MT5821 Advanced Combinatorics Problems 4 4.1. This question is Enigma 1124 from New Scientist. It can be solved by a few applications of the Orbit-counting Lemma, or more easily by a technique we’ll see later, the Cycle Index Theorem. A stained glass window consists of nine squares of glass in a 3 × 3 array. Of the nine squares, k are red, the rest blue. A set of windows is produced such that any possible window can be formed in just one way by rotating and/or turning over one of the windows in the set. Altogether there are more than 100 red squares in the set. Find k. 4.2. A permutation group G on a set X is called transitive if it has only one orbit on X. Show that, if jXj > 1 and G is transitive on X, then G contains a derangement (a permutation with no fixed points). [What is the average number of fixed points of elements of G?] 4.3. Here is a strengthening of the result in the preceding question. We will prove that, if G is transitive on X, with jXj = n > 1, then the number of derangements in G is at least jGj=n. • Show that ∑ fix(g) = jGj. [Easy.] g2G • Consider G acting as a permutation group on X2, the set of all ordered pairs of elements of X. Show that it has at least two orbits, and deduce that ∑ fix(g)2 ≥ 2jGj. g2G • Hence show that ∑ (fix(g) − 1)(fix(g) − n) ≥ jGj: g2G • The summands in this expression are all ≤ 0 except for those with fix(g) = 0. Deduce that there must be at least jGj=n of these. 1 4.4. Some problems on the Petersen graph. The diameter of a graph is the maximum number of steps in a shortest path between any two vertices; the girth is the number of edges in a shortest cycle; and we say that a graph has valency at least (resp., at most) k if every vertex is incident with at least (resp., at most) k edges. • Show that the group of all automorphisms has order 120 (and hence is the group S5 given earlier). • Show that a graph with valency (at most) 3 and diameter (at most) 2 has at most ten vertices, with equality if and only if it is isomorphic to the Petersen graph. • Show that a graph with valency (at least) 3 and girth (at least) 5 has at least ten vertices, with equality if and only if it is isomorphic to the Petersen graph. • Show that it is impossible to colour the edges of the Petersen graph with three colours in such a way that no two edges with the same colour meet at a vertex. 4.5. • Verify the formula given for the chromatic polynomial of the Petersen graph. • Verify the graphs X=g given for the Petersen graph and construct their chro- matic polynomials. 4.6. Show that, if X is the null graph on n vertices (the graph with no edges) and G is the symmetric group Sn, then q + n − 1 OP (q) = : X;G n [You might like to match the colourings up with samples, under appropriate rules, of n things from a set of q.] 2.