## Nonabelian modules

This is a rough overview of my thoughts on a thing I've been thinking about, and as such is incomplete and may contain errors. Proofs have been omitted when writing them out would be at all tedious.

Edit: It has been pointed out to me that near-ring modules have already been defined, and the objects I describe in this post are just near-ring modules where the near-ring happens to be a ring.

### Introduction

As you all know (those of you who have the background for this post, anyway), an $R$-module is an abelian group $M$ (written additively) together with a multiplication map $R\times M\rightarrow M$ such that for all $\alpha,\beta\in R$ and $x,y\in M$, $\alpha\cdot\left(x+y\right)=\alpha\cdot x+\alpha\cdot y$$\left(\alpha+\beta\right)\cdot x=\alpha\cdot x+\beta\cdot x$, $\left(\alpha\beta\right)\cdot x=\alpha\cdot\left(\beta\cdot x\right)$, and $1\cdot x=x$.

What if we don't want to restrict attention to abelian groups? One could attempt to define a nonabelian module using the same axioms, but without the restriction that the group be abelian. As it is customary to write groups multiplicatively if they are not assumed to be abelian, we will do that, and the map $R\times M\rightarrow M$ will be written as exponentiation (since exponents are written on the right, I'll follow the definition of right-modules, rather than left-modules). The axioms become: for all $\alpha,\beta\in R$ and $x,y\in M$, $\left(xy\right)^{\alpha}=x^{\alpha}y^{\alpha}$$x^{\alpha+\beta}=x^{\alpha}x^{\beta}$, $x^{\left(\alpha\beta\right)}=\left(x^{\alpha}\right)^{\beta}$, and $x^{1}=x$.

What has changed? Absolutely nothing, as it turns out. The first axiom says again that $M$ is abelian, because $yx=x^{-1}\left(xy\right)^{2}y^{-1}=x^{-1}\left(x^{2}y^{2}\right)y^{-1}=xy$. We'll have to get rid of that axiom. Our new definition, which it seems to me captures the essence of a module except for abelianness:

A nonabelian $R$-module is a group $M$ (written multiplicatively) together with a scalar exponentiation map $R\times M\rightarrow M$ such that for all $\alpha,\beta\in R$ and $x\in M$, $x^{1}=x$$x^{\alpha+\beta}=x^{\alpha}x^{\beta}$, and $x^{\left(\alpha\beta\right)}=\left(x^{\alpha}\right)^{\beta}$.

These imply that $x^{0}=1$, $1^{\alpha}=1$, and $x^{-1}$ is the inverse of $x$, because $x\cdot x^{0}=x^{1}x^{0}=x^{1+0}=x^{1}=x$$1^{\alpha}=\left(1^{0}\right)^{\alpha}=1^{\left(0\alpha\right)}=1^{0}=1$, and $x\cdot x^{-1}=x^{1-1}=1$.

Just like a $\mathbb{Z}$-module is just an abelian group, a nonabelian $\mathbb{Z}$-module is just a group. Just like a $\mathbb{Z}/n\mathbb{Z}$-module is an abelian group whose exponent divides $n$, a nonabelian $\mathbb{Z}/n\mathbb{Z}$-module is a group whose exponent divides $n$.

### Exponentiation-like families of operations

Perhaps a bit more revealing is what nonabelian modules over free rings look like, since then the generators are completely generic ring elements. Where $A$ is the generating set, a $\mathbb{Z}\left\langle A\right\rangle$-module is an abelian group together with endomorphisms $\left\{ x\mapsto\alpha x\mid\alpha\in A\right\}$, which tells us that modules are about endomorphisms of an abelian group indexed by the elements of a ring. Nonabelian modules are certainly not about endomorphisms. After all, in a nonabelian group, the map $x\mapsto x^{2}$ is not an endomorphism. I will call the things that nonabelian modules are about "exponentiation-like families of operations'', and give four equivalent definitions, in roughly increasing order of concreteness and decreasing order of elegance. Definition 2 uses basic model theory, so skip it if that scares you. Definition 3 is the "for dummies'' version of definition 2.

Definition 0: Let $G$ be a group, and let $A$ be a family of functions from $G$ to $G$ (not necessarily endomorphisms). If $G$ can be made into a nonabelian $\mathbb{Z}\left\langle A\right\rangle$-module such that $x^{\alpha}=\alpha\left(x\right)$ for $x\in G$ and $\alpha\in A$, then $A$ is called an exponentiation-like family of operations on $G$. If so, the nonabelian $\mathbb{Z}\left\langle A\right\rangle$-module structure on $G$ with that property is unique, so define $x^{p}$ to be its value according to that structure, for $p\in\mathbb{Z}\left\langle A\right\rangle$ and $x\in G$.

Definition 1: $A$ is an exponentiation-like family of operations on $G$ if for all $x\in G$, the smallest subgroup containing $x$ which is closed under actions by elements of $A$ (which I will call $\overline{\left\{ x\right\} }$) is abelian, and the elements of $A$ restrict to endomorphisms of it. Using the universal property of $\mathbb{Z}\left\langle A\right\rangle$, this induces a homomorphism $\mathbb{Z}\left\langle A\right\rangle \rightarrow\text{End}\left(\overline{\left\{ x\right\} }\right)^{\text{op}}$. Let $x^{p}$ denote the action of $p$ on $x$ under that map, for $p\in\mathbb{Z}\left\langle A\right\rangle$. By $\text{End}\left(\overline{\left\{ x\right\} }\right)^{\text{op}}$, I mean the endomorphism ring of $\overline{\left\{ x\right\} }$ with composition running in the opposite direction (i.e., the multiplication operation given by $\left(f,g\right)\mapsto g\circ f$). This is because of the convention that nonabelian modules are written as nonabelian right-modules by default.

Definition 2: Let consider the language $\mathcal{L}_{Rings}\sqcup A$, where $\mathcal{L}_{Rings}:=\left\{ 0,1,+,-,\cdot\right\}$ is the language of rings, and each element of $A$ is used as a constant symbol. Closed terms in $\mathcal{L}_{Rings}\sqcup A$ act as functions from $G$ to $G$, with the action of $t$ written as $x\mapsto x^{t}$, defined inductively as: $x^{0}:=1$, $x^{1}:=x$, $x^{\alpha}:=\alpha\left(x\right)$ for $\alpha\in X$, $x^{t+s}:=x^{t}x^{s}$, $x^{-t}:=\left(x^{t}\right)^{-1}$, and $x^{ts}:=\left(x^{t}\right)^{s}$ for closed $\mathcal{L}_{Rings}\sqcup A$-terms $t$ and $s$. $A$ is called an exponentiation-like family of operations on $G$ if $x^{t}=x^{s}$ whenever $T_{Rings}\models t=s$, where $T_{Rings}$ is the theory of rings. If $A$ is an exponentiation-like family of operations on $G$ and $p\in\mathbb{Z}\left\langle A\right\rangle$ is a noncommutative polynomial with variables in $A$, then for $x\in G$$x^{p}$ is defined to be $x^{t}$ where $t$ is any term representing $p$.

Definition 3: Pick a total order on the free monoid on $A$ (e.g. by ordering $A$ and then using the lexicographic order). The order you use won't matter. Given $x\in G$ and $w:=\alpha_{1}...\alpha_{n}$ in the free monoid on $A$, let $x^{w}=\alpha_{n}\left(...\alpha_{1}\left(x\right)\right)$. Where $p\in\mathbb{Z}\left\langle A\right\rangle$ is a noncommutative polynomial, $p=c_{1}w_{1}+...+c_{n}w_{n}$ for some $c_{1},...,c_{n}\in\mathbb{Z}$ and decreasing sequence $w_{1},...,w_{n}$ of noncommutative monomials (elements of the free monoid on $A$). Let $x^{p}=\left(x^{c_{1}}\right)^{w_{1}}...\left(x^{c_{n}}\right)^{w_{n}}$$A$ is called an exponentiation-like family of operations on $G$ if for every $x\in G$ and $p,q\in\mathbb{Z}\left\langle A\right\rangle$$x^{pq}=\left(x^{p}\right)^{q}$ and $x^{p+q}=x^{p}x^{q}$.

These four definitions of exponentiation-like family are equivalent, and for exponentiation-like families, their definitions of exponentiation by a noncommutative polynomial are equivalent.

Facts: $\emptyset$ is an exponentiation-like family of operations on $G$. If $A$ is an exponentiation-like family of operations on $G$ and $B\subseteq A$, then so is $B$. If $G$ is abelian, then $\text{End}\left(G\right)$ is exponentiation-like. Given a nonabelian $R$-module structure on $G$, the actions of the elements of $R$ on $G$ form an exponentiation-like family. In particular, if $A$ is an exponentiation-like family of operations on $G$, then so is $\mathbb{Z}\left\langle A\right\rangle$, with the actions being defined as above.

[The following paragraph has been edited since this comment.]

For an abelian group $A$, the endomorphisms of $A$ form a ring $\text{End}\left(A\right)$, and an $R$-module structure on $A$ is simply a homomorphism $R\rightarrow\text{End}\left(A\right)$. Can we say a similar thing about exponentiation-like families of operations of $G$? Let $\text{Exp}\left(G\right)$ be the set of all functions $G\rightarrow G$ (as sets). Given $\alpha,\beta\in\text{Exp}\left(G\right)$, let multiplication be given by composition: $x^{\left(\alpha\beta\right)}=\left(x^{\alpha}\right)^{\beta}$, addition be given by $x^{\alpha+\beta}=x^{\alpha}x^{\beta}$, negation be given by $x^{-\alpha}=\left(x^{\alpha}\right)^{-1}$, and $0$ and $1$ be given by $x^{0}=1$ and $x^{1}=x$. This makes $\text{Exp}\left(G\right)$ into a near-ring. A nonabelian $R$-module structure on $G$ is a homomorphism $R\rightarrow\text{Exp}\left(G\right)$, and a set of operations on $G$ is an exponentiation-like family of operations on $G$ if and only if it is contained in a ring which is contained in $\text{Exp}\left(G\right)$.

### Some aimless rambling

What are some interesting examples of nonabelian modules that are not abelian? (That might sound redundant, but "nonabelian module'' means that the requirement of abelianness has been removed, not that a requirement of nonabelianness has been imposed. Perhaps I should come up with better terminology. To make matters worse, since the requirement that got removed is actually stronger than abelianness, there are nonabelian modules that are abelian and not modules. For instance, consider the nonabelian $\mathbb{Z}\left[\alpha\right]$-module whose underlying set is the Klein four group (generated by two elements $a,b$) such that $a^{\alpha}=a$, $b^{\alpha}=b$, and $\left(ab\right)^{\alpha}=1$.)

In particular, what do free nonabelian modules look like? The free nonabelian $\mathbb{Z}$-modules are, of course, free groups. The free nonabelian $\mathbb{Z}/n\mathbb{Z}$-modules have been studied in combinatorial group theory; they're called Burnside groups. (Fun but tangential fact: not all Burnside groups are finite (the Burnside problem), but despite this, the category of finite nonabelian $\mathbb{Z}/n\mathbb{Z}$-modules has free objects on any finite generating set, called Restricted Burnside groups.)

The free nonabelian $\mathbb{Z}\left[\alpha\right]$-modules are monstrosities. They can be constructed in the usual way of constructing free objects in a variety of algebraic structures, but that construction seems not to be very enlightening about their structure. So I'll give a somewhat more direct construction of the free nonabelian $\mathbb{Z}\left[\alpha\right]$-module on $d$ generators, which may also not be that enlightening, and which is only suspected to be correct. Define an increasing sequence of groups $G_{n}$, and functions $\alpha_{n}:G_{n}\rightarrow G_{n+1}$, as follows: $G_{0}$ is the free group on $d$ generators. Given $G_{n}$, and given a subgroup $X\leq G_{n}$, let the top-degree portion of $X$ be $\alpha_{n-1}^{k}\left(X\right)$ for the largest $k$ such that this is nontrivial. Let $H_{n}$ be the free product of the top-degree portions of maximal abelian subgroups of $G_{n}$. Let $G_{n+1}$ be the free product of $G_{n}$ with $H_{n}$ modulo commutativity of the maximal abelian subgroups of $G_{n}$ with the images of their top-degree portions in $H_{n}$. Given a maximal abelian subgroup $X\leq G_{n}$, let $\alpha_{n}\restriction_{X}$ be the homomorphism extending $\alpha_{n-1}\restriction_{X\cap G_{n-1}}$ which sends the top-degree portion identically onto its image in $H_{n}$. Since every non-identity element of $G_{n}$ is in a unique maximal abelian subgroup, this defines $\alpha_{n}$. $G:=\bigcup_{n}G_{n}$ with $\alpha:=\bigcup_{n}\alpha_{n}$ is the free nonabelian $\mathbb{Z}\left[\alpha\right]$-module on $d$ generators. If $A$ is a set, the free nonabelian $\mathbb{Z}\left\langle A\right\rangle$-modules can be constructed similarly, with $\left|A\right|$ copies of $H_{n}$ at each step. Are these constructions even correct? Are there nicer ones?

A nonabelian $\mathbb{Z}\left[\frac{1}{2}\right]$-module would be a group with a formal square root operation. As an example, any group of odd exponent $n$ can be made into a $\mathbb{Z}\left[\frac{1}{2}\right]$-module in a canonical way by letting $x^{\frac{1}{2}}=x^{\frac{n+1}{2}}$. More generally, any group of finite exponent $n$ can be made into a $\mathbb{Z}\left[\left\{ p^{-1}|p\nmid n\right\} \right]$-module in a similar fashion. Are there any more nice examples of nonabelian modules over localizations of $\mathbb{Z}$?

In particular, a nonabelian $\mathbb{Q}$-module would be a group with formal $n$th root operations for all $n$. What are some nonabelian examples of these? Note that nonabelian $\mathbb{Q}$-modules cannot have any torsion, for suppose $x^{n}=1$ for some $n\neq0$. Then $x=\left(x^{n}\right)^{\frac{1}{n}}=1^{\frac{1}{n}}=1$. More generally, nonabelian modules cannot have any $n$-torsion (meaning $x^{n}=1\implies x=1$) for any $n$ which is invertible in the scalar ring.

The free nonabelian $\mathbb{Z}\left[\frac{1}{m}\right]$-modules can be constructed similarly to the construction of free nonabelian $\mathbb{Z}\left[\alpha\right]$-modules above, except that when constructing $G_{n+1}$ from $G_{n}$ and $H_{n}$, we also mod out by elements of $G_{n}$ being equal to the $m$th powers of their images in $H_{n}$. Using the fact that $\mathbb{Q}\cong\mathbb{Z}\left\langle \left\{ p^{-1}|\text{primes }p\right\} \right\rangle$, this lets us modify the construction of free nonabelian $\mathbb{Z}\left\langle A\right\rangle$-modules to give us a construction of free nonabelian $\mathbb{Q}$-modules. Again, is there a nicer way to do it?

### Topological nonabelian modules

It is also interesting to consider topological nonabelian modules over topological rings; that is, nonabelian modules endowed with a topology such that the group operation and scalar exponentiation are continuous. A module over a topological ring has a canonical finest topology on it, and the same remains true for nonabelian modules. For finite-dimensional real vector spaces, this is the only topology. Does the same remain true for finitely-generated nonabelian $\mathbb{R}$-modules? Finite-dimensional real vector spaces are complete, and topological nonabelian modules are, in particular, topological groups, and can thus be made into uniform spaces, so the notion of completeness still makes sense, but I think some finitely-generated nonabelian $\mathbb{R}$-modules are not complete.

A topological nonabelian $\mathbb{R}$-module is a sort of Lie group-like object. One might try constructing a Lie algebra for a complete nonabelian $\mathbb{R}$-module $M$ by letting the underlying set be $M$, and defining $x+y=\lim_{\varepsilon\rightarrow0}\left(x^{\varepsilon}y^{\varepsilon}\right)^{\left(\varepsilon^{-1}\right)}$ and $\left[x,y\right]=\lim_{\varepsilon\rightarrow0}\left(x^{\varepsilon}y^{\varepsilon}x^{-\varepsilon}y^{-\varepsilon}\right)^{\left(\varepsilon^{-2}\right)}$. One might try putting a differential structure on $M$ such that this is the Lie algebra of left-invariant derivations. Does this or something like it work?

A Lie group is a nonabelian $\mathbb{R}$-module if and only if its exponential map is a bijection between it and its Lie algebra. In this case, scalar exponentiation is closely related to the exponential map by a compelling formula: $x^{\alpha}=\exp\left(\alpha\exp^{-1}\left(x\right)\right)$. As an example, the continuous Heisenberg group is a nonabelian $\mathbb{R}$-module which is not abelian. This observation actually suggests a nice class of examples of nonabelian modules without a topology: given a commutative ring $R$, the Heisenberg group over $R$ is a nonabelian $R$-module.

The Heisenberg group of dimension $2n+1$ over a commutative ring $R$ has underlying set $R^{n}\times R^{n}\times R^{1}$, with the group operation given by $\left(\boldsymbol{a}_{1},\boldsymbol{b}_{1},c_{1}\right)*\left(\boldsymbol{a}_{2},\boldsymbol{b}_{2},c_{2}\right):=$$\left(\boldsymbol{a}_{1}+\boldsymbol{a}_{2},\boldsymbol{b}_{1}+\boldsymbol{b}_{2},c_{1}+c_{2}+\boldsymbol{a}_{1}\cdot\boldsymbol{b}_{2}-\boldsymbol{a}_{2}\cdot\boldsymbol{b}_{1}\right)$. The continuous Heisenberg group means the Heisenberg group over $\mathbb{R}$. Scalar exponentiation on a Heisenberg group is just given by scalar multiplication: $\left(\boldsymbol{a},\boldsymbol{b},c\right)^{\alpha}:=\left(\alpha\boldsymbol{a},\alpha\boldsymbol{b},\alpha c\right)$.