A NEW PROOF OF

We present a new proof of the famous four colour theorem using algebraic and topological methods. Recent research in physics shows that this proof directly implies the Grand Unification of the Standard Model with Quantum Gravity in its physical interpretation and conversely the existence of the standard model of particle physics shows that nature applies this proof of the four colour theorem at the most fundamental level, giving us a grand unified theory. In particular, we have shown how to use this theory to predict the Higgs Boson Mass [arXiv:0912.5189] with precision. Thus, nature itself demonstrates the logical completeness and consistency of the proof.







This proof was first announced by the Canadian Mathematical Society in 2000. The proof appears as the twelfth chapter of the text book Graph Theory published by Orient Longman and Universities Press of India in 2008. This proof has also been published in the Euroacademy Series Baltic Horizons No. 14 (111) dedicated to Fundamental Research in Mathematics in 2010. Finally, the proof features in an exquisitely illustrated edition of The Four Colour Theorem published by Amazon in 2011. The Endowed Chair of the Institute of Mathematics was bestowed upon Distinguished Professor Ashay Dharwadker in 2012 to honour his fundamental contributions to Mathematics and Natural Sciences.


The famous four colour theorem seems to
have been first proposed by Möbius in 1840, later by DeMorgan and
the Guthrie brothers in 1852, and again by Cayley in 1878. The problem
of proving this theorem has a distinguished history, details of which abound
in the literature. The statement of the theorem may be introduced as follows.
In colouring a geographical map it is customary to give different colours
to any two countries that have a segment of their boundaries in common.
It has been found empirically that any map, no matter how many countries
it contains nor how they are situated, can be so coloured by using only
four different colours. The map of India requires four colours in the states
bordering Madhya Pradesh. The fact that no map was ever found whose colouring
requires more than four colours suggests the mathematical theorem.
FOUR COLOUR THEOREM. For any subdivision of the plane into nonoverlapping regions, it is always possible to mark each of the regions with one of the numbers 0, 1, 2, 3 in such a way that no two adjacent regions receive the same number. STEPS OF THE PROOF: We shall outline the strategy of the new proof given in this paper. In section I on MAP COLOURING, we define maps on the sphere and their proper colouring. For purposes of proper colouring it is equivalent to consider maps on the plane and furthermore, only maps which have exactly three edges meeting at each vertex. Lemma 1 proves the six colour theorem using Euler's formula, showing that any map on the plane may be properly coloured by using at most six colours. We may then make the following basic definitions.

A Steiner system S(t,
k,
v)
is a set P of points together with a set
B
of blocks such that
2. LEMMA. [J. TITS] If there exists a nontrivial Steiner system S(t, k, v) then v ≥ (t+1)(kt+1). PROOF: First show that there exists a set X_{0} of t+1
points that is not contained in any block, as follows. Suppose that for
every set X of t+1 points there is a block B_{X}
that contains it. Then, this block B_{X} must be the unique
block containing X, since X has more than t points.
Let b denote the total number of blocks. Count in two ways the number
of pairs (X, B_{X}) where X is a set of t+1
points and B_{X} is the unique block containing it. One
finds
Count in two ways the number of pairs (Y, B_{Y})
where Y is a set of t points and B_{Y} is
the unique block containing it, by definition of a Steiner system. One
finds
Hence
and it follows that b = 1 and k = v, contradicting the hypothesis that the Steiner system is nontrivial. Now choose a fixed set X_{0} of t+1 points that is not contained in any block. For each set Z of t points contained in X_{0} there is a unique block B_{Z} containing Z. Each such B_{Z} has kt points not in X_{0} and any point not in X_{0} is contained in at most one such B_{Z}, since two such blocks already have t1 points of X_{0} in common. The union of the blocks B_{Z} contains (t+1)+(t+1)(kt) points and this number cannot exceed the total number of points v. ☐ Recall the definition from section I that N is the minimal number of colours required to properly colour any map from the class of all maps on the sphere and m(N) is a specific map which requires all of the N colours to properly colour it. The regions of the map m(N) have been properly coloured using the N colours 0, 1, ..., N1. From the map m(N) and its fixed proper colouring, we shall construct a Steiner system S(N+1, 2N, 6N) by defining the points and blocks in a certain way. The next lemma shows that this construction would force N ≤ 4. 3. LEMMA. Referring to the definition of N in section I, if there exists a Steiner system S(N+1, 2N, 6N) then N ≤ 4. PROOF: Since 4 ≤ N ≤ 6 by definition, the Steiner system is nontrivial if it exists. By lemma 2, 6N ≥ (N+1+1)(2NN1+1) = (N+2)N. Hence 6 ≥ N+2 and it follows that 4 ≥ N. ☐ Now, the goal is to demonstrate the existence of the Steiner system S(N+1, 2N, 6N) based upon the definition of the map m(N). 
Let G be a group with identity
element e and let Z denote the integers. The integral
group algebra (ZG, +, ·) is a ring whose elements are
formal sums
with g in G and n_{g} in Z
such that n_{g} = 0 for all but a finite number of g.
Addition and multiplication in ZG are defined by
The element n of Z is identified with the element
n·e
of ZG and the element g of
G is identified
with the element 1·g of ZG, so that Z
and G are to be regarded as subsets of ZG. The underlying
additive abelian group (ZG, +) is the direct sum of copies
of the integers Z indexed by elements of G. If Q
is a subgroup of G then ZQ is a subring of ZG
in a natural way. For each element g of G, the right multiplication
R(g):
G → G;
x → xg
and the left multiplication
L(g): G → G;
x → gx
are permutations of the set G. Denote the group of all permutations
of the set G by Sym(G). Then
are embeddings of the group G in Sym(G). The images
R(G),
L^{1}(G)
are called the Cayley right and left regular representations of G,
respectively. The subgroup of Sym(G) generated by the set
R(G)∪L^{1}(G)
= {R(g), L(g^{1})g∈G}
is called the combinatorial multiplication group Mlt(G)
of
G. There is an exact sequence of groups
where T(x, y) = R(x)L(y^{1})
and Δc = (c, c) for an
element c of the center C(G) of G. If Q
is a subgroup of G then the relative combinatorial multiplication
group Mlt_{G}(Q)
of Q in G
is the subgroup of Mlt(G) generated by the set R(Q)∪L^{1}(Q)
= {R(q), L^{1}(q)q∈Q}.
The orbits of the action of Mlt_{G}(Q) on G
are the double cosets QgQ of the subgroup Q in G.
The stabilizer of the identity element e is the subgroup of Mlt_{G}(Q)
generated by the set {T(q) = R(q)L^{1}(q)q∈Q}.
A representation of the group Q is usually defined as a module,
i.e. an abelian group (M, +), for which there is a homomorphism
T:
Q → Aut(M,
+) showing how Q acts as a group of automorphisms of the module.
Another approach due to Eilenberg, views a module M for the group
Q
as follows. The set M×Q
equipped with the multiplication
becomes a group M]Q known as the split extension of
M by Q. There is an exact sequence of groups
with ι: M → M]Q;
m → (m,
e)
and π: M]Q → Q;
(m, q) → q split by 0:
Q → M]Q;
q → (0,
q).
The group action
T is recovered from the split extension
M]Q
by mT(q)ι =
mιR((0,
q))L^{1}((0,
q))
for m in M and q in Q. In this context
we shall call M an Eilenberg module for the group Q. For
example, the trivial representation for the group Q is obtained
by defining
T: Q → Aut(M,
+);
q → 1_{M}, the identity
automorphism of (M, +) and the corresponding split extension is
the group direct product M×Q.
The Cayley right regular representation for the group Q is obtained
by defining
Here T(q) = R(q)L^{1}(q)
with L^{1}(q) acting trivially on the module elements
and R(q) acting as the usual right multiplication. The split
extension ZQ]Q has multiplication given by
for m_{1}, m_{2} in ZQ and q_{1}, q_{2} in Q. Referring to the definition in section I, N is the minimal number of colours required to properly colour any map from the class of all maps on the sphere and m(N) is a specific map that requires all of N colours to be properly coloured. Note that m(N) has been properly coloured by using the N colours 0, 1, ..., N1 and this proper colouring is fixed. The set of regions of m(N) is then partitioned into subsets 0, 1, ..., N1 where the subset m consists of all the regions which receive the colour m. Note that the subsets 0, 1, ..., N1 are each nonempty (since m(N) requires all of the N colours to be properly coloured) and form a partition of the set of regions of m(N) (by virtue of proper colouring). Identify the set {0, 1, ..., N1} with the underlying set of the Nelement cyclic group Z_{N} under addition modulo N. Let S_{3} denote the symmetric group on three letters, identified with the dihedral group of order six generated by ρ, σ where ρ = 3 and σ = 2. 4. LEMMA. (Z_{N}, +) is an Eilenberg
module for the group S_{3} with the trivial homomorphism
where1Z_{N}
denotes the identity automorphism of Z_{N}.
The
corresponding split extension Z_{N}]S_{3 }has
multiplication given by
and is a group isomorphic to the direct product Z_{N}×S_{3}. PROOF: Follows from definition. ☐ Referring to section II, the goal is to construct a Steiner system S(N+1, 2N, 6N). We shall take the point set of the Steiner system to be the underlying set of the split extension Z_{N}]S_{3}. The following lemma is used in section V. 5. LEMMA. Let (Z(Z_{N}]S_{3}),
+)
and (ZS_{3}, +) denote the underlying
additive groups of the integral group algebras Z(Z_{N}]S_{3})
and
ZS_{3},
respectively.
Then
(Z(Z_{N}]S_{3}), +)
is
an Eilenberg module for the group (ZS_{3}, +)
with
the trivial homomorphism
where 1Z(Z_{N}]S_{3})
denotes
the identity automorphism of (Z(Z_{N}]S_{3}),
+).
The corresponding split extension Z(Z_{N}]S_{3})]ZS_{3}
has multiplication given by
and is a group isomorphic to the direct product (Z(Z_{N}]S_{3})×ZS_{3}, +). PROOF: Follows from definition. ☐ 
Let Γ be a
bipartite graph with vertex set V = X∪Y
and edge set E (every edge has one end in X and the other
end in Y). A matching from X to Y in Γ
is a subset M of E such that no vertex is incident with more
than one edge in M. A matching M from X to Y in Γ
is called complete if every vertex in X is incident with
an edge in M. If A is a subset of V then let adj(A)
denote the set of all vertices adjacent to a vertex in A.
6. LEMMA. [P. HALL] If adj(A) ≥ A for every subset A of X then there exists a complete matching from X to Y in Γ. PROOF: A matching from X to Y in Γ
with M = 1 always exists by choosing a single edge in E.
Let M be a matching from X to Y in Γ
with m edges, m < X.
Let x_{0}∈X such
that x_{0} is not incident with any edge in M. Since
adj({x_{0}}) ≥ 1,
there is a vertex y_{1} adjacent to x_{0}
by an edge in E\M. If y_{1} is not incident with
an edge in M, then stop. Otherwise, let x_{1} be
the other end of such an edge. If x_{0}, x_{1},
..., x_{k} and y_{1}, ..., y_{k}
have been chosen, then since adj({x_{0}, x_{1},
..., x_{k}}) ≥ k+1,
there is a vertex y_{k}_{+1}, distinct from y_{1},
..., y_{k}, that is adjacent to at least one vertex in {x_{0},
x_{1},
..., x_{k}}. If y_{k}_{+1} is not
incident with an edge in M, then stop. Otherwise, let x_{k}_{+1}
be the other end of such an edge. This process must terminate with some
vertex, say y_{k}_{+1}. Now build a simple path
from y_{k}_{+1} to x_{0} as follows.
Start with y_{k}_{+1} and the edge in E\M
joining it to, say xi_{1}
with i_{1} < k+1. Then
add the edge in M from xi_{1}
to yi_{1}. By
construction, yi_{1}
is joined by an edge in E\M to some xi_{2}
with i_{2} < i_{1}.
Continue adding edges in this way until x_{0} is reached.
One obtains a path y_{k}_{+1}, xi_{1},
yi_{1},
xi_{2},
yi_{2},
..., xi_{r},
yi_{r},
x_{0}
of odd length 2r+1 with the r+1 edges {y_{k}_{+1},
xi_{1}},
{yi_{1},
xi_{2}},
..., {yi_{r}, x_{0}}
in E\M and the r edges {xi_{1},
yi_{1}},
..., {xi_{r}, yi_{r}}
in M. Define
Then M' is a matching from X to Y in Γ, with M' = Mr+r+1 = M+1. Repeating this process a finite number of times must yield a complete matching from X to Y in Γ. ☐ 7. LEMMA. Referring to section III, let Sym(Z_{N}]S_{3}) denote the group of all permutations of the underlying set of the split extension Z_{N}]S_{3 }of lemma 4. Then S_{3 }embeds in Sym(Z_{N}]S_{3}) via the Cayley right regular representation. PROOF: Note that S_{3} = {(0, α)α∈S_{3}} is a subgroup of Z_{N}]S_{3}. Since S_{3} embeds in Sym(S_{3}) via the Cayley right regular representation α → R(α) and Sym(S_{3}) is a subgroup of Sym(Z_{N}]S_{3}), the lemma follows. ☐ 8. LEMMA. By lemma 7, regard S_{3} as a subgroup of Sym(Z_{N}]S_{3}). There exists a common system of coset representatives φ_{1}, ..., φ_{k }such that {φ_{1}S_{3}, ..., φ_{k}S_{3}} is the family of left cosets of S_{3} in Sym(Z_{N}]S_{3}) and {S_{3}φ_{1}, ..., S_{3}φ_{k}} is the family of right cosets of S_{3} in Sym(Z_{N}]S_{3}). PROOF: By Lagrange's theorem the left cosets of S_{3} partition Sym(Z_{N}]S_{3}) into k = [Sym(Z_{N}]S_{3}):S_{3}] disjoint nonempty equivalence classes of size S_{3} = 6. The same is true of the right cosets. Define a bipartite graph Γ with vertices X∪Y where X = {ψ_{1}S_{3}, ..., ψ_{k}S_{3}} is the family of left cosets of S_{3} in Sym(Z_{N}]S_{3}) and Y = {S_{3}ψ'_{1}, ..., S_{3}ψ'_{k}} is the family of right cosets of S_{3} in Sym(Z_{N}]S_{3}) with an edge {ψ_{i}S_{3}, S_{3}ψ'_{j}} if and only if ψ_{i}S_{3} and S_{3}ψ'_{j} have nonempty intersection. Note that we can select representatives of the left cosets that belong to distinct right cosets [see a proof of this fact]. For any subset A = {ψi_{1}S_{3}, ..., ψi_{r}S_{3}} of X, one has ψi_{1}∈ψi_{1}S_{3}, ..., ψi_{r}∈ψi_{r}S_{3} and there exist distinct j_{1}, ..., j_{r} such that ψi_{1}∈S_{3}ψ'j_{1}, ..., ψi_{r}∈S_{3}ψ'j_{r}. Hence, in the graph Γ, adj(A) ≥ A. Hall's hypothesis of lemma 6 is satisfied and there exists a complete matching from X to Y in Γ. This is precisely the statement that a common system of coset representatives φ_{1}, ..., φ_{k} exists. ☐ 
Let C denote the complex
plane. Consider the function C → C;
z → w
= z^{n}, where n ≥ 2.
There is a onetoone correspondence between each sector
and the whole wplane except for the positive real axis. The
image of each sector is obtained by performing a cut along the positive
real axis; this cut has an upper and a lower edge. Corresponding to the
n
sectors in the zplane, take n identical copies of the wplane
with the cut. These will be the sheets of the Riemann surface and
are distinguished by a label k which serves to identify the corresponding
sector. For k = 1, ..., n1 attach the lower edge of the
sheet labeled k with the upper edge of the sheet labeled k+1.
To complete the cycle, attach the lower edge of the sheet labeled n
to the upper edge of the sheet labeled 1. In a physical sense, this is
not possible without selfintersection but the idealized model shall be
free of this discrepancy. The result of the construction is a
Riemann
surface whose points are in onetoone correspondence with the points
of the zplane [see
a geometric model]. This correspondence is continuous in the following
sense. When z moves in its plane, the corresponding point w
is free to move on the Riemann surface. The point w = 0
connects all the sheets and is called the branch point. A curve
must wind n times around the branch point before it closes. Now
consider the nvalued relation
To each w ≠ 0, there correspond
n
values of z. If the wplane is replaced by the Riemann surface
just constructed, then each complex number w ≠ 0
is represented by n points of the Riemann surface at superposed
positions. Let the point on the uppermost sheet represent the principal
value and the other n1 points represent the other values. Then
z
= ^{n}√w becomes
a singlevalued, continuous, onetoone correspondence of the points of
the Riemann surface with the points of the zplane. Now recall the
definition of the map m(N) from section I. The map
m(N)
is on the sphere. Pick a region and deform the sphere so that both 0 and
∞
are two distinct points inside this region when the sphere is regarded
as the extended complex plane. Using the stereographic projection one obtains
the map m(N) on the complex plane C
with the region containing 0 and ∞ forming
a ''sea'' surrounding the other regions which form an ''island''. Put this
copy of C on each sheet of the Riemann surface corresponding
to w = z^{n}. The branch point lies in the ''sea''.
The inverse function z = ^{n}√w
results in n copies of the map m(N) on the
zplane
in
the sectors
The origin of the zplane lies in the n ''seas''.
Referring to section III, the full symmetric group Sym(Z_{N}]S_{3})
acts faithfully on the set Z_{N}]S_{3}.
The action of an element ψ of Sym(Z_{N}]S_{3})
on an element (m, α) of Z_{N}]S_{3}
will be written as (m, α)ψ.
This action extends to the integral group algebra Z(Z_{N}]S_{3})
by linearity
Referring to lemma 8, fix a common coset representative φ_{i} of S_{3} in Sym(Z_{N}]S_{3}) and fix a pair (β, γ)∈S_{3}×S_{3 }= Mlt(S_{3}). There are two cases depending on whether β = γ or whether β ≠ γ. CASE 1. Suppose β ≠ γ. Consider
the composition of the functions
The composite is given by the assignment
There are twentyfour superposed copies of the map m(N)
on the wRiemann surface corresponding to the sectors
on the zplane. These are divided into two sets. The first set
consists of twelve superposed copies of the map m(N)
corresponding to the sectors
of the upper half of the zplane which comprise the upper sheet
of the tRiemann surface. The second set consists of twelve superposed
copies of the map m(N) corresponding to the sectors
of the lower half of the zplane which comprise the lower sheet
of the tRiemann surface.
Label the twelve sectors of the upper sheet of the tRiemann
surface by elements of Z(Z_{N}]S_{3})]ZS_{3}
as shown:
Label the twelve sectors of the lower sheet of the tRiemann
surface by elements of Z(Z_{N}]S_{3})]ZS_{3}
as shown:
Referring to section II, the regions of the map m(N) have been partitioned into disjoint, nonempty equivalence classes 0, 1, ..., N1 and this set of equivalence classes forms the underlying set of the cyclic group Z_{N}. Hence there are twelve copies of Z_{N} on the upper sheet and twelve copies of Z_{N} on the lower sheet of the tRiemann surface. The copies of Z_{N} are indexed by the elements of Z(Z_{N}]S_{3})]ZS_{3} which label the sectors on a particular sheet. The branch point of the tRiemann surface is labeled by the element (0, β+γ) of Z(Z_{N}]S_{3})]ZS_{3} where 0 denotes the zero element of Z(Z_{N}]S_{3}). 9. LEMMA. Referring to lemma 8, fix a common representative
φ_{i
}of
the left and right cosets of S_{3} in Sym(Z_{N}]S_{3}).
Fix
a pair (β, γ)∈S_{3}×S_{3
}with
β ≠ γ.
Referring
to lemma 5,
define a subset T_{(β,
γ)
}of
Z(Z_{N}]S_{3})]ZS_{3
}as
follows:
Referring to the preceding discussion, consider the composite function
of the complex zplane to the wRiemann surface.
There
is a copy of the set T_{(β,
γ)
}on
the upper sheet and a copy of the set T_{(β,
γ)
}on
the lower sheet of the tRiemann surface according to the
labels of the sectors in figures 4 and 5 with the branch
point labeled by the element (0, β+γ)
of
both copies. The rotation of the zplane by π
radians
induces a permutation
given by
for all (m, α)∈Z_{N}]S_{3}, such that each point of the copy of T_{(β, γ) }on the upper sheet moves continuously along a circular curve that winds exactly once around the branch point, to the point superposed directly below it on the copy of T_{(β, γ) }on the lower sheet of the tRiemann surface. PROOF: T_{(β, γ)} is seen to be a welldefined subset of Z(Z_{N}]S_{3})]ZS_{3} by setting the appropriate coefficients to zero in a typical element as described in lemma 5. Each of R(γ)φ_{i}, φ_{i}R(γ), φ_{i}R(β), R(β)φ_{i} are permutations of the set Z_{N}]S_{3} and the rotation of the zplane by π radians clearly induces a permutation p of the set T_{(β, γ)} as described. ☐ 10. LEMMA. Referring to lemma 9, let Sym(T_{(β, γ)}) denote the full permutation group of the set T_{(β, γ)}. Let < p > denote the cyclic subgroup of Sym(T_{(β, γ)}) generated by p. Then < p > is nontrivial and acts faithfully on the set T_{(β, γ)}. PROOF: If p = 1, then ((0,1)R(γ)φ_{i}, β+γ) = ((0,1)R(γ)φ_{i}, β+γ)p = ((0,1)φ_{i}R(γ), β+γ) which implies that (0,1)R(γ)φ_{i}= (0,1)φ_{i}R(γ) in Z(Z_{N}]S_{3}). This is impossible since 1 ≠ 1 in Z. Hence p ≠ 1. Since the full permutation group Sym(T_{(β, γ)}) acts faithfully on T_{(β, γ)}, so does its subgroup < p > . ☐ 11. LEMMA. Referring to lemma 9 and lemma 10, let 1: C → C; z → z denote the identity and π: C → C; z → z denote the rotation through an angle of π radians of the zplane. Then the twoelement cyclic group {1, π} acts faithfully on the set < p > as follows: p^{n}1 = p^{n }and p^{n}π = p^{1n}, for all n in Z. PROOF: The set {1, π} forms a twoelement cyclic group < π > under function composition. To show that {1, π} acts on < p > as defined, observe that (p^{n}π)π = (p^{1n})π = p^{1(1n)} = p^{11+n} = p^{n} = p^{n}1 = p^{n}(ππ), for all n in Z. To show that the action is faithful, let θ∈{1, π}. If θ belongs to the kernel of the action then p^{n}θ = p^{n} for all n in Z so that pθ = p which implies that θ = 1, since p ≠ 1 by lemma 10. ☐ 12. LEMMA. Putting together lemma 9, lemma 10 and
lemma 11, there is a welldefined action of the twoelement cyclic
group {1, π} on the set T_{(β,
γ)
}given
by
and
for all (m, α) in Z_{N}]S_{3}. This action is faithful. PROOF: For each x∈T_{(β,
γ)},
let Orb(x) = {xp^{n}n∈Z}
denote the orbit of x under < p > .
The collection {Orb(x)x∈T_{(β,
γ)}}
forms a partition of the set T_{(β,
γ)}
as follows. Let x, y∈T_{(β,
γ)}.
If z∈Orb(x)∩Orb(y)
then z = xp^{n} = yp^{m} for some
m,
n∈Z.
This implies
xp^{n}^{m }= y ⇒
y∈Orb(x)
⇒
Orb(y)⊆Orb(x)
and yp^{m}^{n }= x ⇒
x∈Orb(y)
⇒
Orb(x)⊆Orb(y).
Hence Orb(x) = Orb(y). Also each x∈T_{(β,
γ)}
belongs to an orbit, namely x∈Orb(x).
Hence the orbits are disjoint, nonempty and their union is all of the set
T_{(β,
γ)}.
For each fixed x∈T_{(β,
γ)}
define
Then π is welldefined, since xp^{n
}=
xp^{m
}⇒
xp^{nm
}=
x
⇒
xp^{m
}=
xp^{n
}⇒
xp^{1m
}=
xp^{1n
}⇒
(xp^{n})π
= (xp^{m})π.
Also, π is a permutation of Orb(x)
with π^{2
}= 1, since for each xp^{n}∈Orb(x)
we have ((xp^{n})π)π
= (xp^{1n})π = xp^{1(1n)}
= xp^{n}. Now define
orbit by orbit. Then, since {Orb(x)x∈T_{(β,
γ)}}
forms a partition of T_{(β,
γ)},
π
is a welldefined permutation of T_{(β,
γ)}
with π^{2
}= 1, the identity permutation
of T_{(β,
γ)}.
Hence, using the definition of p in lemma 9, we obtain an action
of the twoelement cyclic group {1, π} on T_{(β,
γ)}
as follows. For all (m,
α) in
Z_{N}]S_{3}
define
and
Now, using the definition of p in lemma 9, the action of {1,
π}
on T_{(β,
γ)}
may be rewritten
and
for all (m, α) in Z_{N}]S_{3},
as in the statement of this lemma. To verify that the action of {1, π}
on T_{(β, γ)}
is faithful, note that
are permutations of the set T_{(β, γ)}. If θ∈{1, π}and θ belongs to the kernel of the action then xθ = x for all x∈T_{(β,γ)}. Then θ = 1, since π moves ((0, 1)φ_{i}R(γ), β+γ) to ((0, 1)R(γ)φ_{i}, β+γ) which are distinct elements of Z(Z_{N}]S_{3})]ZS_{3}. ☐ CASE 2. Suppose β = γ. Note that in the labeling of the sectors of the sheets of the tRiemann surface in figures 4 and 5, R(β) = R(γ) and β+γ = 2β in the group algebra ZS_{3}. 13. LEMMA. Referring to lemma 8, fix a common representative
φ_{i
}of
the left and right cosets of S_{3} in Sym(Z_{N}]S_{3}).
Fix
a pair (β, β)∈S_{3}×S_{3}.
Referring
to lemma 5, define a subset T_{(β,
β)
}of
Z(Z_{N}]S_{3})]ZS_{3
}as
follows:
Referring to the preceding discussion, consider the composite function
of the complex zplane to the wRiemann surface.
There
is a copy of the set T_{(β,
β)
}on
the upper sheet and a copy of the set T_{(β,
β)
}on
the lower sheet of the tRiemann surface according to the
labels of the sectors in figures 4 and 5. Note that in this
case R(β) = R(γ)
and
β+γ
= 2β
with the branch point labeled by the
element (0, 2β)
of both copies. The
rotation of the zplane by π
radians
induces a permutation
given by
for all (m, α)∈Z_{N}]S_{3}, such that each point of the copy of T_{(β, β) }on the upper sheet moves continuously along a circular curve that winds exactly once around the branch point, to the point superposed directly below it on the copy of T_{(β, β) }on the lower sheet of the tRiemann surface. Then p = p^{1 }so that < p > = {1, p} is a twoelement cyclic subgroup of the full permutation group Sym(T_{(β, β)}) and < p > acts faithfully on the set T_{(β, β)}. PROOF: As in the proof of lemma 9, T_{(β, β)} is seen to be a welldefined subset of Z(Z_{N}]S_{3})]ZS_{3} by setting the appropriate coefficients to zero in a typical element as described in lemma 5. Both φ_{i}R(β), R(β)φ_{i} are permutations of the set Z_{N}]S_{3} and the rotation of the zplane by π radians clearly induces a permutation p of the set T_{(β, β)} as described. Furthermore, it is clear from the definition that p = p^{1} by chasing elements of T_{(β, β)}. Then < p > = {1, p} as a subgroup of Sym(T_{(β, β)}) and < p > acts faithfully on the set T_{(β, β)}. ☐ 14. LEMMA. Referring to lemma 13, let 1: C → C;
z → z
denote the identity and π: C → C;
z → z
denote the rotation through an angle of π radians
of the zplane. Then there is a welldefined action of the
twoelement cyclic group {1, π}
on the
set T_{(β,
β)
}given
by
and
for all (m, α) in Z_{N}]S_{3}. This action is faithful. PROOF: The isomorphism 1 → 1, p → π of the twoelement cyclic groups {1, p} and {1, π} establishes the lemma. ☐ 
RÉSUMÉ. Let us review
the final goal. Recall the definition made in section I. We have defined
N
to be the minimal number of colours required to properly colour any map
from the class of all maps on the sphere. We know that 4 ≤ N ≤ 6.
We have chosen a specific map m(N) on the sphere which
requires all of the N colours 0, 1, ..., N1 to properly
colour it. The map m(N) has been properly coloured
and the regions of m(N) partitioned into disjoint,
nonempty equivalence classes 0, 1, ..., N1
according to the colour they receive. The set {0, 1, ...,
N1}
is endowed with the structure of the cyclic group Z_{N}
under addition modulo N. In section III we have built the split
extension
Z_{N}]S_{3}. The underlying
set Z_{N}]S_{3} of cardinality 6N
is taken to be the point set of a Steiner system S(N+1, 2N,
6N) which will be constructed in this section. We are required to
define the blocks of size 2N and show that every set of N+1
points is contained in a unique block. Once this goal is achieved, lemma
3 shows that N ≤ 4.
15. LEMMA. Let Z_{N}]S_{3
}denote
the split extension defined in lemma 4 and Sym(Z_{N}]S_{3})
denote
the full permutation group on the set Z_{N}]S_{3}.
Define
by
Then μ is a bijection of the set Sym(Z_{N}]S_{3}) with itself. PROOF: Referring to lemma 8, μ is welldefined since each ψ∈Sym(Z_{N}]S_{3}) may be written uniquely as ψ = R(γ)φ_{i} for some γ∈S_{3} and some φ_{i}. Then μ is a surjection because for any ψ∈Sym(Z_{N}]S_{3}) one may also write ψ = φ_{i}R(γ) uniquely for some γ∈S_{3} and some φ_{i}, whence R(γ)φ_{i} → φ_{i}R(γ) = ψ. Since Sym(Z_{N}]S_{3}) is a finite set, μ must be a bijection by counting. ☐ 16. LEMMA. Define the set G as follows:
Define multiplication in G as follows:
i.e.
where R(γ_{3})φi_{3 } is the unique expression for R(γ_{1})φi_{1}R(γ_{2})φi_{2 } according to the right coset decomposition of S_{3} in Sym(Z_{N}]S_{3}). Then G is a group. PROOF: Referring to lemma 8 and lemma 15, the set G is welldefined
by the decomposition of Sym(Z_{N}]S_{3})
into the left and right cosets of S_{3} by the φ_{i}.
Define
Then μ' is a welldefined bijection of the set Sym(Z_{N}]S_{3}) with G since μ is a bijection by lemma 15. The definition of multiplication in G mirrors the multiplication in Sym(Z_{N}]S_{3}) via μ' and is designed to make G a group and μ' an isomorphism. ☐ 17. LEMMA. Consider the set Z_{N}]S_{3
}and
let
Define
Define
Then
Both ↑ and ↓ are welldefined, faithful and Z_{N}]S_{3}transitive right actions of the group G on the set Z_{N}]S_{3}. PROOF: Referring to lemma 12 and lemma 14, put β
= 1. Working in the set T_{(1, γ)},
for each (m, α)∈Z_{N}]S_{3}
we have
using the action of the twoelement cyclic group {1, π}
on the set T_{(1, γ)} according
to lemma 12 and lemma 14. Hence
Since the action ↑ is the usual action of Sym(Z_{N}]S_{3}) on the set Z_{N}]S_{3}, it is faithful and Z_{N}]S_{3}transitive. By the last equality, so is the ↓ action. ☐ 18. LEMMA. Let (m_{1},
α_{1}),
..., (m_{r},
α_{r})
be
any r distinct elements of Z_{N}]S_{3}
and let (n_{1}, β_{1}),
..., (n_{s}, β_{s})
be
any s distinct elements of Z_{N}]S_{3}.
Let
then H_{r, s} is a subgroup of G. PROOF: Note that if ψ = R(γ)φ_{i}=
1 then φ_{i }=
R(γ)^{1}
so that ψ^{μ}= φ_{i}R(γ)
= R(γ)^{1}R(γ)
= 1. Then
since
for i = 1, ..., r and
for j = 1, ..., s. If
then
and
Hence
Since G is finite, H_{r, s} is a subgroup of G. ☐ Note that Z_{N} is embedded as the subgroup {(m, 1)m∈Z_{N}} in Z_{N}]S_{3} and S_{3} is embedded as the subgroup {(0, α)α∈S_{3}} in Z_{N}]S_{3}. Since Z_{N}]S_{3} = Z_{N}×S_{3} is the direct product of groups by lemma 4, both Z_{N} and S_{3} are normal subgroups. Recall the notation
19. LEMMA. Define
Then given
either
or
PROOF: H is a welldefined subgroup of G according to
lemma 18. Let
and (m, α)∈Z_{N}]S_{3}
be given. Referring to lemmas 12 and 14, put β
= γ^{1}αγ.
Working in the set T_{(β, γ)}
we have
using the definition of H and the action of the twoelement cyclic group
{1,π} on the set T_{(β,γ)}.
Hence
Now since
by hypothesis, we have σ = σ^{γ}. Hence γσ = σγ so that either γ = 1 or γ = σ. ☐ 20. LEMMA. Let H be the subgroup of G defined in lemma 19. Then H is a nontrivial group of involutions of the set Z_{N}]S_{3}. In particular, every nontrivial element of H is of order 2. PROOF: Define
Then
and
Now ψ ≠ 1, so
hence H is nontrivial. To show that each nontrivial element of
H
is of order 2, let
Then by the proof of lemma 19, γ = 1 or γ
=
σ.
In particular γ^{2} = 1. Hence, for
any (m, α)∈Z_{N}]S_{3}
= (m, (α^{γ})^{γ})
= (m, α). Since the ↓
action of G on the set Z_{N}]S_{3}
is faithful,
the identity element of G. ☐ 21. LEMMA. Denote the right cosets of Z_{N}
in Z_{N}]S_{3} by
Define
Then Fix↓(H) = {(m, α)∈Z_{N}]S_{3}α = 1 or α = σ}. The ↓ action of a nontrivial element of H transposes the coset Z_{N}ρ with the coset Z_{N}ρ^{2 }and transposes the coset Z_{N}σρ with the coset Z_{N}σρ^{2}. PROOF: By lemmas 19 and 20, the elements
are of two kinds:
in which case
the identity element of H, and
in which case
is an element of order 2 in H. In the second case, compute according
to the cosets of Z_{N} in Z_{N}]S_{3}:
and the lemma follows. ☐ 22. LEMMA. Let Norm_{G}(H) denote the normalizer of H in G. The action ↓ of G on Z_{N}]S_{3 }restricts to an action ↓ of Norm_{G}(H) on Fix↓(H) which is (Z_{N}+1)transitive. PROOF: Let
First show that
as follows:
Now let
Then
showing that the action restricts to an action of Norm_{G}(H)
on Fix↓(H). Now to show that
the action ↓ of Norm_{G}(H)
on Fix↓(H) = Z_{N}∪Z_{N}σ
is
(Z_{N}+1)transitive, label the elements of
Z_{N}
as (m_{1}, α_{1}),
..., (m_{N}, α_{N})
and label (0, σ) = (m_{N}_{+1},
α_{N+1}).
Let (m^{*}_{1},
α^{*}_{1}),
..., (m^{*}_{N+1},
α^{*}_{N+1})
be any Z_{N}+1 distinct points of Fix↓(H).
It is enough to show that there exists
such that
Now there exists
such that
Hence
Note that for every
and for i = 1, ..., N+1:
Hence
23. LEMMA. There exists a Steiner system S(N+1,
2N, 6N ), where the points are the elements of the set
Z_{N}]S_{3
}and
the set of blocks is
PROOF: There are 6N = Z_{N}]S_{3}
points. Each block, for a fixed
contains
points. Label the elements of
Z_{N} as (m_{1},
α_{1}),
..., (m_{N},
α_{N})
and label (0, σ) = (m_{N}_{+1},
α_{N+1}).
Let (m^{*}_{1},
α^{*}_{1}),
..., (m^{*}_{N+1},
α^{*}_{N+1})
be any Z_{N}+1 distinct points of Z_{N}]S_{3}.
Then there exists
such that
Hence, there is at least one block, namely Fix↓(H),
that contains the points (m^{*}_{1},
α^{*}_{1}),
..., (m^{*}_{N+1},
α^{*}_{N+1}).
It remains to show that this is the unique block that contains the points
(m^{*}_{1},
α^{*}_{1}),
..., (m^{*}_{N+1},
α^{*}_{N+1}).
Suppose (m^{*}_{1},
α^{*}_{1}),
..., (m^{*}_{N+1},
α^{*}_{N+1})
are contained in
Then there exist points (m^{**}_{1},
α^{**}_{1}),
..., (m^{**}_{N+1},
α^{**}_{N+1})
in Fix↓(H) such that
By lemma 22, there exists
such that
Hence for i = 1, ..., N+1
= (m^{*}_{i}, α^{*}_{i})
Then by lemma 17
and
Hence
Now H is a subgroup of Norm_{G}(H)
Now, using the first fact in the proof of lemma 22
This establishes the uniqueness of the block. ☐ 24. THEOREM. Any map on the sphere can be properly coloured by using at most four colours. PROOF: Referring to section I, we have defined N to be the minimal number of colours required to properly colour any map from the class of all maps on the sphere. Based on the definition of N, we have selected a specific map m(N) on the sphere which requires no fewer than N colours to be properly coloured. Based on the definition of the map m(N) we have selected a proper colouring of its regions using the N colours 0, 1, ..., N1. Working with the fixed number N, the fixed map m(N), and the fixed proper colouring of the regions of the map m(N), lemma 23 has explicitly constructed a Steiner system S(N+1, 2N, 6N). Now lemma 3 implies that N cannot exceed four. ☐ 

