Could the concept of "finite free groups" be possible?

Question

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

Answer 1

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).

Answer 2

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

Answer 3

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".