Higher Algebra Lecture Notes


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1 Higher Algebra Lecture Notes October 2010 Gerald Höhn Department of Mathematics Kansas State University 138 Cardwell Hall Manhattan, KS USA This are the notes for my lecture Higher Algebra which I taught several times at Kansas State University. Students are expected to have some previous knowledge of algebra, although only basic knowledge of set theory and elementary number theory together with some mathematical maturity is all what is required. Parts of the lecture notes are based on the book Einführung in die Algebra by G. Fischer and R. Sacher. I also used the books Algebra by S. Lang and Abstract Algebra by D. Dummit and R. Foote as well as some randomly found notes on the internet. I tried to resist the temptation to add additional material to the notes. Most of the problems at the end of each section are standard. Some have a computational flavor. Results are numbered continuously in each chapter. For results marked with a star no proof is given. Please send corrections, remarks, etc. to the above address. 1
2 Contents 1 Group Theory Basics Normal subgroups Cyclic and solvable groups Group actions Sylow subgroups Products Examples of finite groups Ring Theory 17 3 Module Theory 18 2
3 Chapter 1 Group Theory 1.1 Basics Definition. Let M be a nonempty set. (a) A map : M M M, (a, b) a b, is called a binary operation or product in M. (b) A product in M is called associative if (a b) c = a (b c) for all a, b, c M and commutative if a b = b a for all a, b M. Remarks: Often one writes ab for a b. If M is finite, say M = {a 1,..., a n }, a product can be described by a product table: a 1 a 2... a j... a n a 1 a 1 a 1.. a i... a i a j. a n... a n a n Let M be a set with product. For every a M one has the maps l a : M M, x ax (left translation), r a : M M, x xa (right translation). M is associative if and only if l a b = l a l b for all a, b M. Definition. A semigroup is a nonempty set M together with an associative product in M. Examples: N = {1, 2, 3,...} with addition or multiplication as product. 3
4 Map(X) = {f : X X} for a set X with composition as product, i.e. (f g)(x) = f(g(x)) for f, g Map(X) and x X. Let a 1,..., a n be elements in a set with product. We write n a i = a 1 a 2... a n = (...((a 1 a 2 )a 3 )...a n 1 )a n. Lemma 1. In a semigroup H one has ( m )( m+n ) a i a i = i=m+1 m+n a i for any a 1,..., a n+m H. Proof: By definition this is true for n = 1. Assume it is true for n = k and consider n = k + 1. ( m )( m+k+1 ) ( m )(( m+k ) ) a i a i = a i a i a m+k+1 i=m+1 = = (( m ( m+k i=m+1 ) ( m+k a i i=m+1 a i ) a n+k+1 = a i ))a m+k+1 m+k+1 Corollary. Let w = [a 1,..., a n ] be a product of a 1,..., a n given by some choice of brackets. Then w does not depend on this choice. Proof: One has w = ([a 1,..., a m ]) ([a m+1,..., a n ]) for some m < n and some choice of brackets for a 1,..., a m and a m+1,..., a n. By induction we may assume [a 1,..., a m ] = m a i, [a m+1,..., a n ] = n i=m+1 a i. The Lemma gives w = n a i and w is therefore welldefined. This allows us to forget about brackets in a semigroup. We write a n for n a. One has a n a m = a n+m, (a m ) n = a nm. Lemma 2. In a commutative semigroup H one has n a i = n a π(i) for any a 1,..., a n+m H and bijective map π : {1,..., n} {1,..., n}. Proof: Exercise. Definition. (a) An identity (or unit element) of a semigroup H is an element e H such that e a = a e = a for all a H. a i. 4
5 (b) A monoid is a semigroup with identity. Remarks: An identity e is unique; for if e is another identity we have e = e e = e. Often one writes just 1 for e; if H is written additively usually the case if H is commutative one writes 0 for e. Examples: (N, ) with identity 1. Map(X) with identity id X : X X, x x. The set of isomorphism classes of ndimensional differentiable manifolds with the commutative connected sum # as product. Identity is the standard nsphere S n. Definition. Let G be a monoid.... (a) An inverse of x G is an element y G such that xy = yx = e. (b) If every element of G has an inverse, G is called a group. Remark. Let G be a group. (a) The inverse of x G is unique. One denotes the inverse of x by x 1. (b) xy = e yx = e for x, y G. (c) (xy) 1 = y 1 x 1 and (x 1 ) 1 = x. (d) For a, b G there exist unique elements x, y G solving the equations ax = b and ya = b. (e) For a, x, y G one has ax = ay x = y and xa = ya x = y. Proof: (a) If y is another inverse of x we have y = y e = y (xy) = (y x)y = ey = y. (b) Let xy = e. Choose z G with zx = e. Then yx = (ey)x = ((zx)y)x = z(xy)x = zex = zx = e. (c) (y 1 x 1 )(xy) = y 1 (x 1 x)y = y 1 y = e. (x 1 ) 1 x 1 = (xx 1 ) 1 = e 1 = e since ee = e. (d) Eq. 1 x = a 1 b and Eq. 2 y = ba 1. (e) Multiplication of eq. with a 1 from left or right. Examples: 5
6 (Z, +) the integers with addition. (C, ) the nonzero complex numbers with multiplication. Perm(X) Map(X) the set of bijective maps of a set X onto itself with composition. Gl n (k) the set of invertible n nmatrices with components in a field k with matrix multiplication. Definition. Let (G, ) and (G, ) be (semi)groups. A map ϕ : G G is called a (semi)group homomorphism if ϕ(a b) = ϕ(a) ϕ(b) for all a, b G. If there is also a homomorphism ψ : G G such that ψ ϕ and ϕ ψ are the identity mappings of G and G, respectively, ϕ is called an isomorphism For G = G a homomorphism ϕ is called an endomorphism, an isomorphism ϕ is called an automorphism. mm mm Remark. Proof: Examples: f a : N G, n a n, for G a semigroup and a G is a semigroup homomorphism. f a : Z G, n a n, for G a group and a G (here a n := (a 1 ) n for n N) is a group homomorphism. For G a group, a G, the map ϕ a : G G, x axa 1 is an (so called inner) automorphism of G. exp : (C, +) (C, ), z exp(z) = n=0 zn n!, is a group homomorphism. Theorem 3. Let H be an abelian semigroup written additively. Then: (a) There exists an abelian group G and a semigroup homomorphism ι : H G such that the following universal property is satisfied: For every semigroup homomorphism ϕ : H G there exists a unique group homomorphism ψ : G G such that the following diagram is commutative...,i.e. ψ ι = ϕ. (b) For x G there exist a, b H such that x = ι(a) ι(b). (c) ι is injective if the cancellation law is valid in H (x + a = y + a x = y). Proof: Lemma 4. Theorem 5 (Lagrange). 6
7 Exercises 1. Prove Lemma 2 from Chapter 1: In a commutative semigroup H one has n a i = n a π(i) for any a 1,..., a n H and bijective map π : {1,..., n} {1,..., n}. 2. A left identity in a semigroup G is an element e such that ea = a for all a G. A left inverse of a G is an element x G such that xa = e, with e a left identity. Prove that a semigroup which has a left identity and every element has a left inverse is in fact a group and the left identity is an identity and a left inverse is an inverse. 3. (i) Is the additive group of integers isomorphic to the additive group of rationals? (ii) Is the additive group of rationals isomorphic to the multiplicative group of nonzero rationals? 4. Prove part (iv) and (v) of the list of subgroup examples: (iv) H is a subgroup of the additive group of integers if and only if there exists a n Z such that H = nz := {nk k Z}. (v) The set Q = {±E, ±I, ±J, ±K} with E = ( ) ( ) ( ) 1 0, I = i 0 0 i, J = , K = ( 0 i i 0 ), is a subgroup of GL 2 (C) of size 8. Furthermore, determine all subgroups of the quaternion group Q. 5. (Calculation problem) Let M be a set with a binary composition. (a) Calculate for n = 6 how many essentially different product expressions w = [a 1, a 2,...,a n ] (i.e., taking into account ordering and bracketing) can be formed with elements a 1, a 2,..., a n M provided that the product in M is assumed to be: (i) associative and commutative, (ii) associative but not commutative, (iii) commutative but not associative, (iv) neither associative nor commutative. (b) List for n = 4 all expressions w as in (a). (Remarks: It is recommended to use a computer. But for any solution by hand, full credit is given. Part (i) and (ii) are worth 0 points but you should try them first. The form in which the products in (b) are listed can be freely chosen. You may solve (b) first.) 1.2 Normal subgroups Theorem 6. Theorem 7 (Homomorphism theorem). Theorem 8 (First isomorphism theorem). Theorem 9 (Second isomorphism theorem)
8 Theorem 10 (JordanHölder). Exercises 1. Prove the following remark from the lecture about the centralizer and normalizer of a subset S of a group G: (a) Cent(S) < Nor(S) < G. (b) H < G implies H Nor(H). (c) K < G and H K implies K < Nor(H), i.e., Nor(H) is the largest subgroup of G in which H is normal. 2. Let c : G Perm(G) be the map defined for a G by c(a) : G G, g aga 1. (a) Show that c is a homomorphism from G into Aut(G) < Perm(G), the subgroup of automorphisms of G. (b) Is the image of c normal in Aut(G)? (c) Determine the kernel of c. 3. The Monster M is a finite simple group of order Can the Monster act nontrivially on a set X of cardinality 48, i.e. does there exist a group homomorphism ϕ : M Perm(X) with trivial kernel? 4. (a) Let G be group and H be a group of finite index. Show that there exists a normal subgroup N of G contained in H and also of finite index. (Hint: If [G : H] = n, find a homomorphism of G into S n (the group of permutations of the set {1,..., n}) whose kernel is contained in H.) (b) Let G be a group and let H 1, H 2 be subgroups of finite index. Prove that H 1 H 2 has finite index. 5. Determine a composition series for the group S 4, the group of permutations of the set {1, 2, 3, 4}. 6. Let G 1,..., G n be finite simple groups. Show that G 1,..., G n are the composition factors of the direct product G 1 G n. 1.3 Cyclic and solvable groups Lemma 11. Theorem 12. Theorem 13. Corollary 14. 8
9 Lemma 15. Theorem 16. Theorem 17. Theorem 18. Theorem 19. Exercises 1. Show that if the automorphism group of the finite cyclic group of order n is itself cyclic that n must be contained in the set S = {2, 4, p k, 2 p k } with p an odd prime and k a positive integer. (Hint: Show Z/nZ = Z/p a 1 1 Z Z/pa l l Z where p a 1 1 pa l l is the prime factorization of n and use this to investigate the ϕ(n) = ϕ(p a 1 ) ϕ(p a l) = (p a (p 1 1)) (p a 1 1 l (p l 1)) generators of Z/nZ. Identify the automorphism group of Z/nZ with the multiplicative group (Z/nZ) of residue classes modulo n of the ϕ(n) nonnegative integers m < n which are relatively prime to n.) 2. If K is a normal subgroup of the group G and K is cyclic, prove that the commutator G is a subgroup of the centralizer C(K) of K. (Hint: Use from Problem 2 that the automorphism group of a cyclic group is abelian.) 3. Let G be a finite solvable group. Prove that there exists a chain G = N 0 > N 1 > > N n = {e} of subgroups of G such that N i is normal in G and N i /N i+1 is abelian for i = 0, 1,..., n 1. (Hint: Prove that a minimal nontrivial normal subgroup M of G is necessarily abelian and use induction. To see that M is abelian, let N be a normal subgroup of M of prime index and show that [x, y] N for all x, y M. Apply the same argument to gng 1, g G, to show that [x, y] lies in the intersection of all Gconjugates of N, and use minimality of M to conclude that [x, y] = e.) 4. Let SL 2 (Z) be the group of 2 2matrices of determinant 1. Show that the commutator subgroup of SL 2 (Z) has index 12. (Hint: Construct a homomorphism from SL 2 (Z) ) to Z/12Z. You can use that SL 2 (Z) is generated by the two matrices S = and T = ( ).) ( (this problem is more number theoretic) Show that the automorphism group of the finite cyclic group of order n is itself cyclic if n is contained in the set S = {2, 4, p k, 2 p k } with p an odd prime and k a positive integer. 1.4 Group actions Lemma 20. Theorem 21. Theorem 22. Lemma 23. 9
10 Lemma 24. Lemma 25. Theorem 26. Lemma 27. Theorem 28. Exercises 1. Let G be a finite group acting on a finite set X. For g G, let X g be the fixpoints of g, i.e., X g = {x X gx = x}. Prove Burnside s formula X/G = 1 G X g for the number of Gorbits. (Hint: Count the subset F = {(g, x) gx = x} G X in two different ways.) 2. Let N be a normal subgroup of a finite group G. Let S G be a conjugacy class of elements in G, and assume that S N. Prove that S is a union of n conjugacy classes in N, all having the same cardinality, where n equals the index [G : N.Cent(x)] of the group generated by N and the centralizer Cent(x) in G of any element x S. 3. Let σ S n such that in the cycle decomposition of σ there are λ i cycles of length i for i = 1,..., n. g G (a) Describe the centralizer of σ in S n and determine its order. (b) Describe the normalizer of the cyclic group σ generated by σ in S n and determine its order. (c) Determine the number of pairs of commuting elements in S n, i.e., the number of pairs (σ, τ) with σ, τ S n and στ = τσ. 4. Let H be a proper subgroup of a finite group G. Show that G g G ghg (the conjugacy classes of A n ) Let T S n be a conjugacy class in S n, i.e., the set of all permutations σ S n which have the same cycle type and assume that T A n. Show that each such conjugacy class T is either also a conjugacy class in A n or decomposes in two conjugacy class of A n of the same size; the second case happens if and only if the cycle type of an element of T consists of distinct odd integers. 6. Prove that A n, the alternating group of degree n, is simple for n 5. (Hint: Use that A n is generated for n 5 by 3cycles. Let σ N \ {e} be an element of a normal subgroup N of A n with a maximal number of fixpoints. Show that σ is conjugated to a 3cycle or an element with more fixpoints. Distinguish the two cases that all orbits of σ have size 1 and 2 or that there is at least one orbit of size 3 or larger.) 10
11 1.5 Sylow subgroups Definition. Let p be a prime. (a) A pgroup is a finite group of order p k for some integer k 0. (b) A psubgroup H of a finite group G is a subgroup of G which is a pgroup. (c) A psylow subgroup H of a finite group G is a psubgroup of G of order p k such that p k is the highest power of p dividing the order of G. Lemma 29. Let G be a finite abelian group and p be a prime divisor of the order of G. Then G has an element of order p. Proof: By induction on G. For G = p, G is cyclic and the Lemma is true. Let a G, not be the identity. If ord(a) is divisible by p, say ord(a) = p t, then b = a t has order p. If otherwise ord(a) is relatively prime to p, then the order of G/ a is divisible by p and smaller than G. Hence the factor group contains an element b a of order p. We claim that the order s of b is divisible by p, for we have (b a ) s = b s a = a. Hence p = ord(b a ) divides s. Since p ord(b), we obtain an element of prime order as before. Theorem 30 (Sylow I). Let G be a finite group and p be a prime number dividing the order of G. Then there exists a psylow subgroup of G. Proof: By induction on G. The case G = p is again clear. Assume therefore that for all subgroups H of a given group G, G > p, with H G the theorem is proven. Consider the class equation G = Z + g i S[G : Cent(g i )]. If p Z, then there is a g i with p [G : Cent(g i )]. Since g i Z we have that Cent(g i ) is a subgroup of G different from G. By induction assumption, Cent(g i ) has a psylow subgroup which is also a psylow subgroup of G and we are done. So we suppose p Z. By Lemma 29, Z contains an element a of order p. If p n is the highest power of p dividing G, then p n 1 is the highest power of p dividing the order of the factor group G/ a. By induction assumption, G/ a contains a psylow subgroup P. The inverse image P under the canonical projection π is a psylow subgroup of G: π P : P P/ a = P is surjective with kernel a, hence P = a P : a = p p n 1 = p n. Lemma 31. Let P be a psylow subgroup of G and H be a psubgroup contained in the normalizer of P. Then H < P. Proof: Since H is a subgroup of Nor(P) and P is a normal subgroup of Nor(P), HP is a subgroup and HP/P = H/H P by the first isomorphism theorem. So if HP P, then HP has a power of p as order, and the order is larger than P, contradicting the hypothesis that P is a psylow subgroup. Hence HP = P and H P. Theorem 32 (Sylow II+III). Let G be a finite group and p be a prime number dividing the order of G. Then: 11
12 (a) Any two psylow subgroups of G are conjugated. (b) The number of psylow subgroups in G is a divisor of the index of any psylow subgroup in G and is 1 (mod p). (c) Any psubgroup is contained in a psylow subgroup. Proof: Let Π be the set of psylow subgroups and let G act on Π by conjugation. Let Σ be one orbit under this action. If P Π we can restrict the action of G on Π to P and decompose Σ into Porbits. Each one of these has cardinality a power of p since the cardinality is a divisor of P. If P Σ then {P } is a Porbit of size 1 and this is the only orbit with this property: If {P } is a Porbit then P Nor(P ) and so, by Lemma 31, P P and hence P = P. This shows Σ 1 (mod p). If P Σ then the same argument shows that all Porbits contained in Σ have size divisible by P. Then Σ is divisible by p. This is a contradiction: The normalizer of a P Σ contains P and has therefore index prime to p. So there are no P Σ and so Σ = Π, which means that G acts transitively on Π. This proves (a) and we also have Π 1 (mod p) proving the second part of (b). For the first, note that Π = [G : Nor(P)] which divides [G : P] since G > Nor(P) > P. For H a psubgroup, restrict the Gaction on Π to H. Since the size of any Horbit is a power of p and since Π 1 (mod p), there exists an orbit {P } containing one element. Then H < Nor(P) and so H < P by Lemma 31. This proves part (c). Applications of Sylow theorems Exercises 1. Prove Cauchy s Theorem: Let p be a prime divisor of the order of a finite group G. Then there exists an element a G of order p. 2. Prove that there is no simple group of order pqr, where p, q and r are distinct primes. 3. Prove that there is no simple group of order Let A be a finite abelian group and let p be a prime. The p th power map is the homomorphism defined by ϕ : A A, a ϕ(a) = a p. Let A p be the image and let A p be the kernel of ϕ, respectively. Prove that A/A p = A p. (Hint: Show that both groups are elementary abelian pgroups of the same order. An elementary abelian pgroup is an abelian group where each element besides the identity has order p.) 5. Prove that there is no simple group of order (Hint: Find all possible values for the numbers n p of psylow subgroups and exclude them one by one. Finally, use a permutation representation of degree 819 to exclude the last case.) 6. (Calculation problem): Use a computer and general results from the lecture and this problem set to eliminate all n smaller than 1000 (or 10000) for which no simple group of order n can exist. Analyze the remaining n further. 12
13 7. (Abelian Groups) (a) Determine the number of nonisomorphic abelian groups of order (b) Determine the number of subgroups of order 1000 of Z/1000Z Z/1000Z. 1.6 Products For G 1, G 2,..., G n, we considered already the direct product G 1... G n, which is the group whose underlying set is the cartesian product and the group structure is defined componentwise. Inside G 1... G n there is for i = 1,..., n the subgroup Lemma 33. bla bla. Proof: H i = {} Next, we generalize the direct product to arbitrary many factors: Theorem 34. Let. Proof: One can reverse the direction of all arrow in Theorem?? and ask if the so obtained universal problem has a solution. This is indeed the case and the corresponding product is called the coproduct or free product of the groups G i. Before we can prove this, we need some preparation: Theorem 35. Let X be a set. Then: (a) There exists a group F(X) and a map ι : X F(X) such that the following universal property is satisfied: For every group G and map f : X G there exists a unique homomorphism ϕ : F(X) G such that the diagram commutes. (b) ι is injective. (c) The image ι(x) generates F(X)....(f = ϕ ι) Proof: Let Y = X {+, } and W(X) = n=0 Y n. The elements of W(X) can be identified with the words x ǫ 1 1 x ǫ 2 2 x ǫn n where n Z and n 0, x 1,..., x n X and ǫ 1,..., ǫ n {+, }. For n = 0 one gets the empty word which we denote by e. A word is called { reduced if there is no i = 1,..., n 1 such that (*) x i+1 = x i and ǫ i+1 = ǫ i = if ǫi = +, Let F(X) W(X) be the set of reduced words. One defines the reduction + if ǫ i =. 13
14 map r : W(X) F(X) by omitting from a word successively beginning from the left all strings x ǫ i i xǫ i+1 i+1 satisfying (*). On F(X) we define now a product by (x ǫ 1 1 x ǫn n ) (y δ 1 1 yδm m ) := r(x ǫ 1 1 x ǫn n y δ 1 1 yδm m ). This binary composition makes F(X) into a group: 1. The empty word e F(X) is the identity. 2. The inverse of x ǫ 1 1 x ǫn n is x ǫn n x ǫ For the associativity law consider for x X {+, } the map σ x : F(X) F(X) defined by { σ x (x ǫ 1 1 x ǫ 2 xx ǫ 1 2 x ǫn n ) = 1 x ǫ 2 2 x ǫn n, if x x ǫ 1 1, x ǫ 2 2 x ǫn n, if x = x ǫ 1 1. Since σ x 1 σ x is the identity map of F(X), σ x is a permutation of F(X). Let A(X) be the subgroup of the permutation group of F(X) which is generated by {σ x x X}. One easily checks that the map σ : W(X) A(X), x ǫ 1 1 x ǫn n σ ǫ x 1 σ 1 x ǫn n restricted to F(X) is a bijection, which respects the binary operation. The point is that σ = σ r and so σ(r(x ǫ 1 1 x ǫn n y δ 1 1 yδm m )) = σ ǫ x 1 σ ǫ 1 x 1 σ δ 1 y 1 σ 1 y δm m = σ(x ǫ 1 1 x ǫn n ) σ(y δ 1 1 yδm m ). Since A(X) is a group, hence associative, so is F(X). We define ι : X F(X) by ι(x) = x + which is clearly injective and ι(x) generates F(X) by construction. This proves part (b) and (c). For the universal property, note that the map ϕ must satisfy ϕ(x ǫ 1 1 x ǫn n ) = f(x 1 ) ǫ 11 f(x n ) ǫn1, hence ϕ is unique. By defining ϕ in this way, one gets indeed a homomorphism since the reduction map does not change the value of ϕ. Definition. The group F(X) is called the free group generated by X. Remark: One has F( ) = {e}, the trivial group; F({x}) = Z, the infinite cyclic group; F({x, y}) is nonabelian. Theorem* (Schreier). Subgroups of a free group are free groups. Definition. Let S be a subset of a group G. The intersection N G S N N of all normal subgroups of G containing S is called the normal hull of S and denoted by S. Remark. (a) S is the smallest normal subgroup containing S. (b) S = {a G a = gxg 1 with g G and x S }. Definition. Let G be a group. A presentation for G is a pair (S, R), where S is a subset of G and R is a subset of F(S) such that F(S)/ R = G and the projection map extends the identity map of S. One writes G = S R. G is said to be finitely generated (resp. finitely presented) if there is a presentation (S, R) of G such that S is finite (resp. S and R are finite). 14
15 It is clear that if G = S R one has G = S, i.e. S generates G. Remark. Every group has a presentation. Proof: Take S = G and consider the identity map S G. By the universal property of the free group F(S) there is a homomorphism ϕ : F(S) G extending this map. Hence ϕ is surjective and G = F(S)/ ker ϕ. So G = G ker ϕ. In fact, one has kerϕ = {a b (ab) 1 a, b G}. Examples: Z/nZ = {x} {x n } = x x n for short. Z Z = x, y [x, y]. Theorem 36. Let I be a nonempty index set and let {G i } i I be a family of groups. Then: (a) There exists a unique group i I G i together with a family {ι i } i I of homomorphisms ι j : G j i I G i such that the following universal property is satisfied: For every group G and every family {α i } i I of homomorphisms α j : G j G there exists a unique homomorphism ϕ : i I G i G such that for all j I the diagram (b) The homomorphisms ι i are injective....(ϕ ι j = α j )commutes. (c) i I G i is generated by the images of the ι i. Proof: Exercises 1. (Presentations) Prove the following group presentations: (a) a, b a 2 b 2, a 1 bab = Q 8, the quaternion group of order 8 defined in Section 1.1. (b) τ 1,..., τ n 1 τ 2 i for 1 i n 1; (τ i τ i+1 ) 3 for 1 i n 2; (τ i τ j ) 2 for 1 i, j n 1 with i j 2 = S n, the symmetric group of degree n. (Hint: τ i is the transposition (1 i+1). Use induction on n.) (c) a, b, c bab 1 a 2, cbc 1 b 2, aca 1 c 2 = {e}, the trivial group with one element. 2. (Coproducts for Abelian Groups) Prove the existence of coproducts for abelian groups, i.e., prove Theorem 39 with each occurrence of the word group replaced by abelian group. 3. (Pushouts) Let F, H and G be groups and f : G F, h : G H be homomorphisms. Prove the existence of a group F G H (called the pushout of f and h) and homomorphisms q 1 : F F G H, q 2 : H F G H with q 1 f = q 2 g such that the following universal property is satisfied: For any group J and homomorphisms j 1 : F J, j 2 : H J with j 1 f = j 2 g there exists a homomorphism ϕ : F G H J such that j 1 = q 1 ϕ and j 2 = q 2 ϕ. 15
16 4. Show that the group S, T S 2, (ST) 3 is isomorphic to the group PSL 2 (Z) = SL 2 (Z)/{±( )}, where SL 2(Z) is the group of 2 2matrices with integer entries and determinant 1. (Hint: Consider the action of SL 2 (Z) by Möbius transformations on the upper halfplane H = {z C I(z) > 0} of complex numbers given by (( a b az+b b c ), z) cz+d ; cf. the book by J.P. Serre: A course in Arithmetic.) 5. Show that there are at most two nonisomorphic groups G of order pq, where p and q are primes with p < q. (Hint: G is either a direct or a semidirect product of its Sylow subgroups.) 6. Let G and F be groups and ϕ : G F be a surjective group homomorphism. Let N = kerϕ be the kernel of ϕ. Prove that G is a semidirect product with normal subgroup N if and only if there is a homomorphism s : F G such that ϕ s = id F, the identity map on F. One says in this case that the short exact sequence splits and s is called a split of ϕ. 1.7 Examples of finite groups {e} N G ϕ F {e} 16
17 Chapter 2 Ring Theory 17
18 Chapter 3 Module Theory 18
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