(math) Equivalence relation properties are just group axioms

06 Nov, 2025

The necessary definitions

We can talk about groups as a collection of symmetries but the formal way is to define them axiomatically. Here are the group axioms for a group $G$ which has multiplication defined by $gh$ for $g,h\in G$:

G1. (Closure): For every $g,h\in G$, $gh\in G$
G2. (Associativity): For every $f,g,h\in G$, we have $f(gh)=(fg)h$.
G3. (Identity): There exists an element $e\in G$ such that for all $g\in G$, we have $eg=ge=g$.
G4. (Inverses): For each $g\in G$, there exists a $g^{-1}\in G$ such that $g{g}^{-1}=g^{-1}g=e$.

We can show that the identity element is unique and the inverses are unique with relative ease. If you have seen groups, then I will take for granted that you have heard about equivalence relations. Given a set $S$, we can define an equivalence relation $R\subset S\times S$ if the following conditions are satisfied:

R1. (Reflexivity): For all $x\in S$, $(x,x)\in R$
R2. (Symmetry): For all $x,y\in S$, if $(x,y)\in R$ then $(y,x)\in R$.
R3. (Transitivity): For all $x,y,z\in S$, if $(x,y)\in R$ and $(y,z)\in R$, then $(x,z)\in R$.

These two notions are more related than they look on the surface and we can formalize that they are the same using group actions. A group action is just a permutation of the elements of some set $X$ arising by multiplication by a group element $g$. For instance, given the set $\{1,3,5,7\}$, the element $g$ can add 2 to each element (modulo 7) and we have $g·X=\{3,5,7,1\}$, which is just the previous set but shuffled. In fact, we can define the action on elements such as $g·1=g(1)=3$.

Using group actions

The main idea is to visualize the pair $(x,y)$ as a process where we start at $x$ and end at $y$. We denote this by $g(x)=y$. So every pair $(x,y)$ is representing a $g$ responsible for sending $x$ to $y$. We call the set of such maps $G$ for foreshadowing reasons.

If we want the set of pairs to be an equivalence relation, we need to have a $g$ that maps all $x$'s to themselves. So, we define $e(x)=x$ for all $x\in S$, which is exactly the identity element of a group!

To have symmetry, for each $g(x)=y$, we must have an $h$ which maps $y$ back to $x$. This is the same as finding inverses within the group!

Transitivity puts up more of a fight as we wrangle the meaning out of it. It says that if I have $(x,y)$ and $(y,z)$ in $R$, then I can find $(x,z)$ in $R$. This sounds a lot like closure, where I can multiply $g$ and $h$ and find $gh$ in $G$. I claim that this is more than just closure, and transitivity actually encodes the action of multiplication itself using pairs. When two pairs of the form $(x,y)$ and $(y,z)$ are lined up together (using the AND operator, say), then multiplication can occur in the universe of groups. Transitivity even comes with associativity built in.

O ye of little faith

We have been a bit sloppy in our exposition above as $(x,y)$ might not necessarily represent a unique element but rather a coset of elements. As an example, both the identity element and the transposition $(24)$ in $S_5$ preserve the set $\{1,3,5\}$. We can solve this by making sure our group action is faithful which means that if $g(x)=h(x)$, then $g=h$. This is not necessary but helps with making the proofs simpler in this case.

We also need that each $(x,y)\in R$ corresponds to some $g\in G$. This is somewhat like transitivity.

What is equality?

The question that stands before us before we do any proofs in this new language is what does $g=h$ in the group universe correspond to in the proposition universe? I am not sure.

(to be continued...)