When considering endomorphisms of a tournament Γ we note any endomor- phism must be an embedding.
Lemma 5.6. Let Γ = (VΓ, EΓ) be a tournament and let f ∈ End(Γ). Then
f is an embedding of Γ into Γ.
Proof. Let u, v ∈ VΓ with u 6= v and assume without loss of generality that (u, v) ∈ EΓ. Now, since f is an endomorphism of Γ, if uf = vf then (vf, vf) ∈ EΓ which is a contradiction. Hence uf 6= vf and f is indeed injective. Furthermore if (uf, vf) 6∈ EΓ then (vf, uf) ∈ EΓ and so it must be the case that (u, v)6∈EΓ since otherwise we would have a contradiction. Thus it follows that if (uf, vf)6∈EΓ then (u, v)6∈EΓ and hence since f was an injective endomorphism it is an embedding.
With Lemma 5.6 in mind, the following result is then of no surprise.
Theorem 5.7. Let f ∈End(T). Then imf is existentially closed and hence imf ∼=T.
Proof. Let U1 and U2 be finite and disjoint subsets in imf. Suppose that
|U1|=m and |U2|=n for m, n∈N. EnumerateU1 as {ui : 1≤i ≤m} and
U2 as {uj : m+ 1 ≤ j ≤ m+n}. Since uk ∈ imf for all 1 ≤ k ≤ m+n,
there exist vertices vk such that vkf = uk. Let V1 = {vi : 1 ≤ i ≤ m} and
subsets of VT. Since T is existentially closed it follows that there exists a
vertex x∈VΓ\(V1∪V2) such that there is an edge from xto every member of V1 and from every member of V2 tox. Now sincef is an endomorphism it follows that xf ∈ imf \(U1∪U2) and that there exists an edge from xf to every member of U1 and from every member of U2 to xf. Since U1 and U2 were arbitrary the result is complete.
Lemma 5.8. Let f ∈E(End(T)). Then f =1.
Proof. If f ∈E(End(T)) then f|imf =1|imf. We also observed that f must
be an injective embedding. So suppose that y ∈VT. Thenyf =x for some
x∈imf. Sincef is idempotentxf =xand hence by injectivity,x=y. Thus
y ∈imf and hence VT \imf =∅. Thus, f =f|VT =1VT as required.
Consequently, we now have the following results on the group H-classes and regular D-classes of End(T).
Theorem 5.9. The only group H -class of End(T) is Aut(T).
Proof. Every group H-class of End(T) contains an idempotent (the sub- group identity). By Lemma 5.8 the only such idempotent is the identity idempotent 1, and H1 = Aut(T) as required.
Corollary 5.10. End(T) has only one regular D-class.
Proof. Every regular D-class contains at least one idempotent. Hence, since Lemma 5.8 told us that the only idempotent in End(T) is the identity, there can only be one regular D-class.
We can now conclude that the only maximal subgroup of End(T) is Aut(T).
Chapter 6
Henson’s Graphs
In this section we discuss Kn-free graphs for n ≥ 3 and introduce Henson’s
graphs, Gn. We will see that, much like the random tournament T, the
graphs Gn are somewhat uninteresting in terms of maximal subgroups.
6.1
Defining Properties and Constructions
Recall that a Kn-free graph is a graph which has no substructure isomorphicto the graph Kn, the complete graph on n vertices. It is not hard to show
that for n≥3, the class of finiteKn-free graphs has the hereditary, joint em-
bedding and amalgamation properties (see [Hen71], for example). Thus, the class of finite Kn-free graphs has a unique homogeneous Fra¨ıss´e limit, which
is known as Henson’s graph, Gn. We will show that Gn can be characterised
by the following property.
We will say that aKn-free graph Γ isexistentially closed (in the class of
Kn-free graphs) if for all finite and disjoint subsets U, V ∈VΓ such thathUi is Kn−1-free, there exists a vertex x∈ VΓ \(U ∪V) such that x is adjacent to every member of U but to no member of V. We will assume for the rest of this chapter that whenever the phrase existentially closed is used for a
Kn-free graph, we mean existentially closed in the class ofKn-free graphs.
It is not hard to see that an existentially closed Kn-free graphs must be
infinite. For suppose that Γ was a such a finite Kn-free graph. Then there
would exists a maximal and finite Kn−1-free set of vertices U from VΓ. But by the existential closure property, there must exist a vertex x∈VΓ\U, such that x is adjacent to no member of U. If |U| =|VΓ|, then such anx cannot exist. On the other hand if |U|<|VΓ| thenU ∪ {x} is a Kn−1-free set which
contradicts the maximality of U.
We can also show that any existentially closed Kn-free graph does not
satisfy the bipartite condition as follows. For suppose that Γ is an existen- tially closedKn-free graph and letv ∈VΓ. Since v by itself is an independent set, there must exist a vertex x1 which is adjacent to v. Another applica- tion of the property ensures the existence of vertices x2 and x3 such that x2 is adjacent to v and not to x1 and such that x3 is adjacent to x1 and not to v nor x2. Finally since {x2, x3} is then an independent set by construc- tion we can find a vertex x4 adjacent to both x2 and x3. Then the path (v, x1),(x1, x3),(x3, x4),(x4, x2),(x2, v) is a cycle of odd length and hence Γ cannot satisfy the bipartite condition.
In some sense the existential closure property forKn-free graphs is ‘equiv-
alent’ to existential closure described for the class of all graphs but which holds only where possible (that is, avoiding the sets for which the property cannot hold due to the graph being Kn-free). Existentially closed Kn-free
graphs also have the following additional property.
Theorem 6.1. Let Γ be a countable existentially closed Kn-free graphs for
some n ≥ 3. Then Γ is homogeneous and every finite Kn-free graph can be
embedded into Γ.
A proof can be found in [Hen71, Theorem 2.3] or alternatively, the theo- rem will follow easily from the construction described in Definition 6.2. The age of an existentially closed Kn-free graph is thus the class of all finite
Kn-free graphs. Since the class of all finite Kn-free graphs has a unique ho-
mogeneous Fra¨ıss´e limit, it follows that if Γ is an existentially closedKn-free
graph, then Γ∼=Gn.
In the case of the random graph and random directed graph, we are able to exhibit a relatively easy probabilistic construction via a finitary method in which edges are chosen one at a time with a set probability. However, to exhibit a random or probabilistic construction of an infinite Kn-free graph is
not so straightforward. For example, consider the triangle-free graph G3. If we start with a countably infinite vertex set V and attempt to construct a triangle-free graph by choosing edges (as symmetric pairs fromV×V\{(x, x) :
x ∈ V}) one at a time with probability 12 say, then we quickly run in to trouble. For example, if we happen to have started the process by choosing the edge (u, v) and then the edge (v, w), we are not allowed to chose the edge (u, w) in order to ensure the graph remains triangle-free. Clearly, the probability that the edge (u, v) is chosen is 1
edge (v, w) is chosen is also 12 and is independent from the choice of (u, v). However the probability that the edge (u, w) is chosen is dependent upon the choices for (u, v) and (v, w) and is thus (using the law of total probability) equal to 0·1 4 + 1 2 · 1 4 + 1 2 · 1 4 + 1 2 · 1 4 = 3
8. Furthermore, this type of procedure provides a graph which is dependent on the order in which we decide to choose edges.
As Cameron discusses in [Cam01, Section 4.10] an attempt to bypass this problem by constructing a random triangle-free graph in a finitary way which is not dependent on the particular ordering of the vertices, still does not have the desired outcome. Any such triangle-free graph which is constructed in this way satisfies, by a result of Erd˝os, Kleitman and Rothschild [EKR76], with probability 1 the bipartite condition, and thus cannot be the Henson graph G3.
At the moment, it is unclear how to construct Henson’s graphG3 using a finitary probabilistic construction. However in [PV10, Section 3], Petrov and Vershik construct a measure-theoretic triangle-free graph on the real num- bers and, by taking countably many independent samples from a probability distribution on the real numbers, use it to produce a graph which is isomor- phic to the Henson graph G3 with probability 1. In effect their method is probabilistic on vertices rather than edges. It has been conjectured that a consequence of some of the stronger results in this paper will show that a finitary probabilistic construction of Henson’s graph G3 (and indeed Gn for
n > 3) is not possible. It is for this reason that the Henson graph Gn not
normally said to be ‘random’ unlike its counterparts R, D and T.
Even with that all said, we can give an explicit construction of Henson’s graphs as follows.
Definition 6.2. Let Γ be a countable Kn-free graph. We will construct a
new graph Ln(Γ) from Γ by adjoining vertices and edges in the following
manner. We will consider the set of finite and Kn−1-free subsets of VΓ. If Γ is countable, then the set of all finite sets of VΓ is countable. Thus the set of all finite and Kn−1-free sets of VΓ is a subset of a countable set and hence countable. We can thus enumerate all such finite Kn−1-free sets from VΓ as
{Ui :i∈N} – replacing the natural numbers by a finite set when necessary.
We createLn(Γ) by adding, for each such finiteKn−1-free setUi, a vertex
More precisely, we let,
VLn(Γ)=VΓ∪ {vi :i∈N}
and
ELn(Γ) =EΓ∪ {(vi, u),(u, vi) :u∈Ui, i∈N}.
If Γ is a finite graph then |VLn(Γ)| ≤ 2
|VΓ|+|V
Γ| and so Ln(Γ) is a finite
graph. Likewise if Γ is countably infinite then the set of all finite Kn−1-free sets is countably infinite (it cannot be finite for then there would exist an infinite subset of vertices such that any two vertices are adjacent – i.e. a complete graph which is impossible since it is not Kn-free) and hence Ln(Γ)
is countably infinite.
We can inductively define a sequence of graphs by setting Γ0 = Γ and Γn+1 =Ln(Γn). Now define Γ∞ to be the limit of this process by letting:
Γ∞= [ n∈N VΓn, [ n∈N VΓn ! .
It should be clear that Γ∞ is Kn-free since it is the union of the Kn-free
graphs Γn where Γn is an induced subgraph of Γn+1.
Example 6.3. [Construction of Γ1 = L3(Γ0) and Γ2 = L3(Γ1) when Γ = ({v},∅)]
Compare with Example 3.3 Γ0 = Γ • Γ1 • • • Γ2 • • • • ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ • ~ ~ ~ ~ ~ ~ ~ • ~ ~ ~ ~ ~ ~ ~ • • • OOOOOO OOOOOOOO JJJJ JJJJJJJJ JJJJ JJJJJJJ
Lemma 6.4. For any countable Kn-free graph Γ, the Kn-free graph Γ∞ is existentially closed and thus Γ∞ ∼=Gn.
Proof. Let U and V be disjoint subsets from Γ∞ such that U is a Kn−1-free set. Then there exists k ∈N such thatU, V ⊂VΓk. By construction of Γk+1,
there exists a vertexv ∈VΓk+1\VΓk adjacent to every member ofU. Moreover
xis adjacent to no other vertices inVΓk. Thusv is adjacent to every member
of U, but to no member of V in Γk+1. Since the construction of Γ∞ makes no change to the edge set of the induced subgraph Γk+1, it follows that v is adjacent to every member of U, but to no member of V in Γ∞.
Since any Kn-free graph Γ can clearly be embedded into Γ∞ and since Γ∞ ∼=Gn by Lemma 6.4, a proof of Theorem 6.1 should now be clear.