Download Math 3121 Lecture 6 ppt97

Math 3121
Abstract Algebra I
Lecture 6
Midterm back over+Section 7
Midterm 1 back and over
• In class
Section 7
• Topics
– Generating Sets
– Cayley Diagraphs
Generating Sets
• Given one element a of a group G, we saw the cyclic
group {an | n is an integer} is the smallest subgroup
of G containing a. This is denoted by <a>.
• What happens if we are given two elements a and b?
In general, we get combinations:
an1 bn2 an3 bn4 …
Where n1, n2, … are integers.
• What happens with more elements?
• Next: a general framework in terms of Set Theory.
Intersections of Sets
Definition: Let T be a set of sets. The intersection of T
is the set whose elements belong to all members of
T. Notation:
T = {x | x in S for all S in T}
Note 1: The axiom of specification guarantees the
existence of T.
Note 2: T is contained in all S in T.
Note 3: Other notation for T includes
S1 S2 … Sn
when T = {S1, S2, …, Sn}
Intersection of Subgroup
Theorem: Let G be a group. The intersection of a set of
subgroups of G is a subgroup of G.
Proof: Let T be a set of subgroups of G, and let H be the
intersection of all members of T. Then h is in H if and only if h
is in K for all K in T. Now we show that H satisfies the
conditions in the theorem for subgroups. 1) We claim that H
is closed under the binary operation of G. Let’s use
multiplicative notation. Let h1 and h2 be in H, then for each K
in T, h1 and h2 are in K, hence the product h1 h2 is in K. Thus
the product h1 h2 is in all K in T, and thus is in H. 2) Similarly,
the identity e of the group is in all K in T, thus is in the
intersection. 3) Finally, for any h in H, h is in all members of T,
thus h-1 is in each member of T because each is a subgroup of
G. So h-1 is in H.
Groups Generated by Subsets
Theorem: Let G be a group, and let S be a subset of G. Then
there is a subgroup H of G such that 1) H contains S and 2) any
subgroup K of G that contain S also contain H.
Proof: Let H be the intersection of all subgroups of G that
contain S. Then H is a subgroup of G and H is contained in all
subgroups of G that contain S.
Definition: For any group G and subset S of G, the group
generated by S is the subgroup of G that satisfies conditions 1)
and 2) in the above theorem. Denote this subgroup by <S>.
Note: Conditions 1) and 2) say that <S> it is the smallest
subgroup of G that contains S.
Explicit Form
Theorem: Let G be a group, and let T be a subset
of G. Then the elements of <T> are finite
products of powers of elements of T.
Proof: The set of all finite products of powers of
elements of T is closed under products and
inverses and contains the identity. Thus it is a
subgroup of G. It contains T. In fact, it must
contain any subgroup containing T.
Cayley Diagraph
• For each generating set of a finite group G, we
can draw a graph whose vertices are elements
of and whose arcs represent right
multiplication by a generator. Each arc is
labeled according to the generator.
• Examples in class: Z6
Properties of Cayley Diagraphs
1.
2.
3.
4.
Can get to any vertex from any other by a path. Reason: Every equation
g x = h has a solution in G and each member of G can be written as a
product of generators and their inverses.
At most one arc goes from any vertex g to a vertex h. Reason: The
solution of g x = h is unique.
Each vertex g has exactly one arc of each type starting at g and exactly
one arc of each type ending at g. Reason: It is constructed this way.
If two different sequences of arc types go from vertex g to vertex h, then
these two sequences applied to any other vertex will go to the same
vertex.
Note: These four properties characterize Cayley diagraphs.
Examples
• Examples:
Write out table of group described by Cayley
diagraph on page 72. Note the inner and outer
squares have different directions.
Try this with triangles - note the directions of inner
and outer triangles have same direction on page
71. Now do them with opposite directions.
What about pentagons?
HW
• Don’t hand in:
Pages 72-73: 1, 3, 5, 9