MAT534 Fall 2013 Practice Midterm I The actual midterm will consist of five problems.

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1 MAT534 Fall 2013 Practice Midterm I The actual midterm will consist of five problems. Problem 1 Find all homomorphisms a) Z 6 Z 6 ; b) Z 6 Z 18 ; c) Z 18 Z 6 ; d) Z 12 Z 15 ; e) Z 6 Z 25 Proof. a)ψ(1) = m Z 6 correctly (check!) defines a homomorphism for any m. b) ψ(1) = m Z 18 has to satisfy 0 ψ(0) ψ(6) 6ψ(1) 6m mod18 m 0 mod3, m {0, 3, 6, 9, 12, 15}c)...d)... e) Any ψ : Z 6 Z 25 is trivial because by Lagrange theorem ψ(z 6 ) 6 and ψ(z 6 ) 25 ψ(z 6 ) = 1. Problem 2 Show that the factor-group of S 4 by the normal subgroup N = {e, (12)(34), (13)(24), (14)(23)} is isomorphic to S 3. Proof. S 4 is the group of orthogonal symmetries S ym( ) of a 3-dimensional tetrahedron = {xe 1 +ye 2 +ze 3 +we 4 x, y, z, w 0, x+y+z+w = 1}/{xe 1 +xe 2 +xe 3 +xe 4 x R} R 4 /R 1 R 3. As in the case of dihedral group it is characterized by its the action on vertices of (fill in the details). We have an ismorphism S ym( ) S 4. The group S 4 acts on the set of unordered pairs of opposite edges X, X = 3. We have a homomorphism S 4 S X S 3, You need to check that it is surjective (find preimages of the generators of S 3 ) and N Ker. Equality N = Ker would follow from Ker = S 4 / S 3 = 4. Problem 3 A subgroup H in the group G is maximal if any subgroup H such that H H G coincides with H or G. Prove that 1

2 1. Show that C C Z(G) if C, C G are distinct maximal subgroups, which are commutative. 2. Let G be finite simple and noncommutative. Then there exists two maximal C, C G such that C C and C C > Let G be finite simple and noncommutative. Then H G, H is noncommutative. Proof. 1. The minimal subgroup < C, C > generated by C, C coincides with G (use maximality). Elements in C C commute with c 1 c 1 c nc n < C, C >= G. Therefor C C Z(G). 2. G is not simple cyclic maximal H G. We have G = H l. Note that N G (H) = H because H is maximal. N G (H) = G is ruled out because G is simple. #{ghg 1 g G} = l. If H ghg 1 > 1 we are done. Otherwise g G ghg 1 = 1 + l( H 1) < G. This means b g G ghg 1. maximal K such that b K. As before G = K l. We assume that none of ghg 1 gkg 1 ghg 1 g Hg 1 is greater then one. Then a = ghg 1 gkg 1 = 1 + l( H 1) + l ( K 1) = g G g G By construction H, K 2, then G /2 G / H = l and l( H 1) G /2. Therefore a 1 + G /2 + G /2 > G. This means that some of subgroups have a nontrivial intersection. 3. If both H, K from previous item were commutative then by the first item 2 H K Z(G), which contradicts simplicity of G. Problem 4 Prove that every group of order a) 15 = 3 5; b) 35 = 5 7; c) 185 = 5 37; d) 255 = ; is commutative. 2

3 Proof. a) n 5 = 1 + 5k, n 5 3 k = 0, Z 5 is normal. Z 3 G/Z 5 Aut(Z 5 ) is trivial because ( Z 3, Aut(Z 5 ) ) = (3, 4) = 1. Z 5 is central G/Z 5 is cyclic G is abelian (prove!). In fact G is cyclic Z 15 (prove!). b)... c)... d)n 17 = k, n k = 0, Z 17 is normal. By a) G/Z 17 is abelian. The homomorphism G/Z 17 Aut(Z 1 7) is trivial because ( G/Z 17, Aut(Z 17 ) ) = (15, 17) = 1. Z 17 is central G/Z 17 is cyclic (see a)) G is abelian. Problem 5 Suppose K G, G = p α, where p is a prime. Prove that K Z(G) > 1 Proof. The group G acts of K by conjugation. K = {1} O(g 1 ) O(g r ) is decomposition into G-orbits. Note O(g i ) = p α i. Then p β = 1 + i p α i. If all α i > 0 we would get a contradiction. As a result g i O(g i ) = {g i } g i K Z(G). Problem 6 Let V be a finite-dimensional vector space over F p, G be a finite p-subgroup in GL(V). Prove that v V such that gv = v g G. Proof. The group G acts of V. V = {0} O(v 1 ) O(v r ) is decomposition into G-orbits. Note O(v i ) = p α i. Then p dim V = 1 + i p α i. If all α i > 0 we would get a contradiction. As a result v i O(v i ) = {v i } gv i = v i g G. Problem 7 Find 1. n p (S p ), where S p is a symmetric group and p is a prime. 2. n p (SL(2, F p )). (Hint: consider upper triangular matrices) Proof. 3

4 1) Note that p p! but p (p 1)! any Sylow is generated by a p-cycle (i 1,..., i p ) S p. The number of such cycles is (p 1)!. Some of the generators generate the same group. The nunber of generators in Z p is p 1. Thus n p (S p ) = (p 1)! p 1 = (p 2)! 2) Note that GL(2, F p = (p 2 1)(p 2 p) SL(2, F p = (p2 1)(p 2 p) p 1 1). The set form a p-subgroup. Its normalizer is prove it! Then n p (SL(2, F p )) = p + 1. Problem 8 Find 1. Z(SL(3, R)). 2. Z(SL(3, C)) 1 a { 0 1 a F p} b a { 0 b 1 a F p, b F p} = p(p 1)(p+ Proof. If a SL(3, K), K = R, C then a is a scalar matrix. Note that Z(SL(3, C)) is defined by algebraic equations. Z(SL(3, R)) is the real locus(subset of points invariant under complex conjugation). The statement follows from the statement for SL(3, C). If g SL(3, C) has no Jordan blocks and two distinct eigenvalues construct h SL(3, C) such that gh hg as a cyclic permutation of eigenvectors. In case when we do have Jordan blocks take h = g t. For scalar matrices equation det(diag(a, a, a)) = a 3 = 1 has one solution a = 1, K = R and three solutions K = C. Problem 9 Let G be a subgroup in S 10 generated by a permutation g =

5 2. g = g = (1, 6, 9)(2, 10)(3, 4, 5, 7, 8) Find all orbits and all stabilizers. Problem 10 Describe all groups that have 1. One conjugacy class. 2. Two conjugacy classes. Proof. a)the group has one element, because 1 Cl G (1) = {1} b) The group has two elements. Indeed G = {1} Cl G (g). Note that Cl G (g) = G 1 = G C G (g) G = C C 1 = C 1 2 Problem 11 Describe the set of left cosets 1. {gso(2, R) g SO(3, R)}, SO(2, R) = {g SO(3, R) that act trivially on x-axis} 2. {gs n 1 g S n }, S n 1 = {g S n that act trivially on k {1,..., n}} Problem 12 Find the centralizer C GL(n,R) (g) where g = diag(λ 1,..., λ n ) if 1. all λ i are distinct. 2. λ 1 = = λ k = a, λ k+1 = = λ n = b, a b Problem 13 Let f be a polynomial in x 1,..., x 4. G f = {σ S 4 f (x σ(1),..., x σ(4) ) = f (x 1,..., x 4 )}. Find G f for a) f = x 1 x 2 + x 3 x 4 ; b) f = x 1 x 2 x 3 ; c) f = x 1 + x 2 ; d) f = x 1 x 2 x 3 x 4 ; e) f = 1 i< j 4(x i x j ). Problem 14 What groups g generated generated by g G are isomorphic 5

6 1. G = C, g = i; G = GL(2, C), g = i 0 ; 3. G = S 6, g = (3, 2, 6, 5, 1); 4. G = C, g = 2 i; 5. G = R, g = 10; 6. G = C, g = cos( 6π 5 ) + i sin( 6π 5 ); 7. G = Z, g = 3. Problem 15 Is the group O(n, Z) = O(n, R) GL(n, Z) finite. If you answer is in the affirmative, then find O(n, Z). Problem 16 Is the the half-interval [0, 1) with a binary operation a b equal to the fractional part of a + b a group. If the answer is yes, identify all finite subgroups of order 5. This is a group isomorphic to R/Z {z C z = 1}. The subgroup is the set of elements {e 2πk 5 k Z} Problem 17 Prove that AutS 3 = S 3 Proof. Let G be the group and G k = {g g = k}. The group Aut(G) acts on G k. In our case G = S 3, G 2 = {(12), (23), (13)}, G 3 = {(1, 2, 3), (1, 3, 2)}. We have a homomorphism ψaut(s 3 ) S G2. The map is surjective because inner automorphisms ψ(inn(s 3 )) already generate S G2. Kerψ consists of all µ Aut(S 3 ) that act trivially on G 2. The set G 2 generate S 3 S 3 = Aut(S 3 ). Problem 18 Find the number of conjugasy classes and number of elements in each class for a noncommutative finite group G of order p 3 where p is a prime. 6

7 Proof. We know that Z(G) {1}. The quotient G/Z(G) is not cyclic because G is not commutative. Z(G) = p. If a Z(G) Z(G), a C G (a) C G (a) = p 2. We conclude that if a Z(G) Cl(a) = p. The class equation becomes p 3 = G = Z(G) + g i Cl(g i ) = p + p g i 1 g i 1 = p 2 1. The total number of conjugasy classes is p Z(G) = p 2 + p 1 Problem 19 Is there an infinite group G such that g G k: g k = 1? Proof. G = Q/Z Problem 20 Find all subgroups of a cyclic group of order a) 24;b)100;c) 360; d)125; e)p n p is a prime n Z 1. 7