Homework 2 /Solutions

MTH 912 Group Theory 1 F18 Homework 2 /Solutions #1. Let G be a Frobenius group with complement H and kernel K. Then K is a subgroup of G if and only if each coset of H in G contains at most one element of K. Proof. Suppose first that K is a subgroup of G and let x, y K with Hx = Hy. Then xy 1 K H = 1 and so x = y. Suppose next that each coset H contains at most one element of K. Let g G. Since K is a normal subset of G we conclude that also each coset of H g contains at most one element of K. Let x, y K. Note that either xy 1 K or xy 1 H g for some g G. In the latter case H g x = H g y and so x = y and xy 1 = 1 K. We proved that xy 1 K for all x, y K. Also 1 K and so K is subgroup of G. #2. Let G be a finite Frobenius group with complement H and kernel K. (a) If H is even, then H contains a unique element t of order 2. In particular, t Z(H). (b) K is a subgroup of G if and only if each coset of H in G contains at least one element of K. (c) If K is a subgroup of G and p is a prime, then H normalizes a Sylow p-subgroup of K. (d) If H is a maximal subgroup of G and K is a subgroup of G, then K is an elementary abelian p-group for some prime p. Proof. (a) Since H has even order, there does exist an element r of order 2 on H. Let t be any element of order 2 in H. By (***) in the proof of Theorem 1.3.7 K = tt S for some S G. Let k K. Then k = tt s with s S and so Lemma 1.3.6 shows that k t = k 1. So also k r = k 1 and k rt = (k r ) t = (k 1 ) t = k Thus K = C K (rt). By Theorem 1.3.7 K is a normal subgroup of G and G is the internal semidirect product of K by H. Thus we can apply 1.3.9f and conclude that C K (h) = 1 for all h H. It follows that rt = 1 and so r = t. Thus t is the unique element of order 2 in H. In particular t H = {t} and so t Z(H). (b) By Lemma 1.3.4 K = G/H. As G/H is finite we see that each coset of H in G contains at least one element of K if and only if each coset of H in G contains at most one element of K, and so by Exercise #1 if and only if K is a subgroup of G. (c) Let P be a Sylow p-subgroup of K. If P = 1 (c) holds. So suppose P 1. Observe that G acts on Syl p (K) be conjugation and by Sylow s Theorem K acts transitively on Syl p (K). Hence the Frattini Argument 1.1.11a shows that G = KN G (P ) Put U := N G (P ). Then P U and G = KU. In particular, exists g G with N G (P ) H g 1. Hence N G (P g 1 ) H = 1. Replacing P by P g 1 we may assume that N G (P ) H 1. Put U := N G (P ). Then P U, G = KU, and U H 1. Since 1 P K we have P H and so also U H. Thus 1.3.6 shows that U is Frobenius group with complement U H and kernel U K. In particular, U K = U/(U H. Note that U K is a subgroup of U and (U K) (U H) K H = 1. Thus U = (U K)(U H). Hence 1

G = KU = K(U K)(U H) = K(U H) and so H = (H K)(U H) = U H. Thus H U = N G (P ) and so H normalizes P. (d) Let p be a prime integer dividing K. By (d) we know that H normalizes a Sylow p subgroup P of K. Since P is non-trivial p-group Z(P ) 1. Put A := {z Z(P ) z p = 1}. Then A is a non-trivial elementary abelian p-subgroup of K. As H normalizes P it also normalizes Z(P ) and A. Thus AH is a subgroup of G. Since A P K we have A H = 1 and so A H. Thus H AH G and since H is maximal subgroup of G we get G = AH. Hence A K AH and so K = A(K H) = A. Thus K is an elementary abelian p-group. #3. Let G be group, H a subgroup of G, A = G/H and S a transversal to H in G. Suppose H acts on a set B and define an action of G on A B as in class. (a) Let B = {H} B. Show that the action of H on B is isomorphic to the action of H on B. (b) Let T be another transversal to H in G. Show that the actions of G on A B defined via S and via T are isomorphic. Proof. (a) Follows from 1.5.7e. (b) Let T 1 = T and T 2 = S. Let {i, j} = {1, 2}. Let Ω i be the G-set obtained from the set A B with the action of G-defined via the transversal T i. Let λ i : B {H} B Ω i be H- isomorphism from 1.5.7e. By Lemma 1.5.8 there exists a unique G-equivariant function ɛ i : Ω i Ω j with λ i ɛ i = λ j. Then λ i (ɛ i ɛ j ) = λ i. Since also λ i id Ωi = λ i the uniqueness assertion in 1.5.8 implies that ɛ i ɛ j = id Ωi. Hence ɛ 1 is a G-isomorphism from Ω 1 to Ω 2 (with inverse ɛ 2 ). #4. Let G be a group acting transitively on the sets A and B. Let a A and b B. Show that A and B are G-isomorphic if and only if G a = G g b for some g G. Proof. Suppose first that σ : A B is a G-isomorphism. If g G, then a = a g aσ = (a g )σ aσ = (aσ) g and so G a = G aσ. Since G acts transitively on B we know that aσ = b g for some g G and so G ασ = G b g = G g b. Thus G a = G g b. Suppose next that G a = G g b for some g G. As Gg b = G b g we get G a = G b g. By Lemma 1.1.9 A = G/G a and B = G/G b g as G-sets. As G b g = G a this shows that A and B are isomorphic G-sets. #5. Let G := Sym(5), A := Syl 5 (G), H := G A and B := Sym(A)/H. (a) G acts faithfully and 3-transitively on A. (b) A = B = 6, Sym(A) acts faithfully on B and Sym(A) B = Sym(B). (c) The action of Sym(A) on A is not isomorphic to the action of Sym(A) on B. 2

Proof. (a): Note first that there are six 5-Sylow subgroups of G. Indeed the non-trivial elements in a Sylow 5 subgroup consists of four 5-cycles, there are 5! 24 5 = 24 five cycles and so 4 = 6 5-Sylow subgroups. Thus A = 6. Let X A. By Sylow s Theorem G acts transitively on A. Hence G/G X = 6 and so G X = 20. A Sylow 5-subgroup cannot normalizes another Sylow 5-subgroup, so X has a unique fixed-point on A, namely itself. Put A = A \ {X}. It follows that any orbit of X on A has size 5. As A = 5 this shows that X acts transitively on A. Hence also G X acts transitively on A. Let Y A. Then G X /G XY = A = 5 and so G XY = 4 and G XY X = 1. Let Z A with Z Y. Since X is transitive on A we have Z = Y x for some x X. Note that G x XY Z Gx XY = G XY x = G XZ. Thus [G XY Z, x] G 1 XY Z, Gx XY Z G XZ On the other hand G XY Z G X = N G (X) and so [G XY Z, x] X G XZ = 1. Thus G XY Z C G (x). Observe that all twenty-four 5 cycles in G are conjugate, so G/C G (x) = 24. Thus C G (x) = 5 and since X C G (x) we get C G (x) = X. Thus G XY Z X and since G XY X = 1 we get G XY Z = 1. In particular, C G (A) G XY Z = 1 and so G acts faithfully on A. Let ω = (X, Y, Z) A 3. Then G ω = G XY Z = 1. Hence ω G = G/G ω = G = 120 = 6 5 4 = A ( A 1)( A 2) = A 3 Thus G acts transitively on A 3 and so 3-transitively on A. (b) We already proved that A = 6. Thus Sym(A) = A! = 6!. Since G acts faithfully on A we have H = G A = G = 5! and so B = Sym(A)/H = 6! 5! = 6. Note that C Sym(A) (B) H and so C Sym(A) (B) H = 120. The only non-trivial proper normal subgroup of Sym(A) is Alt(A) and so has order 360. Thus C Sym(A) (B) = 1. Hence Sym(A) acts faithfully on B and Sym(A) B = Sym(A) = Sym(6) = Sym(B). As Sym(A) B Sym(B) we get Sym(A) B = Sym(B). (c) As H = G A and A acts transitively on A, also H acts transitively on A. On the other hand H fixes the element H of B, and so does not act transitively on B. Thus A and B are not isomorphic as H-sets and so also not isomorphic as Sym(A)-sets. #6. (Comment: The two words in boldface were typos which I corrected) (a) Let G be the internal semidirect product of K by H and let N be a normal subgroup of H. Show that N is normal in G if and only if N centralizes K. If N centralizes K, put H = H/N, observe that H acts on K via k Nh = k h and show that G/N = H K. (b) Let H be a group acting on a set Ω and let K be a group. Let N be a normal subgroup of H acting trivially on Ω. Put H = H/N and observe that H acts on Ω via ω Nh = ω h. Show that N {1} is a normal subgroup of H Ω K and (H Ω K)/(N {1}) = H Ω K. Proof. (a) As G = KH and N H we see that N G if and only of K normalizes N. The latter holds if and only if [K, N] N. Since K G we have [K, N] K and so [K, N] N if and only if [K, N] K N. As K N = 1, this holds if and only if [K, N] = 1, that is if and only if N centralizes K. Thsu the first statement in (a) holds. Suppose now that N centralizes K. Since N H we have KN H = (K H)N = 1N = N and so K H = 1. Also G = KH = K H and so G is the internal semidirect product of K by H. Hence by Homework 1, G = H K with respect to the action of H on K by conjugation. For g G, let g = Ng G. Since K H = 1, the function σ : K K, k k 3

is an isomorphism of groups. Let k K and h = hn H. Then and so σ is also an isomorphism of H-sets. Thus (k h )σ = k h σ = k h = k h H K = H K = G (b) Proof 1: As N acts trivially Ω it also acts trivially on H Ω. Identify H with H {1} and K Ω with {1} K Ω in H Ω K = H K Ω. Then H Ω K is the internal semidirect product of K Ω by H, and N centralizes K Ω in H Ω K. Part (a) now shows that N is normal in H Ω K and H Ω K/N = H K Ω Let h H, f K Ω and ω Ω. With the action (ω, h) ω h of H on Ω as defined in (b) and the action (f, h) f h of H on K Ω as defined in (a) we have ωf h = ωf h = ω h 1 f = ω h 1 f Thus the action of H on K Ω is exactly the action of H on K Ω used in the definition of H Ω K as a external semidirect product of K Ω by H. Hence (b) holds. Proof 2 For h H let h = Nh H. Define φ : H Ω K H Ω K, (h, f) (h, f) We will fist show that φ is a homomorphism. Recall that the action of H on K Ω used in the definition of the wreathed product H Ω K is defined by ωf h = ω h 1 f for all ω Ω, f K ω and h H. If h H with h = Nh, then ω h 1 = ω h 1 = ω h 1 and so ωf h = ω h 1 f = ω h 1 f = ωf h Thus f h = f h Let f 1, f 2 K Ω and h 1, h 2 H. Then ((h 1, f 1 )(h 2, f 2 ))φ = (h 1 h 2, f h2 1 f 2)φ = (h 1 h 2, f h2 1 f 2) = (h 1 h 2, f h2 1 f 2) = (h 1, f 1 )(h 2, f 2 ) = (h 1, f 1 )φ (h 2, f 2 )φ 4

Hence φ is a homomorphism. Observe that φ is surjective and Kerφ = N {1}. The First Isomorphism Theorem nows shows that N {1} is normal in H Ω K and (H Ω K)/(N {1}) = H Ω K #7. Let G be group acting on the sets A and B and let be a G-invariant relation on A and B. Assume that for each a A there exists d B with a d, and that for all b B there exists c A with c b. Let a A and b B. Show that the following three statements are equivalent: (a) G acts transitively on A and G a acts transitively on {d B a d}. (b) G acts transitively on {(c, d) A B c d}. (c) G acts transitively on B and G b acts transitively on {c A c b}. Proof. (a) = (b): : Suppose (a) holds and for i = 1, 2 let a i A and b i B with a i b i. Since G acts transitively on A, there exists g i G with a gi i = a. Put d i = b gi i. Since a i b i and is G equivariant we have a g i bg i, and so a d i. Since G a acts transitively on {d B a d}, there exists g G a with d g 1 = d 2. Put h = g 1 gg2 1. Then and a h 1 = a g1gg 1 2 1 = a gg 1 2 = a g 1 2 = a 2 b h 1 = b g1gg 1 2 1 = d gg 1 2 1 = d g 1 2 2 = b 2 Thus (a 1, b 2 ) h = (a 2, b 2 ) and G acts transitively on {(c, d) A B c d} and (b) holds. (b) = (a): Suppose now that (b) holds. For i = 1, 2 let a i A. Then there exists b i B with a i b i. Moreover, for any such b i, there exists g G with (a 1, b 1 ) g = (a 2, b 2 ). Thus a g 1 = a 2 and b g 1 = b 2. In particular, G acts transitively on A. In the case that a 1 = a 2 = a we have a g = a g 1 = a 2 = a and so g G a. Since b g 1 = b 2 we conclude that G a acts transitively on {d B a d}. We proved that (a) and (b) are equivalent. By symmetry, (b) and (c) are equivalent. #8. Let G be finite group acting n-transitive on a set A and let p be a prime. Let B A with B = n and let P be a Sylow p-subgroup of C G (B). Show that N G (P ) acts n-transitive on C A (P ). Proof. Let A = A n, D = { X A X = n } and B = X D Syl p(c G (X)). Define a relation on A and B by ω P if P G ω. We will now show 1. Let ω A. Then {P B ω P } = Syl p (G ω ). In particular, G ω acts transitively on {P B ω P }. Observe that ω = n and so ω D. Also G ω = C G (ω). So if P Syl p (G ω ), then P ( Syl p CG (ω) ) B and ω P. Conversely let P B with ω P. Then P G ω. Also, since P B, we have P Syl p (C G (X)) for some X D. Pick µ X n. Then µ = X and C G(X) = G µ. Also as G acts n-transitively on A, there exists g G with ω g = µ. Thus C G (X) = G µ = G g ω = G ω. 5

In particular, G ω = C G (X) and since P Syl p (C G (X) and P G ω we conclude that P Syl p (G ω ). 2. C A (P ) n = {ω A ω P }. Note that C A (P ) n An = A and P B. Let ω A. Then ω P iff P G ω iff P C G (ω) iff ω C A (P ) and iff ω C A (P ) n. So (2 ) holds. We now can proof #8. By hypothesis G acts n-transitively on A, that is G acts transitively on A n. By (1 ) G ω acts transitively on {P B ω P } and so Exercise #7 shows that G P acts transitively on {ω A ω P }. Using (1 ) and G P = N G (P ), we conclude that N G (P ) acts transitively on C A (P ) n, that is N G(P ) acts n-transitively on C A (P ). 6