Introduction to vertex-transitive graphs of prime-power order
Dave Witte Morris, University of Lethbridge, March 2013
Abstract. A graph is vertex-transitive if its automorphism group
acts transitively on the set of vertices. (In other words, every vertex
looks exactly like all of the other vertices.) Such graphs can be very
complicated in general, but we will use some group theory to see that
they are easy to describe when the number of vertices is assumed
to be prime. There are similar results when the number of vertices
is the square or cube of a prime, but larger powers are harder to
understand.
X = graph,
V = {vertices of X},
order of X is |V | (finite).
Definition. X is vertex-transitive if Aut X is transitive on V :
∀v, w ∈ V, ∃g ∈ Aut X, gv = w.
0
Example. Circulant graphs are vertex-transitive.
V = cyclic group = Zn = Z7
5
S = {1, 3} ⊆ V .
edge v v ± s for v ∈ V and s ∈ S.
6
• G is transitive on V , and
• |G| = |V |. (I.e., no (nontrivial) g ∈ G fixes any point of V .)
2
4
3
(w + 1).
Exercise. X is (∼
=) circulant ⇐⇒ Aut X contains an n-cycle
(where n = |V |).
Proposition (Turner, 1967). Every vertex-transitive graph of
prime order p is circulant.
Proof. Let G = Aut X, so G is transitive on V , and let v ∈ V .
Orbit-Stabilizer Theorem: |G| = |V | · |G : StabG (v)| = p · (· · · ).
Cauchy’s Theorem (or Sylow): G has an element of order p.
A permutation of order p on a set of p elements
must be a p-cycle (since p is prime).
02
Exercise. X is (∼
=) Cayley ⇐⇒ Aut X has a subgroup G that is
sharply transitive (or “regular”):
1
v w ⇐⇒ (w + 1) − (v + 1) = w − v ∈ ±S ⇐⇒ (v + 1)
So the n-cycle x 7→ x + 1 is in Aut X.
12
22
01 11
V = any group = Z3 × Z3
and S = {(1, 0), (0, 1)} ⊆ V .
00 10
edge v v ± s for v ∈ V and s ∈ S.
This is a Cayley graph on Z3 × Z3 , but it is not circulant.
21
Example. Cayley graphs are vertex-transitive.
Remark. Joy Morris (1999) determined when a circulant graph with
pn vertices is also a Cayley graph on an abelian group that is not
cyclic. The case n = 2 had been done previously by Anne Joseph
(and Andrew Mauer had done part of n = 3, but this was never
published). A shorter proof later appeared in [1]. Mikhail Klin and
Mikhail Muzychuk used linear algebra and elementary Galois theory
of cyclic extensions to give a beautiful proof of part of this theorem.
(It yields the entire theorem when n = 2). This has unfortunately
not been published, but we present the argument in an appendix to
these notes.
20
Proof. (⇒) For g ∈ V :
v w ⇐⇒ (w+g)−(v+g) = w−v ∈ ±S ⇐⇒ (v+g) (w+g).
So the permutation x 7→ x + g is in Aut X.
Let G be the subgroup consisting of these permutations (∼
= V ).
(Cayley: every group is isomorphic to a group of permutations.) Proposition (Marušič?, ∼1980). Every vertex-transitive graph of
order p2 is a Cayley graph.
This will be proved later in the lecture.
Theorem (Marušič, 1985). Every vertex-transitive graph of order p3 is a Cayley graph.
(Independently by Daniel Goldstein,
unpublished.)
Open problem. Find an easy (or short) proof.
Corollary (Marušič (& Goldstein)). Every vertex-transitive graph
of order p3 has a hamiltonian cycle.
Theorem (Marušič, 1985???). Vertex-transitive graphs of order p4
(or larger powers) need not be Cayley graphs.
Exercise. The Petersen graph is vertex-transitive, but not Cayley.
a
0
a
4
0
e
b
d
3
1
d
3
c
1
2
c
2
4
b
e
(Smallest transitive group of automorphisms has order 20, not 10.)
Theorem (Chen, 1998). Every vertex-transitive graph of order p4
has a hamiltonian cycle.
Open problem. Every vertex-transitive graph of order p5 has a
hamiltonian cycle.
Conjecture. Every vertex-transitive graph of order pn has a hamiltonian cycle.
Evidence for the conjecture:
Theorem (Witte, 1985). Every Cayley (di)graph of order pn has a
hamiltonian cycle.
Proof that vertex-transitive graphs of order p2 are Cayley.
Let
• G be a minimal transitive subgroup of Aut X,
• P be a Sylow p-subgroup of G, and
• v ∈V.
The Orbit-Stabilizer Theorem tells us
|G|
|G|p
|P |
p2 = |V | =
=
=
,
| StabG (v)|
| StabG (v)|p
| StabP (v)|
so P is transitive on V .
Minimality: G = P is a p-group (order pk ).
Let M be a maximal subgroup of G.
p-groups are nilpotent (because all p-group have nontrivial centre),
so M / G.
Therefore M · StabG (v) is a subgroup of G.
Suppose StabG (v) 6⊆ M . Then M · StabG (v) ) M .
Maximality: M · StabG (v) = G, so M · StabG (v) is transitive
— it can move v to any other vertex.
But StabG (v) does not help with this (it fixes v),
so M must be transitive.
Since M ( G, this contradicts the minimality of G.
Therefore StabG (v)
T is contained in every maximal subgroup M .
I.e., StabG (v) ⊆
{maximal subgroups of G} = Φ(G).
(“Frattini subgroup” of G)
Exercise. G has more than one maximal subgroup (else G is cyclic).
|G|
So |Φ(G)| ≤ 2 = | StabG (v)| ≤ |Φ(G)|.
p
Therefore StabG (v) = Φ(G) / G.
For all w ∈ V , there is some g ∈ G, such that w = vg, so
w ·StabG (v) = (vg)· g −1 StabG (v) g = v ·StabG (v) ·g = v ·g = w.
This means that StabG (v) is trivial.
Therefore the Orbit-Stabilizer Theorem tells us |G| = p2 .
Remark. The proof actually shows that every transitive permutation group on p2 points has a transitive subgroup of order p2 .
Unfortunately, this is false if you replace p2 with p3 , so a very different argument is required to show that vertex-transitive graphs of
order p3 are Cayley.
References
[1] Brian Alspach and Shaofei Du: Suborbit structure of permutation
p-groups and an application to Cayley digraph isomorphism, Canad.
Math. Bull. 47 (2004), no. 2, 161–167. MR 2059412
[2] Yu Qing Chen: On Hamiltonicity of vertex-transitive graphs and
digraphs of order p4 , J. Combin. Theory Ser. B 72 (1998), no. 1,
110–121. MR 1604701
[3] Dragan Marušič: Vertex transitive graphs and digraphs of order pk ,
in: Cycles in Graphs (Burnaby, B.C., 1982). North-Holland, Amsterdam, 1985, pp. 115–128. MR 0821510
[4] Joy Morris: Isomorphic Cayley graphs on nonisomorphic groups, J.
Graph Theory 31 (1999), no. 4, 345–362. MR1698752
[5] James Turner: Point-symmetric graphs with a prime number of
points. J. Combinatorial Theory 3 (1967) 136–145. MR 0211908
[6] Dave Witte: Cayley digraphs of prime-power order are Hamiltonian,
J. Combin. Theory Ser. B 40 (1986), no. 1, 107–112. MR 0830597
Appendix A. The Klin-Muzychuk Proof
Appendix B. The Converse of Klin-Muzychuk
Theorem (Klin-Muzychuk, 1995, unpublished). If Cay(Zpn , S) is
isomorphic to a Cayley graph on (Zp )n (and S = −S), then
S = (1 + p)S.
Definition. Suppose X and Y are graphs. Their wreath product
(also called the lexicographic product) is the graph X o Y whose vertex set is V (X) × V (Y ), such that
(
x1 x2 in X, or
(x1 , y1 ) (x2 , y2 ) ⇐⇒
x1 = x2 , and y1 y2 in Y .
Remark. The converse is “easy” (see Appendix B).
For convenience, let X = Cay(Zpn , S). The adjacency matrix
of X is the |V | × |V | matrix AX defined by enumerating the vertices
v1 , . . . , vn of X, and letting
(
1 if vi vj ,
(AX )i,j =
0 if vi 6 vj .
Lemma. Every eigenvalue of AX is in the cyclotomic extension
Q[ζp ], where ζp is any primitive pth root of unity (because X is
isomorphic to a Cayley graph on (Zp )n ).
Proof. Any element g of (Zp )n acts on V by a permutation of order p
(or 1), so the corresponding permutation matrix Tg satisfies the polynomial xp − 1 = 0. Therefore, Tg is diagonalizable over Q[ζp ]. Since
(Zp )n is abelian, we know that all Tg ’s commute with each other, so
they are simultaneously diagonalizable over this field. Hence
X
AX =
Tg is diagonalizable over Q[ζp ].
g∈S
This implies that every eigenvalue of AX is in Q[ζp ].
Proof of the theorem. Let ζ be a primitive pn th root of unity.
For any k ≡ 1 (mod p), we have gcd(k, pn ) = 1 and (ζp )k = ζp , so
there exists σk ∈ Gal Q[ζ]/Q[ζp ] , such that σk (ζ) = ζ k . Also, since
n
X is circulant, we know, for every i ∈ Z, P
that (ζ i , ζ 2i , . . . , ζ p i ) is an
eigenvector of AX , with eigenvalue λ = g∈S ζ gi . The lemma tells
us σk (λ) = λ, so
!
X
X
X
X
X
ζ gi = λ = σk (λ) = σk
ζ gi =
σk (ζ gi ) =
ζ kgi =
ζ gi .
g∈S
g∈S
g∈S
pn
g∈S
g∈kS
Since the Vandermonde matrix (ζ gi )g,i=1 is invertible (and i is arbitrary), this implies S = kS, as desired.
Remark. To construct the wreath product, start with the disjoint
union Y1 t Y2 t · · · t Ym of m = |X| copies of the graph Y . Then join
every vertex in Yi to every vertex in Yj iff there is an edge from xi
to xj in X. (In other words, replace each vertex in X with a copy
of Y , and replace an edge in X that joins two vertices with all of the
possible edges joining the two corresponding copies of Y .)
Example. The wreath product of two cycles of length p is isomorphic to both:
(1) the Cayley graph on Zp × Zp with
S = {(1, y) | y ∈ Zp } ∪ {(0, ±1)}, and
(2) the circulant graph on Zp2 with
S = {s ∈ Zp2 | s ≡ 1 (mod p)}.
Exercises.
(KM1) If S = (1 + p)S, then Cay(Zpn , S) is isomorphic to
Xn o Xn−1 o · · · o X1 , where each Xi is a circulant of order p.
(KM2) Suppose N is a normal subgroup of a group G. If
• X is a Cayley graph on G/N , and
• Y is a Cayley graph on N ,
then X o Y is isomorphic to a Cayley graph on G.
Proof of the converse. Exercise KM1 tells us Cay(Zpn , S) is isomorphic to Xn oXn−1 o· · ·oX1 , where each Xi is a circulant of order p.
Also, it is easy to construct an ascending chain
{0} = N0 / N1 / · · · / Nn = (Zp )n
of (normal) subgroups of (Zp )n , such that Ni /Ni−1 ∼
= Zp for each i.
Then Xi is isomorphic to a Cayley graph on Ni /Ni−1 , so Exercise KM2 (and induction) implies that Cay(Zpn , S) is isomorphic
to a Cayley graph on Nn = (Zp )n .
Exercise. Show that if S = (1+p)S, then Cay(Zpn , S) is isomorphic
to a Cayley graph on every group of order pn .