open import Algebra.Group.Cat.FinitelyComplete
open import Algebra.Group.Cat.Base
open import Algebra.Prelude
open import Algebra.Group

open import Cat.Diagram.Equaliser.Kernel

open import Data.Power

open import Order.Instances.Subobjects

import Order.Reasoning as Poset

module Algebra.Group.Subgroup where

private variable
ℓ ℓ' : Level
G : Group ℓ


# Subgroups🔗

A subgroup $m$ of a group $G$ is a monomorphism $H \xrightarrow{m} G$, that is, an object of the poset of subobjects $\operatorname*{Sub}(G)$. Since group homomorphisms are injective exactly when their underlying function is an embedding, we can alternatively describe this as a condition on a predicate $G \to \mathrm{Prop}$.

Subgroup : Group ℓ → Type (lsuc ℓ)
Subgroup {ℓ = ℓ} G = Σ (Group ℓ) λ H → H Groups.↪ G


A proposition $H : G \to \mathrm{Prop}$ of a group $(G, \star)$ represents a subgroup if it contains the group unit, is closed under multiplication, and is closed under inverses.

record represents-subgroup (G : Group ℓ) (H : ℙ ⌞ G ⌟) : Type ℓ where
open Group-on (G .snd)

field
has-unit : unit ∈ H
has-⋆    : ∀ {x y} → x ∈ H → y ∈ H → (x ⋆ y) ∈ H
has-inv  : ∀ {x} → x ∈ H → x ⁻¹ ∈ H


If $H$ represents a subgroup, then its total space $\Sigma H$ inherits a group structure from $G$, and the first projection $\Sigma H \to G$ is a group homormophism.

rep-subgroup→group-on
: (H : ℙ ⌞ G ⌟) → represents-subgroup G H → Group-on (Σ[ x ∈ ⌞ G ⌟ ] x ∈ H)
rep-subgroup→group-on {G = G} H sg = to-group-on sg′ where
open Group-on (G .snd)
open represents-subgroup sg
sg′ : make-group (Σ[ x ∈ ⌞ G ⌟ ] x ∈ H)
sg′ .make-group.group-is-set = hlevel!
sg′ .make-group.unit = unit , has-unit
sg′ .make-group.mul (x , x∈) (y , y∈) = x ⋆ y , has-⋆ x∈ y∈
sg′ .make-group.inv (x , x∈) = x ⁻¹ , has-inv x∈
sg′ .make-group.assoc x y z = Σ-prop-path (λ x → H x .is-tr) associative
sg′ .make-group.invl x = Σ-prop-path (λ x → H x .is-tr) inversel
sg′ .make-group.idl x = Σ-prop-path (λ x → H x .is-tr) idl

predicate→subgroup : (H : ℙ ⌞ G ⌟) → represents-subgroup G H → Subgroup G
predicate→subgroup {G = G} H p = _ , record { mor = map ; monic = ism } where
map : Groups.Hom (el! (Σ _ (∣_∣ ⊙ H))
, rep-subgroup→group-on H p) G
map .hom = fst
map .preserves .is-group-hom.pres-⋆ x y = refl

ism : Groups.is-monic map
ism = Homomorphism-monic map (λ p → Σ-prop-path (λ _ → hlevel!) p)


# Kernels and Images🔗

To a group homomorphism $f : A \to B$ we can associate two canonical subgroups, one of $A$ and one of $B$: $f$’s image factorisation, written $\operatorname*{im}f$, is the subgroup of $B$ “reachable by mapping through $f$”, and $f$’s kernel, written $\ker f$, is the subgroup of $A$ which $f$ sends to the unit.

The kernel can be cheapily described as a limit: It is the equaliser of $f$ and the zero morphism — which, recall, is the unique map $A \to B$ which breaks down as $A \to 0 \to B$.

module _ {ℓ} where
open Canonical-kernels (Groups ℓ) ∅ᴳ Groups-equalisers public

Ker-subgroup : ∀ {A B : Group ℓ} → Groups.Hom A B → Subgroup A
Ker-subgroup f =
ker , record { mor = kernel
; monic = Groups.is-equaliser→is-monic _ has-is-kernel }
where
open Kernel (Ker f)


Every group homomorphism $f : A \to B$ has an image factorisation $\operatorname*{im}f$, defined by equipping its set-theoretic image with a group structure inherited from $B$. More concretely, we can describe the elements of $\operatorname*{im}f$ as the “mere fibres” of $f$: They consist of a point $y : B$, together with (the truncation of) a fibre of $f$ over $y$. We multiply $x$ (in the fibre over $a$) with $y$ (in the fibre over $b$), giving the element $xy$ in the fibre over $ab$.

module _ {ℓ} {A B : Group ℓ} (f : Groups.Hom A B) where
private
module A = Group-on (A .snd)
module B = Group-on (B .snd)
module f = is-group-hom (f .preserves)

Tpath : {x y : image (f #_)} → x .fst ≡ y .fst → x ≡ y
Tpath {x} {y} p = Σ-prop-path (λ _ → squash) p

abstract
Tset : is-set (image (f #_))
Tset = hlevel 2

module Kerf = Kernel (Ker f)


For reasons that will become clear later, we denote the image of $f$, when regarded as its own group, by $A/\ker(f)$, and reserve the notation $\operatorname*{im}f$ for that group regarded as a subgroup of $B$.

The construction of a group structure on $A/\ker(f)$ is unsurprising, so we leave it in this <details> tag for the curious reader.
    T : Type ℓ
T = image (f #_)

A/ker[_] : Group ℓ
A/ker[_] = to-group grp where
unit : T
unit = B.unit , inc (A.unit , f.pres-id)

inv : T → T
inv (x , p) = x B.⁻¹ ,
∥-∥-map (λ { (y , p) → y A.⁻¹ , f.pres-inv ∙ ap B._⁻¹ p }) p

mul : T → T → T
mul (x , xp) (y , yp) = x B.⋆ y ,
∥-∥-elim₂ (λ _ _ → squash)
(λ { (x* , xp) (y* , yp)
→ inc (x* A.⋆ y* , f.pres-⋆ _ _ ∙ ap₂ B._⋆_ xp yp) })
xp yp

grp : make-group T
grp .make-group.group-is-set = Tset
grp .make-group.unit = unit
grp .make-group.mul = mul
grp .make-group.inv = inv
grp .make-group.assoc = λ x y z → Tpath B.associative
grp .make-group.invl = λ x → Tpath B.inversel
grp .make-group.idl = λ x → Tpath B.idl


That the canonical inclusion map $A/\ker(f) \hookrightarrow B$ deserves the name “image” comes from $f$ breaking down as a (regular) epimorphism into $\operatorname*{im}f$ (written A→im), followed by that map:

$(A \xrightarrow{f} B) = (A \xtwoheadrightarrow{f} A/\ker(f) \hookrightarrow B)$

  A→im : Groups.Hom A A/ker[_]
A→im .hom x = f # x , inc (x , refl)
A→im .preserves .is-group-hom.pres-⋆ x y = Tpath (f.pres-⋆ _ _)

im→B : Groups.Hom A/ker[_] B
im→B .hom (b , _) = b
im→B .preserves .is-group-hom.pres-⋆ x y = refl


When this monomorphism is taken as primary, we refer to $A/\ker(f)$ as $\operatorname*{im}f$.

  Im[_] : Subgroup B
Im[_] = _ , record { mor = im→B ; monic = im↪B } where
im↪B : Groups.is-monic im→B
im↪B = Homomorphism-monic im→B Tpath


#### The first isomorphism theorem🔗

The reason for denoting the set-theoretic image of $f : A \to B$ (which is a subobject of $B$, equipped with $B$’s group operation) by $A/\ker(f)$ is the first isomorphism theorem (though we phrase it more categorically): The image of $f$ serves as a quotient for (the congruence generated by) $\ker f$.

Note

In more classical texts, the first isomorphism theorem is phrased in terms of two pre-existing objects $A/\ker(f)$ (defined as the set of cosets of $\ker(f)$ regarded as a subgroup) and $\operatorname*{im}f$ (defined as above). Here we have opted for a more categorical phrasing of that theorem: We know what the universal property of $A/\ker(f)$ is — namely that it is a specific colimit — so the specific construction used to implement it does not matter.

  1st-iso-theorem : Groups.is-coequaliser (Groups.Zero.zero→ ∅ᴳ) Kerf.kernel A→im
1st-iso-theorem = coeq where
open Groups
open is-coequaliser
module Ak = Group-on (A/ker[_] .snd)


More specifically, in a diagram like the one below, the indicated dotted arrow always exists and is unique, witnessing that the map $A \twoheadrightarrow A/\ker(f)$ is a coequaliser (hence that it is a regular epi, as we mentioned above).  The condition placed on $e'$ is that $0 = e' \circ \ker f$; This means that it, like $f$, sends everything in $\ker f$ to zero (this is the defining property of $\ker f$). Note that in the code below we do not elide the zero composite $e' \circ 0$.

    elim
: ∀ {F} {e' : Groups.Hom A F}
(p : e' Groups.∘ Zero.zero→ ∅ᴳ ≡ e' Groups.∘ Kerf.kernel)
→ ∀ {x} → ∥ fibre (f #_) x ∥ → _
elim {F = F} {e' = e'} p {x} =
∥-∥-rec-set ((e' #_) ⊙ fst) const (F .snd .Group-on.has-is-set) where abstract
module e' = is-group-hom (e' .preserves)
module F = Group-on (F .snd)


To eliminate from under a propositional truncation, we must prove that the map $e'$ is constant when thought of as a map $f^*(x) \to F$; In other words, it means that $e'$ is “independent of the choice of representative”. This follows from algebraic manipulation of group homomorphisms + the assumed identity $0 = e' \circ \ker f$;

      const′ : ∀ (x y : fibre (f #_) x)
→ e' # (x .fst) F.— e' # (y .fst) ≡ F.unit
const′ (y , q) (z , r) =
e' # y F.— e' # z  ≡˘⟨ e'.pres-diff ⟩≡˘
e' # (y A.— z)     ≡⟨ happly (sym (ap hom p)) (y A.— z , aux) ⟩≡
e' # A.unit        ≡⟨ e'.pres-id ⟩≡
F.unit             ∎
where


This assumption allows us to reduce “show that $e'$ is constant on a specific subset” to “show that $f(y - z) = 0$ when $f(y) = f(z) = x$”; But that’s just algebra, hence uninteresting:

          aux : f # (y A.— z) ≡ B.unit
aux =
f # (y A.— z)     ≡⟨ f.pres-diff ⟩≡
f # y B.— f # z   ≡⟨ ap₂ B._—_ q r ⟩≡
x B.— x           ≡⟨ B.inverser ⟩≡
B.unit            ∎

const : ∀ (x y : fibre (f #_) x) → e' # (x .fst) ≡ e' # (y .fst)
const a b = F.zero-diff (const′ a b)


The rest of the construction is almost tautological: By definition, if $x : \ker f$, then $f(x) = 0$, so the quotient map $A \twoheadrightarrow A/\ker(f)$ does indeed coequalise $\ker f \hookrightarrow A$ and $0$. As a final word on the rest of the construction, most of it is applying induction (∥-∥-elim and friends) so that our colimiting map elim will compute.

    coeq : is-coequaliser _ _ A→im
coeq .coequal = Forget-is-faithful (funext path) where
path : (x : ⌞ Kerf.ker ⌟) → A→im # A.unit ≡ A→im # (x .fst)
path (x* , p) = Tpath (f.pres-id ∙ sym p)

coeq .universal {F = F} {e′ = e'} p = gh where
module F = Group-on (F .snd)
module e' = is-group-hom (e' .preserves)

gh : Groups.Hom _ _
gh .hom (x , t) = elim {e' = e'} p t
gh .preserves .is-group-hom.pres-⋆ (x , q) (y , r) =
∥-∥-elim₂
{P = λ q r → elim p (((x , q) Ak.⋆ (y , r)) .snd) ≡ elim p q F.⋆ elim p r}
(λ _ _ → F.has-is-set _ _) (λ x y → e'.pres-⋆ _ _) q r

coeq .factors = Forget-is-faithful refl

coeq .unique {F} {p = p} {colim = colim} prf = Forget-is-faithful (funext path)
where abstract
module F = Group-on (F .snd)
path : ∀ x → colim # x ≡ elim p (x .snd)
path (x , t) =
∥-∥-elim
{P = λ q → colim # (x , q) ≡ elim p q}
(λ _ → F.has-is-set _ _)
(λ { (f , fp) → ap (colim #_) (Σ-prop-path (λ _ → squash) (sym fp))
∙ (happly (ap hom prf) f) })
t


## Representing kernels🔗

If an evil wizard kidnaps your significant others and demands that you find out whether a predicate $P : G \to \mathrm{Prop}$ is a kernel, how would you go about doing it? Well, I should point out that no matter how evil the wizard is, they are still human: The predicate $P$ definitely represents a subgroup, in the sense introduced above — so there’s definitely a group homomorphism $\Sigma G \hookrightarrow G$. All we need to figure out is whether there exists a group $H$ and a map $f : G \to H$, such that $\Sigma G \cong \ker f$ as subgroups of $G$.

We begin by assuming that we have a kernel and investigating some properties that the fibres of its inclusion have. Of course, the fibre over $0$ is inhabited, and they are closed under multiplication and inverses, though we shall not make note of that here).

module _ {ℓ} {A B : Group ℓ} (f : Groups.Hom A B) where private
module Ker[f] = Kernel (Ker f)
module f = is-group-hom (f .preserves)
module A = Group-on (A .snd)
module B = Group-on (B .snd)

kerf : ⌞ Ker[f].ker ⌟ → ⌞ A ⌟
kerf = Ker[f].kernel .hom

has-zero : fibre kerf A.unit
has-zero = (A.unit , f.pres-id) , refl

has-⋆ : ∀ {x y} → fibre kerf x → fibre kerf y → fibre kerf (x A.⋆ y)
has-⋆ ((a , p) , q) ((b , r) , s) =
(a A.⋆ b , f.pres-⋆ _ _ ·· ap₂ B._⋆_ p r ·· B.idl) ,
ap₂ A._⋆_ q s


It turns out that $\ker f$ is also closed under conjugation by elements of the enveloping group, in that if $f(x) = 1$ (quickly switching to “multiplicative” notation for the unit), then $f(yxy^{-1})$ must be $1$ as well: for we have $f(y)f(x)f(y^{-1}) = f(y)1f(y^{-1}) = f(yy^{-1}) = f(1) = 1$.

  has-conjugate : ∀ {x y} → fibre kerf x → fibre kerf (y A.⋆ x A.⋆ y A.⁻¹)
has-conjugate {x} {y} ((a , p) , q) = (_ , path) , refl where
path =
f # (y A.⋆ (x A.— y))         ≡⟨ ap (f #_) A.associative ⟩≡
f # ((y A.⋆ x) A.— y)         ≡⟨ f.pres-diff ⟩≡
⌜ f # (y A.⋆ x) ⌝ B.— f # y   ≡⟨ ap! (f.pres-⋆ y x) ⟩≡
⌜ f # y B.⋆ f # x ⌝ B.— f # y ≡⟨ ap! (ap (_ B.⋆_) (ap (f #_) (sym q) ∙ p) ∙ B.idr) ⟩≡
f # y B.— f # y               ≡˘⟨ f.pres-diff ⟩≡˘
f # (y A.— y)                 ≡⟨ ap (f #_) A.inverser ∙ f.pres-id ⟩≡
B.unit                        ∎


It turns out that this last property is enough to pick out exactly the kernels amongst the representations of subgroups: If $H$ is closed under conjugation, then $H$ generates an equivalence relation on the set underlying $G$ (namely, $(x - y) \in H$), and equip the quotient of this equivalence relation with a group structure. The kernel of the quotient map $G \to G/H$ is then $H$. We call a predicate representing a kernel a normal subgroup, and we denote this in shorthand by $H \unlhd G$.

record normal-subgroup (G : Group ℓ) (H : ℙ ⌞ G ⌟) : Type ℓ where
open Group-on (G .snd)
field
has-rep : represents-subgroup G H
has-conjugate : ∀ {x y} → x ∈ H → (y ⋆ x ⋆ y ⁻¹) ∈ H

has-conjugatel : ∀ {x y} → y ∈ H → ((x ⋆ y) ⋆ x ⁻¹) ∈ H
has-conjugatel yin = subst (_∈ H) associative (has-conjugate yin)

has-comm : ∀ {x y} → (x ⋆ y) ∈ H → (y ⋆ x) ∈ H
has-comm {x = x} {y} ∈ = subst (_∈ H) p (has-conjugate ∈) where
p = x ⁻¹ ⋆ ⌜ (x ⋆ y) ⋆ x ⁻¹ ⁻¹ ⌝ ≡˘⟨ ap¡ associative ⟩≡˘
x ⁻¹ ⋆ x ⋆ y ⋆ ⌜ x ⁻¹ ⁻¹ ⌝   ≡⟨ ap! inv-inv ⟩≡
x ⁻¹ ⋆ x ⋆ y ⋆ x             ≡⟨ associative ⟩≡
(x ⁻¹ ⋆ x) ⋆ y ⋆ x           ≡⟨ ap₂ _⋆_ inversel refl ∙ idl ⟩≡
y ⋆ x                        ∎

open represents-subgroup has-rep public


So, suppose we have a normal subgroup $H \unlhd G$. We define the underlying type of the quotient $G/H$ to be the quotient of the relation $R(x, y) = (x - y) \in H$; It can be equipped with a group operation inherited from $G$, but this is incredibly tedious to do.

module _ (Grp : Group ℓ) {H : ℙ ⌞ Grp ⌟} (N : normal-subgroup Grp H) where
open normal-subgroup N
open Group-on (Grp .snd) renaming (inverse to inv)
private
G0 = ⌞ Grp ⌟
rel : G0 → G0 → Type _
rel x y = (x ⋆ y ⁻¹) ∈ H

rel-refl : ∀ x → rel x x
rel-refl _ = subst (_∈ H) (sym inverser) has-unit

    G/H : Type _
G/H = G0 / rel

op : G/H → G/H → G/H
op = Quot-op₂ rel-refl rel-refl _⋆_ (λ w x y z a b → rem₃ y z w x b a) where


To prove that the group operation _⋆_ descends to the quotient, we prove that it takes related inputs to related outputs — a characterisation of binary operations on quotients we can invoke since the relation we’re quotienting by is reflexive. It suffices to show that $(yw - zx) \in H$ whenever $w - x$ and $y - z$ are both in $H$, which is a tedious but straightforward calculation:

      module
_ (w x y z : G0)
(w-x∈ : (w ⋆ inv x) ∈ H)
(y-z∈ : (y ⋆ inv z) ∈ H) where abstract
rem₁ : ((w — x) ⋆ (inv z ⋆ y)) ∈ H
rem₁ = has-⋆ w-x∈ (has-comm y-z∈)

rem₂ : ((w ⋆ (inv x — z)) ⋆ y) ∈ H
rem₂ = subst (_∈ H) (associative ∙ ap (_⋆ y) (sym associative)) rem₁

rem₃ : ((y ⋆ w) — (z ⋆ x)) ∈ H
rem₃ = subst (_∈ H) (associative ∙ ap₂ _⋆_ refl (sym inv-comm))
(has-comm rem₂)


To define inverses on the quotient, it suffices to show that whenever $(x - y) \in H$, we also have $(x^{-1} - y) \in H$.

    inverse : G/H → G/H
inverse =
Coeq-rec squash (λ x → inc (inv x)) λ { (x , y , r) → quot (p x y r) }
where abstract
p : ∀ x y → (x — y) ∈ H → (inv x — inv y) ∈ H
p x y r = has-comm (subst (_∈ H) inv-comm (has-inv r))


Even after this tedious algebra, it still remains to show that the operation is associative and has inverses. Fortunately, since equality in a group is a proposition, these follow from the group axioms on $G$ rather directly:

    Group-on-G/H : make-group G/H
Group-on-G/H .make-group.group-is-set = squash
Group-on-G/H .make-group.unit = inc unit
Group-on-G/H .make-group.mul = op
Group-on-G/H .make-group.inv = inverse
Group-on-G/H .make-group.assoc =
Coeq-elim-prop₃ (λ _ _ _ → squash _ _) λ x y z i →
inc (associative {x = x} {y} {z} i)
Group-on-G/H .make-group.invl =
Coeq-elim-prop (λ _ → squash _ _) λ x i → inc (inversel {x = x} i)
Group-on-G/H .make-group.idl =
Coeq-elim-prop (λ _ → squash _ _) λ x i → inc (idl {x = x} i)

_/ᴳ_ : Group _
_/ᴳ_ = to-group Group-on-G/H

incl : Groups.Hom Grp _/ᴳ_
incl .hom = inc
incl .preserves .is-group-hom.pres-⋆ x y = refl


Before we show that the kernel of the quotient map is isomorphic to the subgroup we started with (and indeed, that this isomorphism commutes with the respective, so that they determine the same subobject of $G$), we must show that the relation $(x - y) \in H$ is an equivalence relation; We can then appeal to effectivity of quotients to conclude that, if $\mathrm{inc}(x) = \mathrm{inc}(y)$, then $(x - y) \in H$.

  private
rel-sym : ∀ {x y} → rel x y → rel y x
rel-sym h = subst (_∈ H) (inv-comm ∙ ap (_⋆ _) inv-inv) (has-inv h)

rel-trans : ∀ {x y z} → rel x y → rel y z → rel x z
rel-trans {x} {y} {z} h g = subst (_∈ H) p (has-⋆ h g) where
p = (x — y) ⋆ (y — z)      ≡˘⟨ associative ⟩≡˘
x ⋆ ⌜ y ⁻¹ ⋆ (y — z) ⌝ ≡⟨ ap! associative ⟩≡
x ⋆ ⌜ (y ⁻¹ ⋆ y) — z ⌝ ≡⟨ ap! (ap (_⋆ _) inversel ∙ idl) ⟩≡
x — z                  ∎

open Congruence
normal-subgroup→congruence : Congruence _ _
normal-subgroup→congruence ._∼_ = rel
normal-subgroup→congruence .has-is-prop x y = hlevel!
normal-subgroup→congruence .reflᶜ = rel-refl _
normal-subgroup→congruence ._∙ᶜ_ = rel-trans
normal-subgroup→congruence .symᶜ = rel-sym

/ᴳ-effective : ∀ {x y} → Path G/H (inc x) (inc y) → rel x y
/ᴳ-effective = equiv→inverse (effective normal-subgroup→congruence)

  private
module Ker[incl] = Kernel (Ker incl)
Ker-sg = Ker-subgroup incl
H-sg = predicate→subgroup H has-rep
H-g = H-sg .fst


The two halves of the isomorphism are now very straightforward to define: If we have $\mathrm{inc}(x) = \mathrm{inc}(0)$, then $x - 0 \in H$ by effectivity, and $x \in H$ by the group laws. Conversely, if $x \in H$, then $x - 0 \in H$, thus they are identified in the quotient. Thus, the predicate $\mathrm{inc}(x) = \mathrm{inc}(0)$ recovers the subgroup $H$; And (the total space of) that predicate is exactly the kernel of $\mathrm{inc}$!

  Ker[incl]≅H-group : Ker[incl].ker Groups.≅ H-g
Ker[incl]≅H-group = Groups.make-iso to from il ir where
to : Groups.Hom _ _
to .hom (x , p) = x , subst (_∈ H) (ap (_ ⋆_) inv-unit ∙ idr) x-0∈H where
x-0∈H = /ᴳ-effective p
to .preserves .is-group-hom.pres-⋆ _ _ = Σ-prop-path (λ _ → H _ .is-tr) refl

from : Groups.Hom _ _
from .hom (x , p) = x , quot (subst (_∈ H) (sym idr ∙ ap (_ ⋆_) (sym inv-unit)) p)
from .preserves .is-group-hom.pres-⋆ _ _ = Σ-prop-path (λ _ → squash _ _) refl

il = Homomorphism-path λ x → Σ-prop-path (λ _ → H _ .is-tr) refl
ir = Homomorphism-path λ x → Σ-prop-path (λ _ → squash _ _) refl


To show that these are equal as subgroups of $G$, we must show that the isomorphism above commutes with the inclusions; But this is immediate by computation, so we can conclude: Every normal subgroup is a kernel.

  Ker[incl]≡H↪G : Ker-sg ≡ H-sg
Ker[incl]≡H↪G = ≤-antisym ker≤H H≤ker where
SubG = Subobjects (Groups ℓ) Groups-is-category Grp
open Poset SubG
open Groups._≅_ Ker[incl]≅H-group

ker≤H : Ker-sg ≤ H-sg
ker≤H .fst = to
ker≤H .snd = Forget-is-faithful refl

H≤ker : H-sg ≤ Ker-sg
H≤ker .fst = from
H≤ker .snd = Forget-is-faithful refl