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.
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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?