasimorphism **homogeneous** if for all $g\in G$ and $n\in\mathbb{N}$ we have that $q(g^n)=nq(g)$. So instead of merely being close to it the quasimorphism behaves well for powers of elements. We can remark that the only bounded homogeneous quasimorphism is the trivial one and that the collection of homogeneous quasimorphisms is again a real vectorspace. A more important result is contained in the following lemma.

**Lemma** A homogeneous quasimorphism $q$ is invariant under conjugation, i.e. we have that $q(hgh^{-1})=q(g)$ for $g,h\in G$.

*Proof:* Take $n>1$, we have

$\left| q(hg^nh^{-1})-(q(h)+q(g^n)+q(h^{-1})) \right|\leq 2\mathrm{D}(q)$

but the left-hand side can be rewritten to $n|q(hgh^{-1})-q(g)|$ by the equality $q(hg^nh^{-1})=q((hgh^{-1})^n)=nq(hgh^{-1})$ and other straightforward applications of homogeneity. Now dividing both sides by $n$ and letting $n$ tend to infinity we get the desired equality. QED

## The main theorem

We can now state what should be considered the main theorem, explaining why the notion of quasimorphism pops up when investigating perfect groups with infinite commutator length.

**Theorem** If $G$ is a perfect group admitting a homogeneous quasimorphism then $\mathrm{cl}_G$ is unbounded.

*Proof:* Take $g\in G$ such that $q(g)\neq 0$. Define $\ell=\mathrm{cl}_G(g)$, then there exist $g_1,\dotsc,g_\ell,h_1,\dotsc,h_\ell\in G$ such that $g=[g_1,h_1]\dotsm[g_\ell,h_\ell]$. This product of commutators can be interpreted as $g_1(h_1g_1^{-1}h_1^{-1})\dotsm g_\ell(h_\ell g_\ell^{-1}h_\ell^{-1})$ and using the bound we obtained before we get
$$
\displaystyle\left| q(g)-\left(\sum_{i=1}^\ell q(g_i)+q(h_ig_i^{-1}h_i^{-1})\right) \right|\leq2\ell\mathrm{D}(q).
$$

Applying the previous lemma on the conjugation we see that the summation is in fact zero, so we get $|q(g)|\leq 2\mathrm{cl}_G(g)\mathrm{D}(q)|$. Now replacing $g$ by $g^n$ we get $|q(g^n)|=n|q(g^n)|\leq 2\mathrm{cl}_G(g^n)\mathrm{D}(q)$ . The left-hand side tends to infinity as $n$ grows, so the right-hand side grows too, and the only factor that can grow is the commutator length. QED

## What's next?

Actually not much remains, it suffices now to give any perfect group $G$ admitting such an unbounded homogeneous quasimorphism. This will be the content of the next post, and will require a little more build-up.