Quotient Groups, Normal Subgroups and Direct Products

Quotient Groups, Normal Subgroups and Direct Products

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia
aram@unimelb.edu.au
and

Department of Mathematics
University of Wisconsin, Madison
Madison, WI 53706 USA
ram@math.wisc.edu

Last updates: 20 April 2010

Quotient subgroups ↔ Normal subgroups

Let H be a subgroup of a group G. We can try to make the set G/H of cosets of H into a group by defining a multiplication operation on the cosets. The only problem is that this doesn't work for the cosets of just any subgroup, the subgroup has to have special properties.

A subgroup N of G is normal if for each n∈N, gn g -1 ∈ N for all g ∈G.

HW:Show that a subgroup N of a group G is normal iff gN=Ng for all g∈G.

Let N be a subgroup of a group G. N is a normal subgroup of G iff G/N with the operation given by aN bN =abN is a group.

Proof.

⇒ ba) Assume N is a normal subgroup of G.
To show:
  1. aN bN = abN is a well defined operation on G/ N .
  2. N is the identity element of G/ N .
  3. g -1 N is the inverse of gN .
  4. (Proof) We want the operation on G/ N given by G/ N×G/ N → G/ N aN bN ↦ abN to be well defined.
  5. To show: If a1 N a2 N , a2 N b2 N ∈ G/ N ×G/ N and a1 N b1 N = a 2 N b2 N then a 1 b1 N= a2 b2 N.
    1. Let a1 N a2 N , a2 N b2 N ∈ G/ N ×G/ N such that a1 N b1 N = a 2 N b2 N .
    2. Then a 1 N = a 2 N and b1 N= b2 N.
    3. To show:
      1. a1 b1 N⊆ a2 b2 N.
      2. a2 b2 N⊆ a1 b1 N.
      3. (Proof) We know a 1 = a 1 · 1 ∈ a 2 N since a1 N = a2 N.
      4. So a 1 = a2 n1 for some n1 ∈ N.
      5. Similarly, b1 = b2 n2 for some n2 ∈ N.
      6. Let k∈ a 1 b1 N.
      7. Then k= a 1 b1 n for some n∈ N. So k= a 1 b1 n = a2 n1 b2 n2 n = a 2 b2 b2 -1 n1 b2 n2 n. Since N is normal, b2 -1 n1 b2 ∈ N, and therefore b2 -1 n1 b2 n2 n ∈ N.
      8. So k = a 2 b2 b2 -1 n1 b2 n2 n ∈ a 2 b2 N.
      9. So a 1 b1 N⊆ a 2 b2 N.
      10. (Proof) Since a 1 N = a2 N , we know a 1 n1 = a 2 for some n1 ∈ N .
      11. Since b 1 N= b2 N , we know b1 n2 = b2 for some n2 ∈ N.
      12. Let k∈ a 2 b2 N.
      13. Then k= a 2 b2 n for some n∈ N. So k= a 2 b2 n = a1 n1 b1 n2 n = a 1 b1 b1 -1 n1 b1 n2 n. Since N is normal, b1 -1 n1 b1 ∈ N, and therefore b1 -1 n1 b1 n2 n ∈ N.
      14. So k = a 1 b1 b1 -1 n1 b1 n2 n ∈ a 1 b1 N.
      15. So a 2 b2 N⊆ a 1 b1 N.
    4. So a 1 b1 N= a 2 b2 N .
    5. So the operation is well defined.
  6. The coset N=1N is the identity since N gN = 1G N =gN = g1 N = gN N , for all g∈ G.
  7. Given any coset gN its inverse is g -1 N since gN g -1 N = g g -1 N =N = g -1 gN = g -1 N gN . for all g∈ G.
So G/ N is a group.
⇐ Assume G/ N is a group with operation aN bN =abN .
To show: If g∈ G and n ∈ N then gn g -1 ∈ N.
  1. First we show: If n∈ N then nN=N.
  2. Assume n∈ N.
  3. To show:
    1. nN∈ N.
    2. N∈ nN.
    3. (Proof) Let x∈ nN .
    4. Then x=nm for some m∈ N.
    5. Since N is a subgroup, nm ∈ N.
    6. So x∈ N.
    7. So nN⊆ N.
    8. (Proof) Assume m∈ N.
    9. Then, since N is a subgroup, m=n n -1 m ∈ nN.
    10. So N⊆ nN.
Now let g∈ G and n∈ N.
Then, by defintion of the operation, gn g -1 N= gN nN g -1 N = gN N g -1 N =g1 g -1 N =N. So gn g -1 ∈ N.
So N is a normal subgroup of G.

□

The quotient group G/N is the set of cosets of normal subgroup N of a group G with the operation given by aN bN =abN

Wow! We actually made this weird set of cosets into a group.

HW: Let N be a subgroup of a group G. Show that N is a normal subgroup of G iff the operation on G/N given by aN bN =abN is well defined.

Homomorphisms

Group homomorphisms are used to compute groups. Let G and H be groups with identities 1 G and 1 H respectively.

A group homomorphism f:G →H is a map between groups G and H such that f gg' =f g f g'   for all  g,g'∈G.

A group isomorphism is a bijective group homomorphism.

Two groups G and H are isomorphic, G≃H, if there exists a group isomorphism f: G → H between them.

Two groups are isomorphic if both the elements of the groups and their operations match up exactly. Think of two groups that are isomorphic as being "the same". When we are classifying groups, we put two two groups in the same class only if they are isomorphic. This is what we mean by classifying groups "up to isomorphism".

HW: Show that if G=N then G/N≃ 1 .

Let f:G→H be a group homomorphism. Let 1 G and 1 H be the identities for G and H respectively. Then

  1. f 1 G = 1 H .
  2. For any g∈G, f g -1 =f g -1 .

Proof.

  1. Multiply both sides of the following equation by f 1G -1 . f 1G =f 1 G · 1G =f 1G f 1G .
  2. Since f g f g -1 =f g g -1 =f 1G = 1H , and f g -1 f g =f g -1 g =f 1G = 1H , then f 1 -1 =f g -1 .

□

The kernel of a group homomorphism f: G→H is the set kerf= g∈G| f g = 1 H , where 1 H is the identity in H .

The image of a group homomorphism f:G→H is the set imf= h∈H| f g =h  for some  g∈G .

Let f:G→H be a group homomorphism. Then

  1. kerf is a normal subgroup of G.
  2. imf is a subgroup of H.

Proof.
To show:
  1. ker f is a normal subgroup of G.
  2. imf is a subgroup of G.
  3. (Proof) To show:
    1. kerf is a subgroup.
    2. kerf is normal.
  4. To show:
    1. aaa) If k1 , k2 ∈ kerf then k1 k2 ∈ kerf.
    2. aab) 1G ∈ kerf.
    3. aac) If k∈ kerf then k -1 ∈ kerf.
    4. aaa) (Proof) Assume k1 , k2 ∈ kerf. Then f k1 = 1H and f k2 = 1H .
    5. So f k1 k2 =f k1 f k2 = 1H .
    6. So k1 k2 ∈ kerf.
    7. aab) (Proof) Since f 1G = 1H , 1G ∈ kerf.
    8. aac) (Proof) Assume k∈ kerf. So f k = 1H .
    9. Then f k -1 =f k -1 = 1H -1 = 1H . So k -1 ∈ kerf.
    So kerf is a subgroup.
  5. (Proof) To show: If g∈ G and k∈ kerf then gk g -1 ∈ kerf.
    1. Assume g∈ G and k∈ kerf. Then f gk g -1 =f g f k f g -1 =f g f g -1 =f g f g -1 =1.
  7. So gk g -1 ∈ kerf.
    8. So kerf is a normal subgroup of G.
  8. To show: im f is a subgroup of H.
    1. To show:
    2. If h1 , h2 ∈ imf then h1 h2 im ∈ f.
    3. 1H ∈ imf.
    4. If h∈ imf then h -1 ∈ imf.
    5. (Proof) Assume h1 , h2 ∈ imf.
    6. Then h1 =f g1 and h2 =f g2 for some g1 , g2 ∈ G.
    7. Then h1 h2 =f g1 f g2 =f g1 g2 since f is a homomorphism.
    8. So h1 h2 ∈ imf.
    9. (Proof) By Proposition 2.1 a), f 1G = 1H so 1H ∈ imf.
    10. (Proof) Assume h∈ imf.
    11. Then h=f g for some g∈ G.
    12. Then, by Proposition 2.1 b), h -1 =f g -1 =f g -1 . So h -1 ∈ imf.
  9. So imf is a subgroup of H.

□

Let f:G→H be a group homomorphism. Let 1 G be the identity in G. Then

  1. kerf= 1 G iff f is injective.
  2. im f=H iff f is surjective.

    • Proof.

    1. Let 1G and 1H be the identities for G and H respectively.
    2. ⇒: Assume kerf= 1G .
    To show: If f g1 =f g2 then g1 = g2 .
    1. Assume f g1 =f g2 .
    2. Then, by Proposition 2.1 b), and the fact that f is a homomorphism, 1H =f g1 f g2 -1 =f g1 g2 -1 . So g1 g2 -1 ∈ kerf.
    3. But kerf= 1G .
    4. So g1 g2 -1 = 1G .
    5. So g1 = g2 .
    6. So f is injective.
    ⇐ : Assume f is injective.
    To show:
    1. 1G ⊆ kerf.
    2. kerf⊆ 1G .
    3. (Proof) Since f 1G = 1H , 1G ∈ kerf.
    4. So 1G ⊆ kerf.
    5. (Proof) Let k∈ kerf. Then f k = 1H . So f k =f 1G . Thus, since f is injective, k= 1G .
    6. So kerf ⊆ 1G .
    So kerf= 1G .
    To show:
    1. imf ⊆ H.
    2. H⊆ imf.
    3. (Proof) Let x∈ imf.
    4. Then x=f g for some g∈ G.
    5. By the definition of f , f g ∈ H.
    6. So x∈ H.
    7. So imf⊆ H.
    8. (Proof) Assume x∈ H.
    9. Since f is surjective there exists a g such that f g =x.
    10. So x∈ imf.
    11. So H ⊆ imf.
    ⇒: Assume im f =H.
    To show: If h∈ H then there exists g∈ G such that f g =h.
    1. Assume h∈ H.
    2. Then h∈ imf .
    3. So there exists some g∈ G such that f g =h.
    So f is surjective.
    ⇐ : Assume f is surjective.
  • So imf=H.

    • □

    Notice that the proof of 2.3 (b) does not use the fact that f:G→H is a homomorphism, only that it is a function.

    HW: Show that if S and T are any two sets and f:S →T is a map then im f=T iff f is surjective.

      2. Let f:G→H be a group homomorphism and let K=kerf. Define f ^ : G/kerf → H gK ↦ f g . Then f ^ is a well defined injective group homomorphism.
    2. Let f:G→H be a group homomorphism and define f': G → imf g ↦ f g . Then f ' is a well defined surjective group homomorphism.
    3. If f:G→H is a group homomorphism then G/kerf ≃imf, where the isomorphism is a group isomorphism.

    Proof.

    1. To show:
      1. f^ is well defined.
      2. f^ is injective.
      3. f^ is a homomorphism.
      4. (Proof) To show:
        1. aaa) If g∈ G then f^ gK ∈ H.
        2. aab) If g1K= g2K then f^ g1 K = f^ g2 K .
        3. aaa) (Proof) Assume g∈ G.
        4. Then f^ gK =f g and f g ∈ H by the definition of f^ and f.
        5. aab) (Proof) Assume g1 K= g2 K .
        6. Then g1 = g2 k for some k∈ K.
        7. To show: f^ g1 K = f^ g2 K , ie,
        8. To show: f g1 =f g2 .
          1. Since k∈ kerf , we have f k =1 and so f g1 =f g2 k =f g2 f k =f g2 .
          2. So f^ g1 K = f^ g2 K .
        9. So f^ is well defined.
        10. To show: If f^ g1 K = f^ g1 K then g1 K= g2 K.
          1. Assume f^ g1 K = f^ g2 K . Then f g1 =f g2 .
          2. So f g1 f g2 -1 =1.
          3. So f g1 g2 -1 =1.
          4. So g1 g2 -1 ∈ kerf.
          5. So g1 g2 -1 =k for some k∈ kerf.
          6. So g1 = g2 k for some k∈ kerf .
          7. To show:
            1. aba) g1 K⊆ g2 K.
            2. abb) g2 K⊆ g1 K.
            3. aba) (Proof) Let g∈ g 1 K. Then g= g 1 k 1 for some k 1 =k.
              So g= g 2 k k 1 ∈ g 2 K, since k k 1 ∈K.
              So g 1 K⊆ g 2 K.
            4. abb) (Proof) Let g∈ g 2 K. Then g= g 2 k 2 for some k 2 =k.
              So g= g 1 k -1 k 2 ∈ g 1 K, since k -1 k 2 ∈K.
              So g 2 K⊆ g 1 K.
            So g 1 K= g 2 K.
          So f ^ is injective.
      5. To show: f ^ g 1 K f ^ g 2 K = f ^ g 1 K g 2 K .
        1. Since f is a homomorphism, f ^ g 1 K f ^ g 2 K = f g 1 f g 2 = f g 1 g 2 = f ^ g 1 g 2 K = f ^ g 1 K g 2 K .
        So f ^ is a homomorphism.
    2. To show:
      1. f' is well defined.
      2. f' is surjective.
      3. f' is a homomorphism.
      4. (Proof) Proved in Ex 2.2.3, Part I ??????????????
      5. (Proof) Proved in Ex 2.2.3, Part I ??????????????
      6. (Proof) Since f is a homomorphism, f' g f' h =f g f h =f gh =f' gh . So f' is a homomorphism.
    3. (Proof) Let K=kerf.
      By a), the function f ^ : G/K → H gK ↦ f g is a well defined injective homomorphism.
      By b), the function f ^ ': G/K → im f ^ gK ↦ f ^ gK =f g is a well defined surjective homomorphism.
      To show:
      1. im f ^ =imf.
      2. f ^ ' is injective.
      3. (Proof) To show:
        1. caa) im f^ ⊆imf.
        2. caa) imf⊆ f^ .
        3. caa) (Proof) Let h∈im f ^ .
          Then there is some gK∈G/K such that f ^ gK =h.
          Let g'∈gK.
          Then g'=gk for some k∈K.
          Then, since f is a homomorphism and f k =1, f g' = f gk = f g f k = f g = f ^ gK = h. So h∈imf.
          So im f ^ ⊆imf.
        4. cab) (Proof) Let h∈imf.
          Then there is some g∈G such that f g =h.
          So f ^ gK =f g =h.
          So h∈im f ^ .
          So imf⊆im f ^ .
      4. (Proof) To show: If f ^ ' g 1 K = f ^ ' g 2 K then g 1 K= g 2 K.
        1. Assume f ^ ' g 1 K = f ^ ' g 2 K .
          Then f ^ g 1 K = f ^ g 2 K .
          Then, since f ^ is injective, g 1 K= g 2 K.
        So f ^ ' is injective.
      Thus we have f ^ ': G/K → im f ^ gK ↦ f g is a well defined bijective homomorphism.

    □

    Direct Products

    Suppose H and K are groups. The idea is to make H×K into a group.

    The direct product H×K of two groups H and K is the set H×K with the operation given by h 1 k 1 h 2 k 2 = h 1 h 2 k 1 k 2 for all h 1 , h 2 ∈H   and   k 1 , k 2 ∈K. We say that the multiplication in H×K is componentwise.

    More generally, given groups G 1 , …, G n , the direct product G 1 ×…× G n is the set given by G 1 ×…× G n with the operation given by h 1 … h i … h n k 1 … k i … k n = h 1 k 1 … h i k i … h n k n where h i , k i ∈ G i and h i k i is given by the operation in the group G i .

    HW: Show that these are good definitions, ie that as defined above, H×K and G 1 × … × G n are groups with identities given by 1 H 1 K and 1 G 1 … 1 G n respectively ( 1 G i denotes the identity in the group G i ).

    Further Definitions

    A group G is abelian if g 1 g 2 = g 2 g 1 for all g 1 , g 2 ∈G.

    The center Z G of a group G is the set Z G = c∈G| cig=gc   for all  g∈G .

    HW: Give an example of a non-abelian group.

    HW: Prove that every subgroup of an abelian group is normal.

    HW: Prove that Z G is a normal subgroup of G.

    HW: Prove that Z G =G iff G is abelian.

    The order G of a group G is the number of elements in G.

    Let G be a group and let g∈G. The order o g of g is the smallest positive integer n such that g n =1. If no such integer exists then o g =∞.

    Let G be a group and let S be a subset of G. The subgroup generated by S,   S , is the subgroup S of G such that

    1. S⊆ S .
    2. If H is a subgroup of G   and S ⊆ H then S ⊆ H.
    3. S is the smallest subgroup of G containing S.

    Think of S as gotten by adding to S exactly those elements of G that are needed to make a group.

    HW: Let G be a group and let S be a subsert of G. Show that the subgroup generated by S,   S , exists and is unique.

    References [PLACEHOLDER]

    [BG] A. Braverman and D. Gaitsgory, Crystals via the affine Grassmanian, Duke Math. J. 107 no. 3, (2001), 561-575; arXiv:math/9909077v2, MR1828302 (2002e:20083)

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