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Solutions to Miscellaneous Problems MT5821 Advanced Combinatorics Solutions to Miscellaneous Problems 1. This exercise outlines a counting proof of the formula n x(x − 1)(x − 2)···(x − n + 1) = ∑ s(n;k)xk k=1 for the generating function of the Stirling numbers of the first kind. (a) Show that it is equivalent to the formula n x(x + 1)(x + 2)···(x + n − 1) = ∑ u(n;k)xk; k=1 where the u(n;k) are the unsigned Stirling numbers of the first kind. (b) Suppose that x is a positive integer, and consider the set of xn functions from f1;:::;ng to a set X of size n. Let the symmetric g g−1 group Sn act on these functions, by the rule that f (i) = f (i ) for i 2 f1;:::;ng. Show that the orbits of Sn correspond to se- lections of n things from a set of size x, with order unimportant and repetitions allowed, and hence that the number of orbits is x + n − 1 . n (c) Show that, if g is a permutation with k cycles, then a function is fixed by g if and only if it is constant on the cycles of g. Deduce that there are xk functions fixed by g. (d) Since u(n;k) is the number of permutations with k cycles, use the Orbit-Counting Lemma to conclude that the formula in part (a) holds. (e) Since it is true for all positive integers x, it is a polynomial iden- tity. 1 (a) Replace x by −x and multiply by (−1)n. Then the coefficient of xk is multiplied by (−1)n−k. (b) An orbit of Sn on functions of the type described is determined once we know how how many times each element of x occurs as a value of the function. (If two functions represent each element of x the same number of times then a permutation of the domain carries one to the other.) Then the correspondence with selections is clear, and the formula was proved in Notes 2, Proposition 2.3. (c) If f is fixed by g, then for any i 2 f1;:::;ng, we have f (i) = f (ig−1) = f (ig−2) = ···; so f is constant on the cycle of g−1 containing i (which, as a set, is the same as the cycle of g). (d) If c(g) is the number of cycles of g, then fix(g) = xc(g), and so x + n − 1 1 n = xc(g) = u(n;k)xk; n n! ∑ ∑ g2Sn k=1 since there are u(n;k) permutations with k cycles. Now multiply by n! to get the result. (e) Quite so. T uT T bb "" '$"b T u" b u " bT " u bT 2. Consider the Fano planeushown&% u above. u (a) What is meant by an automorphism of the Fano plane? (b) Show that the automorphisms of the Fano plane form a group. What is its order? (c) Show that any automorphism fixes equally many points and lines. (d) Identify the points of the Fano plane with the non-zero vectors in a 3-dimensional vector space V over GF(2), so that each line is the set of non-zero vectors in a 2-dimensional subspace of V. 2 Show that any invertible linear transformation induces an auto- morphism of the Fano plane, and so that the group GL(3;2) of all such transformations is a subgroup of the group of automor- phisms of the Fano plane. Is it the full automorphism group? (e) Consider the set D = f1;2;4g ⊆ Z=(7). Show that it is a perfect difference set, in the sense that any non-zero element of Z=(7) has a unique representation in the form x−y for x;y 2 D. Deduce that the translates of D (the sets D + z for z 2 Z=(7)) are the blocks of a structure isomorphic to the Fano plane. Use this to find an element of order 7 in the automorphism group of the Fano plane. (a) An automorphism is a permutation of the points which maps lines to lines. (b) It is clear that the identity is an automorphism, and the composition of two automorphisms is an automorphism. Since the number of lines is finite, the image of the line set under an automorphism consists of all the lines, and so the inverse map also takes lines to lines. It is easy to see that, given any three non-collinear points in a Fano plane, the plane can be completed in a unique way; the uniqueness implies that any map from a non-collinear triple in a Fano plane to another can be extended in a unique way to an isomorphism. So the number of automorphisms is equal to the number of non-collinear triples, which is 7 · 6 · 4 = 168. (c) It is possible to solve this exercise by listing all automorphisms and check- ing; this is not a good solution! Here is a better way. Let g be an automorphism. Count pairs (P;L), where P is a point, L a line, and P;Pg 2 L. If g fixes P, there are three choices of L; otherwise, just one. So if x is the number of fixed points, the number of such pairs is 3x + (7 − x) = 2x + 7. However, the pairs can also be characterised another way, as those for which P 2 L and P 2 Lg−1; by a dual argument, if y is the number of fixed lines of g−1 (which is the same as the number for g), then the number of pairs is 2y + 7. So x = y. Finally here is a linear algebra method of much wider applicability. The per- mutations of points and lines induced by g can be represented by matrices MP and ML, say. If A is the incidence matrix of the Fano plane (with rows and columns indexed by points and lines, with (P;L) entry 1 if P 2 L, 0 otherwise), then one −1 can check that MPA(ML) = A. One can further check that A is invertible (show that AA> = 2I + J, where J is the all-1 matrix), and hence −1 A MPA = ML: 3 Thus the traces of MP and ML are equal. But the trace of MP is the number of points fixed by g, and similarly for ML and lines. 001 T uT T 011 bb "" 101 '$"b111T u" b u " bT " u bT 010 110 100 u&% u u (d) In this picture, 100 means the 1-dimensional subspace spanned by the vec- tor (1;0;0), etc. The picture shows that the lines correspond to 2-dimensional subspaces as in the question. Alternatively, show that the 1- and 2-dimensional subspaces of V form a configuration of seven points with three points on each line and two points together on a unique line, and deduce the result from the unique- ness of the Fano plane. Now any invertible linear map takes subspaces to subspaces, and preserves dimension and inclusion, so induces an automorphism of the Fano plane. Thus GL(3;2) is a subgroup of the automorphism group. But its order is easily seen to be 7 · 6 · 4 = 168, so it is the full automorphism group. (e) The first assertion is verified by a table: − 1 2 4 1 0 6 4 2 1 0 5 4 3 2 0 Now given distinct u;v 2 Z=(7), there are unique x;y 2 D such that u − v = x − y; then D + (u − x) = D + (v − y) is the unique translate of D containing u and v. So we have a Fano plane. Now it is possible to label our Fano plane diagram above to get the correct lines: 1 T uT T 4 bb "" 0 '$"b6 T u" b u " bT " u bT 2 5 3 u&% u u With this labelling, the shift permutation (0;1;2;3;4;5;6) is an automorphism. 4.
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