Some months ago, I found myself wondering about abstract sets and how exactly we can put group structures on them. Groups are often presented as known quantities and there are plenty of examples, e.g. $\mathbb{Z}$ and $\mathbb{R}$. When developing mathematical structures it is of interest to know how common they can be, how broad they are and if you can construct them. This means we must ask the question of when a given set and structure they are compatible.
In the case of finite fields, we know that any finite field must have a specific number of elements, particularly there must be a prime power of them, i.e. $p^k$ for some prime $p$. Thus if we were given an arbitrary finite set, there is no guarantee that we can put a field structure on the set. This tells us that field structures are very restrictive. We could ask a few of weaker questions: given an arbitrary set, can we put a group structure on it? Could we put a ring structure on it? Or are these structures somehow too rigid to accommodate an arbitrary set? Surely if we cannot put a group structure on a particular set, we cannot put a ring structure on that same set. Let us first address the first question.
Putting a group structure on an arbitrary set
Finite sets
Let us first consider the finite setting. Suppose $G = \{g_0, \ldots, g_{n-1}\}$ is a finite set. The elements $g_i$ have no algebraic properties a priori since $G$ is an abstract set. If we were to put a group structure on $G$, the most obvious choice is to associate $G$ to $\mathbb{Z}_n$. To do so, simply map $g_i$ to $i$ and inherit the group operation. In more concrete terms, let $\cdot:G\times G\to G$ be defined by $g_i\cdot g_j = g_{(i+j)\pmod n}$.
It's not hard to check that this does indeed define a group. Associativity is taken care of by noting that $(g_i\cdot g_j)\cdot g_k = g_{(i+j+k)\pmod n} = g_i\cdot (g_j\cdot g_k)$ which follows from elementary properties of modular arithmetic. $g_0$ is an identity element since $g_0\cdot g_i = g_{(0+i)\pmod n} = g_i$. Likewise, inverses exist since we have $g_i\cdot g_{n-i} = g_{(i+n-i)\pmod n} = g_0$.
The case of finite sets is pretty simple! Note that it was not claimed in the above construction that the group structure was unique. In specific examples it is unique, but this isn't the topic of this post. If for instance $|G| = 4$ we could put two different group structures on $G$: the group structure of $\mathbb{Z}_4$ and the Klein four-group.
As is the common theme in set theory, we build up in layers to get an intuition for how things should proceed. After covering the case of finite sets, the next most obvious step is to consider countable sets.
Countable sets
Suppose now that $G$ is a countable set. Let $\{\ldots, g_{-2}, g_{-1}, g_0, g_1, g_2, \ldots\}$ be an enumeration of $G$. Such an enumeration exists since $G$ is countable. Again we can define a binary operation on $G$; this time, let $\cdot:G\times G\to G$ be given by $g_i\cdot g_j = g_{i+j}$.
Again this defines a group structure on $G$. Associativity follows from associativity of the addition of integers: $(g_i\cdot g_j)\cdot g_k = g_{i+j+k} = g_i\cdot(g_j\cdot g_k)$. The identity element is very obviously $g_0$ and the inverse of $g_i$ is clearly $g_{-i}$.
An aside about set theory and the axiom of choice
The next step might be to consider uncountable sets but there are not many uncountable groups that we work with in nature. The most common by far are $\mathbb{R}$ and $\mathbb{C}$ (and subgroups thereof). These are uncountable groups but their cardinality is only $2^{|\mathbb{N}|}$. There are however sets that are "more" infinite than the reals and complexes (for instance the set of all subsets of real numbers has strictly greater cardinality).
Thus we cannot appeal directly to a known group to address this issue in full generality. The next best thing is to come up with some set-theoretic approach to address the problem at hand. At this stage, we must state a little about what model of set theory we are using. By far the most widely accepted model is Zermelo-Fraenkel set theory. Zermelo-Fraenkel (ZF) set theory is often not quite strong enough in mathematics and an additional axiom is imposed: the axiom of choice.
The axiom of choice allows for many nice things in mathematics (e.g. every vector space has a basis and the Cartesian product of nonempty sets is again nonempty) but also introduces some somewhat unsavory consequences (e.g. Banach-Tarski paradox and similar results). Many proofs involving the axiom of choice are existence proofs, i.e. a certain object is proved to exist but there is no way of knowing what that object truly looks like. As such, the axiom of choice is a bit of a contentious piece of mathematics, but most mathematicians accept it as a valid axiom despite its pitfalls.
The axiom of choice formally states that if one has an indexed set of nonempty sets $(S_i)_{i\in I}$, then there is an indexed set $(x_i)_{i\in I}$ such that $x_i\in S_i$ for each $i$. Informally, this can be thought of in the following way: given a collection of nonempty sets, one can pick an element from each set at once. This seems intuitively obvious and in the case of a finite number of sets it is indeed obvious; however in the case of an infinite number of sets, it is not guaranteed. The axiom of choice plays a critical role in what follows.
Arbitrary infinite sets
Suppose now that $G$ is an arbitrary infinite set. Since we cannot appeal to groups we're familiar with, all we have at our disposal in terms of operations are the set operations and subsets of $G$. Somehow we must endeavor to construct a group structure using only these ideas.
Since groups only come with one binary operation, we must choose the correct set operation. For instance, unions and intersections are "one way" processes in some sense. More concretely, given an arbitrary set $B$, $A$ cannot be obtained from $A\cup B$ using only unions. As such, inverses wouldn't exist if we chose our group operation to be set union. Similar logic applies to set intersection. Set difference is not a good option either since it is not associative.
Hence we must look to "higher order" set operations, i.e. operations involving multiple set operations. The simplest higher order set operation is that of symmetric difference: $A\oplus B = (A\setminus B)\cup (B\setminus A)$. Visually, symmetric difference is represented by
where the shaded region represents the symmetric difference. Note that $A\oplus\emptyset = A$ for any set $A$. It is not difficult to prove that $\oplus$ is associative. It is also clear that $A\oplus A = \emptyset$. Thus $\oplus$ seems like a good operation with which to define a group structure on an arbitrary set.
Thus far we have not yet used the axiom of choice. It is at this stage that we use the axiom of choice. It is a consequence of the axiom of choice that the set of all finite subsets of an infinite set has the same cardinality of the set itself. The proof for this is not too difficult and relies on the Cantor-Schroeder-Bernstein theorem and some cardinal arithmetic. The axiom of choice is packed into the cardinal arithmetic that is required.
Denote the set of all finite subsets of the set $X$ by $\mathcal{F}(X)$. By the above, there is a bijection between $G$ and $\mathcal{F}(G)$. Let this bijection be denoted $f$. $\mathcal{F}(G)$ can be turned into a group with the operation of symmetric difference. Note that the symmetric difference of two finite sets is again finite. This however is not a group structure on $G$ but $\mathcal{F}(G)$.
To get a group structure on $G$, note that every $g\in G$ can be written as $f^{-1}(S)$ where $S\in\mathcal{F}(G)$. Let $\cdot$ denote the binary operation on $G$. Suppose that $g_1 = f^{-1}(S_1)$ and $g_2 = f^{-1}(S_2)$. Define $g_1\cdot g_2$ to be $f^{-1}(S_1\oplus S_2)$.
We have to show that this does indeed define a group structure on $G$. Consider $(g_1\cdot g_2)\cdot g_3$ This is nothing more than $f^{-1}(S_1\oplus S_2\oplus S_3)$ which is of course $g_1\cdot(g_2\cdot g_3)$. The identity element is the inverse image of the empty set since if $e = f^{-1}(\emptyset)$, then $e\cdot g = f^{-1}(\emptyset\oplus S) = f^{-1}(S) = g$. Inverses also exist: if $g = f^{-1}(S)$, then $g\cdot g = f^{-1}(S\oplus S) = f^{-1}(\emptyset) = e$.
Consequently $(G,\cdot)$ is indeed a group. In fact it is an abelian group since symmetric difference is, well, symmetric: $f^{-1}(S_1)\cdot f^{-1}(S_2) = f^{-1}(S_1\oplus S_2) = f^{-1}(S_2\oplus S_1) = f^{-1}(S_2)\cdot f^{-1}(S_1)$.
In general we were Abel (haha) to put a group structure on an arbitrary set (assuming the axiom of choice). Not only that, we were able to put an abelian group structure on the set. This is important for the ring portion of the secondary question. If we were not able to guarantee that we would have no hope of addressing the question of putting a ring structure on an arbitrary set.
Putting a ring structure on an arbitrary set
Finite sets
When considering the question of being able to put a group structure on an arbitrary set, we had to differentiate the finite and infinite settings. Unsurprisingly, we must do the same for the ring case. If $R$ is a finite set with $|R| = n$, then we could put an abelian group structure on $R$ by associating it to $\mathbb{Z}_n$. As is well-known, $\mathbb{Z}_n$ is already a unital ring! We don't have any further work to do in this case. The infinite case is much more interesting.
Infinite sets
Above we constructed the abelian group for arbitrary infinite sets so we need only to construct the ring structure for it. Let $R$ be an infinite set. We lifted the group structure of $R$ from $\mathcal{F}(R)$ with symmetric difference. It stands to reason that to construct a ring structure on $R$, we'd need to first put a ring structure on $\mathcal{F}(R)$.
To do this, a multiplication operation must be devised. Way back when when first discussing the elementary set operations, I likened them to usual arithmetic operations. Union is a bit like addition and intersection is a bit like multiplication. For this reason, intersection seems to be the logical choice for the second ring operation of multiplication. Before showing that this does indeed work, let's consider union as our notion of multiplication.
If union is to play the role of multiplication, it must "distribute" over symmetric difference (addition), i.e. $A\cup (B\oplus C) = (A\oplus B)\cup (A\oplus C)$. $A,B,C$ are arbitrary so taking $A = B = C$, the left hand side would become $A$ whereas the right hand side is empty. Thus unions do not distribute over symmetric difference in general. As such, union cannot play the role of multiplication. Intersection, however, does exactly what we want which can be easily verified.
Therefore the natural notion of multiplication on $\mathcal{F}(R)$ is set intersection. Like before, we must pull back the multiplication to $R$. Let $r_1 = f^{-1}(S_1)$ and $r_2 = f^{-1}(S_2)$, where $f$ is the bijection between $R$ and $\mathcal{F}(R)$ as before. Let $\star$ denote the multiplication on $R$ and define it by $r_1\star r_2 = f^{-1}(S_1\cap S_2)$.
Note here that there is no unit since the unit would be played by $R$ itself.
It remains to see that this does indeed define a ring structure on $R$. It must then be checked that the ring axioms are obeyed. This is tantamount to checking that multiplication distributes over addition. Consider $r_1 = f^{-1}(S_1)$, $r_2 = f^{-1}(S_2)$ and $r_3 = f^{-1}(S_3)$, then
\begin{eqnarray*}
r_1\star(r_2\cdot r_3) &=& r_1\star f^{-1}(S_2\oplus S_3) \\
&=& f^{-1}(S_1\cap(S_2\oplus S_3)) \\
&=& f^{-1}((S_1\cap S_2)\oplus (S_1\cap S_3)) \\
&=& f^{-1}(S_1\cap S_2)\cdot f^{-1}(S_1\cap S_3) \\
&=& (r_1\star r_2)\cdot (r_1\star r_3)
\end{eqnarray*}
Hence multiplication distributes over addition and moreover $R$ is endowed with a commutative ring structure since multiplication is commutative (which can be tracked back to the commutativity of set intersection).
We've set out to do exactly what we wanted! We've successfully shown that every set can be endowed with a ring (and a fortiori a group) structure. However one seeming issue is that the ring that we naturally developed in the infinite setting was non-unital. This has its roots in the fact that the unit with regards to set intersection is the universe set. Since our universe set was assumed to be infinite, it is not included in the set of all finite subsets.
It stands to reason then that we should modify our initial construction to include the whole set $R$ to begin with. However if we do this, we immediately lose closure: if $S$ is a finite subset of $R$, then $S\oplus R = R\setminus S$ which is most definitely not a finite set. We could remedy this by considering a union of the sets of finite subsets and cofinite subsets of $R$. This set has the same cardinality as $R$ by cardinal arithmetic. Hence there is a bijection between these sets and the same analysis follows through quite naturally. However this approach gives rise to a unital ring structure on $R$ since now $R$ is included.
The next most logical question to ask is if division rings can be formed on arbitrary sets. This however is beyond my expertise. Based on the construction above, it seems unlikely as intersection is (as stated above) a one-way process in some sense.
In the case of finite fields, we know that any finite field must have a specific number of elements, particularly there must be a prime power of them, i.e. $p^k$ for some prime $p$. Thus if we were given an arbitrary finite set, there is no guarantee that we can put a field structure on the set. This tells us that field structures are very restrictive. We could ask a few of weaker questions: given an arbitrary set, can we put a group structure on it? Could we put a ring structure on it? Or are these structures somehow too rigid to accommodate an arbitrary set? Surely if we cannot put a group structure on a particular set, we cannot put a ring structure on that same set. Let us first address the first question.
Putting a group structure on an arbitrary set
Finite sets
Let us first consider the finite setting. Suppose $G = \{g_0, \ldots, g_{n-1}\}$ is a finite set. The elements $g_i$ have no algebraic properties a priori since $G$ is an abstract set. If we were to put a group structure on $G$, the most obvious choice is to associate $G$ to $\mathbb{Z}_n$. To do so, simply map $g_i$ to $i$ and inherit the group operation. In more concrete terms, let $\cdot:G\times G\to G$ be defined by $g_i\cdot g_j = g_{(i+j)\pmod n}$.
It's not hard to check that this does indeed define a group. Associativity is taken care of by noting that $(g_i\cdot g_j)\cdot g_k = g_{(i+j+k)\pmod n} = g_i\cdot (g_j\cdot g_k)$ which follows from elementary properties of modular arithmetic. $g_0$ is an identity element since $g_0\cdot g_i = g_{(0+i)\pmod n} = g_i$. Likewise, inverses exist since we have $g_i\cdot g_{n-i} = g_{(i+n-i)\pmod n} = g_0$.
The case of finite sets is pretty simple! Note that it was not claimed in the above construction that the group structure was unique. In specific examples it is unique, but this isn't the topic of this post. If for instance $|G| = 4$ we could put two different group structures on $G$: the group structure of $\mathbb{Z}_4$ and the Klein four-group.
As is the common theme in set theory, we build up in layers to get an intuition for how things should proceed. After covering the case of finite sets, the next most obvious step is to consider countable sets.
Countable sets
Suppose now that $G$ is a countable set. Let $\{\ldots, g_{-2}, g_{-1}, g_0, g_1, g_2, \ldots\}$ be an enumeration of $G$. Such an enumeration exists since $G$ is countable. Again we can define a binary operation on $G$; this time, let $\cdot:G\times G\to G$ be given by $g_i\cdot g_j = g_{i+j}$.
Again this defines a group structure on $G$. Associativity follows from associativity of the addition of integers: $(g_i\cdot g_j)\cdot g_k = g_{i+j+k} = g_i\cdot(g_j\cdot g_k)$. The identity element is very obviously $g_0$ and the inverse of $g_i$ is clearly $g_{-i}$.
An aside about set theory and the axiom of choice
The next step might be to consider uncountable sets but there are not many uncountable groups that we work with in nature. The most common by far are $\mathbb{R}$ and $\mathbb{C}$ (and subgroups thereof). These are uncountable groups but their cardinality is only $2^{|\mathbb{N}|}$. There are however sets that are "more" infinite than the reals and complexes (for instance the set of all subsets of real numbers has strictly greater cardinality).
Thus we cannot appeal directly to a known group to address this issue in full generality. The next best thing is to come up with some set-theoretic approach to address the problem at hand. At this stage, we must state a little about what model of set theory we are using. By far the most widely accepted model is Zermelo-Fraenkel set theory. Zermelo-Fraenkel (ZF) set theory is often not quite strong enough in mathematics and an additional axiom is imposed: the axiom of choice.
The axiom of choice allows for many nice things in mathematics (e.g. every vector space has a basis and the Cartesian product of nonempty sets is again nonempty) but also introduces some somewhat unsavory consequences (e.g. Banach-Tarski paradox and similar results). Many proofs involving the axiom of choice are existence proofs, i.e. a certain object is proved to exist but there is no way of knowing what that object truly looks like. As such, the axiom of choice is a bit of a contentious piece of mathematics, but most mathematicians accept it as a valid axiom despite its pitfalls.
The axiom of choice formally states that if one has an indexed set of nonempty sets $(S_i)_{i\in I}$, then there is an indexed set $(x_i)_{i\in I}$ such that $x_i\in S_i$ for each $i$. Informally, this can be thought of in the following way: given a collection of nonempty sets, one can pick an element from each set at once. This seems intuitively obvious and in the case of a finite number of sets it is indeed obvious; however in the case of an infinite number of sets, it is not guaranteed. The axiom of choice plays a critical role in what follows.
Arbitrary infinite sets
Suppose now that $G$ is an arbitrary infinite set. Since we cannot appeal to groups we're familiar with, all we have at our disposal in terms of operations are the set operations and subsets of $G$. Somehow we must endeavor to construct a group structure using only these ideas.
Since groups only come with one binary operation, we must choose the correct set operation. For instance, unions and intersections are "one way" processes in some sense. More concretely, given an arbitrary set $B$, $A$ cannot be obtained from $A\cup B$ using only unions. As such, inverses wouldn't exist if we chose our group operation to be set union. Similar logic applies to set intersection. Set difference is not a good option either since it is not associative.
Hence we must look to "higher order" set operations, i.e. operations involving multiple set operations. The simplest higher order set operation is that of symmetric difference: $A\oplus B = (A\setminus B)\cup (B\setminus A)$. Visually, symmetric difference is represented by
Thus far we have not yet used the axiom of choice. It is at this stage that we use the axiom of choice. It is a consequence of the axiom of choice that the set of all finite subsets of an infinite set has the same cardinality of the set itself. The proof for this is not too difficult and relies on the Cantor-Schroeder-Bernstein theorem and some cardinal arithmetic. The axiom of choice is packed into the cardinal arithmetic that is required.
Denote the set of all finite subsets of the set $X$ by $\mathcal{F}(X)$. By the above, there is a bijection between $G$ and $\mathcal{F}(G)$. Let this bijection be denoted $f$. $\mathcal{F}(G)$ can be turned into a group with the operation of symmetric difference. Note that the symmetric difference of two finite sets is again finite. This however is not a group structure on $G$ but $\mathcal{F}(G)$.
To get a group structure on $G$, note that every $g\in G$ can be written as $f^{-1}(S)$ where $S\in\mathcal{F}(G)$. Let $\cdot$ denote the binary operation on $G$. Suppose that $g_1 = f^{-1}(S_1)$ and $g_2 = f^{-1}(S_2)$. Define $g_1\cdot g_2$ to be $f^{-1}(S_1\oplus S_2)$.
We have to show that this does indeed define a group structure on $G$. Consider $(g_1\cdot g_2)\cdot g_3$ This is nothing more than $f^{-1}(S_1\oplus S_2\oplus S_3)$ which is of course $g_1\cdot(g_2\cdot g_3)$. The identity element is the inverse image of the empty set since if $e = f^{-1}(\emptyset)$, then $e\cdot g = f^{-1}(\emptyset\oplus S) = f^{-1}(S) = g$. Inverses also exist: if $g = f^{-1}(S)$, then $g\cdot g = f^{-1}(S\oplus S) = f^{-1}(\emptyset) = e$.
Consequently $(G,\cdot)$ is indeed a group. In fact it is an abelian group since symmetric difference is, well, symmetric: $f^{-1}(S_1)\cdot f^{-1}(S_2) = f^{-1}(S_1\oplus S_2) = f^{-1}(S_2\oplus S_1) = f^{-1}(S_2)\cdot f^{-1}(S_1)$.
In general we were Abel (haha) to put a group structure on an arbitrary set (assuming the axiom of choice). Not only that, we were able to put an abelian group structure on the set. This is important for the ring portion of the secondary question. If we were not able to guarantee that we would have no hope of addressing the question of putting a ring structure on an arbitrary set.
Putting a ring structure on an arbitrary set
Finite sets
When considering the question of being able to put a group structure on an arbitrary set, we had to differentiate the finite and infinite settings. Unsurprisingly, we must do the same for the ring case. If $R$ is a finite set with $|R| = n$, then we could put an abelian group structure on $R$ by associating it to $\mathbb{Z}_n$. As is well-known, $\mathbb{Z}_n$ is already a unital ring! We don't have any further work to do in this case. The infinite case is much more interesting.
Infinite sets
Above we constructed the abelian group for arbitrary infinite sets so we need only to construct the ring structure for it. Let $R$ be an infinite set. We lifted the group structure of $R$ from $\mathcal{F}(R)$ with symmetric difference. It stands to reason that to construct a ring structure on $R$, we'd need to first put a ring structure on $\mathcal{F}(R)$.
To do this, a multiplication operation must be devised. Way back when when first discussing the elementary set operations, I likened them to usual arithmetic operations. Union is a bit like addition and intersection is a bit like multiplication. For this reason, intersection seems to be the logical choice for the second ring operation of multiplication. Before showing that this does indeed work, let's consider union as our notion of multiplication.
If union is to play the role of multiplication, it must "distribute" over symmetric difference (addition), i.e. $A\cup (B\oplus C) = (A\oplus B)\cup (A\oplus C)$. $A,B,C$ are arbitrary so taking $A = B = C$, the left hand side would become $A$ whereas the right hand side is empty. Thus unions do not distribute over symmetric difference in general. As such, union cannot play the role of multiplication. Intersection, however, does exactly what we want which can be easily verified.
Therefore the natural notion of multiplication on $\mathcal{F}(R)$ is set intersection. Like before, we must pull back the multiplication to $R$. Let $r_1 = f^{-1}(S_1)$ and $r_2 = f^{-1}(S_2)$, where $f$ is the bijection between $R$ and $\mathcal{F}(R)$ as before. Let $\star$ denote the multiplication on $R$ and define it by $r_1\star r_2 = f^{-1}(S_1\cap S_2)$.
Note here that there is no unit since the unit would be played by $R$ itself.
It remains to see that this does indeed define a ring structure on $R$. It must then be checked that the ring axioms are obeyed. This is tantamount to checking that multiplication distributes over addition. Consider $r_1 = f^{-1}(S_1)$, $r_2 = f^{-1}(S_2)$ and $r_3 = f^{-1}(S_3)$, then
\begin{eqnarray*}
r_1\star(r_2\cdot r_3) &=& r_1\star f^{-1}(S_2\oplus S_3) \\
&=& f^{-1}(S_1\cap(S_2\oplus S_3)) \\
&=& f^{-1}((S_1\cap S_2)\oplus (S_1\cap S_3)) \\
&=& f^{-1}(S_1\cap S_2)\cdot f^{-1}(S_1\cap S_3) \\
&=& (r_1\star r_2)\cdot (r_1\star r_3)
\end{eqnarray*}
Hence multiplication distributes over addition and moreover $R$ is endowed with a commutative ring structure since multiplication is commutative (which can be tracked back to the commutativity of set intersection).
We've set out to do exactly what we wanted! We've successfully shown that every set can be endowed with a ring (and a fortiori a group) structure. However one seeming issue is that the ring that we naturally developed in the infinite setting was non-unital. This has its roots in the fact that the unit with regards to set intersection is the universe set. Since our universe set was assumed to be infinite, it is not included in the set of all finite subsets.
It stands to reason then that we should modify our initial construction to include the whole set $R$ to begin with. However if we do this, we immediately lose closure: if $S$ is a finite subset of $R$, then $S\oplus R = R\setminus S$ which is most definitely not a finite set. We could remedy this by considering a union of the sets of finite subsets and cofinite subsets of $R$. This set has the same cardinality as $R$ by cardinal arithmetic. Hence there is a bijection between these sets and the same analysis follows through quite naturally. However this approach gives rise to a unital ring structure on $R$ since now $R$ is included.
The next most logical question to ask is if division rings can be formed on arbitrary sets. This however is beyond my expertise. Based on the construction above, it seems unlikely as intersection is (as stated above) a one-way process in some sense.
