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 {a n | n is an integer} is the smallest subgroup of G containing a. This is denoted by. What happens if we are given two elements a and b? In general, we get combinations: a n1 b n2 a n3 b n4 … 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 S 1  S 2  …  S n when T = {S 1, S 2, …, S n }

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. Note: Conditions 1) and 2) say that 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 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: Z 6

Properties of Cayley Diagraphs 1.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. 2.At most one arc goes from any vertex g to a vertex h. Reason: The solution of g x = h is unique. 3.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. 4.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