Mon 3/6/00

 Old business:

   (1) determinants, column swaps
   (2) boosts; Lorentz group (6 parameter)
   (3) Show D_2n = D_n X Z_2 for n odd.
         - example:  n=3 =>  D_6 = C_6 X Z_2 = (C_3 X Z_2) X Z_2 = D_3 X Z_2.

           Note that factoring C_6 depends on fact that 3,2 are relatively prime.
           Otherwise, of elements (i,j) i=0,1,2; j=0,1 something like (1,1) 
           doesn't generate the whole group.
           
           even counterexample: n=2 => D_4 = C_4 X Z_2 but C_4 != C_2 X C_2 because
           (1,1) doesn't generate whole of C_4 as (i,j) with i=0,1 and j=0,1.

 New business:

   *   read Joyner, chap 5, 6, 9

   *   define "homomorphism" between groups
	  map G -> H which preserves multiplication

   *   define "kernel" of the homomorphism
	  {g} in G s.t. {g}->identity

   *   define a "normal subgroup" J
          - contains entire conjugacy classes, OR
          - g J G' = J for all g in G,  OR
          - right and left cosets same, xJ=Jx

   *   define G/J = set of left cosets of subgroup J of a group G.

   THEN, given a homomorphism G->H,
	(1) the kernel is a normal subgroup, and 
        (2) G/kernel is isomorphic to the group H. called the "quotient group"
              This is a "quotient group", or G mod kernel
   ALSO,
         if J is normal -> G/J is a group with group operation (aJ)(bJ)=(abJ)

   examples and counterexamples. 

   * normal subgroups of D4 and D5.
   * what happens if H is not normal?

  THEOREM:
     (1) The commutator subgroup [G,G] is normal.
     (2) G/[G,G] is abelian.
     (3) If H is a normal subgroup, and G/H is abeliean, then [G,G] is contained in H.

   * commutator subgroup of A4. (Armstrong 15.8)


  definition SIMPLE GROUP:
     Only normal subgroups are identity and entire group.

=============================

In class:

 * C6 has normal subgroups C2 and C3 (all subgroups of an abelian group are normal),
   so we have both C6/C2 = C3 and C6/C3=C2, as well as C2 x C3 = C6 (direct product).

 * S3 = {I,r,l,X,Y,Z} has normal subgroup {I,r,l}, so S3/C3=C2.  
   But there is no normal C2 subgroup!  Note that {I,X} is a subgroup,
   but since {x,y,Z} is one of conjugacy classes, {I,X} is not normal.
   Questions -> Why can't we divide S3/C2?  [i.e. what makes C2 here not normal?]
                why isn't there a 6-element non-abelian group G s.t. G/C2 = C3?
                Can we multiply C2 and C2 to get S3? [Yes, with "semi-direct" product.]
 
                What does all this have to do with classification of 
                 of groups given C2, C3 as simple groups and their products?