1. Math 3121 Abstract Algebra I Lecture 7: Finish Section 7 Sections 8

2. Finish Section 7 Examples in class of Cayley Digraphs

3. Cayley Diagraph For each generating set of a finite group G, we can draw a graph whose vertices are elements of G and whose arcs represent right multiplication by a generator. Each arc is labeled according to the generator. Examples in class: Z 6

4. 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.

5. 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?

6. HW for Section 7 Don’t hand in: Pages 72-73: 1, 3, 5, 9 Hand in (Due Tues, Oct 28): page 73: 12, 14, 16

7. Section 8 Section 8: Groups of Permutations – Definition and Notation of Permutation – Theorem: Permutations on a set form a group with composition as binary operation. – Definition: Symmetric Group on n letters – Definition: Dihedral group – Cayley’s Theorem

8. Permutations Definition: A permutation of a set A is a function from A to A that is one-to-one and onto. Examples: 1)Let A = {a, b, c}, and let f: A  A such that f(a) = b f(b) = c f(c) = a 2)Let A = the set of real numbers, and let f: A  A such that f(x) = 2 x. Does this work if A is the set of integers?

9. Permutation Groups Theorem: Let A be a nonempty set, and let Perm[A] be the set of permutations of A. Then Perm[A] is a group with composition as the binary operation. Proof: Composition is a well defined binary operation on Perm[A]. It satisfies: 1) It is associative because composition is associative. 2) The identity map from A to itself acts as an identity for composition. Hence Perm[A] has an identity. 3) Every permutation is one-to-one onto, and thus has an inverse function. The inverse is also a 1-1 function from A onto itself. Hence Perm[A] is closed under inverses.

10. Permutation Notation Write for f: {x 1, x 2, …, x n }  {x 1, x 2, …, x n } Note: Most of the time, the set will be {1, 2, …,n}

11. Composition in Permutation Notation Composition in permutation notation is represented by multiplicatively. Note the order is right to left in this textbook (and hence in this class).

12. Composition in Permutation Notation Procedure: Fill in each column at a time. For each column of the result, the top row determines a column of the right system. In that column, the entry in the second row determines a column of the left system. In that column, the entry in the second column determines the entry of the selected column of the result.

13. Example Try

14. The Symmetric Group on n Letters S n denotes the permutation group on the set {1, 2, …, n} and is called the symmetric group on n letters. Note: S n has n! elements. Why? Look at S 1, S 2, S 3

15. S 3 = Symmetric Group on 3 Letters ρ0ρ0 ρ1ρ1 ρ2ρ2 μ1μ1 μ2μ2 μ3μ3 ρ0ρ0 ρ0ρ0 ρ1ρ1 ρ2ρ2 μ1μ1 μ2μ2 μ3μ3 ρ1ρ1 ρ1ρ1 ρ2ρ2 ρ0ρ0 μ3μ3 μ1μ1 μ2μ2 ρ2ρ2 ρ2ρ2 ρ0ρ0 ρ1ρ1 μ2μ2 μ3μ3 μ1μ1 μ1μ1 μ1μ1 μ2μ2 μ3μ3 ρ0ρ0 ρ1ρ1 ρ2ρ2 μ2μ2 μ2μ2 μ3μ3 μ1μ1 ρ2ρ2 ρ0ρ0 ρ1ρ1 μ3μ3 μ3μ3 μ1μ1 μ2μ2 ρ1ρ1 ρ2ρ2 ρ0ρ0

16. Symmetries of the Equilateral Triangle ρ0ρ0 1 3 2 μ3μ3 1 3 2 μ2μ2 1 3 2 μ1μ1 1 3 2 ρ1ρ1 1 3 2 ρ2ρ2 1 3 2

17. S 3 =D 3 = 3 rd Dihedral Group ρ0ρ0 ρ1ρ1 ρ2ρ2 μ1μ1 μ2μ2 μ3μ3 ρ0ρ0 ρ0ρ0 ρ1ρ1 ρ2ρ2 μ1μ1 μ2μ2 μ3μ3 ρ1ρ1 ρ1ρ1 ρ2ρ2 ρ0ρ0 μ3μ3 μ1μ1 μ2μ2 ρ2ρ2 ρ2ρ2 ρ0ρ0 ρ1ρ1 μ2μ2 μ3μ3 μ1μ1 μ1μ1 μ1μ1 μ2μ2 μ3μ3 ρ0ρ0 ρ1ρ1 ρ2ρ2 μ2μ2 μ2μ2 μ3μ3 μ1μ1 ρ2ρ2 ρ0ρ0 ρ1ρ1 μ3μ3 μ3μ3 μ1μ1 μ2μ2 ρ1ρ1 ρ2ρ2 ρ0ρ0 1 3 2

18. S 4 =D 4 = Dihedral Group ρ0ρ0 ρ1ρ1 ρ2ρ2 ρ3ρ3 μ1μ1 μ2μ2 δ1δ1 δ2δ2 ρ0ρ0 ρ0ρ0 ρ1ρ1 ρ2ρ2 ρ3ρ3 μ1μ1 μ2μ2 δ1δ1 δ2δ2 ρ1ρ1 ρ1ρ1 ρ2ρ2 ρ3ρ3 ρ0ρ0 δ1δ1 δ2δ2 μ2μ2 μ1μ1 ρ2ρ2 ρ2ρ2 ρ3ρ3 ρ0ρ0 ρ1ρ1 μ2μ2 μ1μ1 δ2δ2 δ1δ1 ρ3ρ3 ρ3ρ3 ρ0ρ0 ρ1ρ1 ρ2ρ2 δ2δ2 δ1δ1 μ1μ1 μ2μ2 μ1μ1 μ1μ1 δ2δ2 μ2μ2 δ1δ1 ρ0ρ0 ρ2ρ2 ρ3ρ3 ρ1ρ1 μ2μ2 μ2μ2 δ1δ1 μ1μ1 δ2δ2 ρ2ρ2 ρ0ρ0 ρ1ρ1 ρ3ρ3 δ1δ1 δ1δ1 μ1μ1 δ2δ2 μ2μ2 ρ1ρ1 ρ3ρ3 ρ0ρ0 ρ2ρ2 δ2δ2 δ2δ2 μ2μ2 δ1δ1 μ1μ1 ρ3ρ3 ρ1ρ1 ρ2ρ2 ρ0ρ0 1 3 2 4

19. Symmetries of the Square 3 ρ0ρ0 4 21 3 ρ1ρ1 4 21 3 ρ2ρ2 4 21 3 ρ3ρ3 4 21 3 μ1μ1 4 21 3 μ2μ2 4 21 3 δ1δ1 4 21 3 δ2δ2 4 21

20. Cayley’s Theorem Theorem (Cayley’s Theorem): Every group is isomorphic to a group of permutations. Proof: Let G be a group. We show that G is isomorphic to a subgroup of the permutations on the set G. For each a in G, let ρ a be the map from G to itself given by left multiplication by a. That is, ρ a (g) = a g. Then ρ a is one- to-one and onto. In fact, it has an inverse. So ρ a is a permutation of the set G. The map ρ that takes a to ρ a is a homomorphism, because ρ a b (g) = a b g = a ρ b (g) = ρ a (ρ b (g)) = ( ρ a ρ b )(g) ρ is one-to-one and it is onto its image f[G]. It is straightforward to show that f[G] is a subgroup of Perm[G]. Thus f induces an isomorphism between G and f[G].

21. Right and Left Regular Representations ρ x (g) = g x as in the theorem defined the left regular representation f of G. Multiplication on the right gives the property that is not homomorphism: μ x y = μ y μ x. This is sometimes called the antihomomorphism property. The inverse map reverses the order. So μ x (g) = g x -1 defines a map that has the correct order to be a homomorphism. This is called the right regular representation f of G.

22. Examples Find the left and right representations of Z 2, Z 3, Z 4, S 3

23. HW for Section 8 Don’t hand in: pages 83-84: 1, 3, 5, 7, 9, 11, 13 Hand in (Due, Tues, Oct 28): Page 84: 18