Title: Cayley Digraphs of Groups
1Cayley Digraphs of Groups
2A directed graph (or digraph) is a finite set of
points, called vertices, and a set of arrows,
called arcs, connecting some of the
vertices. Definition Cayley digraph Let G be a
finite group and let S be a set of generators for
G. We define a digraph Cay(SG), called the
Cayley digraph of G with generating set S, as
follows 1. Each element of G is a vertex of
Cay(SG) 2. For some x and y in G, there is an
arc from x to y iff xsy for some sÎS, the
generating set.
3Examples!
Cay(1Z5)
Cay(1Z2)
0 4 1 3 2
0 1
4What about groups with generating set, S1?
Cay((12), (123) S3)
(123) (12)
5And different generating sets?
(12) (13)
Cay((12), (13) S3)
6Around the World
In 1859, Irish mathematician, Sir William
Hamilton (father of the Hamiltonian numbers)
invented this puzzle Take any dodecahedron you
happen to have lying around and label each of the
twenty vertices with a different city name. Start
at any city and follow the edges (arcs) in a path
such that each city is visited exactly once
before returning to the original starting city.
7Hamiltonian paths and circuits
A solution to the Around the World problem is a
Hamiltonian circuit. In general, a Hamiltonian
circuit is a sequence of arcs in a digraph
passing through each vertex exactly once before
returning to the starting vertex. A Hamiltonian
path is a sequence of arcs in a digraph passing
through each vertex exactly once without
returning to the starting vertex.
8What groups have Hamiltonian circuits or paths?
Our Around the World dodecahedron group has a
Hamiltonian circuit, but what other groups
do? And what about the generator? Can different
digraphs of the same groups be all that different?
9Some sufficient conditions
All cyclic groups have at least one Cayley
digraph with a Hamiltonian circuit.
Theorem Cay((1,0),(0,1) Zm x Zn ) has a
Hamiltonian circuit when n divides m.
10Cumbersome?
Agreed. There is a shorthand notation to
describe circuits and paths. We can list the
sequence of arcs with respect to a starting
position (s1, s2, s3, s3, s4, s5,, sn) where
si is a generator. We can show the repetition of
a sequence k (s1, s2, s3, s3,, sn)
concatenates the sequence k-times.
11Back to our theorem
The shorthand proof for our theorem is
simple. Begin at (0,0). Apply m((n-1)(0,1),
(1,0)). Q.E.D.
12When can we be sure we dont have a Hamiltonian
circuit?
Theorem Cay((1,0), (0,1) Zm x Zn) does not
have a Hamiltonian circuit when m and n are
relatively prime.
13but wait
What about Cay((1,1) Zm x Zn) where m and n
are relatively prime?
14How about Hamiltonian paths?
Theorem Let G be a finite Abelian group, and
let S be any nonempty generating set for G. Then
Cay(SG) has a Hamiltonian path.
15Homework
- Find a solution for Hamiltons Around the World
problem. - Show that if (x1, x2, x3, , xk) is a sequence of
generators constituting a Hamiltonian circuit for
some vertex, then the same sequence determines a
Hamiltonian circuit for any starting vertex.