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.