Groups

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

Last updates: 10 July 2011

Groups and group homomorphisms

HW: Show that the identity 1G is unique.
HW: Show that if gG then the inverse g-1G is unique.
HW: Why isn't {1,2,3,4, 5} a group?

Important examples of groups are:

(a)   The integers with the operation of addition,
(b)   The integers mod n, /n, with operation addition,
(c)   The symmetric group Sn,
(c)   The general linear group of invertible n×n matrices GLn().

Group homomorphisms are for comparing groups.

Let G and H be groups with identities 1G and 1H, respectively.

  • A group homomorphism from G to H is a function f:GH such that
    (a)   If g1,g2 G then f(g1g2) = f(g1) f(g2) ,
    (b)   f(1G) = 1H.
  • A group isomorphism is a bijective group homomorphism.
  • Two groups G and H are isomorphic, GH, if there exists a group isomorphism f:GH between them.

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

Let f:GH be a group homomorphism. Let 1G and 1H be the identities in G and H, respectively. Then

(a)   f(1G) =1H.
(b)   If gG then f(g-1) = f(g)-1.

Proof.

(a)   Multiply both sides of the following equation by f (1G) -1 . f( 1G ) =f( 1G· 1G) =f( 1G ) f(1G).
(b)   Since f(g) f(g-1) =f(g g-1 ) =f(1G) =1H and f(g-1) f(g) =f(g-1g) =f(1G) =1H, f(g) -1 =f( g-1).

  • A subgroup H of a group G is a subset HG such that
    (a)   If h1,h2 H then h1h2 H,
    (b)   1H,
    (c)   If hH then h-1H.
  • The trivial group {1} is the set containing only 1 with the operation given by 11 =1.

Cosets

Let G be a group and let H be a subgroup of G. We will use the subgroup H to divide up the group G.

  • A left coset of H in G is a set gH= {gh | hH}, where gG.
  • G/H (pronounced "G mod H") is the set of left cosets of H in G.
  • A right coset of H in G is a set Hg= {hg | hH}, where gG.
  • H\G is the set of right cosets of H in G.

Unless specified otherwise we will always work with left cosets and just call them cosets.

HW: Let G be a group and let H be a subgroup of G. Let x and g be two elements of G. Show that xgH if and only if gH=xH.

g1H g4H g5H g6H g7H g8H H g3H g2H G

Let G be a group and let H be a subgroup of G. Then the cosets of H in G partition G.

Proof.
  1. To show: (a)   If gG then there exists gG such that gg'H.
  2. To show: (b)   If g1H g2H then g1H = g2H.
  3. (a)   Let gG.
  4. (a)   Then g= g·1gH, since 1H.
  5. (a)   So ggH.
  6. (b)   Assume g1H g2H .
  7. (b)   To show: (ba)   g1H g2H .
  8. (b)   To show: (bb)   g2H g1H.
  9. (b)   Let k g1Hg2H .
  10. (b)   Suppose k =g1h1 =g2h2 , where h1, h2H .
  11. (b)   Then g1 = g1 h1 h1-1 =k h1-1 = g2 h2 h1-1 ,and g2 = g2h2 h2-1 = k h2 -1 = g1h1 h2 -1.
    1. (ba)   Let gg1H.
    2. (ba)   Then there exists hH such that g =g1h.
    3. (ba)   Then g = g1h = g2h2 h1-1 h g2H ,since h2 h1-1 hH.
    4. (ba)   So g1H g2H.
  12. (b)   (bb)   Let gg2H.
    1. (bb)   Then there exists hH such that g=g2h.
    2. (bb)   Then g=g2h =g1h1 h2-1h g1H ,since h1 h2-1 h H.
    3. (bb)   So g2H g1H.
  13. So g1H =g2H .
So the cosets of H in G partition G.

Let G be a group and let H be a subgroup of G. If g1, g2 G then Card(g1H) =Card(g2H) .

Proof.
  1. To show: There is a bijection from g1H to g2H.
  2. Define a map φ by φ: g1H g2H x g2 g1-1 x.
  3. To show: (a)   φ is well defined.
  4. To show: (b)   φ is a bijection.
  5. (a)   To show: (aa)   If xg1H then φ(x) g2H.
  6. (a)   To show: (ab)   If x=y then φ(x) =φ(y).
    1. (aa)   Assume xg1H.
    2. (aa)   Then x=g1h for some hH.
    3. (aa)   So φ(x) = g2 g1-1 g1h = g2 hg2H .
    4. (ab)   This is clear from the definition of φ.
  7. So φ is well defined.
  8. (b)   By virtue of Theorem 2.2.3 Part I, ???????????????????? we want to construct an inverse map for φ.
  9. (b)   Define ψ: g2H g1H y g1 g2-1 y.
  10. (b)   HW: Show (exactly as in (a) above) that ψ is well defined. Then ψ(φ(x) = g1 g2-1 φ(x) = g1 g2-1 g2 g1-1 x =x, and φ(ψ)y) =g2 g1-1 φ(y) =g2 g1-1 g1 g2-1 y =y. and
  11. (b)   So ψ is an inverse function to φ.
  12. So φ is a bijection.

Let H be a subgroup of a group G. Then Card(G) =Card(G/H) Card(H).

Proof.
  1. By Proposition 2.2, all cosets in G/H are the same size as H.
  2. Since the cosets of H partition G, the cosets are disjoint subsets of G, and G is a union of those subsets.
  3. So G is the union of Card(G/H) disjoint subsets all of which have size Card(H).

The above results show that the cosets of a subgroup H divide the group G into equal size pieces, one of these pieces being the subgroup H itself.

  • A set of coset representatives of H in G is a set of distinct elements {gi} of G such that
    1. each coset of H is of the form giH for some gi and
    2. giH gjH unless gi=gj.
  • The index of a subgroup H in a group G, |G:H| , is the number of cosets of H in G, |G:H| =Card(G/H).

HW:Show that |G:{1}| =Card(G).

Quotient groups Normal subgroups

Let H be a subgroup of a group G. We can try to make the set of cosets of H into a group by defining an 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 normal subgroup N is a subset of a group G such that if nN and gG then gng-1 N.

HW: Show that a subgroup N of a group G is normal if and only if N satisfies: if gG then gN=Ng.

Let N be a subgroup of a group G. Then N is a normal subgroup of G if and only if G/N with operation given by (aN) (bN) =(abN) is a group.

Proof.
Proof
  1. Assume N is a normal subgroup of G.
  2. 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-1N is the inverse of gN.
  3. We want the operation on G/N given by G/N×G/N G/N (aN,bN) abN to be well defined.
  4. To show: If (a1N,b1N ), (a2N,b2N ) G/N×G/N and (a1N, b1N ) =(a2N, b2N ) then a1 b1N =a2b2N.
  5. Assume (a1N ,a2N ), (a2N ,b2N) G/N×G/N and (a1N ,b1N )=( a2N, b2N ).
  6. Then a1N =a2N and b1N =b2N .
  7. To show: aa. a1b1 N a2b2 N.
  8. To show: ab. a2b2N a1b1N.
    1. Since a1N = a2N , we know a1 =a1·1 a2N.
    2. So there exists n1N such that a1 = a2n1 .
    3. Similarly, there exists n2N such that b1 =b2n2.
    4. Let ka1 b1N.
    5. Then there exists nN such that k= a1 b1n.
    6. So k = a1b1n = a2n1 b2 n2n = a2 b2 b2-1 n1b2 n2 n.
    7. Since N is normal, b2-1 n1b2 N, and therefore ( b2-1 n1b2 ) n2n N.
    8. So k= a2 b2 ( b2-1 n1 b2 ) n2n a2 b2N.
    9. So a1 b1N a2 b2N .
    10. To show: a2b2N a1b1N.
    11. Since a1N =a2N , there exists n1N such that a1n1 = a2.
    12. Since b1N = b2N , there exists n2N such that b1 n2 =b2 .
    13. Let k a2b2N.
    14. Then there exists nN such that k=a2 b2n .
    15. So k = a2b2n = a1n1 b1n2n = a1b1 b1-1 n1b1 n2n.
    16. Since N is normal, b1-1 n1b1 N, and therefore (b1-1 n1b1 ) n2nN .
    17. So k =a1b1 ( b1-1 n1b1 ) n2n a1 b1N .
    18. So a2b2N a1b1N.
  9. So (a1 b1)N = (a2 b2)N.
  10. So the operation is well defined.
  11. The coset N=1N is the identity since if gG then (N) (gN) = (1G)N =gN = (g1)N = (gN)(N),
  12. If gN is a coset then its inverse is g-1N since (gN) (g-1N) = (gg-1)N = N = g-1gN = (g-1N) (gN).
  13. So G/N is a group.

  1. Assume G/N is a group with operation (aN) (bN)=abN .
  2. To show: If gG and nN then gng-1 N .
    1. First we show: If nN then nN=N.
    2. Assume nN.
    3. To show: a. nNN.
    4. To show: b. NnN.
    5. Assume xnN.
    6. Then there exists mN such that x=nm.
    7. Since N is a subgroup, nmN.
    8. So xN.
    9. So nNN.
    10. Assume mN.
    11. Then, since N is a subgroup, m=n n-1 mnN.
    12. So NnN.
    Now let gG and nN.
    Then, by definition of the operation, gng-1N = (gN) (nN) (g-1N) = (gN) (N) (g-1N) = g1g-1N = 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 the normal subgroup N of the group G with operation given by (aN) (bN) =(abN) .

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

Let N be a subgroup of a group G. Then N is a normal subgroup of G if and only if G/N with operation given by (aN) (bN) =(abN) is well defined.
HW: Show that if G=N then G/N{1}.

Kernel and image of a homomorphism

  • The kernel of a group homomorphism f:GH is the set kerf={gG | f(g)=1H}, where 1H is the identity in H.
  • The image of a group homomorphism f:GH is the set imf={f(g)H | gG }.

Let f:GH be a group homomorphism. Then

(a)   kerf is a normal subgroup of G.
(b)   imf is a subgroup of H.

Proof.
    To show:
    1. kerf is a normal subgroup of G.
    2. imf is a subgroup of G.
  1. To show: aa. kerf is a subgroup.
  2. To show: ab. kerf is normal.
    1. To show: aaa. If k1,k2 kerf then k1 k2kerf .
    2. To show: aab. 1Gkerf.
    3. To show: aac. If kkerf then k-1 kerf .
      1. Assume k1,k2 kerf . Then f(k1) =1H and f(k2) =1H.
      2. So f(k1k2) =f(k1) f(k2) =1H .
      3. So k1k2 kerf.
      4. Since f(1G) =1H, 1G kerf.
      5. Assume kkerf. So f(k)=1H.
      6. Then f(k-1) =f (k)-1 =1H -1 =1H . So k-1 kerf .
      So kerf is a subgroup.
    4. To show: If gG and kkerf then gk g-1 kerf.
    5. Assume gG and kkerf.
    6. Then f( gkg-1) = f(g)f(k) f(g-1) = f(g) f(g-1) = f(g) f(g) -1 = 1.
    7. So gkg-1 kerf.
  3. So kerf is a normal subgroup of G.
  4. To show: imf is a subgroup of H.
  5. To show: ba. If h1, h2imf then h1h2 imf.
  6. To show: bb. 1Himf.
  7. To show: bc. If himf then h-1 imf.
    1. Assume h1,h2 imf .
    2. Then there exist g1,g2 G such that h1 =f(g1) and h2 =f(g2).
    3. Then h1h2 =f(g1) f(g2) = f(g1g2) , since f is a homomorphism.
    4. So h1h2 imf.
    5. By Proposition 2.1 (a), f( 1G)=1H .
    6. So 1H imf .
    7. Assume himf .
    8. Then there exists gG such that h=f(g).
    9. Then, by Proposition 2.1 (b), h-1 =f(g) -1 =f( g-1). So h-1 imf .
  8. So imf is a subgroup of H.

Let f:GH be a group homomorphism. Let 1G be the identity element of G. Then

(a)   kerf={1G} if and only if f is injective.
(b)   imf=H if and only if f is surjective.

Proof.

  1. Let 1G and 1H be the identities for G and H, respectively.
    1. Assume kerf=(1G) .
    2. To show: If f(g1) =f(g2) then g1 =g2 .
    3. Assume f(g1) =f(g2) .
    4. Then, by Proposition 2.1 (b), and the fact that f is a homomorphism, 1H =f(g1) f(g2) -1 =f( g1 g2 -1 ) .
    5. So g1 g2 -1 kerf.
    6. Since kerf =(1G), then g1 g2 -1 =1G.
    7. So g1=g2.
    8. So f is injective.
  2. :
    1. Assume f is injective.
    2. To show: aa. (1G) kerf.
    3. To show: ab. kerf (1G) .
      1. Since f(1G) =1H ,1G kerf .
      2. So (1G)kerf.
      3. (Proof) Let kkerf. Then f(k)=1H . So f(k) =f(1G) . Thus, since f is injective, k=1G.
      4. So kerf(1G) .
    4. So kerf =(1G).
    1. Assume imf=H.
    2. To show: If hH then there exists gG such that f(g)=h.
      1. Assume hH.
      2. Then himf .
      3. So there exists gG such that f(g)=h.
    3. So f is surjective.
    1. Assume f is surjective.
    2. To show: ba. imfH.
    3. To show: bb. Himf.
      1. Let ximf .
      2. Then there exists gG such that x=f(g).
      3. By the definition of f,f(g) H.
      4. So xH.
      5. So imfH .
      6. (Proof) Assume xH.
      7. Since f is surjective there exists a g such that f(g) =x.
      8. So ximf.
      9. So Himf.
    4. So imf=H.

Notice that the proof of Proposition (gpinjsur)(b) does not use the fact that f:GH is a homomorphism, only the fact that f:GH is a function.

HW: Show that if S and T are sets and f:ST is a functions then imf=T if and only if f is surjective.

(a)   Let f:GH 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.
(b)   Let f:GH be a group homomorphism and define f: G imf g f(g). Then f is a well defined surjective group homomorphism.
(c)   If f:GH is a group homomorphism then G/kerfimf, where the isomorphism is a group isomorphism.

Proof.

  1. To show: aa. f^ is well defined.
  2. To show: ab. f^ is injective.
  3. To show: ac. f^ is a homomorphism.
    1. To show: aaa. If g G then f^(gK) H.
    2. To show: aab. If g1K =g2K then f^(g1K) = f^(g2K) .
      1. Assume gG.
      2. Then f^(gK) =f(g) and f(g)H by the definition of f^ and f.
      3. Assume g1K=g2K.
      4. Then there exists kK such that g1=g2k.
      5. To show: f^(g1K) =f^(g2K) .
      6. To show: f(g1) =f(g2).
      7. Since kkerf, then f(k)=1 and so f(g1) =f(g2k) =f(g2) f(k) =f(g2).
      8. So f^(g1K) = f^(g2K).
    3. So f^ is well defined.
    4. To show: If f^(g1K) =f^(g1K) then g1K =g2K.
    5. Assume f^(g1K) =f^(g2K) .
    6. Then f(g1) =f(g2).
    7. So f(g1) f (g2) -1 =1.
    8. So f(g1 g2-1 )=1.
    9. So g1 g2-1 kerf.
    10. So g1 g2-1 =k for some kkerf.
    11. So g1=g2k for some kkerf.
    12. To show: aba. g1K g2K.
    13. To show: abb. g2K g1K.
      1. Let gg1K.
      2. Then there exists k1K such that g=g1k1.
      3. So g =g2kk1 g2K, since kk1K .
        So g1K g2K.
      4. Let gg2K. Then g=g2k2 for some k2=k.
        So g =g1 k-1 k2g1K , since k-1k2 K.
        So g2K g1K.
      So g1K =g2K.
    14. So f^ is injective.
    15. To show: f^(g1K) f^(g2K) = f^( (g1K) (g2K)) .
    16. Since f is a homomorphism, f^(g1K) f^(g2K) = f(g1) f(g2) = f(g1g2) = f^ (g1g2K) = f^( (g1K) (g2K)).
    17. So f^ is a homomorphism.
  4. To show: ba. f is well defined.
  5. To show: bb. f is surjective.
  6. To show: bc. f is a homomorphism.
    1. This is proved in Ex 2.2.3, Part I ??????????????
    2. This is proved in Ex 2.2.3, Part I ??????????????
    3. Since f is a homomorphism, f(g) f(h) = f(g)f(h) = f(gh) = f(gh).
    4. So f' is a homomorphism.
  7. Let K=kerf.
  8. By (a), the function f^: G/K H gK f(g) is a well defined injective homomorphism.
    By (b), the function f^: G/K imf^ gK f^(gK) =f(g) is a well defined surjective homomorphism.
  9. To show: ca. imf^ =imf.
  10. To show: cb. f^' is injective.
    1. To show: caa. imf^ imf.
    2. To show: cab. imff^.
      1. Let himf^ .
      2. Then there exists gKG/K such that f^(gK) =h.
      3. Let ggK.
      4. Then there exists kK such that g=gk.
      5. Then, since f is a homomorphism and f(k)=1, f(g) = f(gk) = f(g)f(k) = f(g) = f^(gK) = h.
      6. So himf.
      7. So imf^imf .
      8. Let himf.
      9. Then there exists gG such that f(g)=h.
      10. So f^(gK) =f(g) =h .
      11. So himf^ .
      12. So imfimf^.
    3. To show: If f^'(g1K) =f^' (g2K) then g1K =g2K.
      1. Assume f^ (g1K) =f^ (g2K) .
      2. Then f^(g1K) = f^(g2K) .
      3. Then, since f^ is injective, g1K =g2K.
    4. So f^ is injective.
  11. Thus f^: G/K imf^ 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 (h1,k1) (h2,k2) = (h1h2, k1k2) , for h1, h2H, k1, k2K.
  • More generally, given groups G1, G2,, Gn , the direct product G1× G2×× Gn is the set G1× G2×× Gn with the operations given by (h1,, hi,, hn) + (k1,, ki,, kn) = (h1k1 ,, hiki,, hnkn) where hi,ki Gi, and hiki is given by the operation in the group Gi. The operation in the direct product is just the operations from the original groups acting componentwise.

HW: Show that these are good definitions, i.e. that, as defined above, H×K and G1× G2×× Gn are groups with identities given by (1H,1K) and (1G1, , 1Gn), respectively (1Gi denotes the identity in the group Gi).

Further Definitions

  • An abelian group is a group G such that if g1,g2 G then g1g2 =g2g1.
  • The center Z(G) of a group G is the set Z(G) ={cG | if gG then cg=gc }.

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 if and only if G is abelian.

  • The order |G| of a group G is the number of elements in G.
  • Let G be a group and let gG. The order o(g) of g is the smallest positive integer n such that gn =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 is the subgroup S of G such that
    (a)   S S,
    (b)   If H is a subgroup of G and S H then SH.
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 subset of G. Show that the subgroup generated by S exists and is unique.

Notes and References

These notes are written to highlight the analogy between groups and group actions, rings and modules, and fields and vector spaces.

References

[Ram] A. Ram, Notes in abstract algebra, University of Wisconsin, Madison 1993-1994.

[Bou] N. Bourbaki, Algèbre, Chapitre 9: Formes sesquilinéaires et formes quadratiques, Actualités Sci. Ind. no. 1272 Hermann, Paris, 1959, 211 pp. MR0107661.

[Ru] W. Rudin, Real and complex analysis, Third edition, McGraw-Hill, 1987. MR0924157.

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