Chapter 3 Coset Graphs We have seen that every transitive action is equivalent to a coset action, and now we will show that every vertex-transitive graph or digraph can be represented as a so called coset graph. 3.1 Definition and properties For a group G and a subgroup H G, denoteby[g:h] the set of right cosets of H in G, thatis [G:H]={Hx x G}. For any subset S G, we may define a di-graph on [G : H] as follows: Definition 3.1 Let G be a group, H a subgroup of G, ands a subset of G. Define the coset graph of G with respect to H and S to be the directed graph with vertex set [G : H] and such that, for any Hx,Hy V, Hx is connected to Hy if and only if yx 1 HSH, and denote the di-graph by Γ(G, H, HSH). Example 3.2 Let G = a, z a n = z 2 =1, z 1 az = a 1 = D2n, a dihedral group of order 2n where n 2 is a positive integer. Then each element of G may be written as a i or a i z for i =0,1,...,n 1. Let H = z,andlets={a}.setγ=γ(g, H, HSH). 25
26 Coset Graphs Then the vertex set of Γ equals [G : H] ={ z, z a,..., z a n 1 }, and the vertex z a i is connected to the vertex z a j if and only if a j i = a j (a i ) 1 HSH = z a z. There are two possibilities: a j i = a or zaz. Further, a j i = a if and only if j i = 1; while, as zaz = a 1,wehavea j i =zaz if and only if j i = 1. It follows that the neighborhood of the vertex H is {Ha,Haz}. Therefore, the graph Γ is an undirected cycle of size n. Exercise 3.3 Let G = a, z a n = z 2 =1,z 1 az = a 1 = D2n,andletH= z. Let S 1 = {a}, S 2 = {a 1 }, S 3 = {a, a 1 },ands 4 ={a, a 1,z}. Prove that Γ(G, H, HS i H), where i {1,2,3}, are all the same graph, but they are not isomorphic to Γ(G, H, HS 4 H). Observations: (i) In the case H = 1, the coset graph Γ(G, H, HSH) is the Cayley graph Cay(G, S,). So coset graphs are a generalisation of Cayley graphs. (ii) In the coset graph Γ(G, H, HSH), the neighborhood of the vertex H is equal to {Hsh s S, h H}, which is a partition of HSH. The following lemma collects some basic properties of coset graphs. Lemma 3.4 Let Γ=Γ(G, H, HSH) with vertex set V =[G:H].Then (1) Γ is vertex-transitive; (2) Γ is undirected if and only if HSH = HS 1 H; Proof. (1). For each g G and each Hx V, define (Hx) g = Hxg V, (right multiplication). Then, for any two vertices Hx,Hy V,wehavethat Hx is connected to Hy yx 1 HSH (yg)(xg) 1 HSH Hxg is connected to Hyg (Hx) g is connected to (Hy) g.
Groups & Graphs 27 Thus, g is an automorphism of Γ. Since the group G acts transitively on V by right multiplication (coset action), AutΓ is transitive on V and Γ is vertex-transitive. (2). Suppose that Γ is undirected. For any two vertices Hx,Hy V,wehave yx 1 HSH Hx is connected to Hy Hy is connected to Hx xy 1 HSH yx 1 (HSH) 1 = HS 1 H It follows that HSH = HS 1 H. Suppose now that HSH = HS 1 H. For any two vertices Hx,Hy V,wehave Hx is connected to Hy yx 1 HSH xy 1 (HSH) 1 = HS 1 H = HSH Hy is connected to Hx. 2 The next lemma enables us to exclude some trivial cases. Lemma 3.5 Let G be a group, and let Γ=Γ(G, H, HSH) with vertex V.Then (1) G acts faithfully on V =[G:H] if and only if H is core-free in G; (2) Γ contains no self-loops if and only if H S =. Proof. (1). Let C = Core G (H),thecoreofH in G, that is the largest normal subgroup of G contained in H.LetKbe the kernel of G acting on V, that is, any element of K fixes every vertices of Γ. Then K is a normal subgroup of G. For any k K,wehaveHk = H k = H. Hence k H,sothatK H. By definition, K C. On the other hand, for any c C and any Hx V,wehave (Hx) c = Hxcx 1 x = Hc x = Hx, sothatcfixes Hx. Thus C K,andsoC=K. Now G is faithful on V K =1 C =1 H is core-free in G.
28 Coset Graphs (2). Since Γ is vertex-transitive, Γ contains no self-loop if and only if there is no self-loop at the vertex H, that is, the vertex H is not connected to H.Now the vertex H is connected to itself 1 HSH 1=h 1 sh 2 for some h 1,h 2 H and some s S s = h 1 1 h 1 2 H,sothats S H. Therefore, we will assume that, for a coset graph Γ = Γ(G, H, HSH), H is core-free in G, and H S=. Under this assumption, Γ is a simple di-graph, and G is a subgroup of Aut Γ. Example 3.6 Let G = A 4,actingonΩ={1,2,3,4}. Write a = (12)(34), b = (23)(14), c = (13)(24), and z = (123). Then ab = ba = c, bc = cb = a, andca = ac = b; z 1 az = b, z 1 bz = c, and z 1 cz = a. Let H = a, and S = {z}. Set Γ=Γ(G, H, HSH). Then V Γ=[G:H]={H, Hz, Hz 2,Hb,Hbz,Hbz 2 },andadjacency relation is as follows: H is connected to Hz and Hbz; Hz is connected to Hz 2 and Hbz 2 ; Hz 2 is connected to H and Hb; Hb is connected to Hz and Hbz; Hbz is connected to Hz 2 and Hbz 2 ; Hbz 2 is connected to H and Hb. Then Γ is a directed graph of out-valency 2 and degree 6. Exercise 3.7 Using notation given in Example 3.6, let T = {z,z 1 }. Prove that Γ(G, H, HT H) is undirected and G-edge-transitive. 2
Groups & Graphs 29 3.2 Coset graph representations If Γ is a di-graph and G is a vertex-transitive group of automorphisms of Γ, then each vertex of Γ can be thought of as a right coset of G α in G for some vertex α. One can naturally ask the question: is Γ isomorphic to a coset graph? The affirmative answer is provided by the following theorem. Theorem 3.8 Let Γ=(V,E) be a di-graph and G a vertex transitive group of automorphisms. Then there exist a subgroup H and a subset S of G, such that Γ is isomorphic to the coset graph Γ(G, H, HSH). Proof. Let α be an arbitrary vertex of Γ and let H = G α, the stabiliser of α in G, S = {s G α s Γ(α)}, where Γ(α) is the neighborhood of α, that is the set of vertices to which α is conected. Let Σ=Γ(G, H, HSH). We claim that Γ = Σ. Write V = {α 0 = α, α 1,...,α n }.SinceGis transitive on V,foreachα i, there exists some element g i G such that α g i = α i. Further, it is easy to show that an element g G maps α to α i if and only if g Hg i, that is, Hg i = {g G α g = α i }. Define a map ϕ from V to [G : H]: ϕ : α i Hg i. We claim that ϕ is an isomorphism from Γ to Σ. First we show that HSH = {g G α g Γ(α)}. (3.1) Suppose that g HSH,thatis,g=h 1 sh 2 for some h 1,h 2 Hand some s S.Then β:= α s Γ(α), and as h 1 fixes α 0 and h 2 permutes Γ(α), we obtain α g = α h 1sh 2 = α sh 2 = β h 2 Γ(α).
30 Coset Graphs Hence HSH {g G α g Γ(α)}. Now suppose that g is such that α g Γ(α), that is, α g = α i = α s i for some s i S. Then α gs 1 i = α, andsogs 1 i G α = H. Therefore, g Hs i HSH, and thus {g G α g Γ(α)} HSH, so that HSH = {g G α g Γ(α)}. For any vertices α i,α j V,wehave α i is connected to α j α = α g i 1 i g j gi 1 is connected to α g 1 i j = α g jg 1 i 0 HSH (Hg i,hg j )isanarcinγ(g, H, HSH). Thus, noting that α ϕ i = Hg i and α ϕ j = Hg j, α i is connected in Γ to α j if and only if α ϕ i is connected in Γ(G, H, HSH) to α ϕ j,sothatϕisan isomorphism from Γ to Γ(G, H, HSH). 2 We now study an example. For a graph Γ with vertex set V and edge set E,the complement of Γ is the graph with vertex V such that vertices α, β are adjacent if and only if they are not adjacent in Γ, denoted by Γ. It is easily shown that Aut Γ = Aut Γ. Example 3.9 Let Γ be the complement of the Petersen graph. Identify vertices of Γ with 2-subsets of Ω = {1, 2, 3, 4, 5}, andchooseα={1,2}. Then neighbors of α in Γ are {1, 3}, {1, 4}, {1, 5}, {2, 3}, {2, 4}, {2, 5}. Let G be the subgroup of Aut Γ such that G = A 5,andletH be the stabilizer of α in A 5.ThenH= (345), (12)(45) = D 6.Let S={(23)(45), (24)(35), (25)(34), (13)(45), (14)(35), (15)(34)}. It is easy to see that for each neighbor β of α, there exists g S such that α g = β. Thus Γ = Γ(G, H, HSH). Exercise 3.10 Using notation given in Example 3.9, prove that Γ has valency 6, and for any element g S,wehave Γ=Γ(G, H, HgH).
Groups & Graphs 31 3.3 Edge-transitive graphs and digraphs In the following we examine what conditions the input G, H,andS has to satisfy, so that the coset graph has some prescribed properties. Theorem 3.11 Let G be a group, and let Γ=Γ(G, H, HSH) for some subgroup H and some subset S of G. ThenΓis G-edge-transitive if and only if HSH = HsH for some s G; further, if Γ is G-edge-transitive, then each g HSH is such that HSH = HgH. In particular, each G-orbital can be written as Γ(G, H, HgH) for some element G G. Proof. Denote by α the vertex of Γ corresponding to H. Then the neighborhood of α is {Hsh s S, h H}. Take an arbitrary vertex β Γ(α). Then β = Hs 0 for some s 0 S. Recall that, as G is transitive on the vertex set V,thatGis transitive on the edge set of Γ if and only if G α is transitive on Γ(α), that is, Γ(α)=β G α.now G α is transitive on Γ(α) H acts on {Hsh s S, h H} transitively HSH = {Hsh s S, h H} =(Hs 0 ) H = Hs 0 H. 2 To obtain such a result for undirected graphs, we first prove a lemma. Lemma 3.12 Let G be a group, and let Γ=Γ(G, H, HsH) for some element s G. Then Σ:=Γ(G, H, Hs 1 H) is the paired orbital graph of Γ. Further, Γ(G, H, H{s, s 1 }H) is undirected and G-edge-transitive. Proof. By definition, for any Hx,Hy [G : H], Hx is connected in Γ to Hy yx 1 HsH xy 1 Hs 1 H Hy is connected in Σ to Hx. It is now easily shown that Γ(G, H, H{s, s 1 }H) isg-edge-transitive. 2 The next theorem describes vertex- and edge-transitive graphs in terms of coset graphs. Theorem 3.13 Let G be a group, and let Γ=Γ(G, H, HSH) for some subgroup H and some subset S of G. ThenΓ is undirected and G-edge-transitive if and only if HSH = H{s, s 1 }H for some s G.
32 Coset Graphs Proof. It is easily shown that Γ is G-edge transitive if and only if Γ is an edge-disjoint union of two paired G-orbitals, hence G is transitive on EΓ E Γ= for some G-orbital Γ(α) = (α) (α) HSH =(Hs 0 ) H (Hs 1 0 ) H = H{s 0,s 1 0 }H. 2 3.3.1 s-arc transitive graphs For a graph Γ = (V,E), a sequence α 0,α 1,...,α s of vertices in V is called an s-arc if α i 1 is adjacent to α i and α i 1 α i+1. A graph Γ is called a (G, s)-arc transitive graph if G Aut Γ is transitive on V andonthesetofs-arcs of Γ. Clearly, (G, 1)-arc transitive graphs are exactly G-arc transitive; (G, s)-arc transitive graphs are (G, s 1)- arc transitive. Theorem 3.14 Let G be a group, and let Γ=Γ(G, H, HSH). Then (i) Γ is undirected and G-arc-transitive if and only if HSH = HgH for some g G such that g 2 H ; (ii) Γ is (G, 2)-arc transitive if and only if HSH = HgH for some g G such that g 2 H and H is 2-transitive on [H : H H g ]. Moreover, Γ(G, H, HgH) has valency H / H H g. Proof. (i) Suppose that Γ = (V,E) is undirected and G-arc-transitive. Let α, β V be adjacent. Then there exists g G which interchanges α and β. Choose α = H,so that β = Hg. By Theorem 3.13, HSH = HgH. NowHg 2 =(Hg) g = β g = α = H,and so g 2 H. Conversely, suppose that HSH = HgH and g 2 H. By Theorem 3.11, Γ is G-arc-transitive. Further, g 2 = h for some H H, and hence g = hg 1,sothat HgH = Hg 1 H. Thus Γ is undirected.
Groups & Graphs 33 (ii). Let α be the vertex corresponding to H,and β the vertex corresponding to Hg. Recall that the neighborhood of the vertex α is Γ(α) ={Hgh h H}. SinceΓisG-arc transitive, H is transitive on {Hgh h H}. NowHg {Hgh h H}, and hence the H -action on {Hgh h H} is equivalent to the coset action of H on [H : H β ]. Since G β = G g α = G g α,wehavethath β =H G β =H G g α=h H g.thus Γis(G, 2)-arc transitive G α is 2-transitive on Γ(α) H is 2-transitive on {Hgh h H} H is 2-transitive on [H : H H g ], as required. 2