Group Theory Problems

Transcription

1 Grou Theory Problems Ali Nesin 1 October 1999 Throughout the exercises G is a grou. We let Z i = Z i (G) and Z = Z(G). Let H and K be two subgrous of finite index of G. Show that H K has also finite index in G. Show that H has finitely many conjugates in G. Conclude that if a grou has a subgrou of finite index, then it has a normal subgrou of finite index. (See also Exercise 49). 1. (P. Hall) a. Show that for x, y, z G, [x, yz] = [x, z][x, y] z and [xy, z] = [x, z] y [y, z]. Conclude that if H, K G, then H and K normalize the subgrou [H, K]. Conclude also that if A G is an abelian subgrou and if g N G (A), then ad(g) : A A is a grou homomorhism whose kernel is C A (g). b. Let x, y, z be three elements of G. Show that [[x, y 1 ], z] y [[y, z 1 ], x] z [[z, x 1 ], y] x = 1. Conclude that if H and K are two subgrous of a grou G and if [[H, K], K] = 1, then [H, K'] = 1. c. (Three Subgrou Lemma of P. Hall) Let H, K, L be three normal subgrous of G. Using art b, show that [[H, K], L] [[K, L], H][[L, H], K]. d. Conclude from art (c) that [G i, G j ] G i+j+1, G (i) i G 2, [G i, Z j ] Z j i 1, [Z i+1, G i ] = 1. e. Show that a nilotent grou is solvable. Show that the converse of this statement is false. 2. Let A G be an abelian subgrou and let g G. By Exercise 1.a, ad(g): A A is a grou homomorhism. Assume that [G', A] = 1. Show the following: a. C A (g) G. b. [g, A] = ad(g)(a). c. [g, A] G. 3a. Let G be nilotent of class n. Show that G n i Z i. Conclude that G = Z n. b. Conversely, assume that G = Z n. Show that G i Z n i. Conclude that G is nilotent of class n. c. Show that G is nilotent of class n if and only if Z n = G and Z n 1 G. 4. Let H G and K, L G. a. Show that [KH/H, LH/H] = [K, L]H/H. 1 From Borovik-Nesin, Grous of Finite Morley Rank, chater 1.

2 b. Conclude that if G is solvable (res. nilotent), then so are H and G/H. c. Show that if G/H and H are solvable, then so is G. d. Find an examle where the revious result fails if we relace the word solvable by nilotent. e. Deduce from art c that if A and B are solvable subgrous of G and if one of them normalizes the other, then A, B = AB is also solvable. 5. Let X Z n be a normal subgrou of G. Show that G is nilotent if and only if G/X is. Let i be fixed integer. Show that G is nilotent of class n if and only if G/Z i is nilotent of class n i. Show that Z i is nilotent of class i. Find a (nilotent) grou where Z 2 Z and Z 2 is abelian. (See also Exercise 41). 6. Show that a nilotent grou G satisfies the normalizer condition (i.e. if H < G then H < N G (H)). 7. (Hirsch) Let G be a nilotent grou. Show that if G = HN ' for some H G, then H = G. 8. (Hirsch) Let G be a nilotent grou. Show that if 1 H G, then H Z Let A and B be two normal nilotent subgrous of G. Show that the subgrou A, B = AB is also normal and nilotent. 10. a. Show that the subgrou G n is generated by the elements of the form [x 1, [x 2,..., [x n, x n+1 ]...]], where x i G. Find a similar statement for G (n). b. Show that an abelian grou is locally finite if and only if it is a torsion grou. Conclude that a solvable grou is locally finite if and only if it is a torsion grou. c. Let be a rime. Show that a nilotent-by-finite -grou is solvable and hence locally finite. 11. Show that for x, y G and n a ositive integer, [x n, y] = n 1 n 2 x x [ x, y] [ x, y]...[ x, y]. 12a. Let g G and H G be such that [g, H] Z. Show that the ma ad(g): H Z is a grou homomorhism. Show that for all h H, n Z, [g, h] n = [g n, h] = [g, h n ]. b. Using Exercise 11, show that if z Z 2 and z n Z, then [z, G] is a central subgrou of finite exonent and that ex([z, G]) divides n. c. (Mal'cev, McLain) Use art b to rove, by induction on the nilotency class, that if a nilotent grou has an element of order where is a rime, then it has central elements of order. d. Let G be a nilotent grou and D a -divisible subgrou of G. Show that D commutes with all the -elements of G. Deduce that in a divisible nilotent grou, elements of finite order form a central subgrou Divisible Nilotent Grous. (Chernikov) Let be a rime and let G be a -divisible nilotent grou. a. Show that if g Z, then g Z.

3 b. Conclude that Z is -divisible, contains all the -elements and that G/Z is - torsion-free and -divisible. c. Show that G/Z i is -torsion-free for all i 1. d. Conclude that Z i+1 /Z i is -torsion-free and -divisible for i Let G be a nilotent grou. a. Let i 1 be an integer. Show that G/G i is -divisible if and only if G/G i+1 is - divisible. b. Conclude that G is -divisible if and only if G/G' is -divisible. c. Show that G has a unique maximal -divisible subgrou D. d. Assume that for some D G, D and G/D are -divisible. Show that G is - divisible. 15. (Dixmier). Let G be nilotent and assume that ex(g/g') = n. a. Show that ex(g i /G i+1 ) n for all i. b. Conclude that ex(g) n c where c is the nilotency class of G. 16. Let P be a Sylow -subgrou of G. Show that P is characteristic in N G (P). Conclude that N G (N G (P)) = N G (P). By Exercise 6, if G is nilotent, N G (P) = G, i.e. P G. Conclude that, for a given rime, a nilotent grou G has a unique Sylow - subgrou, and that if G is torsion, then G is the direct sum of its Sylow -subgrous. 17. Let t G be an involution. Let X = {[t, g]: g G} a. Show that for x X, x t = x 1 and that t X. Conclude that the elements of tx are involutions. b. Show that the ma ϕ : G/C G (t) X defined by ϕ(gc G (t)) = [t, g 1 ] is a welldefined bijection. c. Assume from now on that G is finite and that C G (t) = {1, t}. We will show that X is an abelian 2'-subgrou and G = X {1, t}. By art b, X = G /2. By art a and by assumtion, X has no involutions. Therefore X tx =. Conclude that G = X tx and that X is the set of elements of order 2 of G. Therefore, X is a characteristic subset of G. Let x X \ {1} be a fixed element. Conclude that t x inverts X as well (relace t by t x ). Conclude that 1 x 2 = tt x centralizes X. Therefore X = C G (x 2 ) G. Since t inverts X, X is an abelian grou without involutions. 18. Let G be a finite grou with an involutive automorhism α without nontrivial fixed oints. Show that G is inverted by α. 19a. Let G be a grou of rime exonent. Show that for g G *, no two distinct elements of g can be conjugated in G. b. Show that if ex(g) =, then G has at least conjugacy classes. c. (Reineke) Let G be a grou and assume that for some x G of finite order, we have G = x G {1}. Show that G = 1 or Let G be an arbitrary torsion grou without involutions. Note that G is 2- divisible (see Exercise 36). Assume G has an involutive automorhism α that does

4 not fix any nontrivial elements of G. We will show that G is abelian and is inverted by α. a. Show that for a, b G, if a 2 = b 2 then a = b. Let g G. Let h G be such that h 2 = g α g. b. Show that (h α ) 2 = (h 1 ) 2. Conclude that h α =h 1. c. Show that (gh 1 ) α = gh 1. Deduce that g = h. This roves the result. 21. (I. Schur 2 ). Assume that G/Z is finite. We will show that G' is finite. Let G/Z = n. a. Show that the set X = {[g, h]: g, h G} has cardinality n 2. n1 nk b. Let X = {x 1,..., x k }. Show that G' = { x... x : n i N}. c. Show that for all g, h G, [g, h] n+1 = g 1 [g, h] n g[g, h] = g 1 [g, h] n 1 [ ] 2, h g 1 g g. d. Conclude from arts (b) and (c) that every element of G' is a roduct of at most n 3 elements of X and so G' is finite. 22. (R. Baer). Let A, B be subgrous of G that normalize each other. Assume that the set X = {[a, b]: a A, b B} is finite. We will show that [A, B] is finite. Note first that, without loss of generality, we may assume that G = AB. With this assumtion A and B are normal subgrous of G. Let U = [A, B] A B. Clearly U G. a. Show that C G (X) is a normal subgrou of finite index in G. Show that C G (X) centralizes U. b. Deduce from art (a) that C G (X) U is a central subgrou of U and has finite index in U. Exercise 21 imlies that U' is finite. c. Show that, without loss of generality, we may assume that U' = 1. d. Clearly the subset {[a, u] : a A, u U} of X is finite and these elements commute with each other. Show that [a, u] 2 = [a, u 2 ]. Conclude that [A, U] is finite. Show that, without loss of generality, we may assume that [A, U] = 1. Conclude that, without loss of generality U is central in G. e. Show that X is closed under the squaring ma x x 2. Conclude that [A, B] is finite. 23. Comletely Reducible Grous. A grou is said to be comletely reducible if it is the direct sum of finitely many nonabelian simle grous. A subgrou H of a grou G is called subnormal if there is a finite chain H = H 1... H _n = G. We assume in this exercise that G is a comletely reducible grou and we let G = A i where each A i is a simle nonabelian grou. i =1,...,n a. We first want to show that any simle, nontrivial and normal subgrou H of G is one of the subgrous A i. This will show that the subgrous A i are uniquely determined. Let 1 h = a 1... a n H where a i A i. Assume a i 1. Let b i A i \ C a ). Show that 1 [h, b i ] A i H, conclude that A i = H. A ( i i b. Show that every normal subgrou of G is a direct sum of the subgrous A i. Conclude that every normal subgrou of G is comletely reducible and has a comlement. 2 If I am not mistaken, this exercise, the way it intends to lead to the result, contains a mistake. 1 k

5 c. Conclude from (a) that a subnormal subgrou of G is a normal subgrou. 24. Let G be a finite grou. Let A be a minimal normal subgrou of G. a. Show that if A has a nontrivial normal solvable subgrou, then A is an elementary abelian subgrou. From now on we assume that A has no nontrivial normal solvable subgrou. We will show that A is comletely reducible. Let B be a minimal A-normal subgrou of A. b. Show that for any g G, B g is also a minimal A-normal subgrou of A. Conclude that if C A, then either B g C or C B g g1 = 1. Deduce that A = B... gn B for some g 1,..., g n G. c. Show that B is simle. 25. (Generalized Quaternions) Let G be the grou generated by the elements x and y subject to the relations x m = y 2 and x y = x 1 where m > 0. Show that x G. Note that x m = x my = y 2y = y 2 = x m. Conclude that G 4m. In fact G = 4m. When m is even, G is called a generalized quaternion grou. When m = 2, G is called the quaternion grou. Semidirect Products Let U and T be two grous and let ϕ: T Aut(U), t ϕ t be a grou homomorhism. We will construct a new grou denoted by U ϕ T, or just by U T for short. The set on which the grou oeration is defined is the Cartesian roduct U T, and the oeration is defined as follows: (u, t)(u', t') = (u.ϕ t (u'), tt'). The reader will have no difficulty in checking that this is a grou with (1, 1) as the identity element. The inverse is given by the rule: (u, t) 1 1 = ( ϕ ( u ), t 1 ). Let G denote this grou. G is called the semidirect roduct of U and T (in this order; we also omit to mention ϕ). U can be identified with U {1} and hence can be regarded as a normal subgrou of G. T can be identified with {1} T and can be regarded as a subgrou of G. Then the subgrous U and T of G have the following roerties: U G, T G, U T = 1 and G = UT. Conversely, whenever a grou G has subgrous U and T satisfying these roerties, G is isomorhic to a semidirect roduct U ϕ T where ϕ : T Aut(U) is given by ϕ t (u) = tut 1. When G = U T, one says that the grou G is slit 3 ; then the subgrous U and T are called each other's comlements. We also say that T (or U) slits in G. Note that T is not the only comlement of U in G: for examle, any conjugate of T is still a comlement of U. When the subgrou U is abelian, it is customary to denote the grou oeration of U additively. In this case, it is suggestive to let tu = ϕ t (u). Then the grou oeration can be written as: (u, t)(u', t') = (tu' + u, tt'). The reader should comare this with the following formal matrix multilication: t 0 u t' t u' tt' tu' + u = This is an abuse of language: every grou G is slit, for examle as G = G {1}. When we use the term slit, we have either U or T around.

6 Examles. 1. Let V be a vector sace and GL(V) be the grou of all vector sace automorhisms of V. The grou V GL(V) (where ϕ = Id) is a subgrou of Sym(V) as follows: (v, g)(w) = gw + v. 2. The subgrou B n (K) that consists of all the invertible n n uer triangular matrices over a field K is the semidirect roduct of UT n (K) (uer-triangular matrices with ones on the diagonal) and T n (K) (invertible diagonal matrices). Exercises. 26. Let K be any field. Show that the grou t u * G = : t K, 0 1 u K is a semidirect roduct of the form G' T for some subgrou T. This grou is called the affine grou. 27. Show that the direct roduct of two grous is a secial case of semidirect roduct. 28. Let G = U T. a. Let U H G. Show that H = U (H T). b. Let T H G. Show that H = (U H) T. c. Show that if T is abelian then G' U. d. Show that if T 1 T, then N U (T 1 ) = C U (T 1 ). 29. Let G = U T. Let t T and x U. Show that xt is G-conjugate to an element of T if and only if xt is conjugate to t if and only (xt) u = t for some u U if and only if x [U, t 1 ]. 30. Let G = U T and let V U be a G-normal subgrou of U. Show that G/V U/V T in a natural way. 31. Let G = U T and let V U be a G-normal subgrou of U. By Exercise 30, G/V U/V T. Let t T be such that V = ad(t)(v) and U/V = ad(t)(u/v). Show that U = ad(t)(u). 32. Let K be a field and let n be a ositive integer. For t K * and x K, let ϕ t (x) = t n x. Set G = K + ϕ K *. What is the center of G? Show that Z 2 (G) = Z(G). What is the condition on K that insures G' K +? Show that G is isomorhic to a subgrou of GL 2 (K). Abelian Grous We will need the following fact several times:

7 Fact 1. Let G be an abelian grou. Let D be a divisible subgrou of G. Then D has a comlement in G, i.e. G = D H for some H G. Furthermore every subgrou disjoint from D can be extended to a comlement of D. Sketch of the roof: It is enough to rove the second statement. Let K be a subgrou disjoint from D. Using Zorn's Lemma, find a subgrou H containing K, disjoint from D and maximal for these roerties. The maximality of H insures that G = D H. From this fact it follows that, for some subgrou H, G = D(G) H where D(G) is the unique maximal divisible subgrou of G. Clearly H has no nontrivial, divisible subgrous. We will also make use of the following elementary result: Fact 2. A finitely generated abelian grou is a direct sum of finitely many cyclic grous. Prüfer -grou. Let be any rime and consider the subset n Z = {x C: x = 1 for some n N} of comlex numbers of norm 1. With the usual multilication of comlex numbers, Z is an infinite countable abelian grou. It is called the Prüfer -grou. Every element of Z exactly n elements of e n 2kπi / has finite order n for some n. Given a natural number n, there are Z that satisfy the equation n x = 1 (namely the elements where k = 0,..., n 1). Note that Z is the union of the ascending chain of finite subgrous n {x C: x = 1} which are isomorhic to the cyclic grous Z/ n Z. Thus every finite subset of Z generates a finite cyclic grou (isomorhic to Z/ n Z for some n N), i.e. Z is a locally cyclic grou. Exercises. 33. Let G and H be two abelian grous of the same rime exonent (such grous are called elementary abelian -grous) and of the same cardinality. Noting the fact that G and H are vector saces over the field F, show that these grous are isomorhic to each other. (See also Exercise 34). 34. Let G and H be two torsion-free abelian divisible grous of the same uncountable cardinality. Noting the fact that G and H are vector saces over Q, show that G H. (Comare this with Exercise 35). 35. Show that the grou Q has no roer, nontrivial divisible subgrous. Conclude that Q and Q Q are not isomorhic. Generalize this to Q. i =1,..., n 36. Show that a (not necessarily abelian) torsion grou that has no elements of order where is a rime is -divisible. Show that a grou which is -divisible for all rimes is divisible. Deduce that Z is a divisible abelian grou.

8 37. Show that if a divisible abelian grou contains an element of order, then it contains a subgrou isomorhic to Z. 38. Let G be a divisible abelian grou and a rime. Let G be the set of elements of order of G together with 1. G is an elementary abelian -grou and so it can be considered as a vector sace over the field F of elements. Let κ be the dimension of G over F. Show that G contains a direct sum of κ coies of called the Prüfer -rank of G. Z. κ is 39. Let G be a divisible abelian grou. Show that G = T(G) F where T(G) is the set of torsion elements and F is some divisible torsion-free subgrou. Conclude that a divisible abelian grou is isomorhic to a grou of the form: ( Z ) ( Q) for some sets I and I. rime I 40. Let K be an algebraically closed field. First assume that char(k) = 0. Show that * K Z ( Q) = rime for some I. Now assume that char(k) = > 0. Show that * K Z ( Q) for some I. = q, q rime 41. Let G = Z Z/2Z where Z/2Z acts on Z I I I by inversion (i.e. if 1 i Z/2Z then ϕ i (g) = g 1 for all g Z. Show that G is solvable of class 2, nonnilotent but that the chain (Z _n (G)) n N is strictly increasing. Show that G is isomorhic to a Sylow 2-subgrou of PSL 2 (K) where K is an algebraically closed field of characteristic 2. (Recall that SL 2 (K) is the grou consisting of 2 2 matrices of determinant 1 over K, and PSL 2 (K) is the factor grou of SL 2 (K) modulo its center that consists of the two scalar matrices ±1). What is the Sylow 2-subgrou of SL 2 (K) when char(k) = 2? 42. Let G be the direct sum of finitely many coies of Z. Show that if H G is an infinite subgrou then H contains a nontrivial divisible subgrou. 43. This exercise will show the advantages of the additive notation over the multilicative one. Let G be a grou and let A G be an abelian subgrou. Let g N G (A). Thus g acts on A by conjugation. Let ğ Aut(A) denote the automorhism of A induced by g. We can view ğ as an element of the ring End(A) and, denoting A additively, we can consider the endomorhism ğ 1. (For a A, (ğ 1)(a) translates into [g, a 1 ] when the grou oeration of A is denoted multilicatively). Let be a rime number and assume that g C G (A). Show that either g C G (A) or ğ is an automorhism of order. Assume now that ex(a) =. Show that (ğ - 1) = 0. Conclude that C A (g) 0. Conclude also that if A is infinite then C A (g) is also infinite.

9 44a. Conclude from the receding exercise that if H G is a normal subgrou of finite index with a nontrivial center and if G is a -grou for a rime, then G has a nontrivial center. Deduce that a nilotent-by-finite -grou has a nontrivial center. b. Let G be a nilotent-by-finite -grou. Let 1 H G. Show that H Z(G) 1. c. Show that if G is a nilotent-by-finite -grou and X < G, then X < N G (X). This roerty is called the normalizer condition. Permutation Grous. Let G be a grou and X a set. We say that G acts on X or that (G, X) is a ermutation grou if there is a ma G X X (denoted by (g, x) g * x or gx) that satisfies the following roerties: 1 For all g, h G and all x X, g(hx) = (gh)x. 2 For all x X, 1x = x. This is saying that there is a grou homomorhism ϕ: G Sym(X) where Sym(X) is the grou of all bijections of X. The kernel of ϕ is called the kernel of the action. When ϕ is one-to-one, the action is called faithful. In other words, G acts faithfully on X when gx = x for all x X imlies g = 1. Note that G/ker(ϕ) acts on X in a natural way: ğx = gx, and this action is faithful. Two ermutation grous (G, X) and (H, Y) are called equivalent if there are a grou isomorhism f : G H and a bijection ϕ : X Y such that for all g G, x X we have ϕ(gx) = f(g)ϕ(x). Let (G, X) be a ermutation grou. For any Y X, we let G Y = {g G : gy = y for all y Y}. G Y is called the ointwise stabilizer of Y. Note that G Y G is a subgrou. When Y = {x 1,..., x n }, we write G x,..., x instead of G 1 n Y. Clearly G Y is the intersection of the subgrous G y for y Y. For g G and Y X we define gy = {gy: y Y} and the setwise stabilizer G(Y) = {g G : gy = Y} of Y. We have G Y G(Y). Finally for A G, we define F(A) = {x X: ax = x for all a A}, the set of fixed oints of A. Exercise. 45. Let A, B G and Y, Z X. Then the following hold: i. A G F(A). ii. Y F(G A ). iii. If A B then F(B) F(A). iv. If Y Z, then G Z G Y. v. F(G F(A) ) = F(A). vi. G F G ) = G Y. ( Y We say that G acts n-transitively on X if X n and if for any airwise distinct x 1,..., x n X and any airwise distinct y 1,..., y n X, there is a g G such that gx i = y i for all i = 1,..., n. Transitive means 1-transitive. We say that (G, X) is sharly n- transitive if it is n-transitive and if the stabilizer of n distinct oints is reduced to {1}; in other words, if for any distinct x 1,..., x n X and any distinct y 1,..., y n X, there is a unique g G such that gx i = y i X for all i = 1,..., n. Sharly 1-transitive actions

10 are also called regular actions. U to equivalence, each grou has only one regular action (see Exercise 46). Clearly, for every n and X = n, (Sym(X), X) is sharly n and also sharly (n 1)-transitive. If for g G, x X, gx = x imlies g = 1, we say that the action of G is free or that G acts freely on X. Let X be a grou and G Aut(X). Then (G, X) is a ermutation grou. By abuse of language, one says that G acts freely (res. regularly) on X if G acts freely (res. regularly) on X *. Now we give the most imortant and, u to equivalence, the only examle of transitive grou actions: Left-Coset Reresentation. Let G be a grou and B G a subgrou. Set X = G/B, the left-coset sace. We can make G act on X by left multilication: h(gb) = hgb. This action is called the left-coset action, or the the left-coset reresentation. The kernel of this action is the core g G B g of B in G, which is the maximal G- normal subgrou of B. Exercises 46. Let (G, X) be a transitive ermutation grou. Let x X be any oint and let B = G x. Then the ermutation grou (G, X) is equivalent to the left-coset reresentation (G, G/B). (Hint: Let f = Id G and ϕ: G/B X be defined by ϕ(gb) = gx.) 47. If N G (B) = B, then the left-coset action of G on G/B is equivalent to the conjugation action of G on {B g : g G}. 48. Let (G, X) be a 2-transitive grou and B = G x. Then G = B BgB for every g G \ B. In articular B is a maximal subgrou of G. Conversely, if G is a grou with a roer subgrou B satisfying the roerty G = B BgB for every (equivalently some) g G \ B, then the ermutation grou (G, G/B) is 2-transitive. (Hint: Assume G is 2-transitive, and let x and B as in the statement. Let g G \ B be a fixed element of G. Let h G \ B be any element. Since G is 2-transitive, there is an element b G that sends the air of distinct oints (x, gx) to the air of distinct oints (x, hx). Thus b B and bgx = hx, imlying h 1 bg B and h BgB.) 49. Let G be a grou and let H G be a subgrou. Assume [G:H] = n. By considering the coset action G Sym(G/H) show that [G : g G H g ] divides n!. The subgrou g G H g is called the core of H in G. 50. Let (G, X) be a ermutation grou. Assume G has a regular normal subgrou A (i.e. the ermutation grou (A, X) is regular). Show that G = A G x for any x X. Show that (G, X) is equivalent to the ermutation grou (G, A) where G = A G x acts on A as follows: For a A, h G x and b A, (ah).b = if and only if C H (A) = 1. 1 h ab. Show that G is faithful 51 Let (G, X) be a ermutation grou. Show that g = G g x x for any x X. Show that if G is an n-transitive grou, then for any 1 i n, all the i-oint stabilizers are conjugate to each other. G 1

11 52. Let (G, X) be a transitive ermutation grou. Show that if G is abelian then, for any x X, G x is the kernel of the action and (G/G x, X) is a regular ermutation grou. 53. Let n 2 be an integer. Show that (G, X) is n-transitive if and only if (G x, X \ {x}) is (n 1)-transitive for any (equivalently some) x X. State and rove a similar statement for sharly n-transitive grous. 54. Let (G, X) be a ermutation grou. A subset Y X is called a set of imrimitivity if for all g, h G, either gy = hy or gy hy =. If the only sets of imrimitivity are the singleton sets and X, then (G, X) is called a rimitive ermutation grou. Show that a 2-transitive grou is rimitive. Assume that (G, X) is transitive. Show that (G, X) is rimitive if and only if G x is a maximal subgrou for some (equiv. all) x X. Conclude that if G is a 2-transitive grou, then G x is a maximal subgrou. (This also follows from Exercise 48). 55. Let G be a grou and B < G be a roer subgrou with the following roerties: There is a g G such that G = B BgB and if agb = a'gb' for a, a', b, b' B then a = a' and b = b'. Show that (G, G/B) is a sharly 2-transitive ermutation grou. 56. Let G = A H be a grou where H acts regularly on A by conjugation (i.e. on A * ). Show that G is a sharly 2-transitive grou. 57. Let (G, X) be a sharly 2-transitive ermutation grou, and for a fixed x X, set B = G x. Show that for any fixed g G \ B, G = B BgB and if agb = a'gb' for a, b, a', b' B, then a = a' and b = b'. Show also that the conjugates of B are disjoint from each other. Show that there are involutions that swa given any two oints. Conclude that there are involutions outside of B. 58. Show that the grou t u * G = : t K, u K 0 1 acts sharly 2-transitively on the set x X = : x K Show that G = PGL 2 (K) = GL 2 (K)/Z where Z is the set of scalar matrices (which is exactly the center of GL 2 (K)) acts sharly 3-transitively on G/B where B = B 2 (K). Show that there is a natural corresondence between G/B and the set K {}. Transort the action of G on K {} and describe it algebraically. 60. Let V be a vector sace over a field K. Show that V GL(V) acts 2- transitively on V (see Examle 1). Show that, when dim K (V) = 1, we find the examle of Exercise 58.