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

module Algebra.Group.Subgroup where


A subgroup mm of a group GG is a monomorphism HmGH \xto{m} G, that is, an object of the poset of subobjects Sub(G)\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 GPropG \to \prop.

Subgroup : Group   Type (lsuc )
Subgroup { = } G = Precategory.Ob (Subobj (Groups ) G)

A proposition H:GPropH : G \to \prop of a group (G,)(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 ) (H :  (G .fst)) : Type  where
  open Group-on (G .snd)

    has-unit : unit  H
    has-⋆    :  {x y}  x  H  y  H  (x  y)  H
    has-inv  :  {x}  x  H  x ⁻¹  H

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

  : (H :  (G .fst))  represents-subgroup G H  Group-on (Σ[ x  G .fst ] x  H)
rep-subgroup→group-on {G = G} H sg =
    (Σ-is-hlevel 2 has-is-set λ x  is-prop→is-set (H x .is-tr))
    (unit , has-unit)
     { (x , xin) (y , yin)  x  y , has-⋆ xin yin} )
     { (x , xin)  (x ⁻¹ , has-inv xin) })
     x y z  Σ-prop-path  x  H x .is-tr) (sym associative))
     x  Σ-prop-path  x  H x .is-tr) inversel)
     x  Σ-prop-path  x  H x .is-tr) inverser)
     x  Σ-prop-path  x  H x .is-tr) idl)
  where open Group-on (G .snd)
        open represents-subgroup sg

predicate→subgroup : (H :  (G .fst))  represents-subgroup G H  Subgroup G
predicate→subgroup {G = G} H p = cut map , ism where
  map : Groups.Hom (Σ (∣_∣  H) , rep-subgroup→group-on H p) G
  map .fst = fst
  map .snd .Group-hom.pres-⋆ x y = refl

  ism : map
  ism g h p = Forget-is-faithful
    (embedding→monic (Subset-proj-embedding (is-tr  H)) (fst g) (fst h) (ap fst p))

Kernels and Images🔗

To a group homomorphism f:ABf : A \to B we can associate two canonical subgroups, one of AA and one of BB: ff’s image, written imf\im f, is the subgroup of BB “reachable by mapping through ff”, and ff’s kernel, written kerf\ker f, is the subgroup of AA which ff sends to the unit.

The kernel can be cheapily described as a limit: It is the equaliser of ff and the zero morphism — which, recall, is the unique map ABA \to B which breaks down as A0BA \to 0 \to B.

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

  Ker-subgroup :  {A B : Group }  Groups.Hom A B  Subgroup A
  Ker-subgroup f = cut kernel ,→is-monic _ has-is-kernel where
    open Kernel (Ker f)

Every group homomorphism f:ABf : A \to B has an image imf\im f, defined by equipping its set-theoretic image with a group structure inherited from BB. More concretely, we can describe the elements of imf\im f as the “mere fibres” of ff: They consist of a point y:By : B, together with (the truncation of) a fibre of ff over yy. We multiply xx (in the fibre over aa) with yy (in the fibre over bb), giving the element xyxy in the fibre over abab.

For reasons that will become clear later, we denote the image of ff, when regarded as its own group, by A/ker(f)A/\ker(f), and reserve the notation imf\im f for that group regarded as a subgroup of BB.

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

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

    inv : T  T
    inv (x , p) = x B.⁻¹ ,
      ∥-∥-map  { (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  _ _  squash)
         { (x* , xp) (y* , yp)
            inc (x* A.⋆ y* , f.pres-⋆ _ _  ap₂ B._⋆_ xp yp) })
        xp yp

    grp : Group-on T
    grp = make-group Tset unit mul inv
       x y z  Tpath (sym B.associative))
       x  Tpath B.inversel)  x  Tpath B.inverser) λ x  Tpath B.idl

That the canonical inclusion map A/ker(f)BA/\ker(f) \mono B deserves the name “image” comes from ff breaking down as a (regular) epimorphism into imf\im f (written A→im), followed by that map:

(AfB)=(AfA/ker(f)B) (A \xto{f} B) = (A \xepi{f} A/\ker(f) \mono B)

  A→im : Groups.Hom A A/ker[_]
  A→im .fst x = f .fst x , inc (x , refl)
  A→im .snd .Group-hom.pres-⋆ x y = Tpath (f.pres-⋆ _ _)

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

When this monomorphism is taken as primary, we refer to A/ker(f)A/\ker(f) as imf\im f.

  Im[_] : Subgroup B
  Im[_] = cut im→B , im↪B where
    im↪B : im→B
    im↪B = injective-group-hom im→B Tpath

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

Note: In more classical texts, the first isomorphism theorem is phrased in terms of two pre-existing objects A/ker(f)A/\ker(f) (defined as the set of cosets of ker(f)\ker(f) regarded as a subgroup) and imf\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)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 : (→ ∅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 AA/ker(f)A \epi A/\ker(f) is a coequaliser (hence that it is a regular epi, as we mentioned above).

The condition placed on ee' is that 0=ekerf0 = e' \circ \ker f; This means that it, like ff, sends everything in kerf\ker f to zero (this is the defining property of kerf\ker f). Note that in the code below we do not elide the zero composite e0e' \circ 0.

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

To eliminate from under a [propositional truncation], we must prove that the map ee' is constant when thought of as a map f(x)Ff^*(x) \to F; In other words, it means that ee' is “independent of the choice of representative”. This follows from algebraic manipulation of group homomorphisms + the assumed identity 0=ekerf0 = e' \circ \ker f;

      const′ :  (x y : fibre (f .fst) 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 fst p)) (y A.— z , aux) 
        e' A.unit      ≡⟨ e'.pres-id 

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

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

      const :  (x y : fibre (f .fst) x)  (e' (x .fst))  (e' (y .fst))
      const a b = (const′ a b)

The rest of the construction is almost tautological: By definition, if x:kerfx : \ker f, then f(x)=0f(x) = 0, so the quotient map AA/ker(f)A \epi A/\ker(f) does indeed coequalise kerfA\ker f \mono A and 00. 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 .fst)  A→im .fst A.unit  A→im .fst (x .fst)
      path (x* , p) = Tpath (f.pres-id  sym p)

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

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

    coeq .universal = 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 .fst x  elim p (x .snd)
        path (x , t) =
            {P = λ q  colim .fst (x , q)  elim p q}
             _  F.has-is-set _ _)
             { (f , fp)  ap (colim .fst) (Σ-prop-path  _  squash) (sym fp))
                           sym (happly (ap fst prf) f) })

Representing kernels🔗

If an evil wizard kidnaps your significant others and demands that you find out whether a predicate P:GPropP : G \to \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 PP definitely represents a subgroup, in the sense introduced above — so there’s definitely a group homomorphism ΣGG\Sigma G \mono G. All we need to figure out is whether there exists a group HH and a map f:GHf : G \to H, such that ΣGkerf\Sigma G \cong \ker f as subgroups of GG.

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 00 is inhabited, and they are closed under multiplication and inverses, though we shall not make note of that here).

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

  kerf : Ker[f].ker .fst  A .fst
  kerf = Ker[f].kernel .fst

  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 kerf\ker f is also closed under conjugation by elements of the enveloping group, in that if f(x)=1f(x) = 1 (quickly switching to “multiplicative” notation for the unit), then f(yxy1)f(yxy^{-1}) must be 11 as well: for we have f(y)f(x)f(y1)=f(y)1f(y1)=f(yy1)=f(1)=1f(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 .fst (y A.⋆ (x A.— y))             ≡⟨ ap (f .fst) A.associative 
      f .fst ((y A.⋆ x) A.— y)             ≡⟨ f.pres-diff 
      f .fst (y A.⋆ x) B.— f .fst y        ≡⟨ ap (B._⋆ _) (f.pres-⋆ y x) 
      (f .fst y B.⋆ f .fst x) B.— f .fst y ≡⟨ ap (B._⋆ _) (ap (_ B.⋆_) (ap (f .fst) (sym q)  p)  B.idr) 
      f .fst y B.— f .fst y                ≡˘⟨ f.pres-diff ≡˘
      f .fst (y A.— y)                     ≡⟨ ap (f .fst) A.inverser  f.pres-id 

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

record normal-subgroup (G : Group ) (H :  (G .fst)) : Type  where
  open Group-on (G .snd)
    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 ≡⟨ ap₂ _⋆_ refl (sym associative)   i  x ⁻¹  x  y  inv-inv {x = x} i) 
        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 HGH \unlhd G. We define the underlying type of the quotient G/HG/H to be the quotient of the relation R(x,y)=(xy)HR(x, y) = (x - y) \in H; It can be equipped with a group operation inherited from GG, but this is incredibly tedious to do.

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

    op : G/H  G/H  G/H
    op = Quot-op₂ rel-refl rel-refl _⋆_  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 (ywzx)H(yw - zx) \in H whenever wxw - x and yzy - z are both in HH, which is a tedious but straightforward calculation:

        _ (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 (xy)H(x - y) \in H, we also have (x1y)H(x^{-1} - y) \in H.

    inverse : G/H  G/H
    inverse =
      Coeq-rec squash  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 GG rather directly:

    Group-on-G/H : Group-on G/H
    Group-on-G/H = make-group squash (inc unit) op inverse
      (Coeq-elim-prop₃  _ _ _  squash _ _)
        λ x y z i  inc (associative {x = x} {y} {z} (~ i)))
      (Coeq-elim-prop  _  squash _ _) λ x i  inc (inversel {x = x} i))
      (Coeq-elim-prop  _  squash _ _) λ x i  inc (inverser {x = x} i))
      (Coeq-elim-prop  _  squash _ _) λ x i  inc (idl {x = x} i))

  _/ᴳ_ : Group _
  _/ᴳ_ = G/H , Group-on-G/H

  incl : Groups.Hom Grp _/ᴳ_
  incl .fst = inc
  incl .snd .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 GG), we must show that the relation (xy)H(x - y) \in H is an equivalence relation; We can then appeal to effectivity of quotients to conclude that, if inc(x)=inc(y)\id{inc}(x) = \id{inc}(y), then (xy)H(x - y) \in H.

    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 (x ⋆_) associative 
          x  ((y ⁻¹  y)  z) ≡⟨ ap (x ⋆_) (ap (_⋆ _) inversel  idl) 
          x  z                

  /ᴳ-effective :  {x y}  Path G/H (inc x) (inc y)  rel x y
  /ᴳ-effective = equiv→inverse
    (effective  _ _  H _ .is-tr) (rel-refl _) rel-trans rel-sym)

The two halves of the isomorphism are now very straightforward to define: If we have inc(x)=inc(0)\id{inc}(x) = \id{inc}(0), then x0Hx - 0 \in H by effectivity, and xHx \in H by the group laws. Conversely, if xHx \in H, then x0Hx - 0 \in H, thus they are identified in the quotient. Thus, the predicate inc(x)=inc(0)\id{inc}(x) = \id{inc}(0) recovers the subgroup HH; And (the total space of) that predicate is exactly the kernel of inc\id{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 .fst (x , p) = x , subst (_∈ H) (ap (_ ⋆_) inv-unit  idr) x-0∈H where
      x-0∈H = /ᴳ-effective p
    to .snd .Group-hom.pres-⋆ _ _ = Σ-prop-path  _  H _ .is-tr) refl

    from : Groups.Hom _ _
    from .fst (x , p) = x , quot (subst (_∈ H) (sym idr  ap (_ ⋆_) (sym inv-unit)) p)
    from .snd .Group-hom.pres-⋆ _ _ = Σ-prop-path  _  squash _ _) refl

    il = Forget-is-faithful $ funext λ x  Σ-prop-path  _  H _ .is-tr) refl
    ir = Forget-is-faithful $ funext λ x  Σ-prop-path  _  squash _ _) refl

To show that these are equal as subgroups of GG, 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 ) Groups-is-category Grp
    open Poset SubG
    open Groups._≅_ Ker[incl]≅H-group

    ker≤H : Ker-sg  H-sg
    ker≤H ./ = to
    ker≤H ./-Hom.commutes = Forget-is-faithful refl

    H≤ker : H-sg  Ker-sg
    H≤ker ./ = from
    H≤ker ./-Hom.commutes = Forget-is-faithful refl