I. Group Theory. Proof. a. Since. n 11 1 (mod 11), n so n 11 = 1. So H has a normal Sylow 11-subgroup. b. Since

I. Group Theory Jan 2000 #1 Let G be a group of order 2 4 5 3 11 and let H be a group of order 5 3 11. a. Show that H has a normal Sylow 11-subgroup. b. If the number of Sylow 5-subgroups of G is (strictly) less than 16, prove that G has a proper normal subgroup of order divisible by 5. c. If G has exactly sixteen Sylow 5-subgroups, show that G has a normal Sylow 11-subgroup. Proof a. Since So H has a normal Sylow 11-subgroup. b. Since n 11 1 (mod 11), n 11 5 3 so n 11 = 1 n 5 1 (mod 5), n 5 2 4 11 so n 5 = 1, 11, 16, 176 But n 5 < 16, so n 5 = 1 or 11. If n 5 = 1 we are done. If n 5 = 11, let P = Syl 5. G : N G (P ) = n 5 = 11. Now G acts by conjugation on its 11 Sylow 5-subgroups, so its action affords a permutation representation ϕ : G S 11. The kernel K = ker ϕ of this action is the subgroup of G which normalizes all Sylow 5-subgroups of G. In particular, K N G (P ). Since G/K = Imϕ S 11. Now, since 5 3 S 11, K must be divisible by 5 3. So K is the desired normal subgroup. c. If n 5 = 16. Let P = Syl 5 (G). Let H = N G (P ). Since G : H = 16, we have H = 5 3 11. By (a) H has a unique, hence normal Sylow 11-subgroup Q. So N H (Q) = H (since n 11 (H) = N G (P ) / N H (Q) = 1). Since Q is also a Sylow 11-subgroup of G and H = N H (Q) N G (Q), we have Q H N G (Q) G Now, n 11 = G G divides N G (Q) H = 16. But n 11 1 mod 11 and n 11 5 3 2 4, so n 11 = 1. This implies that the Sylow 11-subgroup Q is unique, hence is normal in G. Aug 2000 #1 Suppose that a group G is the (internal) direct product of subgroups S and T. Let H be a subgroup of G such that SH = G = T H. a. Prove that S H and T H are normal subgroups of G. b. If S H = 1 = T H, prove that S and T are isomorphic. c. If S H = 1 = T H and H is normal in G, show that G is abelian. Proof a. By definition of internal direct product (p172), S, T G and S T = 1. So [S, T ] [S, G] S by Proposition 7(2) in Sec 5.4. Similarly, [S, T ] T, hence [S, T ] S T = 1. This implies that S, T commute elementwise. Also, S H H and T H H by Exercise 3.1.24. Thus, T centralizes S H (because every element in T commutes with elements S, hence with elements S H). Want to show that N G (S H) = G, but G = HT = T H by assumption and T G. 1

It is enough to show that HT N G (S H). Indeed, for ht HT and a S H, we have (ht)a(ht) 1 = h(tat 1 )h 1 = hah 1 S H since T centralizes S H. Hence, S H G. Similarly we can prove that T H G. b. Since G/H = SH/H = S/S H = S and G/H = T H/H = H/T H = T. Hence S = T. c. If H G, then [S, H] [S, G] S and [S, H] [H, G] H. So [S, H] S H = 1. Similarly, [T, H] T H = 1. Thus, H centralizes S and T, so H C G (S). It follows that HT C G (S). This is because in (a) we proved that S, T commute elementwise. So for ht HT, we have (ht)s(ht) 1 = htst 1 h 1 = hsh 1 = s. Hence G = HT C G (S). Similarly, we can prove that G = HS C G (T ). This implies that G = ST Z(G). Hence, G is abelian. Jan 2001 #1 Let X and Y be a distinct subgroups of a finite group G. We say that X and Y are a weird pair if X = Y and if no subgroup of G other than X and Y has this same order. a. If G is a group having a weird pair of subgroups, show that some subgroup of G has a weird pair of normal subgroups. b. If G = A B is a direct product of solvable groups, show that the subgroups A 1 and 1 B cannot be a weird pair. c. Show that a solvable group cannot obtain a weird pair of subgroups. Aug 2001 #1 Let G be a finite group of order 504 = 2 3 3 2 7. a. Show that G cannot be isomorphic to a subgroup of the alternating group Alt 7. b. If G is simple, determine the number of Sylow 3-subgroups of G. Proof a. Suppose G is a subgroup of A 7. Then G has index A 7 / G = 5. By Exercise 4.2.8, there is a normal subgroup K of G with K A 7 and A 7 : K 5!. This is a contradiction since A 7 is simple. b. Since n 3 1 (mod 3), n 3 2 3 7 so n 3 = 1, 4, 7, 28 G is simple, so n 3 1. If n 3 = 4, G acts by conjugation on its four Sylow 3-subgroups, so its actions affords a permutation representation ϕ : G S 4. Let K = ker ϕ, then K N G (P ) and G/K = Imϕ S 4. Thus, G/K 4!. K must be divisible by 21. This is a contradiction since G is simple. If n 3 = 7, by the same reason above, G/K = Imϕ S 7. But G is simple, K = 1. So G S 7. By (a) G cannot be a subgroup of A 7, we have the following diagram: S 7 2 m A 7 G m A 7 G 2 2

Note that S 7 : A 7 = 2. So S 7 : G = m with 2 m. Hence (A 7 G) G because G : A 7 G = 2. Since G is simple, we must have A 7 G = G which implies that G A 7. This contradicts to (a). Therefore, n 3 = 28. Jan 2002 #1 Let N be a normal subgroup of the finite group G. A subgroup H of G is said to be a complement for N in G if NH = G and N H = 1. a. Show that all complements for N in G are isomorphic. b. If N has a complement in G that is a p-group for some prime p, prove that every Sylow p-subgroup of G contains a complement for N. c. Assume that the center of N is trivial, that is equal to the identity subgroup, and that every automorphism of N is inner. Prove that N has a unique complement H that is normal in G. Proof a. Since G/N = NH/N = H/N H = H. Thus, every complement for N in G is isomorphic to G/N hence all complements are isomorphic. b. Let H be a complement of N in G, and H is a p-group. Suppose S is the Sylow p-subgroup containing H. Let P be any Sylow p-subgroup. Then P = gsg 1 for some g G. But H S, so P ghg 1 for some g G. We claim that ghg 1 is a complement for N, i.e. want to show that G = NgHg 1. Since G = ggg 1 = gnhg 1 = gng 1 ghg 1 = NgHg 1. Furthermore, N ghg 1 = gng 1 ghg 1 = g(n H)g 1 = g1g 1 = 1, as desired. c. Since N G, G acts by conjugation on N as automorphisms of N (Proposition 13 in Sec 4.4) i.e. There is a homomorphism ϕ : G Aut(N) defined for each g G by g (n gng 1 ). The kernel K = ker ϕ = C G (N) G. Want to prove that K is the complement for N. Indeed, by assumption Inn(N) = Aut(N), so ϕ(n) = Aut(N). Thus, NK contains K and maps onto Aut(N) = ϕ(g). But Aut(N) = Inn(N) = N/Z(N) = N by Corollary 15 in 4.4 and because Z(N) = 1 by assumption. So NK/K = N, but G/K = Aut(N). Hence NK = G. Also, K N = C G (N) N = Z(N) = 1. Hence K is a normal complement for N in G. Finally, let H be a normal complement for N. Since H and N are disjoint normal subgroup, we have NH = HN by Proposition 14 and Corollary 15 in Sec 3.2. Hence H C G (N), so C G (N) = NH C G (N) = (N C G (N))H = H. Hence H = C G (N) is unique. Aug 2002 #1 For any finite group G and prime p, we let n p (G) denote the number of Sylow p-subgroups of G. Now suppose K G, and let P be a Sylow p-subgroup of G. a. Show that n p (G/K) divides n p (G). b. Prove that n p (G/K) = n p (G) if and only if P P K. Proof 3

a. Consider the following correspondence diagram: G/K G P K/K P K 1 K Since P K/K is the Sylow p-subgroup of G/K with normalizer N G/K (P K/K) (refer Exercise 34 in Sec 4.5). By definition, n p (G) = and n p (G/K) = = G G/K N G (P ) N G/K (P K/K) G / K N G (P K) / K = G N G (P K) because N G/K (P K/K) = {ḡ G/K ḡ P K K ḡ 1 P K K } = {g G gp Kg 1 P K}/K = N G (P K)/K Since N G (P ) N G (P K), thus N G (P ) divides N G (P K). Thus, n p (G/K) divides n p (G). b. It is enough to show that N G (P ) = N G (P K). We already know that N G (P ) N G (P K). If P P K, then P char P K (Corollary 20 in Sec 4.5) since P is clearly the Sylow p-subgroup of P K. Since P K N G (P K), actually P K N G (P K). By (3) on page 135, we have P N G (P K). Thus N G (P K) = N G (P ). Conversely, suppose N G (P K) = N G (P ). Since P P K N G (P K) = N G (P ), but P N G (P ) (by definition of normalizer), we have N G (P ) = P. All inequalities become equalities. In particular, P K = N G (P ) hence P P K, as desired. Jan 2003 #1 Let N be a normal subgroup of the finite group G and suppose that G/N is a p-group for some prime p. a. If N Z(G), the center of G, show that the commutator subgroup G of G is a p-group. b. Now assume that N is cyclic (but not necessarily central in G). Prove that N G Z(G ) and deduce that G is a p-group. Proof a. Let P be a Sylow p-subgroup of G. Then NP/N is a Sylow p-subgroup of G/N by the following diagram G/N G NP/N NP 1 N But G/N is a p-group, so NP/N = G/N. Hence G = NP and G = [NP, NP ] = [ax, by] a, b N, x, y P. Note that (ax) 1 (by) 1 (ax)(by) = x 1 y 1 xy since N Z(G). So 4

G = P P is a p-group. b. By Proposition 13 in Sec 4.4, G acts by conjugation on N as automorphisms of N. And G/C G (N) is isomorphic to a subgroup of Aut(N). Since N is cyclic, by Proposition 16 Aut(N) = (Z/nZ) is abelian. So G/C G (N) is abelian and G C G (N) (Proposition 7(4) in Sec 5.4). G centralizes N so N commutes with G, hence N G Z(G ). Now, since G /(N G ) G/N. The latter is a p-group by assumption, so G /(N G ) is also a p-group. Then we apply (a) on N G G (because N G Z(G ) and center is a normal subgroup), so G is a p-group. Aug 2003 #1 Same as Aug 2001 Jan 2004 #1 Let G be a finite group and let H G be a subgroup of index G : H = n. a. Show that H : (H H g ) n for all g G. b. If H is a maximal subgroup of G and H is abelian, show that (H H g ) G for all g H. c. Now suppose that G is simple. If H is abelian and n is a prime, prove that H = 1. Proof a. Let K = ghg 1. Then we have H H K = HK K G ghg 1 = G H = n. b. Since H is abelian, so is H g. Thus H and H g centralize H H g, hence H H g C G (H H g ). But H is a maximal subgroup of G, either G = C G (H H g ) or H = C G (H H g ). In the first case, H H g Z(G) G, so H H g G. In the second case, we have H g C G (H H g ) = H N G (H). It is clear that H H g, hence H = H g and g N G (H). In this case, H H g = H and we want to show that H G. Since H is the maximal subgroup, G = N G (H). Hence H G. c. By (b), (H H g ) G. Since G is simple, H H g = 1. By (a), H : (H H g ) = H n. H has index n, so G n 2. But n is prime and G is simple, the equality cannot hold by Theorem 1(3) in 6.1. So G < n 2 and actually G = mn where m < n. Thus, G has Sylow n-subgroup of order n and the number of Sylow n-subgroups divides m. But m < n, so the number of Sylow n-subgroup must be 1. But G is simple, so G = n and H = 1. Aug 2004 #1 Let G be a finite group of order pm, where p is a prime that does not divide m, and let n denote the number of Sylow p-subgroups of G. a. Show that there exists a homomorphism θ from G to the symmetric group Sym(n) such that, for all x G of order p, the image θ(x) has exactly one fixed point. b. Now suppose that G is simple and contains an element y of order pq, for some prime q p. If θ is as in part (a), show that θ(y) must contain a cycle of length pq in its cycle decomposition. c. Now let p = 5 and suppose that G is a simple group of order 660. Show that G has no element of order 15. 5

Proof (a) G acts by conjugation on n Sylow p-subgroups, so this action affords a permutation representation θ : G S n. For x G, θ(x) is an element in the symmetric group S n. Denote these Sylow p-subgroups by {P 1, P 2,, P n }. Let x G be an element of order p, since p m, x Syl p (G). Let P = x. Want to show that P is the only one fixed point. Suppose P i is another fixed point, so P i is sent, by conjugation by all elements in G, to itself. In particular, xp i x 1 = P i. So P = x N G (P i ). On the other hand, since P i N G (P i ) G, actually P i N G (P i ). Then P i is the unique Sylow p-subgroup of N G (P i ). But as shown earlier that P = x is also a Sylow p-subgroup of G contained in N G (P i ). Hence P i = P. (b) Let K = ker θ. Then G/K = Im(θ) S n. Since G is simple, K = 1 and θ is injective. Since θ is a homomorphism. If y G is an element of order pq, θ(1) = θ(y pq ) = θ(y) pq = 1. To show that θ(y) S n contains a cycle of order pq, let σ 1 σ m be the cycle decomposition of θ(y). If none of the σ i s has length pq there must be at least one of the length q, otherwise all would have length p. Let σ 1 have length q. Then θ(y q ) fixes at least q components in σ 1 (why?). This contradicts to (a) since θ(y q ) has order p and should have only one fixed point. (c) Since G = 660 = 5 2 2 3 11, so n 5 1 (mod 5), n 5 2 2 3 11 so n 5 = 1, 6, 11, 66 Suppose that there is g G of order 15. Let x = g 3 and y = g 5, then [x, y] = 1 and y C G (P ) where P = x Syl 5 (G). Since C G (P ) N G (P ), we must have 3 (the order of y) divides N G (P ). Now, since G is simple, n 5 1. If n 5 = 6, G acts by conjugation on these 6 Sylow 5-subgroups, the action affords a permutation representation ϕ : G S 6. The kernel K = ker ϕ = {1} since G is simple, so ϕ is injective and G = ϕ(g) S 6. But this is a contradiction since G = 660 does not divides S 6 = 6! = 720. If n 5 = 11, again there is ϕ : G S 11. But by (b) this implies that there is a cycle of length 15 in S 11 which is impossible. Finally, if n 5 = 66 = G N G (P ), so N G(P ) = 10, which is not divisible by 3, a contradiction. Therefore, there s no such g exists. Jan 2005 #1 Let G be a finite group with G = 660 = 2 2 3 5 11 and suppose that E G is a subgroup of order 11. Assume that C G (E) = E. a. Prove that N G (E) = 55. b. If M G, show that either E M or M 1 mod 11. c. Show that every minimal normal subgroup of G contains E. Proof a. E is actually a Sylow 11-subgroup since 11 2 2 3 5. Since If n 11 = 12, n 11 1 (mod 11), n 11 2 2 3 5 so n 11 = 1, 12 G N G (E) = 12. So N G(E) = 55. If n 11 = 1, then G = N G (E). So we have E = C G (E) N G (E) = G But N G (E)/C G (E) is isomorphic to a subgroup of Aut(E) = Z/10Z (Corollary 15, 16 in Sec 4.4). This implies that N G (E)/C G (E) = 660/11 = 60 divides 10, a contradiction! Therefore 6

n 11 = 12 and N G (E) = 55. b. If E M, then 11 M. We claim that C G (E) M = {1}. Suppose y C G (E) M, since y E, the order of y divide E = C G (E). And by Lagrange o(y) = 11 since y {1}. But M cannot contain an element of order 11. A contradiction. Hence, C G (E) M = E M = {1}. The result follows from the conjugacy class equation. c. Let M be a nontrivial minimal normal subgroup. Suppose E M. By (b), M = 12. By Example on page 144, a group of order 12 either has a normal subgroup Sylow 3-subgroup or M = A 4. If M has a normal Sylow 3-subgroup, this contradicts to the minimality of M. If M = A 4. By Example (4) on page 142, A 4 has a unique Sylow 2-subgroup, again a contradiction. Hence, M must contain E. Jan 2005 #5 An additive abelian group U is said to be uniform if, for every two nonzero subgroups X and Y, we have X Y 0. Let us also say that U is max-uniform if U is uniform and if U is not contained in any properly larger uniform group. Jan 2006 #1 Let A, B and K be minimal normal subgroups of the group G with K A, K B and K AB. a. Show that KA = AB = KB. b. Prove that A = K = B. c. Show that AB is abelian. Proof Lemma: Dedekind modular law: If A, B, C are subgroups of a group G with A B, then A(B C) = B AC. (I guess the proof is: A(B C) = AB AC = B AC since A B). a. Since K AB, A KA AB. By Lemma, we have A(KA B) = KA. Since B is minimal, KA B = 1 or B. If KA B = 1, KA = A, so K A, a contradiction. If KA B = B, then KA = AB. Similarly, KB = AB. b. Since A, B and K are minimal normal subgroups of G with K A and K B, we have K A = K B = 1. Furthermore, K AB and K A and K B and A B = 1. By Theorem 9 in Sec 5.4, K A = KA = AB = A B, hence K = KA/A = AB/B = B. Similarly, we have K = B. Therefore, A = K = B. c. Since A, B G, by Proposition 7 in Sec 5.4, [A, B] [A, G] A and [A, B] [B, G] B, so [A, B] A B = 1. Similarly, [A, K] A K = 1. So A centralizes both K and B, hence A centralizes KB. But A AB = KB, so A centralizes itself, i.e. A is abelian. Since A = B, B is abelian. Thus AB = A B is abelian. Aug 2006 #1 Let M be a minimal normal subgroup of the finite group G and let N/M be a nontrivial normal subgroup of G/M. Assume that M is a p-group and that N/M is a q-group for some primes p and q, not necessarily distinct. a. Show that G = MH where H is a subgroup of G having a nontrivial normal 7

q-subgroup. b. If M is self-centralizing in G, prove that p q. c. If M is self-centralizing and if H is as in part a, prove that M H = 1. Proof a. Let Q be a Sylow q subgroup of N so that N = MQ. Consider the following diagram, G/M G N/M N 1 M Since N/M is a normal subgroup of G/M, N is a normal subgroup of G by correspondence theorem (?). By Frattini s Argument (Proposition 6 in Sec 6.1), G = NH where H = N G (Q). So G = NH = (MQ)H = M(QH) = MH (since Q H), with Q H. b. If p = q, then N is a p-group. We have 1 Z(N) = C G (M) = M. But M is a minimal normal subgroup of G and Z(N) G, so M = Z(N). This implies that M = Z(N) = C G (M) N, hence N = M, a contradiction since N/M is a q-group. c. Want to show that M H G and draw a contradiction. Since M is normal, M H H (by Exercise 24 in Sec 3.1). On the other hand, since M is a minimum normal subgroup and M Z(M), then M = Z(M). So M is abelian. This implies that M H M (because every subgroup of an abelian group is normal). Thus, M H MH = G, hence M H = 1 or M. If M H = M, then G = MH = H has a nontrivial normal q-subgroup Q and p q by (a) and (b). Since Q G and M G, we have [M, Q] [G, Q] Q and [M, Q] [M, G] M. Thus, [M, Q] M Q = 1, hence Q centralizes M, a contradiction since M = C G (M). Therefore, M H = 1. Jan 2007 #1 Let G be a finite group and let Syl p (G) denote its set of Sylow p-subgroups. a. Suppose that S and T are distinct members of Syl p (G) chosen so that S T is maximal among all such intersections. Prove that the normalizer N G (S T ) has more than one Sylow p-subgroup. b. Show that S T = 1 for all S, T Syl p (G), with T S, if and only if N G (P ) has exactly one Sylow p-subgroup for every nonidentity p-subgroup P of G. Proof Aug 2007 #1 Let G be a finite group of order G = 504 = 2 3 3 2 7. a. If G has a normal subgroup N of order 8, show that G has at most 8 Sylow 7-subgroups, that is Syl 7 (G) 8. b. If Syl 7 (G) 8, prove that G has an element of order 21. c. If G is isomorphic to a subgroup of Sym 9, the symmetric group of order 9, show 8

that G cannot have a normal subgroup of order 8. Proof a. Consider the following diagram G C G (P ) N C G (P ) N Note that N is the unique Sylow 2-subgroup since it is normal. Let P = Syl 7, then P is abelian hence P C G (P ) N G (P ). So C G (P ) is divisible by 7. Thus, N C G (P ) = 1. This means that C G (P ) is odd. Now, G = 504 = 2 3 3 2 7, so n 7 1 (mod 7), n 7 2 3 3 2 so n 7 = 1, 8, 36 G If n 7 = 36, then 36 = N G (P ) = 22 3 2 implies that N G (P ) = 14. In this case, N G (P ) is abelian by Example on page 181, hence N G (P ) = C G (P ), but this is a contradiction since C G (P ) is odd. Thus, n 7 36 so n 7 = 1 or 8. Probably a better solution: (I think above is done! - 12/28/2015) Let P = Syl 7, then P is abelian hence P C G (P ) N G (P ). Since N G (P )/C G (P ) is isomorphic to a subgroup of Aut(P ), N G (P ) / C G (P ) divides 6. Suppose n 7 = G N G (P ) = 36, then N G (P ) = 14. This implies that C G (P ) can only be 7. Then what? b. If n 7 = G N G (P ) = 1, 8, then N G(P ) = 2 3 3 2 7 or 3 2 7. Since N G (P )/C G (P ) is isomorphic to a subgroup of Aut(P ), a cyclic group of order 6. In either case, C G (P ) is divisible by 3. Choose x C G (P ) of order 3 and y P of order 7, we obtained an element of order 21. c. Since S 9 has no element of order 21. The result follows from (a) and (b). Jan 2008 #1 Let G be a finite nonabelian group with center Z. a. If G/Z is a p-group, for some prime p, show that G has a normal Sylow p- subgroup and that p divides Z. b. If G/Z is solvable, show that G has a nonidentity normal p-subgroup for some prime dividing G : Z. Proof G/Z a. Let P = Syl p (G). By Third Isomorphism Theorem, P Z/Z = G. Since P P Z, P Z G/Z P Z/Z = G P Z divides G. But G/P is relatively prime to p. And by assumption P G/Z is a p-group, so G/P Z can only be 1. i.e. G = P Z. Now, since P N G (P ) and 9

Z = Z(G) N G (P ), we have G = P Z N G (P ). This implies that P G. Now, since Z(P ) centralizes P and clearly it centralizes Z, we have 1 Z(P ) = Z P. Since p-groups have non-trivial center, Z(P ) is non-trivial. By Class Equation, p divides Z. b. Recall that every solvable group contain a normal abelian subgroup (by Exercise 11 in Sec 3.4). Here G/Z is a p-group, so the subgroup is a normal abelian p-group. By Correspondence Theorem, there exists H G such that Z H and H/Z is an abelian p-group. G/Z G H/Z H 1 Z So we have two cases: (a) If H is abelian, then H has a unique Sylow p-subgroup Q (Example (2) on p142). So Q char H. But H G, hence Q G. (b) If H is nonabelian, since Z Z(H) H, we have H/Z(H) H/Z. Since H/Z is abelian, so is H/Z(H). By (a) H has a unique Sylow p-subgroup Q. Since Q char H and H G, we have Q G. Aug 2008 #1 In this problem, we prove that a Sylow 2-subgroup of a simple group of order 168 is its own normalizer. a. If G is a group of order 24 and G has a normal Sylow 2-subgroup, show that G contains an element of order 6. b. If G is a simple group and H is a subgroup of G with G : H = 7, show that H contains no element of order 6. c. Let G be a simple group with G = 168 and let P be a Sylow 2-subgroup of G. Prove that N G (P ) = P. Proof a. Let P be the unique Sylow 2-subgroup and Q the Sylow 3-subgroup. Since n 3 1 (mod 3), n 3 2 3 so n 3 = 1, 4 If n 3 = 1, then G is the internal direct product of P and Q since P, Q G and P Q = 1. In particular, they centralize each other ([P, Q] [G, Q] Q and [P, Q] [P, G] P, so [P, Q] P Q = 1). Since P is a p-group, let x Z(P ) 1. In fact x Z(G). Let y Q be an element of order 3. Let z = xy, since [x, y] = 1, z has order 6. (Note: we pick an element from Z(G) so that x commutes with y when computing the degree of z = xy) G If n 3 = 4, n 3 = 4 = N G (Q) = 24 N G (Q), so N G(Q) = 6. G acts by conjugation on these four Sylow 3-subgroups, this action affords a permutation representation ϕ : G S 4. And K = ker ϕ N G (Q). Since G/K = Im(ϕ) S 4. If K = 1, ϕ is injective, but they have the same order, so G = S 4. This is a contradiction since n 2 (S 4 ) = 3 (Example on page 142). If K = N G (Q), then we have Q N G (Q) = K G. Since Q is the unique Sylow subgroup of 10

N G (Q) (or it has index 2, hence is normal), it is characteristic in N G (Q). But N G (Q) = K G thus Q G, which is a contradiction to the assumption that n 3 = 4. Therefore, K = 2 is the only possibility. Note that a normal subgroup of order 2 is central, so pick x K and and y P, then z = xy has order 6, as in (a). b. G acts by left multiplication on the set of left coset G/H. Let ϕ be the associated permutation representation afforded by this action ϕ : G S 7. But G is simple, we might well just let ϕ : G A 7 because S 7 is not simple. By Theorem 3 in Sec 4.2, K = ker ϕ = x G xhx 1 and K is the normal subgroup of G contained in H. By first isomorphism Theorem, G/K = ϕ(g) S 7. But G is simple, so K = 1 and ϕ : G A 7 is injective. Now, since the stabilizer in G of the point 1H G/H is the subgroup H. Namely, any element in H fixes at least one element corresponding to 1 H. On the other hand, any element of order 6 in S 7 fixing one point must a product of 2-cycle and 3-cycle. But none of them is a even permutation. Therefore H cannot contain an element of order 6. c. Since G = 168 = 2 3 3 7. And n 2 1 (mod 2), n 2 3 7 so n 2 = 1, 3, 7, 21 Since G is simple, n 2 1. If n 2 = 3, there is a homomorphism ϕ : G S 3 and this is not possible. If n 2 = 7. n 2 = 7 = G N G (P ), so N G(P ) = 24. By (a) N G (P ) has an element of order 6. This is a contradiction to (b). Hence, n 2 = 21 and N G (P ) = 8 which implies that P = N G (P ). Jan 2009 #1 Let G be a finite group of order p(p + 1), where p is an odd prime, and assume that G does not have a normal Sylow p-subgroup. a. Find (without proof) the number of elements of G with order different from p. b. Show that each nonidentity conjugacy class of elements with order different from p has size at least p, and conclude that there is precisely one such conjugacy class. c. Prove that p + 1 is a power of 2. Proof a. n p 1 (mod p), n p p + 1 so n p = 1, p + 1 Since G does not have a normal Sylow p-subgroup, n p = p + 1. Since n p = p + 1 = G N G (P ), we have N G (P ) = p. Furthermore, P is abelian, P C G (P ). Note that C G (P ) N G (P ). So we have P = C G (P ) = N G (P ). Since the number of elements in G with order p is (p 1)(p + 1). Hence the number of elements with order different form p is p(p + 1) (p 1)(p + 1) = p + 1. b. Let x G be an element of order p, from above we see that C G (x) cannot contain any element of order p. Now, C G (x) p + 1 because by (a) there are p + 1 elements in G of order p, so G : C G (x) = G p(p + 1) = p. Since there are p + 1 elements of order p, C G (x) p + 1 this means that there are p non-identity elements of order p. All these elements are in the same conjugacy class. Thus, there is precisely one such conjugacy class. 11

c. Since p + 1 is even (because p is an odd prime). Suppose p + 1 is not power of 2, it must be divisible by another prime, say q p. Since G is even there are x, y of order 2 and q respectively. But this is a contradiction to (b) since elements of different order can t be conjugated and can t be in one conjugacy class. Aug 2009 #1 Let H be a maximal subgroup of finite group G and let X be the set of normal subgroups X of G such that X 1 and X H = 1. a. Show that all members of X are minimal normal subgroups of G of the same order. b. If some member of X is abelian, show that all members of X are abelian p-groups for some prime p. c. Let U, V X be distinct and assume that X contains at least one additional member different U and V. Show that (UV H) G and conclude that (UV H) Z(UV ). Jan 2010 #1 Let S 7 denote the symmetric group on seven points, and let A 7 be the corresponding alternating group. a. Find the number of elements of order 7 in S 7, and find the order of the centralizer in S 7 of one of these elements. b. Find the order of the normalizer of a Sylow 7-subgroup in A 7. c. Prove that S 7 does not contain a simple subgroup G of order 504 = 2 3 3 2 7. Proof a. The element in S 7 of order 7 is 7-cycle. And the number of 7 cycle is 7!/7 = 6!. Note that the elements of the same cycle type are in the same conjugacy class. And Proposition 6 in Sec 4.3 says that the number of conjugates of an element s G is the index of the centralizer of s, [G : C G (s)]. So C G (s) = 7!/6! = 7. b. Since n 7 = [G : N G (P )]. Also, we know that the number of elements of order 7 in S 7 is the number of non-identity elements in one Sylow 7-subgroup (which is 6) times the number of Sylow 7-subgroups, i.e. 6 n 7. i.e. n 7 6 = # of elements of order 7. From (a), the number of elements of order 7 in S 7 is 6!, so n 7 = 6!/6 = 5!. Hence N G (P ) = 7!/5! = 42. c. If S 7 contains a simple subgroup G of order 504. We have G A 7 S 7. Since S 7 : A 7 = 2 and S 7 : G = 10, this implies that A 7 : G = 5. But by Exercise 4.2.8, there is a normal subgroup K of A 7 contained in G such A 7 : K 5!. This is a contradiction since A 7 is simple. Aug 2010 #1 Let G be a finite group and let N be a minimal normal subgroup of G. Suppose N = S 1 S 2 S r, where each S i is a simple subgroup and where S 1 is not abelian. a. Show that Z(N) = 1, where Z(N) is the center of N, and deduce that each S i is nonabelian. b. If g G, show that (S 1 ) g = S i for some i = 1, 2,, r. c. Prove that G has a subgroup of index r. 12

Proof a. Since Z(N) N, Z(N) char N. But N G, so Z(N) G. Since N is the minimal normal subgroup of G, Z(N) = N or 1. If Z(N) = N, this is a contradiction since S 1 is not abelian. Thus, Z(N) = 1. Similarly, each S i is not abelian. b. Since each S j is simple, it is easy to see that the normal subgroup of N has the form of A 1 A r where A i is either 1 or S i. Furthermore, the only minimal normal subgroups of N are the S i s. Thus, gs 1g 1 = S j for some j. c. The subgroup group S1 G generated by gs 1g 1 is normal in G and contained in N, from (b) we have S1 G = N (Note that S i can not be normal in G since N is minimal normal subgroup of G. So in general S g i S i. And different g may send S 1 to different S i. Furthermore, since N is normal, this implies that conjugating by g will only send N to itself, not elsewhere). Thus, the size of the orbit (of the action of G on S i ) is r, so this gives a subgroup of index r because there are only r possibilities to send. This subgroup is the stabilizer of S i. Orb(x) = G Stab(x) Jan 2011 #1 (a) By definition of internal direct product, H and K are normal subgroup. So HK G, hence HK = KH. To prove that in particular U is a subgroup, given h 1 k 1, h 2 k 2 U for h 1, h 2 H and k 1, k 2 K with θ(h 1 ) = φ(k 1 ), θ(h 2 ) = φ(k 2 ). Note that (h 1 k 1 )(h 2 k 2 ) 1 = h 1 k 1 k2 1 h 1 2 = h 1 h 1 2 k 1k2 1 HK, we want to show that θ(h 1 h 1 2 ) = φ(k 1k2 1 ) so that U is a subgroup. Indeed, θ(h 1 h 1 2 ) = θ(h 1)θ(h 2 ) 1 = φ(k 1 )φ(k 2 ) 1 = φ(k 1 k2 1 ). Hence U is a subgroup. Show that G = UK, it is enough to show that G = HK UK. Let hk G. Since θ(h) X and φ is surjective, there is l K such that θ(h) = φ(l). Thus, g = hk = hl l 1 k UK. Similarly, we can prove that G = UH. Now, if g = h 1 U H, then we must have θ(h) = φ(1) = 1, so g = h ker θ. Conversely, if h ker θ, then θ(h) = 1 = φ(1). So h U. Thus, U H = ker θ. Similarly, U K = ker φ. (b) By (a), G = UH. So it is enough to show that V H UH. Since V H consists of elements in H only, we only need to show that V H is normalized by H (why?). Let h 1 V H and h 2 H. Since V contains U, h 2 1 U and φ is surjective, there exists k K such that θ(h 2 ) = φ(k). So (h 2 k)(h 1 )(h 2 k) 1 V since h 2 k U. After simplifying we get h 2 h 1 h 1 2 V H. Similarly we can show that V K G. (c) First, note that U cannot contain all of H since by (a) U H = ker θ, so θ would be a zero map. Similarly, U cannot contain all of K. Now, let V be a subgroup of G with U V and not containing H or K. By (b), we have V H G, so V H H. This implies that θ(v H) X (the image of a normal subgroup is normal), a contradiction since X is simple. Aug 2011 #1 (a) Let K be a normal p-complement of G. We claim that H K is a normal p-complement in H. Since [H : H K] = HK G K K = pm 13

for some m. Thus, [H : H K] is a power of p. Furthermore, H K K and p K, so p H K. (b) Let K be a normal p-complement of G. Consider the diagram, G/N G NK/N NK 1 N We claim that NK/N is a normal p-complement of G/N. First, N, K G implies NK G. G/N Since NK/N = G KN divides G / K, so it is a power of p. Finally, since = K KN N K N, we conclude that p cannot divide KN/N since p K. (c) Let S, T be the normal p-complements for U and V respectively. Claim that ST is the a normal p-complement for UV. Since UV U V / U V = = U V ST S T / S T S T S T UV. Thus, U V ST divides U V, hence is a power of p. Finally, p S and p T implies that p ST since S T ST divides S T. Jan 2012 #1 (a) Since G = 4312 = 2 3 7 2 11. n 11 1 (mod 11), n 11 2 3 7 2 so n 11 = 1, 56 If n 11 = 1, then P 11 G. Let Q be a subgroup of order 7, then P 11 Q is a subgroup of order 77. If n 11 = 56, since n 11 = G/N G (P 11 ), then the normalizer of P 11 in G, N G (P 11 ), has order 77. (b) Let H be the subgroup of order 77 in (a), let P 7 be a Sylow 7-subgroup of H. Since H is abelian, we have P 7 C H (P 7 ) N H (P 7 ) H. Since P 7 is the unique Sylow 7-subgroup of H as follows: n 7 1 (mod 7), n 7 11 so n 7 = 1 So P 7 H. i.e.p 7 is normalized by H, so H N G (P 7 ). On the other hand, the Sylow 7- subgroup of G, denoted by P 49, contains P 7 and is also abelian, we have P 49 is contained in N G (P 7 ). So any Sylow 7-subgroup of G containing P 7 normalizes P 7. N G (P 7 ) P 49 H P 7 14

So we see that N G (P 7 ) contains a subgroup of order 49 and a subgroup of order 77. Thus, N G (P 7 ) is divisible by 7 2 11, hence has index dividing 8. (c) If G is simple. The permutation representation of G on the cosets of the subgroup of index 8, i.e. ϕ : G S 8, in injective. But S 8 contains no element of order 11. So this is a contradiction. Aug 2012 #1 (a) First, show that B(G) is normal. Note that every Sylow p-subgroup is a p-group, every element of Sylow p-subgroup has order power of p. Since conjugating an element of order p-power will get an element of order p-power. Same is true for the product of elements of order p-power. Hence B(G) is normal. Let H be a normal subgroup of G whose index is not divisible by p. Then its order must be divisible by p a, so it contains a Sylow p-subgroup of G. Since all Sylow p-subgroups of G are conjugate, H contains all Sylow p-subgroups, hence contains B(G). (b) Let h = g 1 g k B(L) where g i are element of each Sylow p subgroup of L. Since conjugating each g i by G lies in L since L is normal. Actually, it is not only lying in L, it was sent to some other Sylow p-element of L (why?). So by the same reason as in (a), we see that ghg 1 B(L). Thus, B(L) is normal in G. Now, if G : L is not divisible by p, L must be divisible by p a. Then L contains a Sylow p-subgroups which is also a Sylow p-subgroup of G. As in (a), L contains all Sylow p-subgroups of G hence contains B(G). By minimality property of (a), B(L) = B(G). (c) If H is a subgroup such that G : H = p, then H = p a 1 m. If L is the largest normal subgroup of G contained in H, we have the field extension L H G. Now, suppose H : L is divisible by p, then L = p a k n for some k > 1 and n m. But this implies that L can not be a normal subgroup of H since L is not the Sylow p-subgroups of H, a contradiction. (Not sure if the above was correct, but I tried another way to prove this as follows: By Theorem 3, G acts by left multiplication on the cosets of H in G. Let ϕ : G S p be the associated permutation representation, then K = ker ϕ is the largest normal subgroup of G contained in H. So we must have L = K. Since G/K = Imϕ S p, we have G/K = G/L divides S p = p!. Suppose p H/L. But G/L = p k m n for k > 1, hence pk m n divides p!. This is a contradiction since p k does not divide p! if k > 1.) Finally, since L is normal in G, L is normal in H. So by (b), we have that B(H) = B(L). Since B(L) is characteristic in L. This is because by (b) B(L) is normal in H, hence is normal in L. And by (a), B(L) is the unique normal subgroup of L. So it is characteristic in L. But L is normal subgroup of G. Hence, B(H) = B(L) is normal in G (by statement (3) on page 135 D&F). Jan 2013 #1 (a) The symmetric group S k has Property C. For any element σ C k, the order of σ is the l.c.m. of the lengths of its cycles. Two permutations of S k are conjugate if and only if they have the same cycle type. So if n is relatively prime to the order of σ, then n is relatively prime to each cycle length of σ. Since the n th power of a k-cycle is again a k-cycle if n is relatively to k (Example: (1234)(1234) = (13)(24), but (1234) 3 = (1432)). Hence the cycles in σ n have the same lengths as in σ. Therefore, σ and σ n are conjugate. 15

(b) Any abelian group that is not an abelian group of order 2 does not have property C, since no two elements are conjugate in an abelian group and every element has a power relatively prime to and strictly less than its order, unless its order is 1 or 2. (c) If G has property C and ρ : G GL m (C) is a homomorphism. If g has order k, then its image of ρ has order k, i.e., ρ(g) k = 1. So eigenvalues of ρ(g) are k th -roots of unity. Since the matrix has dimension m, let ξ be the primitive m th -root of unity for m k, with multiplicity a j. Recall that Galois conjugates of ξ are ξ j where j runs through {1,, m} relatively prime to m. By assumption, g and g j are conjugate, we see that ξ and ξ j are eigenvalues of ρ(g) with equal multiplicities. Now, since the sum of all m th -roots of unity is in Q, the sum of all eigenvalues of ρ(g) is rational. This implies that the trace of ρ(g) lines in Q for every g G. Aug 2013 #1 (a) Let B = (v 1, v 2, v 3 ) be an ordered basis for B. Define a map ϕ by sending v i to the i th column in a GL 3 (Z/p). Then ϕ is clearly a bijection. For v 1, there are p 3 1 choices (3 positions and each position has Z/p options, but minus the case where v 1 = (0, 0, 0)). For v 2, there are p 3 p choices since we need to subtract those in the span of v 1 = {av 1 a Z/p}. Similarly, for v 3, there are p 3 p 2 choices since we need to subtract those in the span of v 1 and v 2, i.e.{av 1 + bv 2 a, b Z/p}. Therefore, the order is (p 3 1)(p 3 p)(p 3 p 2 ). (b) To find the kernel of ϕ, for example p = 3, the natural homomorphism ϕ will send 1 1, 2 2, 3 0, 4 1, 5 2, 6 0, 7 1, 8 2, 9 = 0 0. The homomorphism is simply the reduce entries mod p. So a matrix with entries 0, 3, 6 will be the kernel of ϕ. In general, the kernel of ϕ is an element in GL 3 (Z/p 2 Z) with entries 0, p, 2p,, (p 1)p, i.e. ker ϕ = I + pm where M GL 3 (Z/p). Hence, the kernel of ϕ is simply M 3 3 (Z/p) of order p 9. By the first isomorphism theorem, GL 3 (Z/p 2 ) = p 9 (p 3 1)(p 3 p)(p 3 p 2 ). Remark: Exponent means the smallest integer n such that x n = 1 for all x G. 16

II. Linear Algebra Jan 2001 #4 (a) Let λ be an eigenvalue of A and x the corresponding eigenvector. Then λ is the eigenvalue of A. So we have Ax = λx and A x = λ x. This implies that A Ax = A λx = λa x = λλ x = λ 2 x. (b) Note that the determinant of a matrix is the product of its eigenvalues. Also, let c be any scalar, then in general if λ is an eigenvalue of A, then λ + c is an eigenvalue of A + ci. From (a) we proved that the eigenvalues of A A are real and nonnegative. The eigenvalues of I+A A is eigenvalues of A A plus 1. Hence the product of these eigenvalues is real and positive. Aug 2001 #4 Let A be a real n n matrix. We say that A is a difference of two squares if there exist real n n matrices B and C with BC = CB = 0 and A = B 2 C 2. (a) If A is a diagonal matrix, show that it is a difference of two squares. (b) If A is a symmetric matrix that is not necessarily diagonal, again show that it is a difference of two squares. (c) Suppose A is a difference of two squares, with corresponding matrices B and C as above. If B has a nonzero real eigenvalue, prove that A has a positive real eigenvalue. Proof (a) Let B be a matrix with b i = a i if a i > 0 b i = 0 if a i 0, and let C be a matrix with c i = 0 if a i > 0 and c i = a i if a i 0. Then BC = CB = 0 and A = B 2 C 2. (b) If A is a symmetric matrix, there is a invertible matrix P such that D = P 1 AP is a diagonal matrix (Corollary on page 369 in Hoffman and Kunze). By (a) there are matrices U and V such that UV = V U = 0 and D = U 2 V 2. Now, let B = P UP 1 and C = P V P 1. Then we see that BC = CB = 0. Furthermore, we have B 2 C 2 = (P UP 1 ) 2 (P V 1 ) 2 = P U 2 P 1 P V 2 P 1 = P (U 2 V 2 )P 1 = P DP 1 = A (c) Let λ be a nonzero eigenvalue of B and x the corresponding eigenvector. Then Bx = λx for x 0. So we have x = λ 1 Bx, hence Cx = C(λ 1 Bx) = λ 1 CBx = 0. This implies that x is an eigenvector for C with eigenvalue 0. Thus, Ax = (B 2 C 2 )x = B 2 x C 2 x = λ 2 x 0x = λ 2 x. So λ 2 > 0 is an eigenvalue for A. Aug 2002 #4 (a) Since the field is algebraic closed, the minimal polynomial is the product of linear factors (see Theorem 5 and 6 in Sec 6.4 in H&K). So the matrix is triangulable. Since triangular matrix can be decomposed into diagonal matrix, say B, and strictly triangular matrix, say C. And C is nilpotent. Apparently, BC = CB. (b) Let n be such that C n = 0 since C is nilpotent. Then choose t such that p t > n, then we have A pt = (B + C) pt = B pt + C pt = B pt. Since B is diagonalizable, B pt is diagonalizable. 17

(c) Since C n = 0, we have Hence we have Exp(A) = A k k! k=0 = A k k! (B + C)k k! = n 1 k=0 i=0 n 1 = i=0 B k i C i (k i)! i! The result follows since the last term exists! Jan 2003 #4 Aug 2003 #4 See Aug 2001 #4. ( ) n 1 k B k i C i /k! = i i=0 = k=0 B k i n 1 C i (k i)! i! i=0 B k i C i (k i)! i! n 1 = Exp(B) Jan 2004 #4 (a) Since f(x) is the minimal polynomial, it is irreducible. Let g(x) be a polynomial factor of degree m, then the null space of g(t ) is an invariant subspace of V of dimension m with T (W ) W. (b) Refer Lemma on page 200 in H&K. See Jan 2010 #4. Aug 2004 #4 Let V be an n-dimensional vector space over K spanned by v 0,, v n where v 0 + v 1 + + v n = 0. Let W be a second K-vector space and let w 0,, w n W. Find necessary and sufficient conditions on w 0,, w n so that there exists a linear transformation T : V W with T (v i ) = w i for i = 0,, n. Proof We claim that w 0 + + w n = 0 is the necessary and sufficient condition. Since 0 = T (0) = T (v 0 + + v n ) = T (v 0 ) + + T (v n ) = w 0 + + w n, this is a sufficient condition. On the other hand, since v 0 = (v 1 + + v n ), B = {v 1,, v n } spans V. But V has dimension n, B is actually a basis. Define T (v i ) = w i for i = 1,, n, then extending linearly to all of V. Then we have So we are done. w 0 = T (v 0 ) = T ( v 1 v n ) = T (v 1 ) T (v n ) = w 1 w n Jan 2005 #4 Let A = {Y M n (F ) Y X = XY } and B = {X Y A Y is diagonalizable}. We claim that X B if and only if the characteristic polynomial has no repeated roots. Suppose X B, since X commutes with itself, X is diagonalizable. So X is similar to a diagonal matrix D. We claim that the entries of D are distinct. If not, say, d i = d j for i j and let E ij = 1 at (i, j) and zero otherwise. Then DE ij = E ij D but E ij is not diagonalizable, a contradiction. Thus, D has distinct diagonal entries, hence the characteristic polynomial of X has no repeated roots. i=0 C i i!. 18

Conversely, if the characteristic polynomial of X has no repeated roots, X is diagonalizable. So X is similar to a diagonal matrix with entries consisting of distinct eigenvalues. Let Y be a matrix commuting with X, prove that Y is diagonalizable (http://www.math.lsa.umich.edu/ tfylam/math217/proofs10-sol.pdf). First, if Xv = λv, then XY v = Y Xv = Y λv = λy v So Y v is also an eigenvector of X with eigenvalue λ. Since X has distinct eigenvalues, all eigenspaces are 1-dimensional. Both v an Y v lie in the one-dimensional eigenspace of X, they must be linearly dependent. Hence, Y v = µv for some µ. So v is an eigenvector of Y corresponding to the eigenvalue µ. This shows that every eigenvector for X is also an eigenvector for Y. Finally, since X has distinct eigenvectors, it has linearly independent eigenvectors. These vectors are also eigenvectors of Y and are linearly independent. This implies that Y is diagonalizable. Aug 2005 #4 (a) Note that 8det(A) = det(2a) = det(b 1 AB) = det(b 1 )det(a)det(b) = det(a). 7det(A) = 0. Since det(a) 0, F has characteristic 7. So (b) Similar matrices have the same trace. To see this, note that tr(ab) = tr(ba). So if P 1 AP = B, then tr(b) = tr(p 1 AP ) = tr(p P 1 A) = tr(a). Now, since (B 1 AB) n = B 1 A n B = (2A) n. Hence, A n and (2A) n are similar. So tr(a n ) = tr(2 n A n ) = 2 n tr(a). This implies that (2 n 1) tr(a n ) = 0. But 3 n, 2 n 1 0, so tr(a n ) = 0. (c) Let F be an algebraic closed filed of F and let λ 1, λ 2, λ 3 F be eigenvalues of A. Since tr(a) = λ 1 + λ 2 + λ 3, tr(a 2 ) = λ 2 1 + λ2 2 + λ2 3. So The characteristic polynomial of A is λ 1 λ 2 + λ 1 λ 3 + λ 2 λ 3 = (λ 1 + λ 2 + λ 3 ) 2 (λ 2 1 + λ2 2 + λ2 3 ) 2 (x λ 1 )(x λ 2 )(x λ 3 ) = X 3 tr(a)x 2 + 1 2 (tr2 (A) tr(a 2 ))X det(a). By (b), tr(a) = tr(a 2 ) = 0. It follows that X 3 det(a). Jan 2006 #4 (a) If dim R V is odd, then the degree of characteristic polynomial is odd. The roots in C are in pair. So there exists at least one root in R. (b) Suppose W is a proper subspace that is invariant under all A i. Since the eigenvalues of A i on W and on V/W are eigenvalues of A i on V, the hypotheses are satisfied when A i are restricted to W or V/W. Assume that dim R W and dim R V/W are even. Since dim V = dim W + dim V/W, by induction dim R V is even. Suppose there is no invariant subspace. We prove by contradiction. Suppose dimension of V is odd, from (a) each A i has at least one real eigenvalue. Let W i be the corresponding eigenspace. Since A i commutes with A j, W i is invariant under A j (see example 8 in Sec 6.4 in Hoffman and Kunze). By assumption (there is no invariant subspace), W i = V. So A i v = λ i v for v V. This implies that A i acts on V is like λ i I acts on V. Now, since I = i A i = ( i λ i)i, we have i λ i = 1, this is a contradiction since none of A i has a negative real eigenvalues. 19

Thus, dim R V is even. Aug 2006 #4 (a) First, let W be T -cyclic subspace of V generated by v. i.e. W = span({v, T v, T 2 v, }). Since v 0, the set {v} is linearly independent. V is finite dimensional, let n be such that B = {v, T v, T 2 v,, T n 1 v} are linearly independent. Let U = span(b), then B is a basis for U. Furthermore, T n v U (because B T n v is linearly dependent if and only if T n v span(b).) Want to show that U is T -invariant, so that U = W. Let w U, then w is a linear combination of B, i.e. w = b 0 v + b 1 T v + + v n 1 T n 1 v. Hence, T (w) = b 0 T v + b 1 T 2 v + + v n 1 T n v. Thus, T (w) U. So U is T -invariant. Furthermore, v U and W is the smallest T -invariant subspace of V that contains v. So W U. Clearly we have U W. Hence W = U. It follows that B is a basis for W and dim(w ) = n. Characteristic polynomial f(x) has degree n. By Cayley-Hamilton, the minimal polynomial g(x) divides f(x). If deg g(x) = k < n, then g(t ) = 0, so g(t )v = 0. This implies that {v, T v,, T k v} is linearly dependent, which means that k n. So both f(x) and g(x) must have degree n. (b) Let matrix of T be ( ) A B 0 C then A is a matrix for T W and C a matrix for T V/W. Let f 1 be the characteristic polynomial for T W and g 1 be minimal polynomial. And let f 2 be the characteristic polynomial for T V/W and g 2 the minimal polynomial. Thus, g 1 f 1 and g 2 f 2. Since g 2 annihilates V/W, we have g 2 (T )V W, so g 1 (T )g 2 (T )V g 1 (T )W = 0. Thus, the minimal polynomial g(x) divides g 1 g 2. But f = f 1 f 2, g = g 1 g 2 and f = g, we must have f 1 = g 1 and f 2 = g 2. (could also refer Dummit s solution.) Aug 2006 #5 (a) Let λ be the eigenvalue of A in the algebraic closure of F. Then tr(a) = 2λ F and det (A) = λ 2 F since they are coefficients of characteristic polynomial (x 2 tr(a)x+det(a) = 0, so x 2 2λx + λ 2 = 0). If F has characteristic 2, then λ = 2λ/2 F because 2λ, as a coefficient of characteristic polynomial, is in F. If char(f ) = 2, since F is perfect (Corollary 36 in Sec 13.5), every element of F is square of an element of F. So λ 2 F implies that λ F. (b) Every nilpotent matrix in M 2 (F ) is conjugate to the Jordan form ( ) 0 1 N =. 0 0 By Orbit-Stabilizer Theorem, the size of the orbit equals to the index of stabilizer G N in G = GL 2 (F ). Recall that GL 2 (F ) = (q 2 1)(q 2 q). It is enough to compute G N. That is, ( ) ( ) ( ) ( ) a b 0 1 0 1 a b = c d 0 0 0 0 c d We get a = d, c = 0, so the stabilizer G N has the form ( ) a b 0 a So there are q(q 1) elements. It follows that GL 2 (F ) / G N = (q 2 1)(q 2 q)/(q 2 q) = q 2 1. 20

Jan 2007 #4 (a) Define T : V (V/R) by x ϕ(x) : V/R F defined by v + R x, v. To see ϕ is well-defined, let v 1 + R = v 2 + R, this implies v 1 v 2 R. Then for all x V, we have x, v 1 v 2 = 0, i.e. x, v 1 = x, v 2. So ϕ(x) is well-defined. And ker(t ) = L is clear. (b) Let (a 1,, a n ) be an ordered basis for V. Consider the matrix A = ( a i, a j ) for 1 i, j n. First show that R = ker(a) and L = ker(a T ). Let x V, x = x 1 a 1 + + x n a n. It suffices to work on the basis element a i V. If x R, for each i, we have a i, x = a i, x 1 a 1 + + x n a n = x 1 a i, a 1 + + x n a i, a n = 0 a 1, a 1 a 1, a n x 1 x 1 a 1, a 1 + + x n a 1, a n 0...... =. =. a n, a 1 a n, a n x 1 a n, a 1 + + x n a n, a n 0 x n So x R if and only if x ker(a). i.e. R = ker(a). Similarly, we can show that L = ker(a T ). a 1, a 1 a n, a 1 x 1 x 1 a 1, a 1 + + x n a n, a 1 0...... =. =. a 1, a n a n, a n x 1 a 1, a n + + x n a n, a n 0 x n Now, since the dimension of the row and column space are the same, their codimensions are also the same. In other words, dim K (ker(a)) =dim K (ker(a T )). Equivalently, dim K (R) =dim K (L). Finally, dim(t (V )) =dimv dim(ker(t )) =dimv diml =dimv dimr =dimv/r =dim(v/r). So T is surjective. Aug 2007 #4 (a) Define a linear map f : V W by v (v, ). It is well-defined since (, ) is bilinear. So ker f = W. Thus, dim(v ) dim(w ) dim(w ) = dim(w ). b) Let f : V V as in (a). By assumption f is injective and both spaces have the same dimension, so f is an isomorphism. Given a basis A = {a 1,, a n } there is a standard dual basis A = {a 1,, a n} such that a i (a j) = δ ij. Let a i = f 1 (a i ). Since f is an isomorphism, A = {a 1,, a n} is a basis for V. So we have (a i, a j ) = (a j, a i) = f(a j )(a i) = a j (a i) = δ ij. (c) Jan 2008 #4 (a) Since T p = I, the polynomial x p 1 = 0 satisfies T. So the minimal polynomial divides x p 1. Hence the eigenvalue is 1 since x p 1 = (x 1) p. And we have (T I) p = 0. So on main diagonal of the Jordan form, the values are either 1 or 0. The algebraic multiplicity of eigen value 1 (i.e. = p) is greater than its geometric multiplicity (which is the nullity of the matrix T I and indeed by assumption dim F W = 1). (To find the dimension of V is to find the dimension of generalized eigenspace.) Since (T I) p = 0, the dimension of the generalized eigen space is p (refer wikipedia for generalized eigenspace). But dimension V is the degree of characteristic polynomial that has degree this implies that generalized eigenspace for eigenvalue 1 has dimension 1. Since there is only one eigenvalue (b) Since minimal polynomial divides x p 1 = (x 1) p, it is (x 1) k for some k p. By assumption the degree of characteristic polynomial dim F V is < p, so we have k dim F V < p. 21

This implies that k p 1. Therefore, (T I) p 1 = 0. (c) Since (x 1) p 1 = xp 1 x 1 = xp 1 + + x 2 + x + 1 If there is v V such that (I + T + T 2 + + T p 1 )v 0, this implies that v ker(t I) p 1. By (b), we have that dim F V p, together with (a) proved that dim F V = p. Aug 2008 #4 (a) Suppose S n 1 M = MT n 1, then S n M = S(S n 1 M) = S(MT n 1 ) = (SM)T n 1 = (MT )T n 1 = MT n. So by induction on n we see that S n M = MT n for all nonnegative integer n. So p(s)m = Mp(T ) for all polynomials p(x) in C[x]. In particular, f(s)m = Mf(T ). Since f(x) is the minimal polynomial of T, we have f(s)m = Mf(T ) = 0. (b) Since M 0, Mv 0, this implies that f(s) in singular (Note: A is non-singular if the only solution to Ax = 0 is x = 0), i.e. det(f(s)) = 0. Since f(x) is the minimal polynomial of T, we can write f(x) = i (x λ i) n i where λ i are eigenvalues of T with multiplicities n i. Hence, 0 = det(f(s)) = i det(s λ ii) n i. So det(s λ i I) = 0 for some λ i, this implies that λ i are eigenvalues for both S and T. (c) Let ( ) a b M = c d Set SM = MT. By direct computation, we get ( ) 1 0 M = 1 0 Clearly M is not invertible. Jan 2009 #4 (a) If W is an invariant subspace for T. The characteristic polynomial for the restriction operator T W divides the characteristic polynomial for T. Same for minimal polynomial. Recall that T is diagonalizable if and only if the minimal polynomial for T has distinct roots. Thus, T is diagonalizable on W. (b) The characteristic polynomial M xi is f(x) = x 2 (x 1). So the eigenvalues are λ = 0 and 1. When λ = 0, the eigenvector is [0, 0, 1] T. When λ = 1, the eigenvector is [0, 1, 0] T. The eigenspace has only two dimension, so cannot span 3-dimensional space V. So T is not diagonalizable. Aug 2009 #4 Let V be a vector space over a field F and let (, ) : V V F be a bilinear form. For each x V define A(x) = {y V (x, y) = (y, x)}. Now suppose v is a fixed element of V with (v, v) 0. (a) For all x V, show that A(x) is a subspace of V of codimension at most 1. (b) If the characteristic of F is different from 2, prove that A(v) is a subspace of V of codimension precisely 1. (c) If F is algebraically closed and has characteristic different from 2, show that 22