quotient monoids, or how i became universal-algebra-pilled
The goal of this blog post is to offer the "universal algebra pill" to readers already familiar with abstract algebra but not its more exotic structures.
After all, you probably know how to take group quotients, ring quotients, etc.
However, usually these constructions are taught in a way that does not generalize to more general algebraic structures.
In a nutshell, kernels are much nicer for group homomorphisms than in general.
But before I get ahead of myself...
reframing quotient groups
Let's start with the familiar.
Recall how quotient groups are constructed.
Well, you may be thinking of normal subgroups and their friends right now.
This is not what we want.
We want an equivalence relation that respects the group operation.
Normal subgroups and their friends are results, not definitions.
Let $\sim$ be our equivalence relation over our group $(G, \cdot)$.
Then we define our equivalence relation to respect $\cdot$ if for all $a, b, c \in G$,
$a \sim b \rightarrow ca \sim cb$ and $a \sim b \rightarrow ac \sim bc$
From there it is possible to prove that cosets constructed via a normal subgroup are precisely these types of equivalence relations.
However, let's pretend we did not know that.
Define $G/\sim$ as G's multiplication applied over the cosets (i.e. partitions) induced by $\sim$.
This is well-defined because of our equivalence relation's respect for $\cdot$.
It also induces a natural projection $G \rightarrow G/\sim$ by mapping elements of G onto their cosets.
Even with this bare-bones treatment of quotient groups, we still have a version of the Homomorphism Theorem!
(Also known as the First Isomorphism Theorem, or a lot of other names, because this theorem has way too many names.)
To see why, consider any surjective group homomorphism $\varphi: G \rightarrow H$.
We can define $\sim$ over $G$ by letting $a \sim b$ iff $\varphi(a) = \varphi(b)$.
It is quick to verify this is an equivalence relation.
Critically, $\sim$ also respects $\cdot$; given any $a, b, c \in G$,
$a \sim b \rightarrow \varphi(a) = \varphi(b) \rightarrow \varphi(c) \varphi(a) = \varphi(c) \varphi(b) \rightarrow \varphi(ca) = \varphi(cb) \rightarrow ca \sim cb$
and because $\varphi$ is surjective, for any $h \in H$ there exists at least one $c$ such that $\varphi(c) = h$.
This, along with the argument applied to right-multiplication instead of left-multiplication, proves that $\sim$ respects G's $\cdot$.
Let's (suggestively) rename $\sim$ to $\ker \varphi$.
Then we have constructed a natural isomorphism between $G/\ker \varphi$ and $H$, and further, the natural projection mentioned earlier factors through it!
hold on, you called that equivalence relation what?
Okay so hear me out.
Recall the traditional definition of a kernel: that which a map sends to identity.
Define the set $\{a \in G \text{ s.t. } a \sim \text{e}\}$ where $\text{e}$ designates the identity element of the group G.
I claim this set is the kernel and thus $\sim$ encodes all the information that the kernel encodes.
Indeed, applying our definition of $\sim$ yields $\varphi(a) = \text{e}$, which is precisely the definition of a kernel!
In fact, our induced equivalence relation from $\varphi$ contains more information than its traditional kernel.
So, why do we not use this definition of kernel when working with groups?
As it turns out, for "nice" algebraic structures, all you need is the traditional kernel to recover the entire equivalence relation.
Additionally, traditional kernels tend to be nice substructures too; groups have normal subgroups, rings have ideals, and modules have submodules.
However, if we want to take the quotient of a monoid, we need to forget all of this niceness and focus on the true definition of a kernel:
$\ker \varphi = \;\sim$ where $a \sim b$ iff $\varphi(a) = \varphi(b)$.
let's actually take a quotient of a monoid now
Recall that a monoid is a group but without the requirement that all elements have inverses.
That is, it is a set equipped with an associative binary operation with identity.
One easy example of a monoid is $(\mathbb{N}, +)$, that is, the set of naturals (including 0) over addition.
In order to take the quotient of a monoid, we need to specify an equivalence relation over that monoid.
Here is a weird but illustrative one:
$a \sim b$ iff $\begin{cases}
a = 0 \text{ and } b = 0 \\
a > 0 \text{ and } b > 0 \text{ and } 2 \mid a - b
\end{cases}$
In other words, $0 \sim 0$, $1 \sim 3 \sim 5 \sim \ldots$, and $2 \sim 4 \sim 6 \sim \ldots$.
Importantly, this equivalence relation respects $+$.
It is a good exercise to convince yourself of this.
Now we have that $(\mathbb{N}, +)/\sim \; = (\{[0], [1], [2]\}, +)$, where [n] indicates a coset represented by n.
Here is the Cayley table:
$\begin{array}{c | c c c c c}
+ & [0] & [1] & [2] \\
\hline
[0] & [0] & [1] & [2] \\
[1] & [1] & [2] & [1] \\
[2] & [2] & [1] & [2] \\
\end{array}$
By the Homomorphism Theorem, it is possible to view this quotient an entirely different way.
We can view this $+$ respecting equivalence relation as the kernel of a surjective monoid homomorphism.
For example, we could consider $\varphi: (\mathbb{N}, +) \rightarrow (\{1,2,4\}, \cdot_6)$ where $\cdot_6$ is integer multiplication mod 6.
We define $\varphi$ as follows:
$\varphi(a) = \begin{cases}
1 & a = 0 \\
4 & a > 0 \text{ and } 2 \mid a \\
2 & \text{otherwise}
\end{cases}$
Note the similarity between the definition of this homomorphism and our equivalence relation.
This is not a coincidence!
In fact, I have engineered this scenario such that $\ker \varphi = \;\sim$ and proving this is yet again an exercise to the reader.
conclusion
I sincerely hope you enjoyed learning about more exotic quotients of algebraic structures.
I think they are really neat, anyway.
Here is a final exercise: find all quotients of $(\mathbb{N}, +)$.
Or in other words, find all surjective (monoid) homomorphisms (up to isomorphism) with $(\mathbb{N}, +)$ as their domain.
Enjoy! (Hint: you will need two different natural-number parameters to describe them all.)