advertisement

Solution Sketch for Homework Nine Part I (of II) (1) A permutation is a bijection of a finite sets, as such we need to find how many bijections there are from S to S. Solution: Let S = {s1 , s2 , . . . , sn }. If f : S → S is a bijection, there are n possible choices for f (s1 ), n−1 possible choices for f (s2 ), and so on. As such, there are n! possible bijections. (2) Describe a group structure on the set of moves of a Rubik’s cube. If you are unsure what these moves are, you might find the first few seconds of this video helpful. Your description should include your chosen identity element, chosen binary operation and a brief explanation for why the group axioms hold. Solution: Your operation here would be something that combines your moves together to form other moves. If your moves are permutations of the facets of the cube, then two permutations can be combined by performing them one after another. This is associative because it corresponds to function composition and we know that function composition is associative. The tricky thing here is that it’s not necessarily commutative, meaning that doing the moves in different orders might yield different results. But there’s nothing in the description of a group that requires g1 ∗ g2 to coincide with g2 ∗ g1 . You have already seen this kind of thing happen, for instance, in the context of matrix multiplication. (3) Let G be a finite group (i.e. a group with finitely many elements) with binary operation ? and identity eG . Prove that for every g ∈ G there is some n ∈ N such that (†) g ? g ? g ? . . . ? g = eG . n times It might be helpful to figure this out concretely for your favourite permutation group before tackling the case of an abstract group. What does (†) mean in the context of the Rubik’s cube? Solution: This boils down to observing that the action of G on itself by left multiplication (c.f. piazza discussion) is a bijection of the elements of G. As such, since G is finite, after finitely many iterations it will send eG back to eG . In the context of the Rubik’s cube, (†) means that if we start with a solved cube and perform a fixed sequence of moves, say LRU , then if we perform this sequence over and over again we will eventually return to the solved cube, i.e., (LRU )(LRU ) . . . (LRU ) = Identity. 1 2 (4) Let Γ and Ξ be graphs and suppose that there is a surjective graph homomorphism ϕ : Γ → Ξ. Prove that if Γ is connected then Ξ is also connected. Solution: The key is to observe that any two vertices v1 , v2 in Ξ have pre-images w1 , w2 in Γ (why?) which are connected by a path in Γ. The image of this path under ϕ must then be a path in Ξ joining v1 to v2 (why?). (5) Prove or disprove: there is a graph homomorphism from a square graph to a graph consisting of a single edge (the answer is not what you initially think it is). Solution: The “twist” here is simply to view the square as a bipartite graph to produce a homomorphism. (6) Prove that a graph whose vertices and edges can be drawn as such is not isomorphic to a graph whose vertices and edges can be drawn as such (here, as in the previous question, it might be helpful to label the vertices before trying to write anything down). Solution: Proceeding by contradiction, this boils down to the fact that isomorphisms preserve adjacencies between vertices as well as the degrees of vertices. As such, the contradiction arises by analyzing the neighbours of the neighbours of the unique vertex of degree three in each graph. (7) Find all the automorphisms of your favourite graph. In other words, describe its automorphism group. Solution: This could have been the trivial graph with a single vertex and no edges. Its automorphism group is the trivial group with a single element: the identity.