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THE FRIENDSHIP THEOREM
Craig Huneke
The ‘friendship theorem’ can be stated as follows ([1, p. 183]):
Suppose in a group of at least three people we have the situation that any pair of persons
have precisely one common friend. Then there is always a person who is everybody’s friend.
The first proof of this theorem was due to Paul Erd˝os, Alfred R´enyi, and Vera S´os [3].
Translating the theorem into graph theory yields the following theorem:
Theorem. If Gis a graph in which any two distinct vertices have exactly one common
neighbor, then Ghas a vertex joined to all others.
As a consequence, such graphs are completely determined; they consist of edge-disjoint
triangles around a common vertex. The best known and simplest proof is based on com-
puting the eigenvalues (and their multiplicities) of the square of the adjacency matrix of
the graph. Such a proof is given in [1]. In [6], a similar idea is used, but phrased in terms
of showing the graph has the structure of a ‘projective plane’. (See also [2].) In his recent
book [5 p. 466], West writes, “It is startling that such a combinatorial-sounding result
seems to have no short combinatorial proof. There do exist proofs avoiding eigenvalues (see
Hammersley, [4]), but they require complicated numerical arguments to eliminate regular
graphs.” The goal of this note is to provide one proof which is more combinatorial, and
another proof which does not explicitly use eigenvalues (though it does use traces), and in
some sense combines the combinatorics with the linear algebra.
I first heard of this problem from William Lang as a graduate student in 1975; he chal-
lenged me to solve it. I used the idea of counting walks of length p, where pis a carefully
chosen prime number, to construct a proof that same year. I recently showed this proof to
my colleague Fred Galvin, who then significantly simplified it; this resulted in our first proof
following. The first three paragraphs of the proof are standard; they reduce the problem to
a regular graph. However, we make this reduction in a slightly different way than in the
references to this paper.
Proof of Theorem. We first claim that if x, y ∈Gand are not adjacent, then they have
the same degree (i.e., the same number of adjacent vertices). Let N(x) denote the vertices
adjacent to x(the ‘neighborhood’ of x). Define a map α:N(x)→N(y) by sending a vertex
z∈N(x) to the common neighbor of zand y. This common neighbor cannot be x, as x
and yare not adjacent. The map is one-to-one since z∈N(x) is uniquely determined as
the common neighbor of xand α(z). By symmetry the map is onto.
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2 CRAIG HUNEKE
Suppose there is a vertex of degree k > 1. We claim that all vertices have degree k, unless
there is a universal friend. Let Abe the set of all vertices of degree k, let Bbe the set of all
vertices of degree different from k, and assume for a contradiction that Bis nonempty. By
the first claim, every vertex in Ais adjacent to every vertex in B. If Aor Bis a singleton,
then that singleton is a universal friend; otherwise, there are two different vertices in A, and
they have two common neighbors in B, contradicting the hypothesis. It follows that Gis
k-regular, i.e., the degree of every vertex is k.
Next we claim that the number nof vertices in Gis exactly k(k−1) + 1. This follows by
counting the paths of length two in G: by assumption there are n
2such paths. For each
vertex v, there are exactly k
2paths of length two having vin the middle, giving n·k
2total
paths of length two. Equating both counts permits us to conclude that n=k(k−1) + 1.
We can assume that k≥3, else n= 3 and the theorem is clear.
Awalk of length non Gis an ordered sequence v0v1...vnof vertices such that viand vi+1
are neighbors. We say the walk is closed if vn=v0. A closed walk is considered to have a
starting point and an orientation, always returning to the starting vertex; thus, if uvwu is
a closed walk in G, then uvwu,vwuv,vuwv, etc. are considered distinct closed walks. It
follows that, if pis a prime number, then the number of closed walks of length pis divisible
by p.
For a fixed vertex v, let f(n) be the number of walks from vto vof length n. If n > 1,
then the number of closed n-walks v0...vn−2vn−1vnfrom v=v0=vnwith vn−2=vis
kf (n−2), and the number of such walks with vn−26=vis kn−2−f(n−2). The total
number of walks v0v1....vn−2from a fixed vertex v=v0is kn−2as Gis k-regular. Thus
f(n)=(k−1)f(n−2) + kn−2. Let pbe a prime divisor of k−1; then f(p)≡1 (mod
p). Finally, the total number of closed walks of length pis [k(k−1) + 1]f(p)≡1 (mod p),
contradicting the fact that the number of such walks is divisible by p.
Although the preceding proof requires no linear algebra, one can compare this proof to
the proof of [1], which uses eigenvalues. Thinking about it gives another simple proof,
provided one knows a little linear algebra in characteristic p.
Second Proof of the Friendship Theorem. We again reduce to the case in which the graph is
k-regular, i.e., each vertex has exactly kadjacent vertices and the total number of vertices
is n=k(k−1) + 1, with k≥3. Let G={v1, ..., vn}. We let Abe the adjacency matrix
of G, whose (i, j) entry is 1 if viand vjare adjacent, and 0 otherwise. The matrix Ahas
zeroes on the diagonal, so the trace of Ais 0. We let Bbe the nby nmatrix having a 1 in
every entry. The trace of Bis n.
By assumption and the fact that Gis k-regular, A2= (k−1)I+B, and AB =kB,
where Iis the identity matrix of size nby n. We now pass to the field Zp, where pis a
prime dividing k−1. We continue to call the matrices Aand B, though we now think of
them with entries in Zp. Observe that both nand kare now equal to 1. Hence A2=B,
and furthermore AB =kB =B. It follows that for all l≥2, Al=B. Let trC denote the
trace of a square matrix C. In characteristic p,trAp= (trA)p. We reach a contradiction:
1 = n=trB =trAp= (trA)p= 0.
The relation between the first proof and the second is simply that the trace of Apcounts
THE FRIENDSHIP THEOREM 3
the closed walks of length pin the graph. The relationship between the second proof and
the usual proof is clear: in characteristic 0, one computes the eigenvalues of A2and then
proves that Acould not have trace 0. The second proof takes advantage of the fact that
(a+b)p=ap+bpin characteristic p. This allows us to avoid the actual computation of the
eigenvalues, to push the calculation of the trace out to the pth power.
Bibliography
[1] M. Aigner and G. Ziegler, Proofs from the Book, Springer-Verlag, Berlin, 1999.
[2] J. Brunat, Una demostraci´o del teorema de l’amistat per m`etodes elementals, Butllet´
i Societat
Catalana de Matem`atiques 7(1992), pp. 75–80.
[3] P. Erd˝os, A. R´enyi, and V. S´os, On a problem of graph theory, Studia Sci. Math. 1(1966), pp.
215–235.
[4] J. Hammersley, The friendship theorem and the love problem, in Surveys in Combinatorics, London
Math. Soc. Lec. Notes 82 (1983), Cambridge Univ. Press, Cambridge, pp. 31–54.
[5] D. B. West, Introduction to Graph Theory, 2nd edition, Prentice Hall, Upper Saddle River, NJ,
2001.
[6] H. Wilf, The friendship theorem, Combinatorial Mathematics and its Applications, Proc. Conf.
Oxford, 1969 (1971), Academic Press, London and New York, pp. 307–309.
University Of Kansas, Lawrence, KS 66045, USA
E-mail address:huneke@math.ukans.edu
