Download Notes on Cayley Graphs for Math 5123

Notes on Cayley Graphs for Math 5123
Matt Day
Fall 2012
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Cayley graphs
The purpose of these notes is to define and explain Cayley graphs, which are combinatorial objects
closely related to groups. All the statements in these notes are well known. Since they do not
appear in Dummitt & Foote, I am providing these supplementary notes.
1.1
Definitions and examples
By an alphabet, we simply mean a set whose elements we think of as being symbols (or letters).
From here on, when X is mentioned without introduction, it is assumed to be an alphabet for an
X-digraph.
Definition 1.1. Let X be an alphabet. An X-digraph (short for X-labeled directed graph) is a
directed graph where each edge is labeled by an element from X. More carefully, an X-digraph Γ is
the following structure: Γ has a vertex set V (Γ) and an edge set E(Γ), and for each edge e ∈ E(Γ),
the edge has an origin o(e) ∈ V (Γ) and a terminus t(e) ∈ V (Γ) and a label µ(e) ∈ X. Parallel
edges, anti-parallel edges and edges from a vertex to itself (called loops or slings) are all allowed.
There is a natural way to turn a group together with a generating set into an X-digraph.
Definition 1.2. Suppose G is a group and X is a generating set for G. The Cayley graph Γ(G, X)
of G with respect to X is the following X-digraph: the vertex set of Γ(G, X) is the set of elements
in G, and for each s ∈ X and g ∈ G, there is a single directed edge from g to gs labeled by s.
Here we use multiplication on the right to determine edges in the Cayley graph. It is possible to
define essentially the same object using multiplication on the left. However, one must be consistent
with the choice of convention.
The Cayley graph is a combinatorial picture of a group. The following examples illustrate that
the Cayley graph depends very much on the given generating set as well as on the group. We can
draw the full Cayley graph when G and X are finite, but the definition makes sense even when
they are not.
Examples 1.3. We can easily construct the Cayley graphs for the following G and X:
• G = Zn , X = {x} where G = ⟨x⟩.
• G = D2n , X = {s, r}, where s and r are the usual generators with |s| = 2 and |r| = n.
• G = D2n , X = {s, sr}, with s and r as above.
1
x3
x4
x2
x5
x
1
Figure 1: Cayley graph of Z6 = ⟨x⟩ with respect to X = {x}.
• G = Znm , X = {xm , xn }, where n and m are relatively prime positive integers and G = ⟨x⟩.
• G = Z and X = {1}; this example has an infinite underlying graph but we know how to draw
as much of it as we like.
• G = Z × Z, X = {(0, 1), (1, 0)}.
• G = Q, X = {1, 1/2, 1/3, 1/4, 1/5, . . . }; this example has infinitely many edges at each vertex
but it still exists in a mathematical sense.
We illustrate a few of these examples in the figures.
1.2
Basic properties
Following paths in a Cayley graph corresponds to taking products of generators in a group. To
express this more carefully, we make the following definitions.
Definition 1.4. A finite edge path in an X-digraph Γ is a finite sequence of edges in Γ, with each
edge followed forwards or backwards, so that the far endpoint of each edge in the sequence matches
the near endpoint of the next edge in the sequence. Γ is connected if for any two vertices, there is
an edge path starting at one vertex and ending at the other.
The edge label sequence of a finite edge path is a sequence of elements from X ∪ X −1 (where
X −1 is the set of formal inverses of elements of X), where the ith element in the label sequence is
the label on the ith edge in the edge path, with that label inverted if the edge is read backwards
in the sequence.
Lemma 1.5. There is an edge path in Γ from the vertex 1 to the vertex g with label sequence xϵ11 ,
xϵ22 , . . . , xϵkk (with xi ∈ X and ϵi ∈ {+1, −1}) if and only if the following equation is true in G:
g = xϵ11 xϵ22 · · · xϵkk .
2
x4
x2
x
x5
x3
x2
x3
1
Figure 2: Cayley graph of Z6 = ⟨x⟩ with respect to X = {x2 , x3 }.
r2
r
sr
sr2
s
r
1
s
Figure 3: Cayley graph of D6 with respect to X = {r, s}.
3
sr2
r2
r
sr
s
sr
s
1
Figure 4: Cayley graph of D6 with respect to X = {s, sr}.
Proof. This is an exercise using induction and the definitions.
At this point we can get a clearer picture of what Cayley graphs look like in general:
Proposition 1.6. Suppose G is a group with generating set X and Γ is the Cayley graph of G with
respect to X. Then each vertex of Γ has exactly one edge with each label from X leaving it and
exactly one edge with each label from X arriving at it. Also, Γ is connected.
Proof. For each element g ∈ G and each s ∈ X, there is exactly one element h1 ∈ G with h1 = gs
and exactly one element h2 ∈ G with g = h2 s (this is because cancellation is possible in groups).
This means that there is exactly one edge labeled with s leaving g (going from g to h1 ) and exactly
one edge labeled with s arriving at g (going from h2 to g).
Now we show that Γ is connected. Since X generates G, for each g ∈ G, there is an expression
for g as a product of elements of X and their inverses. Then by Lemma 1.5, there is a path from
the vertex 1 to the vertex g. So to get a path from a vertex g to a vertex h, we concatenate a path
from g to 1 with a path from 1 to h.
In fact, a Cayley graph is a complete picture of a group, allowing us to reconstruct the original
group.
Proposition 1.7. Suppose we are given an X-digraph Γ that is the Cayley graph of a group G with
respect to a generating set X. Then we can reconstruct the group product on G.
Proof. The underlying set of G must be the vertex set of Γ, so we know what G is as a set. Now
suppose that g and h are in G. Since G is connected, there is a path from 1 to g and a path from
1 to h. These paths give us expressions for g and h as products of generators and their inverses
as in Lemma 1.5. We multiply these expressions together to get an expression for gh. Then by
Lemma 1.5, this expression gives us a path from 1 to the vertex gh. This allows us to identify
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the vertex for a product gh from the vertices g and h, and therefore to reconstruct the product
operation in G.
1.3
Symmetries of Cayley graphs
Definition 1.8. A symmetry (or an automorphism) of an X-digraph Γ is a pair of bijective maps,
one from the edge set E(Γ) to itself, and one from the vertex set V (Γ) to itself, such that
• each edge is mapped to an edge with the same label,
• the initial vertex of an edge maps to the initial vertex of the corresponding edge, and
• the terminal vertex of an edge maps to the terminal vertex of the corresponding edge.
Proposition 1.9. Suppose G is a group and Γ is the Cayley graph of G with respect to a generating
set X. Then G acts by symmetries on Γ. The action on the vertices of Γ is faithful and transitive.
Here “by symmetries” means that for any g in G, the maps on edges and vertices by e 7→ g · e and
v 7→ g · v constitute a symmetry of Γ.
Proof. We define the action as follows. For g ∈ G and h a vertex of Γ, g · h is the vertex gh of
Γ. For e an edge of Γ labeled by s ∈ X starting at h ∈ G, we send e to the edge of Γ labeled by
s starting at gh. Since there is a unique such edge, this map is well defined. It is easy to check
that this satisfies the axioms for an action (really there are two actions, one on vertices and one on
edges, and both satisfy the axioms).
We want to verify that G acts on Γ by symmetries. First we note that the action on vertices is
bijective: the action function for g −1 ∈ G is the inverse of the action function for g ∈ G. The action
on edges is bijective for the same reason. It is also obvious that the action on edges preserves labels.
Further, it should be clear that the action sends the initial vertex of an edge to the initial vertex of
the corresponding edge. The only nontrivial thing to show is the statement for terminal vertices.
Suppose h is a vertex, s ∈ X and e is the edge labeled by s starting at h. Then by definition, the
terminal vertex of e is hs. Action by g ∈ G sends e to the edge g · e starting at gh and labeled by s.
By definition, the terminal vertex of g · e is ghs. However, ghs is also the vertex g · hs, the image
of the terminal vertex of e under the action of g.
Next we show that the action on vertices is transitive. If h1 and h2 are two different vertices,
we take g = h2 h−1
1 ∈ G. Then g · h1 = h2 .
Finally, we show that the action on vertices is faithful. Suppose g ∈ G and g acts on the vertices
of Γ trivially, that is, sending each vertex to itself. Then g sends the vertex for 1 to the vertex for
g: g · 1 = g. However, since the action is trivial, the vertex for g and the vertex for 1 are the same
vertex, and therefore g and 1 are the same element of G. This means that the kernel of the action
contains only the trivial element.
Lemma 1.10. Suppose Γ is the Cayley graph of a group G with respect to a generating set X.
Then if α and β are symmetries of Γ sending a vertex v1 to a vertex v2 , then α = β.
Proof. Suppose α, β, v1 and v2 are as above. To show α = β, we show that for each vertex v of Γ,
α(v) = β(v). It then will follow that α and β agree on edges by the uniqueness of labeled edges at
a vertex.
So suppose v is a vertex of Γ. Then by connectedness, there is an edge path from v1 to v. It
is an exercise using induction on the length of the edge path and uniqueness of edges at a given
vertex to then show that α(v) = β(v).
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Proposition 1.11. Suppose Γ is the Cayley graph of a group G with respect to a generating set
X. Then the set of all symmetries Aut(Γ) of Γ is a group isomorphic to G.
Proof. Let Ω = E(Γ) ∪ V (Γ), the union of the set of edges and the set of vertices of Γ. The set of
symmetries Aut(Γ) is a nonempty subset of SΩ , and it is a straightforward exercise to show that it
is a subgroup.
We recall the action of G on Γ from Proposition 1.9. An element g ∈ G acts on Γ by sending
the vertex h to the vertex gh and sending each edge at h to the edge with the same direction and
label at gh. Of course, this defines a homomorphism from G to SΩ , and by Proposition 1.9, the
image is a subset of Aut(Γ). Proposition 1.9 showed that the action of G is faithful, and therefore
the map G → Aut(Γ) is injective. To prove the current proposition, it is enough to show that the
map is surjective.
Suppose α is a symmetry of Γ. Let g = α(1). The action of g on Γ is a symmetry sending 1 to
g. However, α is also a symmetry sending 1 to g. By Lemma 1.10, this means that α and action
by g represent the same symmetry of Γ, and therefore α is in the image of G → Aut(Γ). Since α
was arbitrary, the map is surjective.
The upshot of these notes is the following: if we have a group G and we want to share it with
someone, all we need to do is share instructions for constructing the Cayley graph Γ of G with
respect to some generating set. The other person can recover G from Γ using Proposition 1.7, or
recover the isomorphism class of G by finding Aut(Γ).
Exercises
1. Draw the Cayley graph of the quaternion group Q8 with respect to the generators i and j.
2. Draw the Cayley graph of the subgroup of S4 generated by {(123), (234)}, with respect to
that generating set. Bonus points for making your drawing highly symmetric.
3. Finish the proof of Lemma 1.5.
4. Finish the proof of Lemma 1.10.
5. Show that the symmetry group of a Cayley graph Γ is a subgroup of SΩ , where Ω = V (Γ) ∪
E(Γ).
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