interesting, nonstandard explanation of quotient groups

Stream: maths

[Ben Selfridge]

Hello -- I've recently stumbled on a slightly different way of defining/motivating the concept of a quotient group. I'd be interested to see what people think -- it's probably obvious to most here, but I found it very illuminating, and I would not have stumbled upon this other way of thinking about it without trying to formalize my own understanding in Lean.

Instead of first defining left and right cosets, and then talking about "well-definedness" of multiplication and so forth (which I always found confusing), this explanation starts by asking the very general question: "How can we make a new group out of subsets of an existing group G?"

Let G be a group. We can extend the definition of multiplication to
expressions of the form α * β, where α and β either elements of G or sets
containing elements of G. In particular, we have a natural definition for
multiplication on subsets of G: A * B = { a * b | a ∈ A, b ∈ B }. We also
have a natural definition of "inverse" on subsets: A⁻¹ = { a⁻¹ | a ∈ A }.

These extended operations induce a group-like structure on the subsets of
G, but the set of all subsets of G clearly doesn't form a group; no
matter what identity you try to pick, general subsets will never be
invertible for non-trivial groups. In a sense, there are "too many"
subsets.

Therefore, let's pick a subcollection Γ of nonempty subsets of G, and we
will do it in a way that guarantees Γ forms a group under setwise
multiplication and inversion as defined above. Note that we can always do
this in at least two ways -- we can pick the singleton sets of elements of
G, which is isomorphic to G, or we can pick the lone set G, which is
isomorphic to the trivial group.

If Γ forms a group, it must have an identity. Call that identity N. Then
certainly

N * N = N

and

N⁻¹ = N

owing to the fact that it is the identity element of Γ. It also contains
the identity of G, since it is nonempty and closed under * and ⁻¹.
Therefore, N is a subgroup of G.

What about the other elements of Γ? Well, we know that for every A ∈ Γ, we
have N * A = A * N = A and A⁻¹ * A = A * A⁻¹ = N. Let's define a *coset of
N* to be ANY subset A ⊆ G satisfying this relationship with N. Then, as it
happens, the cosets of N are closed under multiplication and inversion,
and form a group.

It is easy to prove that the cosets all satisfy A = aN = Na for all a ∈ A,
and cosets are disjoint. [edited]

Note that it is possible that not all elements of G are contained in a
coset of N. If it happens that every element is contained in some coset,
we say that N is a normal subgroup of G.

[Edward van de Meent]

Ben Selfridge said:

It is easy to prove that the cosets all satisfy A = aN = Na for all a ∈ A,
and form a partition of G.

Note that it is possible that not all elements of G are contained in a
coset of N.

you seem to be contradicting yourself. is it a partition or not?

[Edward van de Meent]

perhaps you meant to say "and form disjoint subsets of G"?

[Ben Selfridge]

Yep, that is exactly what I meant.

[Antoine Chambert-Loir]

To motivate the introduction of quotient groups, I prefer saying that in some cases, I wish to identify some elements of the group, and still get a group in a natural way, will all identifications that follow my requests. One is lead to the concept of an equivalence relation ~ on G compatible with a law *, and for groups, everything boils down to what is the equivalence class of 1.

• If you wish for a group law on G/~, then the equivalence class of 1 is a normal subgroup, and conversely.

• If you just wish for a left-action of G on G/~, then the equivalence class of 1 is a subgroup, and the equivalence classes are the cosets (the side of which depends on your convention).

• Idem for right-actions.

[Patrick Massot]

Ben, your motivation focuses entirely on the implementation of quotient groups (in set theory based foundations). But this is not what is relevant in math. The usual path goes as follows: a quotient of a group $G$ is group whose elements come from $G$ or, as Antoine puts it, a group obtained by gluing some elements of $G$. How to make the precise? We want the quotient to be also a group, say $G'$. And we want the “coming from” to be in the group theory sense, so we want a group morphism $π$ from $G$ to $G'$. Saying that all elements of $G'$ come from $G$ means we want $π$ to be surjective. So you have your definition: a group quotient of $G$ is a pair $(G', π)$ where $G'$ is a group and $π$ is a surjective group morphism from $G$ to $G'$. Next comes the question: how to understand all quotients of a given $G$? How to build them? Here you have a series of exercises to work on.l First prove that the kernel of a group morphism is a normal subgroup. Then prove that any two elements $x$ and $y$ have the same image under $π$ if and only if $xy^{-1} ∈ \ker π$. Then prove the converse: given a normal subgroup of $G$, define a relation by $x \sim y$ if $xy^{-1} ∈ \ker π$, prove this is an equivalence relation, prove there is a unique group structure on any quotient of $G$ by this relation so that the projection is a group morphism.

[Maja Kądziołka]

As far as interesting motivations for quotient groups go, I really like the one set out in this post: https://gowers.wordpress.com/2011/11/20/normal-subgroups-and-quotient-groups/

[Antoine Chambert-Loir]

On the other hand, Patrick, group theorists (at least some of them) are interested in the monoid of subsets of a monoid $M$, setting $S \bullet T=\{s\cdot t; s\in S; t\in T\}$ for $S,T\in\mathfrak P(M)$. The empty subset is absorbing., of course, and $S=\{1\}$ is the neutral element. This monoid $\mathfrak P(M)$ contains the initial monoid $M$ as a submonoid, given by singletons. If $M$ is a group, then those singletons are the unit elements of $\mathfrak P(M))$. That remark shows that $\mathfrak P(M)$ determines $M$ when $M$ is a group, but it doesn't seem to be known whether that holds in general.

[suhr]

Why one needs to mystify a straightforward concept (making it look like an arbitrary nonsense), when one can explain it with a single picture: https://q.uiver.app/#q=WzAsMyxbMCwwLCJHIl0sWzIsMCwiRyciXSxbMSwxLCJHL0giXSxbMCwxLCJcXHZhcnBoaSJdLFswLDIsIlxccGkiLDJdLFsyLDEsIlxcZXhpc3QhXFx0aWxkZVxcdmFycGhpIiwyXV0=?

If you believe that ad-hoc constructions are easier for beginners to understand than a simple universal property, then I'm eager to tell you that it is not. Abstract algebra did not make much sense to me until I had luck to find Aluffi's Algebra Chapter 0. I'm very grateful that this textbook exists.

[Ben Selfridge]

The motivation for this explanation came from a gap in my own understanding. It was helpful for me to focus on the implementation details.

Setwise multiplication is a very intuitive idea, and it was illuminating to me to realize that without making any assumptions about how to do it, one could recover all the "rules" about quotient groups by merely assuming that we were making "groups out of subsets" of a group.

As an undergraduate, this kind of explanation would have helped me a great deal. I don't think it's mystifying a straightforward concept at all, in some sense it's a more general and tangible way of conjuring up cosets and normal subgroups out of a very minimal set of requirements.

@suhr, your simple picture is also extremely illuminating, but it obscures a lot of gory details, and without mucking around in the details a bit, it's hard for someone who is just learning these concepts for the first time to really "trust" the foundations all the way down.

[Edward van de Meent]

suhr said:

Why one needs to mystify a straightforward concept (making it look like an arbitrary nonsense), when one can explain it with a single picture: https://q.uiver.app/#q=WzAsMyxbMCwwLCJHIl0sWzIsMCwiRyciXSxbMSwxLCJHL0giXSxbMCwxLCJcXHZhcnBoaSJdLFswLDIsIlxccGkiLDJdLFsyLDEsIlxcZXhpc3QhXFx0aWxkZVxcdmFycGhpIiwyXV0=?

because reading that picture requires a whole lot of math background which students typically don't have when the concept of a group quotient is explained

[Edward van de Meent]

even right now i'm not sure exactly how i'm supposed to interpret this picture

[Edward van de Meent]

(so good job, you've mystified a straightforward concept by claiming it is explained by a picture)

[Ben Selfridge]

Antoine Chambert-Loir said:

On the other hand, Patrick, group theorists (at least some of them) are interested in the monoid of subsets of a monoid $M$, setting $S \bullet T=\{s\cdot t; s\in S; t\in T\}$ for $S,T\in\mathfrak P(M)$. The empty subset is absorbing., of course, and $S=\{1\}$ is the neutral element. This monoid $\mathfrak P(M)$ contains the initial monoid $M$ as a submonoid, given by singletons. If $M$ is a group, then those singletons are the unit elements of $\mathfrak P(M))$. That remark shows that $\mathfrak P(M)$ determines $M$ when $M$ is a group, but it doesn't seem to be known whether that holds in general.

This is exactly the kind of viewpoint that inspired me in all of this. Realizing that the subsets themselves form a monoid, but not a group, and asking the question -- "how do we make them into a group? How many different ways to do that are there?" and realizing that the answer to that question is "There's as many different ways to do that as there are subgroups of G."

[Ben Selfridge]

Antoine Chambert-Loir said:

To motivate the introduction of quotient groups, I prefer saying that in some cases, I wish to identify some elements of the group, and still get a group in a natural way, will all identifications that follow my requests. One is lead to the concept of an equivalence relation ~ on G compatible with a law *, and for groups, everything boils down to what is the equivalence class of 1.

• If you wish for a group law on G/~, then the equivalence class of 1 is a normal subgroup, and conversely.

• If you just wish for a left-action of G on G/~, then the equivalence class of 1 is a subgroup, and the equivalence classes are the cosets (the side of which depends on your convention).

• Idem for right-actions.

This is a totally reasonable and probably better way to introduce the topic initially. But I've been mulling over these basic notions for a long time, and this other way of looking at it was helpful for me.

[Ben Selfridge]

Patrick Massot said:

Ben, your motivation focuses entirely on the implementation of quotient groups (in set theory based foundations). But this is not what is relevant in math. The usual path goes as follows: a quotient of a group $G$ is group whose elements come from $G$ or, as Antoine puts it, a group obtained by gluing some elements of $G$. How to make the precise? We want the quotient to be also a group, say $G'$. And we want the “coming from” to be in the group theory sense, so we want a group morphism $π$ from $G$ to $G'$. Saying that all elements of $G'$ come from $G$ means we want $π$ to be surjective. So you have your definition: a group quotient of $G$ is a pair $(G', π)$ where $G'$ is a group and $π$ is a surjective group morphism from $G$ to $G'$. Next comes the question: how to understand all quotients of a given $G$? How to build them? Here you have a series of exercises to work on.l First prove that the kernel of a group morphism is a normal subgroup. Then prove that any two elements $x$ and $y$ have the same image under $π$ if and only if $xy^{-1} ∈ \ker π$. Then prove the converse: given a normal subgroup of $G$, define a relation by $x \sim y$ if $xy^{-1} ∈ \ker π$, prove this is an equivalence relation, prove there is a unique group structure on any quotient of $G$ by this relation so that the projection is a group morphism.

I mean, yeah, that seems like the standard way of teaching it, and it's a good one.

My motivation does indeed focus on the implementation of quotient groups, because that's a more constructive, and to me, intuitive way of motivating it. I also disagree that it's "not relevant in math" -- who are you (or who am I) to decide what's universally relevant in all of math? If the implementation is wrong, the theory doesn't matter. That seems relevant. For me, these low-level details have always felt tricky and subtle, and any additional insight gained is valuable, especially implementation-level insight.

[suhr]

because reading that picture requires a whole lot of math background

But it doesn't. It's page 93 of Algebra Chapter 0, and this book requires basically no prerequisites at all (might require a bit of mathematical maturity).

(so good job, you've mystified a straightforward concept by claiming it is explained by a picture)

Yeah, I definitely forgot some words.

Theorem 7.12. Let H be a normal subgroup of a group G. Then for every group homomorphism ϕ : G → G' such that H ⊆ ker ϕ there exists a unique group homomorphism ϕ̃ : G/H → G' so that the diagram ... commutes.

And this is the middle of the chapter which starts with

Recall that we have the notion of a quotient of a set by an equivalence relation (§I.1.5) and that this notion satisfies a universal property (clumsily stated in §I.5.3). It is natural to investigate this notion in Grp.
We consider then an equivalence relation ∼ on (the set underlying) a group G; we seek a group G/∼ and a group homomorphism π : G → G/∼ satisfying the appropriate universal property, that is, initial with respect to group homomorphisms ϕ : G → G' such that a ∼ b =⇒ ϕ(a) = ϕ(b).
It is natural to try to construct the group G/∼ by defining an operation • on the set G/∼. The situation is tightly constrained by the requirement that the quotient map π : G → G/∼ (as in §I.2.6) be a group homomorphism: for if [a] = π(a), [b] = π(b) are elements of G/∼ (that is, equivalence classes with respect to ∼), then the homomorphism condition forces
[a] • [b] = π(a) • π(b) = π(ab) = [ab].
But is this operation well-defined? This amounts to conditions on the equivalence relation, which we proceed to unearth.

[suhr]

By the way, https://xenaproject.wordpress.com/2025/02/09/what-is-a-quotient/ explains quotients with the same universal property.

[Ben Selfridge]

Algebra Chapter 0 is a great book, and is an awesome way to learn algebra, even for a beginner.

[Edward van de Meent]

for some reason i keep hoping for a full categorytheory phrasing of "the map acts the same on related elements" when people claim to talk about the "universal property" of quotients

[Edward van de Meent]

but i always get disappointed

[Aaron Liu]

What is disappointing?

[Edward van de Meent]

that they don't give such a phrasing

[Aaron Liu]

Just use coequalizers then

[Edward van de Meent]

of which maps?

[Aaron Liu]

That depends on how you implement your relation

[Aaron Liu]

But for quotient groups it's the inclusion from the normal subgroup and zero

[Adam Topaz]

If you want to mimic a general relation, you consider a type X with two maps f and g to a group G. This induces two maps from the free group on X to G, and the coequalizer of those two maps is the quotient of G by the congruence relation generated by the relation that identifies f x and g x for any x : X.

[Edward van de Meent]

right, but i was wondering how one would go about constructing the relevant equivalence relation in the category of groups

[Edward van de Meent]

or wait, is what you said actually the equivalence relation? :thinking:

[Adam Topaz]

What is the definition of “an equivalence relation in the category of groups” that you want to use?

[Edward van de Meent]

an equivalence relation on $G$ is a subobject $X$ of $G \times G$ (with map $m$) such that:

• there exists a map $d:G \rightarrow X$ with $\pi_1 m d = \text{Id}_G = \pi_2 m d$ (meaning the relation is reflexive)
• there exists a map $s : X \rightarrow X$ such that $\pi_1 m s = \pi_2 m$ and $\pi_2 m s = \pi_1 m$ (meaning the relation is symmetric)
• for a pullback $\overline{X}$ of $\pi_1 m$ and $\pi_2 m$, there is a map $t:\overline{X} \rightarrow X$ such that $\pi_1 m t = \pi_1 m \overline{\pi_1 m}$ and $\pi_2 m t = \pi_2 m \overline{\pi_2 m}$
  (i guess this definition is easier if you later prove it is a subobject, and just assume the two maps into $G$)

[Adam Topaz]

FWIW any quotient G -> Q in groups is a coequalizer, either as Aaron describes, or more canonically by looking at the two projections from G x_Q G to G.

[Edward van de Meent]

right

[Aaron Liu]

Adam Topaz said:

If you want to mimic a general relation, you consider a type X with two maps f and g to a group G. This induces two maps from the free group on X to G, and the coequalizer of those two maps is the quotient of G by the congruence relation generated by the relation that identifies f x and g x for any x : X.

How do I get the f and g if I have r : G → G → Prop?

[suhr]

For a full category theory phrasing one can always consult nLab: https://ncatlab.org/nlab/show/quotient+object. I'm not sure if using https://ncatlab.org/nlab/show/congruence instead of "equivalence relation" makes things easier.

[Adam Topaz]

@Aaron Liu given such an r look at the subtype of G x G associated to it, call it X, then the two projections are your f and g

[Edward van de Meent]

suhr said:

For a full category theory phrasing one can always consult nLab: https://ncatlab.org/nlab/show/quotient+object. I'm not sure if using https://ncatlab.org/nlab/show/congruence instead of "equivalence relation" makes things easier.

that definition of congruence looks to be the same as what i had in mind, untill transitivity, where i'm not entirely clear if it's equivalent

[Edward van de Meent]

it is

[Edward van de Meent]

i got confused where they mentioned fiber products, but TIL that's another way to say pullback

[Edward van de Meent]

And then it's just the same, I think

[Aaron Liu]

What's your $\overline{\pi_1 m}$ and $\overline{\pi_2 m}$?

[Edward van de Meent]

it's the respective pullbacks of $\pi_1 m$ and $\pi_2 m$ along eachother

[Edward van de Meent]

i.e. they are the maps such that $\pi_2 m \overline{\pi_1 m} = \pi_1 m \overline{\pi_2 m}$

[Aaron Liu]

that works

[Aaron Liu]

I've always called them "projections"

[Edward van de Meent]

which ones?

[Edward van de Meent]

(this might be due to a lack of experience in this field, but to me projections are only the maps out of products that are part of a limiting cone)

[Aaron Liu]

Edward van de Meent said:

(this might be due to a lack of experience in this field, but to me projections are only the maps out of products that are part of a limiting cone)

This might be due to a lack of experience in this field, but I have no idea what you mean by this

[Aaron Liu]

Edward van de Meent said:

which ones?

The ones that go $X \times_{f, g} Y \to X$ and $X \times_{f, g} Y \to Y$

[Edward van de Meent]

i mean that to me, only the canonical maps out of true products like $\pi_1 : X \times Y \rightarrow X$ and $\pi_2 : X \times Y \rightarrow Y$ are projections

[Johan Commelin]

@Edward van de Meent I think it is custom to generalize that from "product" to "limit". At least, in Mathlib all those maps are called $\pi$.

[Andrew Yang]

Unfortunately we have docs#CategoryTheory.Limits.equalizer.ι and docs#CategoryTheory.Limits.coequalizer.π