# Could the concept of “finite free groups” be possible?

Is it possible to define "finite free groups" ? could that make it easier to deal with group presentations ?

• @user82770 Then how would you multiply two words whose length was already at this limit (unless the limit is zero)? – Trevor Wilson Nov 25 '13 at 17:09
• By zero I assume you mean the identity of the group? If you did that then that would be a relation between the generators and the group would not be free. – Jim Nov 25 '13 at 17:14
• You can say whatever you want, @user82770, but then that finite group would definitely not be free in any even more or less close sense to what we all now know as "free groups"...But there are free solvable, free nilpotent, free abelian groups (free objects in the corresponding category), but all these non-trivial things are way non-finite... – DonAntonio Nov 25 '13 at 18:40
• I do not understand why this question has been closed. You get free groups, you get free abelian groups. You get free nilpotent groups. Why not free finite groups? As the answer shows, this question can be -essentially- formalised and answered. – user1729 Nov 26 '13 at 9:21
• Meta-discussion: meta.math.stackexchange.com/questions/11789/… – Willie Wong Nov 26 '13 at 11:09

## 3 Answers

The question is ill-posed. But let me mention one way how to make it precise.

If $\mathcal{C}$ is a concrete category, i.e. a category equipped with a forgetful functor $U : \mathcal{C} \to \mathsf{Set}$, then a free $\mathcal{C}$-object is usually defined to be one of the form $F(X)$, where $X$ is a set and $F : \mathsf{Set} \to \mathcal{C}$ is left adjoint of $U$. In other words, we have $\hom_{\mathcal{C}}(F(X),G) \cong \hom_{\mathsf{Set}}(X,U(G))$, natural in $G \in \mathcal{C}$. For example we get the usual notions of free groups, free $R$-modules, and if $R$ is commutative a free commutative $R$-algebra is a polynomial algebra over $R$. Taking $\mathcal{C}=\mathsf{FinGrp}$, we obtain the notion of a free (finite group). I've put the brackets here because I don't mean (free and finite) groups. For example, the trivial group is the free (finite group) on $\emptyset$ (in general, a free object on $\emptyset$ is the same as an initial object). I claim that no others exist:

Assume that $X$ is a non-empty set and that $F(X)$ is a finite group with the property $\hom_{\mathsf{FinGrp}}(F(X),G) \cong \hom_{\mathsf{Set}}(X,U(G))$, naturally in $G \in \mathsf{FinGrp}$, where $U(G)$ denotes the underlying set of a finite group $G$. Then we have a map $\iota : X \to U(F(X))$ which induces the bijection (Yoneda). Choose some $x_0 \in X$, and choose some finite cyclic group $G$ with generator $g$ whose order is larger than the order of $\iota(x_0)$. Define $f : X \to U(G)$ to be the map which is constant $g$. Then there is a homomorphism $\tilde{f} : F(X) \to G$ such that $\tilde{f} \circ \iota = f$. In particular, the order of $\tilde{f}(\iota(x_0))=g$ divides the order of $\iota(x_0)$, a contradiction. $\square$

We have proven that the functor $\mathsf{FinGrp} \to \mathsf{Set},~ G \mapsto \hom_{\mathsf{Set}}(X,U(G))$ is not representable. But it has a property related to that. Say for example $X=\{x_0\}$, so that we consider just $U$. The underlying set $U(G)$ of a finite group is the directed union of the $n$-torsion subsets $U_n(G) = \{g \in G : g^n = 1\}$ for $n \in \mathbb{N}^+$. For $n|m$ we have $U_n \subseteq U_m$. It follows that $$U = \varinjlim_n U_n \cong \varinjlim_n \hom_{\mathsf{FinGrp}}(\mathbb{Z}/n\mathbb{Z},-)$$ is a directed colimit of representable functors, i.e. ind-representable. One can show that the category of ind-representable functors on $\mathsf{FinGrp}$ is equivalent to the category of pro-finite groups. Here, $U$ corresponds to the pro-finite completion $\widehat{\mathbb{Z}} = \varprojlim_n \mathbb{Z}/n\mathbb{Z}$ of $\mathbb{Z}$, which is in fact the free pro-finite group on one generator. Although we haven't got free finite groups, we have free pro-finite groups (at least on finitely many generators).

The Burnside group $B(n,m)$ is in some sense the "free $n$-generator group with exponent $m$". Here "exponent $m$" means $x^m=1$ for all $x \in B(n,m)$. Interestingly some of these are finite and some aren't: $B(1,n)$ is cyclic of order $n$, but I believe it's still unknown whether or not $B(2,5)$ is finite. You can read about these groups in the wiki article http://en.wikipedia.org/wiki/Burnside_group

• That $m=2$ implies finite is a classic undergrad question, while that $m=3$ implies finite isn't much harder. Also, one can gain an analogue $B_f(n, m)$ of $B(n, m)$ for finite groups, by quotienting out the intersection of all subgroups of finite index. This is perhaps more relevant here. In fact, this is really close to an answer! The restricted Burnside problem asks whether this group $B_f(n, m)$ is finite. Efim Zelmalov solved this problem, and won a fields medal for his efforts! – user1729 Nov 26 '13 at 10:45

Groups have a first-order axiomatization, and one could repeat for groups the definition of pseudo-finite fields, which are the infinite fields satisfying the same sentences as (all) finite fields in the first-order theory of commutative fields.

[Edit. Search shows that the theory exists as expected, but is much more complicated than the theory of pseudofinite fields.]

The free $n$-generator pseudo-finite group would be the group containing $n$ distinct elements with no other properties except the ones that follow from the first-order theory of pseudofinite groups and the axiom "there exist $n$ distinct non-identity elements".

• Your last paragraph is unclear to me — what kind of properties are you considering, when you say “no other properties”? If you mean “first-order properties”, then this is impossible: the theory of pseudo-finite groups is clearly not a complete theory, but the set of sentences satisfied by any specific structure is always complete. If you take it to mean just “algebraic equations”, then it makes sense (and is equivalent I think to defining it as an initial object in the category of “psuedo-finite groups with n distinguished elements”), but it is not obvious that such a group exists. – Peter LeFanu Lumsdaine Dec 4 '13 at 23:16
• A first-order theory (complete or not) does not usually define an infinite structure up isomorphism, though there are exceptions in countable cardinality, like random graph or dense orders. However, one could hope for a parametrization of the elementary equivalence classes as something like $GL_n(F)$ for pseudofinite $F$, because (a quick skimming of online literature found that...) natural classes of pseudofinite groups seem to be closely tied to algebraic groups over pseudofinite fields. So to get actual uniqueness perhaps one wants "free $n$-generator group algebra". @PeterLeFanuLumsdaine – zyx Dec 7 '13 at 23:48