Basic Group Theory. labeled trees with N i vertices. More precisely, there are n!/n i labeled trees isomorphic to T i.

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1 Basic Group Theory Let us consider the problem of counting all labeled trees of size n By Cayley s theorem, the number of such trees is n n, but suppose we didn t know this We could still efficiently count the labeled trees by means of the following result If Γ is a (simple) graph, by an automorphism of Γ we mean a map f : Γ Γ, or more precisely on the set of vertices, such that if x, y Γ then x and y are adjacent if and only if f(x) and f(y) are adjacent For example, the following graph has two automorphisms: A B C D The first automorphism is the identity map 1 Γ The other one is the map that interchanges A and D and also B and C Proposition 1 Let T 1,, T k be a complete set of nonisomorphic trees on n vertices Let N i be the number of automorphisms of T i Then there are ( 1 n! ) N 1 N N k labeled trees with N i vertices More precisely, there are n!/n i labeled trees isomorphic to T i We will prove this statement as an application of group theory But let is illustrate it with a couple of examples If n = 4, there are two nonisomorphic trees with 4 vertices, so we may take T 1 and T to be the following trees T i N i 6 1

2 So we expect 4!/ = 1 labeled trees isomorphic to T 1, and indeed, these are: (We re not listing 431 since it is the same as 134, etc) Similarly there are 4!/6 = 4 labeled trees isomorphic to T, or = 16 labeled trees altogether, as predicted by Cayley s theorem In order to prove Proposition 1, we need a few easy concepts from group theory A group is a set G with a unit element 1 = 1 G and a multiplication, which is a map G G G The image of (g, h) under the multiplication is denoted as gh or g h As part of the definition of a group, it is assumed that the associative law is satisfied, so g (hk) = (gh) k, and also that 1 g = g 1 = g Finally G has inverses, so if g G there is an element g 1 with gg 1 = g 1 g = 1 One thing that is not assumed is gh = hg If this is true, the group is called abelian Lemma 1 The identity element of G is unique The inverse g 1 is unique Finally, we have (gh) 1 = h 1 g 1 Proof If 1 is another identity element, that is, if 1 x = x 1 = x, then 1 = 1 1 = 1 If g 1 is another inverse of g, that is, if gḡ 1 = 1, then g 1 = g 1 1 = g 1 gḡ 1 = 1 ḡ 1 = ḡ 1 Finally, we have (gh)h 1 g 1 = g 1 g 1 = 1, so h 1 g 1 is another inverse of gh, that is, (gh) 1 Another useful concept is that of a group action Suppose we have a group G and a set X together with a multiplication G X X As with the multiplication in G, we denote this map as (g, x) g x or gx It is assumed that if x X then 1 x = x, and if g, h G, then g(hx) = (gh)x

3 For example, let Γ be a graph Let G be the set of automorphisms of Γ Then G is a group, where the group law is composition of mappings: if f, g G, then fg is the automorphism f g, that is, (fg)(x) = f(g(x)) for x X It is clear that the composition of two automorphisms is an automorphism, and the inverse of an automorphism is an automorphism, and that G is indeed a group It acts on Γ, if we consider gx = g(x) to be a multiplication G Γ Γ If G is a group acting on a set X, we call the action transitive if for every u, v X there exists a g G such that gu = v For example, consider the following graph w x z y (1) The action is transitive, since there is an automorphism τ defined by τ(x) = w, τ(w) = z, τ(z) = y and τ(y) = x If u and v are any pair of vertices, some power of t then takes u to v, proving transitivity On the other hand, the automorphism group of the tree is not transitive on the set of vertices since no automorphism interchanges 1 and However this automorphism group induces an action on the subset {, 3, 4} and that action is transitive As another example, let S n be the symmetric group on n letters, that is, the group of permutations (bijections) of the set X n = {1,, 3,, n} The group S n acts on X n, namely if σ S n and x X, then we write σx for σ (x) and we think of (σ, x) σx as a group action This action is obviously transitive Let us introduce a couple of common notations for permutations The permutation σ of {1,, 3, 4} such that σ (1) = 3, σ () = 1, σ (3) = 4 and σ (4) = is denoted ( ) 3

4 Thus we write x above σ (x) The same permutation is denoted more succinctly in cycle notation as (134) The permutation ( 1 3 ) is written in cycle notation as (13) (4) Here (13) means that 1 3 and 3 1, while (4) means 4 and 4, which is what this permutation does If G acts on a set X, and if x X, let G x = {g Ggx = x} It is easy to see that G x is a group, the isotropy subgroup or stabilizer of x For example, in the graph (1), the isotropy subgroup of x has two elements, namely the identity automorphism, and an automorphism θ such that θ(x) = x, θ(w) = y, θ(y) = w and θ(z) = z Clearly any automorphism g of this graph such that gx = x must be either θ or the identity, so G x = {1, θ} The following statement is very powerful We will denote by X the cardinality of a set X, that is (if it is finite) the number of elements (The notion of cardinality extends to infinite sets, but for our purposes you may assume that G and X are finite in the following Proposition) Proposition Let G be a group acting transitively on a set X Let x X, and let G x be the isotropy subgroup of x Then G = ΓG x As an example, let Γ be the graph above with four elements We showed that G acts transitively on Γ and that G x has order This proves that the group of automorphisms of G has order 4 = 8 Proof If y Γ let φ y G such that φ y x = y Such an automorphism exists because G acts transitively It may not be unique, but fix such a φ y We will show that every element g of G has the form φ y θ for a unique y Γ and θ G x To prove that such y Γ and θ G x exist, let y = gx and let θ = φ 1 y g Clearly φ y θ = g We need to check that θ G x Indeed, θx = φ 1 y gx = φ 1 y y = x since φ y x = y This proves the existence of y and θ To prove uniqueness, suppose that φ y θ = g with y Γ and θ G x We must show y = y and θ = θ Since θ G x we have y = gx = φ y θ x = φ y x = y, and g = φ 1 y g = θ We see that the map (y, θ) φ y θ is a bijection from Γ G x to G, and so G = Γ G x then θ = φ 1 y Proposition 3 Let T be a tree with n vertices Then the number of labeled trees isomorphic to T equals n!/n, where N is the order of the automorphism group of T 4

5 Proof Let Γ be the set of all labeled trees isomorphic to T The labels are thus 1,,, n Let us fix one labeling, which we call x For example, if T is the graph: we can choose the labeling x = Denote this labeled tree as x Now let G be the group S n This acts on the labeled trees isomorphic to T For example, the permutation (4) which sends 1 1, 4, 3 3 and 4 sends the labeled tree x to this one: The action is clearly transitive Since S n = n! it follows from Proposition that n! = N G x The proposition is proved if we check that the group G x is isomorphic to the group of automorphisms of T But this is obvious For example the permutation (3) turns x into the labeled tree 1 4, 3 corresponding to the automorphism that switches the right and bottom nodes of T Summing over all isomorphism types of trees, it is clear that Proposition 3 implies Proposition 1 5