Algebra I Notes. Clayton J. Lungstrum. July 18, Based on the textbook Algebra by Serge Lang

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1 Algebra I Notes Based on the textbook Algebra by Serge Lang Clayton J. Lungstrum July 18, 2013

2 Contents Contents 1 1 Group Theory Basic Definitions and Examples Subgroups Homomorphisms Lagrange s Theorem and Its Applications Normal Subgroups Sequences Towers of Subgroups Commutator Subgroups Groups Acting on Sets Applications of Group Actions Advanced Topics in Group Theory Sylow Theorems Free Abelian Groups Ring Theory Basic Definitions and Examples Operations on Ideals Homomorphisms of Rings Localization Modules Noetherian and Artin Rings Field Theory Introduction to Field Theory More on Field Extensions Extensions of Field Homomorphisms

3 Chapter 1 Group Theory 1.1 Basic Definitions and Examples Definition A monoid is a set M together with a composition law such that it is (i) associative (i.e., (xy)z = x(yz)) for all x, y, z M; (ii) There exists e M such that ex = x = xe for all x M. Proposition In a monoid, the identity element e is unique. Proof. To see that the identity is unique, suppose that there are 2 identities, say e 1 and e 2. Then we have: e 1 = e 1 e 2 = e 2 since being an identity implies it is a left identity as well as a right identity. Definition A subset N of a monoid M is a submonoid if it is itself a monoid under the same operation and having the same identity as M. Note Being a subset and a monoid is not enough! The identity must also be the same. Example Z Z is a monoid under multiplication with identity (1, 1). Let N = Z {0}, which is clearly a subset of Z Z and a monoid under multiplication, but the identity is (1, 0, therefore N is not a submonoid of Z Z. Example Let M = (Z,, 1) and N = ({0},, {0}). N is certainly a subset and a monoid under the same operation, but it is not a submonoid because the identity is 0 and not 1. Following shall be several examples of monoids with the following conventions: N = {0, 1,...} and P = {1, 2,...}. We shall also denote the power set of a set S by P(S) and the set of all transformations (maps) from S to S by M(S).. (i) (N,, 1) 2

4 CHAPTER 1. GROUP THEORY (ii) (N, +, 0) (iii) (P,, 1) (iv) (Z,, 1) (v) (Z, +, 0) (vi) (P(S),, ) (vii) (P(S),, S) (viii) (M(S),, id S ) Example Let S = {1, 2}. Then we have the following transformations of S: ( ) ( ) Id: α : ( ) ( ) β : γ : This gives us the following multiplication table: Id α β γ Id Id α β γ α α Id γ β β β β β β γ γ γ γ γ More generally, if S = {1, 2,..., n}, then S n is the subgroup of M(S) consisting only of permutations. Note that we get the following properties: S n = n! and M(S) = n n. Definition A semigroup is an algebraic structure consisting of a set together with an associative binary operation. Example Let S be a set with a product such that ab = b for all a, b S. Then this is a semigroup that is associative, but is a monoid only if S = 1. Example Let M be all strings of 8 letter words with a product of taking the first 4 letters of the first word followed by the last 4 letter of the last word. We can check associativity, thus it is a semigroup (since there is no identity element). If we reverse the product (i.e., take the last 4 letters of the last word followed by the first 4 letters of the first word), we lose associativity. Example Let (M, P, 1) be a monoid, where the product is P. Let m M and define a new product P m such that P m (x, y) = xmy. We can check associativity; however, it is only a monoid if m has a 2-sided inverse. Definition A group is a monoid in which every element has a 2-sided inverse. Equivalently, a group is a set with an operation satisfying: 3.

5 CHAPTER 1. GROUP THEORY (i) associativity; (ii) identity; (iii) invertible. 1.2 Subgroups Definition We say H is a subgroup of G if H G and H is a group under the same operation as G. When checking if a subset is actually a subgroup, we need to show three things: (i) the set is closed under the operation; (ii) the identity is in the set (more generally, that the set is nonempty); (iii) if x H, then x 1 H. As an exercise, show that if H is finite, it is enough to check that H is closed under the operation and nonempty. Example Let M be a monoid. u(m) is the set of invertible elements (also called units; sometimes denoted U) in M. u(m) is always a group. Example We know M(S) is a monoid, but u (M(S)) = Perm(S) = S n is a group under composition. Example The group D 2n = σ, τ, where O(σ) = n and O(τ) = 2. When n = 4, this is also known as the symmetries of the square. Note that it is always the case that D 2n = 2n. Definition The center of a group G is the set of all elements in G that commute with all of the other elements, i.e., Z(G) = {x G : ax = xa for all a G}. Now we shall classify all groups of order less than or equal to 8 (up to isomorphism). Groups of order: 1: {id}; 2: {id,τ}; 3: C 3 ; 4: C 4 or C 2 C 2 = V 4 (also called the Klein 4-group); 5: C 5 ; 6: C 6 = C2 C 3 or S 3 = D6 ; 4

6 CHAPTER 1. GROUP THEORY 7: C 7 ; 8: C 8, D 8, Quaternion (denoted Q 8 ), C 4 C 2, or C 2 C 2 C 2. The quaternion group is Q 8 = i, j, where O(i) = O(j) = 4. Define k = ij. Then ascribe the properties: i 4 = j 4 = k 4 = 1; i 2 = j 2 = k 2 = 1; ij = ( j)i. This implies O( 1) = 2, 1 commutes with i.j, and k, and ji = k. Remark. All subgroups of this non-abelian group are abelian. Definition Suppose G is a group and S G. We say that S generates G and write G = S if G is the smallest subgroup of G containing S. Equivalently, S generates G if Another equivalent definition is if G = (all subgroups of G containing S). G = {s 1 s 2 s n : s i S or s 1 i S}. An important special case is when S = {a}, then we write G = a = {a n : n Z}. Our multiplicative notation will be a n = (a 1 ) n = (a n ) 1, whereas our additive notation will be na = a + a + + a }{{} n times and ( n)a = (na) = n( a). Observe that 0a = 0, though the 0 on the left is from the integers (we re adding 0 multiples of a). Definition A group G is cyclic if G = a for some a G. If we have a cyclic group G, then G has two possible scenarios: (i) a i a j for all i j, then G = ; or, (ii) a i = a j for some i j, and without loss of generality, j > i. Then a j i = e and let d be the smallest positive integer such that a d = e and a = {e = a d, a 1,..., a d 1 }. Let G = a, f : Z G defined by f(n + m) = f(n)f(m). We can check this is a surjective homomorphism, so this gives rise to two possible cases: (i) ker(f) = {0}, thus G = ( Z, + ) ; or, (ii) ker(f) = dz, thus G = Z/dZ. 5

7 CHAPTER 1. GROUP THEORY Theorem Any subgroup H of a cyclic group G is cyclic. Moreover, except in the case where G = and H = {e}, we can always write H = a d, where d is the smallest positive integer such that a d H, and in that case, a s H if and only if d s. Proof. If G is infinite and H = {e}, then H = e, so H is cyclic and we are done in that case. In all other cases, there exists a positive integer m such that a m H, so we can choose d to be the smallest positive integer such that a d H. Note that a d H implies a d H. To show the reverse containment, let x H and we will show that x a d, i.e., x = a dq for some q Z. Since x H G = a, thus x = a s for some s Z. Let s = dq + r where q Z and 0 r < d. Then x = a s = a dq+r = a dq a r, thus a dq a r H, which means ( ) a dq 1 x = a r. }{{} H Hence a r H. If r 0, then a r H for a smaller positive integer than d, contradicting the minimality of d. So r = 0, which means s = dq, i.e., x = a s = (a d ) q a. Corollary The only subgroups of Z are {0}, Z, 2Z,.... Corollary Let H be a subgroup of G = a and let d be as above (the smallest positive integer such that a d H). Then d n where n = O(a), i.e., the smallest positive integer such that a n = e. We may also write G = n. Proof. a n = e and e H, thus d n. Corollary Let n be the smallest positive integer n such that a n = e. Then a s = e if and only if n s. Proof. Let H = {e}. Then n here is d in the previous theorem, so a s H if and only if n s. In finite cases, one can deduce that H = n d = q. Corollary If G is a cyclic group of order n, then G has exactly 1 subgroup of order q for each q n. Example Suppose G = a and O(a) = 45. There are 6 subgroups: (i) a = {e, a, a 2,..., a 44 } (ii) a 3 = {e, a 3, a 6,..., a 42 } (iii) a 5 = {e, a 5, a 10,..., a 40 } (iv) a 9 = {e, a 9, a 18,..., a 36 } (v) a 15 = {e, a 15, a 30 } 6

8 CHAPTER 1. GROUP THEORY (vi) a 45 = {e} A natural question is to ask about the subgroups whose order doesn t divide the order of the group. Let us take a 25 from above. The most natural first step is to list the elements generated, that is a 25 = {a 25, a 5, a 30, a 10, a 35, a 15, a 40, a 20, e} = a 5. Theorem Let n = O(a). Then O(a i ) = n. gcd(n,i) Corollary a i = a j if and only if gcd(i, n) = gcd(j, n). Corollary If G = a and n = O(a), then G = a i if and only if i and n are relatively prime. Corollary If G = a and n = O(a), then G has ϕ(n) generators, where ϕ is Euler s ϕ function or Euler s totient function. Definition ϕ(n) is the number of positive integers less than or equal to n and relatively prime to n. Notice we require the equal part so we can include 1. Theorem If G has at most one subgroup of order q for each q n, then G is cyclic. We ll present two proofs of the preceding theorem; one will rely on G being abelian, the other proof will not assume abelian from the beginning. Proof 1. For the abelian case, we ll make use of the Fundamental Theorem of Finite Abelian Groups, thus G = Z p r 1 Z p r 2 Z p rs. Assume G has at most 1 cyclic group of order q for any q n. Then the p i must all be distinct as otherwise G contains a copy of Z p s Z p s. But that implies G has more than 1 subgroup of order p, so all the p i are distinct. Then G is isomorphic to a direct sum of cyclic subgroups of order q and are pairwise relatively prime. Hence, G is cyclic. In the general case, don t assume G is abelian to start. We ll use the fact the G = n and thus, for all x G, x divides n; this is known as Lagrange s Theorem. We ll also use Euler s ϕ-function, and the following lemma: Lemma Let n Z +. Then n = d n ϕ(d). Proof. Let G be a cyclic group of order n. Then there are ϕ(d) elements of order d for each d n, and there are 0 elements of order d for each d n. Every element of G has some order (since it is cyclic and of order n), so n = G = d = d n number of elements in G of order d ϕ(d). 7

9 CHAPTER 1. GROUP THEORY Proof 2. Now suppose G is any group of order n such that G has at most 1 cyclic subgroup of order d for each d n. Then if d n, the number of elements in G of order d is either 0 or ϕ(d) by the lemma and G has no elements of order d if d does not divide n. Then G = d n ϕ(d), so this equals n. In particular, the number of elements of order n is ϕ(n) 1, so G has at least 1 element x such that O(x) = n = G, thus x = G implies G is cyclic. Corollary Any finite subgroup G of the multiplicative group of any field must be cyclic. In particular, if K is a finite field, then the set of nonzero elements of K, denoted by K, is cyclic under multiplication. Also, in C, the group of all n th roots of unity is cyclic for all n. Proof. Let G = n. The equation x d 1 has at most d roots in any field. This implies G can have at most d elements x satisfying x d = 1. This implies G can have at most 1 cyclic subgroup of order d. Example Z p is cyclic for all primes p. Take p = 11; then 2 = {2, 4, 8, 16 = 5, 10, 20 = 9, 18 = 7, 14 = 3, 6, 12 = 1}. Note Z n is not always cyclic. It is cyclic if and only if n = 2, 4, an odd prime p, a power of an odd prime p, or twice the power of an odd prime p. Note Z p 2, where p is an odd prime, Z 25 is cyclic, Z 25 = 20, but isn t a multiplicative subgroup of a field. 1.3 Homomorphisms Definition A map f : M M between 2 monoids is called a monoid-homomorphism if (i) f(xy) = f(x)f(y) for all x, y M. (ii) f(1 M ) = 1 M. Definition A map f : G G between two groups is called a group-homomorphism if f(xy) = f(x)f(y) for all x, and y in G. Notice that for groups, this condition is sufficient to imply that f(1 G ) = 1 G. Example An example that the second condition is necessary for monoids. Let f : Z Z Z. Both sets are monoids under multiplication, but if we define f(a) = (a, 0), we satisfy the first condition, but not the second. 8

10 CHAPTER 1. GROUP THEORY (i) If a homomorphism f : G G is one-to-one, we call it a monomorphism and we say it is an embedding, denoting it by f : G G. (ii) If a homomorphism f : G G is onto we call it an epimorphism and say it is surjective, denoting it by f : G G. (iii) A homomorphism f : G G is an isomorphism if it is bijective, that is, one-to-one and onto. Definition An endomorphism on G is a homomorphism f from G into itself. If f is bijective, we call it an automorphism. Example An endomorphism which isn t an automorphism is given by ϕ : Z Z where ϕ(z) = 2z. Example ϕ g : Z j + G j defined by ϕ g (n) = g n. The image is {g n : n Z} = g. Example Let G be an abelian group. Fix n Z and define ϕ : G G by ϕ(x) = x n. Example G = G 1 G 2 G n, define the projection ϕ i : G G i by ϕ i ((x 1,..., x i,..., x n )) = x i. The projection is onto the i th component and the kernel is G 1 {0} G n, where the {0} is the i th component. If G = S, then any homomorphism f : G G 2 is determined by its action on S. Note A special case is when G = a. Then any homomorphism f : G G 2 is determined by f(a), but f(a) cannot be arbitrary in general. Another special case is when G = D 2n = σ, τ. Here, any homomorphism is determined by f(σ) and f(τ). Theorem The composition of two homomorphisms is always a homomorphism. Theorem A map f : G G 2 is a bijection if and only if there exists f 1 such that f 1 : G 2 G such that f f 1 = id G2 and f 1 f = id G. In this case, f is a homomorphism if and only if f 1 is a homomorphism. Let f : G G 2 be a homomorphism and define ker(f) = {x G : f(x) = e G2 }. The kernel of f turns out to be a normal subgroup of G. The image under f is and is a subgroup of G 2. Im(f) = {y G 2 : y = f(x) for some x G} Theorem A homomorphism f : G G is one-to-one if and only if ker(f) = {e}. 9

11 CHAPTER 1. GROUP THEORY Proof. Suppose f is one-to-one. We know f(e) = e from the properties of group homomorphisms, so {e} ker(f). Since f is one-to-one, nothing else can map to e. Conversely, suppose ker(f) = {e}. Then let f(x) = f(y) so that e = f(x) (f(y)) 1 = f(xy 1 ). Since ker(f) = {e}, we have xy 1 = e, i.e., x = y. Corollary A surjective homomorphism is an isomorphism if and only if its kernel is trivial. Let f : G G be a homomorphism. Then ker(f) = K = {g G : f(g) = e } Ka = {ka : k K} ak = {ak : k K}. Theorem If g G, then Kg is the inverse image of f(g), i.e., the set {x G : f(x) = f(g)}; the same is true for gk. Proof. It is clear that Kg {x G : f(x) = f(g)}; if x Kg, then x = kg for some k K. Thus, f(g) = f(kg) = f(k)f(g) = e f(g) = f(g). Conversely, let x {x G : f(x) = f(g)}. Then f(x) = f(g), which means f(x)[f(g)] 1 = e. Using the property of homomorphisms, we have f(xg 1 ) = e, so xg 1 K, i.e., xg 1 = k for some k K, thus x = kg and we have x Kg. This finishes the proof. Corollary gk = Kg for all g G. Proposition Suppose H and K are subgroups such that (i) hk = kh for all h H and k K (ii) H K = {e}. Then HK is a subgroup and f : H K HK defined by f((h, k)) = hk is an isomorphism. In particular, HK = H K. Proof. It is easy to show HK is a subgroup (true whenever H and K normalize each other). So look at the map f : H K HK. Clearly it is onto since e H K. Also, it is clear it is a homomorphism (note this uses the fact that the elements commute with each other). Now we will show that the kernel is trivial, which implies the injectivity of the function. To that end, suppose (h, k) ker(f). Then we have f((h, k)) = e = hk. Thus, h = k 1. Therefore, h K and h H (since H and K are subgroups), thus h H K = {e} and we have h = e = k. Thus ker(f) = {(e, e)}, and so f is injective. Note We have to use (a) to show that f is a homomorphism, though we can change it to H and K being normal subgroups. This implies the original (a). Just notice that hkh 1 k 1 = (hkh 1 )k 1 K since K is normal. Similarly, hkh 1 k 1 = h(kh 1 k 1 ) H since H is normal. Hence, hkh 1 k 1 = e, i.e., hk = kh. 10

12 CHAPTER 1. GROUP THEORY Proposition Let H 1, H 2,..., H n be subgroups such that (i) H i G for all i (ii) H 1 H 2 H i 1 H i = {e} for all 1 i n. Then H 1 H 2 H n is a group and it is isomorphic to n i=1 H i. Let G be a group and a G, then define ah = {ah : h H} and Ha = {ha : h H}. Proposition a bh if and only if ah = bh. Proof. If a bh, then a = bh for some h H. Therefore, we have ah = (bh)h = bh, thus ah = bh. Corollary Distinct cosets are disjoint. Definition Any element of gh is called a representative of gh. Proposition We have ah = H if and only if a H (since e H). Proposition The order of ah is the same as the order of H, i.e., ah = H. 1.4 Lagrange s Theorem and Its Applications Theorem (Lagrange s Theorem). Let G be a group of order n. If G has a subgroup H of order d, then d n. Proof. If we take one element from each coset and let that be the representative for that coset, we have G = gh. Then we have G = gh = (#ofcosets)( H ), thus the order of H divides the order of G. Corollary A group G with G > 1 has no nontrivial proper subgroups if and only if G = p, a prime, in which case G is cyclic. Proof. If G = p, a prime, and H is a subgroup, then either (i) H = p, thus H = G; or (ii) H = 1, thus H = {e}. Conversely, suppose the only subgroups are H and {e}. Since G > 1, let a G with a e. Let H = a, a subgroup of G and not e. Thus, G = a, hence G is cyclic. Clearly G is not of infinite order since infinite cyclic groups have infinitely many subgroups, and G is not a composite number as cyclic groups have at least 1 subgroup of order d, hence G is prime. 11

13 CHAPTER 1. GROUP THEORY Corollary All groups of prime order are cyclic. Corollary If G is a group and G = n, then a n = e for all a G. Proof. Let a G and let d = O(a). Then let H = a and note that H = d. By Lagrange s Theorem, d n, so say n = dr. Then a n = a dr = (a d ) r = e r = e. Corollary (Fermat s Little Theorem). Let p be a prime. If a Z such that a p, then a p 1 1 (mod p). Note that we can state an equivalence in the following fashion; let p be a prime and a Z, then a p a (mod p). Proof. Denote the nonzero elements of Z p by Z p. Then Z p is a group under multiplication of order p 1. Also, note that Z p = (Z/pZ). Theorem (Euler s Theorem). Suppose m Z + and a Z such that gcd(a, m) = 1. Then a ϕ(m) 1 (mod m). Proof. Done! Question: Is the converse of Lagrange s theorem true? Answer: The converse of Lagrange s Theorem is true under restrictions, i.e., it s true if G is cyclic, abelian, or nilpotent, but it is not true in general. Example A 4 = 12, but there is no subgroup of order 6. This is because if there were such a subgroup N, then N would have to contain at least one element of order 3, call this element a. Then we have in A/N that (an) 3 = N, but A/N only has two elements, hence we must have a N. This tells us that N contains all of the elements of order 3, but there are 8 elements of order 3, hence N has order greater than 6, a contradiction to the assumption. Therefore, A 4 must not have a subgroup of order 6. Definition [G : H] is the index of H in G, i.e., the number of left cosets of H. Proposition If G is a group, and K H G, then [G : K] = [G : H][H : K]. Let G = S n, H = {σ G : σ(n) = n}. Then [G : H] = n. Let σ i = (n, i), that is σ i (n) = i, σ i (i) = n, and σ i (k) = k for all k i, n. Notice, then that σ n is the identity. Claim σ i H = {σ G : σ(n) = i} Proof. Clearly σ i H {σ G : σ(n) = i}. Now, σ i H = H = (n 1)! = {σ G : σ(n) = i}. An alternate proof follows. Again, clearly the left-hand side is contained in the right. Conversely, if σ {σ G : σ(n) = i}, then σ(n) = i implies σ 1 i (σ(n)) = σ i (σ(n)) = n, thus σ 1 σ H, i.e., σ σ i H. 12

14 CHAPTER 1. GROUP THEORY So, {σ 1, σ 2,..., σ n } is a complete set of left coset representatives and also a complete set of right coset representatives. In general, if {a 1, a 2,..., a n } is a complete set of left coset representatives and also a complete set of right coset representatives as x a i H if and only if x 1 Ha 1 i. In our case, σ i = σ 1 i. Let K H G, with G and H as before. Let K = {σ S n : σ(n) = n and σ(n 1) = n 1}. The left coset representatives for H in G are σ i = (n, i) and the left coset representatives for K in H are τ j = ((n 1), j). The left coset representatives for K in G are σ i τ j since Thus, we have σ i τ j if i j, i n 1; (n, i)(n 1, j) σ i τ j if i = n 1 j; σ i τ j if i = j; n 1 n j. n n 1 j n σ i τ j K = {σ G; σ(n) = i and σ(n 1) = j} σ i τ j K = {σ G; σ(n) = n 1 and σ(n 1) = j} σ i τ j K = {σ G; σ(n 1) = n and σ(n) = j}. Let f : G G with kernel K. Recall ak = {x G : f(x) = f(a)} and Ka = {x G : f(x) = f(a)}. Corollary ak = Ka. 1.5 Normal Subgroups Theorem Let G be a group and H G a subgroup of G. The following are equivalent: (i) every left coset of H in G is also a right coset of H in G; (ii) ah = Ha for all a G; (iii) aha 1 = H for all a G; (iv) aha 1 H for all a G; (v) H is the kernel of a group homomorphism. Proof. We have established equivalence between the first three, as well as the third condition implies the fourth and the fifth condition implies the fourth. Let us consider how the fourth condition implies the third. Suppose aha 1 H for all a G. Then since a 1 G, we have a 1 Ha H for all a G, thus H aha 1 for all a G. Thus aha 1 = H. 13

15 CHAPTER 1. GROUP THEORY Definition H G, that is, H is a normal subgroup, if any one the first four conditions above is true. Corollary Any subgroup of index 2 is normal. Let G = D 4 = {σ, σ 2, σ 3, id, στ, σ 2 τ, σ 3 τ, τ}. The subgroups of order 4 are σ and {σ 2 τ, τ, σ 2, id}. All of these are normal subgroups, being of index 2. If G is abelian, then all subgroups are normal; however, the converse of this statement is not true in general. Consider Q 8 = {i, j, k, i, j, k, 1, 1}, with o(i) = o(j) = o(k) = 4 and i 2 = j 2 = k 2 = 1. Now, let us shift the focus to factor groups. Consider G/N = {an : a N}. Our approach will be such that for any subsets S 1, S 2 of N, define S 1 S 2 = {s 1 s 2 : s 1 S 1, s 2 S 2 }. So, (xn)(yn) = x(ny)n = x(yn)n = xyn. We could approach this from a second standpoint: let G and S be subsets and define x y if and only if x 1 y S. It is an exercise to show that is an equivalence relation if and only if S is a subgroup. So, suppose S is a subgroup. For x G, define x = [x] = {g G : g x}. Our goal is to define xy = xy. We must issue a word of caution here, though. We need to check the operation is well-defined, i.e., we need to show that if x = x and y = y, then xy = x y. In order for this to be true, we need S to be normal. We can check x = xh. Then, note here, the approach gives the same multiplication as the first approach, that is, xhyh = x y xy = xyh. It is easy to check when the above multiplication is well-defined, thus we have G/N is a group under x y = xy when x = xn. Example Take the additive group of the integers and the subgroup nz. Then Z/nZ = {0,..., n 1} = Z n. In G/N, x = e if and only if xn = N if and only if x N. If N G, then G/N is a group under multiplication. Define ϕ : G G/N by x x = xn. We can check it is a homomorphism: Also, we have the kernel as follows: ϕ(xy) = xyn = (xn)(ny) = (xn)(yn) = ϕ(x)ϕ(y). ker(ϕ) = {x G : ϕ(x) = e} = {x G : x = e} = {x G : x N} = N. An important property of homomorphisms is that if f : G G is a homomorphism and ker(f) = K, then G/K = Im(f). Theorem Let f : G G be a homomorphism. Let K = ker(f), f(g) = Im(f). Then there exists a unique f that makes the diagram commute. 14

16 CHAPTER 1. GROUP THEORY Proof. Define f (x) = f (xk) = f(x). We can check f is a well-defined homomorphism, and it is injective as f (gk) = e if and only if f(g) = e if and only if g K, i.e., g = e. Let G be a group, S G. Definition The normalizer of S in G is defined to be N G (S) = {g G : gsg 1 = S}. Note N G (S) is a subgroup. If H is a subgroup of G, then H N G (H). As an exercise, verify the above note. In general, N G (H) is the biggest subgroup of G in which H is normal. If H G, then N G (H) = G. If H K and K N G (H), then H K. Even if H K, if K is a subgroup of G and K N G (H), then H K. We have a special case when S = {a}. Then N G (S) = C G (a) = {g G : ga = ag}, where C G (a) is called the centralizer of a. More generally, if S G, then Z S = {g G : gs = sg for all s S}. Note that Z G is called the center of G. Example (i) G = D 8 and G = 8. The center of G is Z G = {id, σ 2 }. (ii) G = Q 8, then Z Q8 = {1, 1}. (iii) G = D 2n, then Z G = {id, σ n 2 when n is even, and Z G = {id} when n is odd. (iv) S n when n 3, then Z G = {id}. 1.6 Sequences Definition A sequence G f G g G is exact if Im(f) = ker(g). Some special cases include the following: (i) To say that {0} G f G is exact means f is injective. (ii) To say that G g G {0} is exact means g is surjective. 15

17 CHAPTER 1. GROUP THEORY (iii) To say that {0} G f G g G {0} is exact means G f G (injective), so we can identify G with a subgroup of G; G g G (surjective). Since G = ker(g), we have G = G/G. Definition We say that f i f i+1 f i+2 G i Gi+1 Gi+2 is exact if and only if all of the short exact sequences are exact. f i+1 {0} f i (G i ) G i+1 fi+1 (G i+1 ) {0} f 1 f 2 f 3 {0} G 1 G2 G3 G4 {0} We have f 1 is injective and f 3 is surjective; if we identify G 1 with its image in G 2, then G 1 = ker(f 2 ). Also, G 4 = G3 / ker(f 3 ) = G 3 /Im(f 2 ) = coker(f 2 ). Observe that a homomorphism f is injective if and only if ker(f) = {0} while f is surjective if and only if coker(f) = {0}. Conversely, if f : G G is a homomorphism, More generally, if f : G G, {0} ker(f) G f G coker(f) {0}. {0} ker(f) G G/ ker(f) {0}. (i) G f f[g] ϕ G/ ker(f) f (ii) N G and N ker(f) G f f[g] G/N f (iii) If H G, K G, and K H, then K H. There exists a natural homomorphism G/K G/H given by xk xh. We need to check the map is well-defined, i.e., check if xk = yk implies xh = yh. 16

18 CHAPTER 1. GROUP THEORY Proof. We know xk = yk if and only if x 1 yk = K. This only happens if x 1 y K, if and only if x 1 yh = H. From this it follows yh = xh. It is easy to check it is a homomorphism and surjective. Observe that kernel = {xk : xh = H} = {xk : x H} = H/K. Thus, we can conclude that / G/K H/K = G/H. (iv) Let G be a group with subgroups H and K where H N G (K). Then HK = KH is a group and H K H. Moreover, ϕ : HK HK/K where K HK, H HK HK/K, by ϕ(h) = hk. Observe that ker(ϕ) = {h H : hk = K} = {h H : h K} = H K. Corollary H / H K = HK / K. We can write the above corollary more generally. Theorem Let H, K be subgroups of a finite group G. Then HK = H K / H K. Proof. HK is a subset of G. We can still write it as a union of cosets by noting (i) K = n i=1 (H K)k i where n = [K : H K] = K H K (ii) HK = n i=1 H(H K)k i = n i=1 Hk i. H K So, we will be done if we can show cosets are distinct, as then HK = H n =. But H K this is easy. Suppose Hk i = Hk j. Then k i k 1 j H but also k i k 1 j K, so k i k 1 j (H K)k i if and only if i = j. Let f : G G be a group homomorphism and let H be a subgroup of G. Define H to equal f 1 (H ). It is easy to check H is a subgroup. Theorem H G implies H G. Proof. f(xhx 1 ) = f(x)f(h)f(x 1 ) = H, therefore xhx 1 f 1 (H ) = H. 17

19 CHAPTER 1. GROUP THEORY Theorem (i) Let f : G G be a surjective homomorphism. Then there exists a one-to-one correspondence between subgroups H of G and subgroups H of G containing ker(f) given by H = f 1 (H) and H = f(h). (ii) Moreover, H G if and only if H G, and in that case, G/H = G/H. Proof. (i) From the homework, we know that f 1 (H) = HK, where K = ker(f). (ii) Similar in nature to the homework exercise. 1.7 Towers of Subgroups Definition A tower of subgroups is G = G 0 G 1 G m. A normal tower 1 is a tower such that G i 1 G i. An abelian normal tower is a normal tower in which all of the factor groups G i /G i+1 are abelian. A refinement of a tower is just a tower gotten by inserting a finite number of subgroups into a given tower. The refinement of a normal tower is completely analogous. Note Observe that G i+1 G i does not imply G i+1 G (consider D 8, for example). Proposition Refinement of cyclic (respectively, abelian) towers are cyclic (respectively, abelian). Proof. Suppose G i G i+1, so that when you refine, you get G i K G i+1. Observe there is an injective homomorphism from K/G i+1 into G i /G i+1, and this is enough. Definition A group is called solvable if it has an abelian normal tower ending in {e}. Example (i) G = D 2n σ {e} with G/ σ = Z 2 and σ /{e} = C n, where C n is the cyclic group with n elements. (ii) S 1, S 2, S 3, and S 4 are solvable, but S n is not solvable for n 5. (iii) All abelian groups are solvable. (iv) W. Feit and J. Thompson proved that all groups of odd order are solvable. (v) A Galois group is solvable if and only if the associated polynomial is solvable by radicals. 1 Some books use subnormal series (such as Thomas W. Hungerford s textbook, Algebra). 18

20 CHAPTER 1. GROUP THEORY (vi) Subgroups and homomorphic images of solvable groups are solvable. (vii) If N G, then G is solvable if and only if both N and G/N are solvable. Definition A composition series is a normal tower admitting no nontrivial refinements. Equivalently, G = G 0 G 1 G n = {e}, each G i+1 is the maximal normal proper subgroup of G i. Another equivalent statement is to say that each G i /G i+1 is a simple group. Proposition Any normal tower for a finite group G with no repetitions can be refined to get a composition series. Proof. If all G i /G i+1 are simple, we are done. If some G i /G i+1 is not simple, say we start with the first one, then G i contains a normal subgroup N i containing G i+1. Set G i N i G i+1, where G i /N i is simple. If N i /G i+1 is simple, we stop and go to the next i where G i /G i+1 is not simple. If N i /G i+1 is not simple, then repeat the above process as many times as necessary until the desired result is achieved. We know this process will eventually terminate since the order of G is finite. Proposition Any simple abelian group is cyclic of prime order. Proof. Abelian implies that all subgroups are normal and simple implies that the only normal subgroups are G and {e}. Now, let x G, then x with x e would be isomorphic to Z, thus x 2 would be a proper subgroup, a contradiction. Thus, x has finite order, say n. If n is composite with d n, then x d is a proper subgroup of x, again a contradiction. Thus, we deduce that G is cyclic of prime order. Corollary A finite group is solvable if and only if it has a cyclic tower ending in {e}. Proof. A cyclic tower implies that the tower is also abelian. Conversely, if G has an abelian tower ending in {e}, refine this tower to get a composition series (which we can do since G is finite). The factors G i /G i+1 for the composition series are simple and abelian, which, from the above, implies they are cyclic. Example (i) S 3 C 5 A 3 C 5 {id} C 5 {e} is a composition series. All of the factors are cyclic (thus abelian) as (S 3 C 5 )/(A 3 C 5 ) = C 2 (A 3 C 5 )/({id} C 5 ) = C 3 ({id} C 5 )/{e} = C 5. 19

21 CHAPTER 1. GROUP THEORY (ii) S 3 C 5 S 3 {id} A 3 {id} {e} is a composition series distinct from above, but observe (S 3 C 5 )/(S 3 {id}) = C 5 (S 3 {id})/(a 3 {id}) = C 2 (A 3 {id})/{e} = C 3 Thus, at least in this case, even though the composition series are different, their factors are just a permutation of each other. The following theorem generalizes this statement. Theorem (Jordan-Hölder). Let G be a finite group and let G = G 0 G s = {e} and G = H 0 H r = {e} be composition series for G. Then r = s and there exists a permutation i i of 0, 1,..., s 1 such that G i /G i+1 = Hi /H i +1. Proof. We will use induction on the order of G. Theorem Let G be a group, N G. Then G is solvable if and only if N and G/N are solvable. Proof. If G is solvable, start with an abelian tower for G ending in {e}. Derive from it an abelian tower for both N and G/N. Lang (our author) goes through the details for N (many details to check). Conversely, if N and G/N are both solvable, start with an abelian tower for G/N ending in {e} and lift back to get an abelian tower for G ending in N (Lang does something similar a page or two prior to this theorem in the book). Glue this tower together with an abelian tower for N ending in {e} and get an abelian tower for G ending in {e}. Details left for students (me). 1.8 Commutator Subgroups Definition If x, y G, then we define the commutator of x and y by (x, y) = xyx 1 y 1. Think of (x, y) as the obstacle keeping x and y from commuting as xy = (x, y)yx. Definition The commutator subgroup of G, denoted G C, is the subgroup of G generated by all the x, y G, i.e., the set of all finite products of all commutators. Note G C may not equal the set of all commutators as that set is closed under inverses, but may not be closed under products! Proposition (i) Any subgroup containing G C is a normal subgroup. In particular, G C G. 20

22 CHAPTER 1. GROUP THEORY (ii) More generally, any time N G, then N C G. (iii) If N G, then G/N is abelian if and only if G C N. In particularly, G/G C is abelian. Note G C = {e} if and only if G is abelian. Proof. (i) Suppose H is a subgroup of G containing G C. Then let g G and h H; we have ghg 1 = (ghg 1 )h 1 h = (ghg 1 h 1 )h where ghg 1 h 1 is in the commutator subgroup, and thus in H, while h H by hypothesis, hence their product is in H, as H is a subgroup. Thus, H is normal. See Homework Set 4 for an alternate proof. (ii) Let xyx 1 y 1 N C for x, y N. Argue that g(xyx 1 y 1 )g 1 N C and explain why that is enough. This says suppose (x, y) N C. Show that g(x, y)g 1 N C, but this is easy to show since g(x, y)g 1 = (gxg 1, gyg 1 ) which is a commutator of N. You can fill in the remainder of the details. (iii) Let N G. G/N is abelian if and only if xy = yx if and only if xyx 1 y 1 = e for all x, y G/N if and only if xyx 1 y 1 = e for all x, y G if and only if G C N. An alternate approach to the first part is to prove directly that G C G using an argument similar to the argument in (ii). Then also show directly G/G C is abelian and then argue G G/G C is onto. G abelian implies all subgroups of G are normal by one-to-one correspondence between normal subgroups of G/K and normal subgroups of G containing K, this implies all subgroups of G containing G C are normal. We will introduce some new notation here. We will let G (0) = G, G (1) = G C, G (2) = (G (1) ) C, etc., i.e., for any i 1, let G (i) = (G (i 1) ) C. Then, by part (i) above, G = G (0) G (1) and by part (iii), G (i) /G (i+1) is abelian and G (s) = {e} for some s G implies G is solvable. Theorem Let G be a finite group, G (i) be as above for every (i), then G is solvable if and only if G (s) = {e} for some s. Note In this case, part (ii) from above implies G (i) G for each i. Proof. Suppose G (s) = {e}. Then we have G = G (0) G (1) G (s) = {e} where the factors are abelian, thus G is solvable. 21

23 CHAPTER 1. GROUP THEORY Conversely, suppose G is solvable. This means there exists a normal tower G = G 0 G 1 G s = {e}, where all G i /G i+1 are abelian. Then, since G/G 1 is abelian, we know that G 1 contains G C = G (1). Now, G 1 /G 2 is abelian, thus (G 1 ) C G 2, but we know G (1) G 1, which implies G (2) (G 1 ) C G 2. In particular, then, we have G (2) G 2. Continuing in this manner, we see that G (s) G s, but G s = {e}, hence G (s) = {e}. Corollary If G is solvable, we can always find an abelian tower where G i G. G = G 0 G 1 G s = {e} Note Not all abelian towers have this property; the corollary simply states that such an abelian tower exists, not that all abelian towers have the property. 1.9 Groups Acting on Sets Definition We say a group G acts from the left on a set S if for all s S and all g G, then gs S such that (i) g 1 (g 2 s) = (g 1 g 2 )s; (ii) es = s for all s S, where e is the identity in G. Claim A group G acts on a set from the left if and only if there is a grouphomomorphism from G into the permutation set of S. Proof. Given a homomorphism ϕ : G Perm(S), where g f g, we can define an action of G on S by gs = f g (s). Lets check that we have the properties from the above definition: (i) g 1 (g 2 s) = f g1 (f g2 (s)) = f g1 g 2 (s) = (g 1 g 2 )s; (ii) es = f e (s) = Id(s) = s, where Id is the identity permutation. Conversely, suppose we re given an action of G on S. Then we must show there exists a corresponding homomorphism from G to the permutation set of S. We can do this by simply defining f g (s) = gs, but we need to check some things. It is clear this gives a map from G to M(S), the maps from S to S, but we have to check that f g Perm(S), i.e., f g is one-to-one and onto and we need to check that g f g is a homomorphism. We ll show that latter statement first. We begin by showing g f g is a homomorphism of monoids, i.e., showing that f g1 f g2 = and f e is the identity map. So, f g1 g 2 f g1 f g2 (s) = f g1 f g2 (s) = f g1 (g 2 s) = g 1 (g 2 s) = (g 1 g 2 )s = f g1 g 2 (s), 22

24 CHAPTER 1. GROUP THEORY and since s S was arbitrary, we have that it is a homomorphism. Likewise, f e (s) = es = s = Id(s), thus, e is mapped to the identity map. It remains to show that f g is a permutation of S, not just a map, i.e., we need to show f g is one-to-one and onto; equivalently, show f has a 2-sided inverse. But this is clear, since f g f g 1 = f e = f g 1f g, thus f g is a permutation of the elements of S. We could also define an action on the right of G on S by g taking s to sg such that (sg 1 )g 2 = s(g 1 g 2 ) and se = s for every s S. If we try to define a map G into the permutation set of S, we cannot define it by g fg (think about inverses to see why this is true). To fix it, instead we map g to sg 1. Example (i) S = G, where the action is conjugation, i.e., f g (s) = gsg 1 = g s. Then ϕ : G Perm(S) = Aut(G), where the kernel is precisely the center of G, all of the elements in G that commute with every element in G. (ii) S = G, where the action is left translation, i.e., f g (s) = gs, then ϕ : G Perm(G). (iii) S is the set of left cosets of H in G, i.e., S = {G/H : H is a subgroup of G, where the action is left translation. Then f g (ah) = g(ah). This gives a homomorphism ϕ : G Perm(G/H where the kernel can be calculated as follows: ker(ϕ) = {g G : gah = ah} = {g G : a 1 gah = H for all a G} = {g G : a 1 ga H for all a G} = {g G : g aha 1 for all a G} = a G aha 1, which is the intersection of all of the conjugates of H. Note then that K = ker(ϕ) has the properties (a) K G; (b) H K; (c) K contains any normal subgroup N of G contained in H. To see the last claim, let N H. Then N = ana 1 aha 1, which means N K, so K is the largest normal subgroup of G contained in H. Corollary Suppose G is a simple group and H is any proper subgroup of index d. Then G divides d!. 23

25 CHAPTER 1. GROUP THEORY Proof. Let G act on G/H by left translation. Let ϕ : G Perm(G/H) be the canonical homomorphism. Then ker(ϕ) is a proper normal subgroup (since it is contained in H), thus ker(ϕ) = {e}, which implies ϕ is one-to one, so G is isomorphic to a subgroup of Perm(G/H), which implies G divides d!. Note the contrapositive of the above corollary: if H is a subgroup of G with index d and if G d!, then H contains a nontrivial normal subgroup of G. Let G be a group and S, S be two G-sets, a map f : S S is called a G-map if f(gs) = gf(s) for all g G and s S. If, in addition, f is a bijection, it is called an equivalence of actions. Now suppose G acts on S. Define the isotropy subgroup G x = {g G : gx = x}. Observe that G x need not be normal. Example (i) G acts on G by conjugation, i.e., S = G where the action is conjugation. Here, we have the centralizer of x in G. G x = {g G : gxg 1 } = {g G : gx = xg} = C G (x), (ii) G acts on subgroups of G by conjugation, thus G H = {g G : ghg 1 = H} = N G (H). Now suppose G acts on S with s, s S and gs = s. What is the connection between G s and G s? Claim G s = gg s g 1. Proof. Clearly gg s g 1 G s. Also, g 1 s = s. So, by symmetry, g 1 G s g G s, thus G s gg s g 1. Definition We say G acts faithfully on S if ker(ϕ) = {e}. Observe that ker(ϕ) = s S G s. In general, if K = ker(ϕ) and S is a G-set and also S is a G/K-set in a natural way; gs = gs. Definition If G acts on S, a fixed point of G is an element x S such that G x = G; equivalently, such that gx = x for all g G. Definition Suppose G acts on S, s S. Define the orbit of s under the action of G to equal [s] = {gs : g G}. Check that if denotes being in the same orbit, then is an equivalence relation. (i) Clearly s [s], thus is reflexive. (ii) If s 1 [s 2 ], then s 1 = gs 2 for some g G, so that g 1 s 1 = s 2, i.e., s 2 [s 1 ], thus, is symmetric. 24

26 CHAPTER 1. GROUP THEORY (iii) If s 1 [s 2 ] and s 2 [s 3 ], then s 1 [s 3 ], hence is transitive. Hence, we can conclude that S is the disjoint union of its orbits under the action of G. We have a special case, though: if there is only 1 orbit, then given any s and s in S, there is a g G such that gs = s. In this case, we say G is transitive. Example (i) S n acts transitively on S = {1, 2,..., n}. (ii) If S is the set of subgroups of G, and G acts on S by conjugation. A natural question arises, what is the orbit of H under the above action? The answer is the set of all conjugate subgroups of H (sometimes called conjugacy class of H); that is, the set [H] = {ghg 1 : g G}. Now we can ask another question: what does it mean if [H] = 1? It means that H is normal, since for all g, H = ghg 1 (observe from the above that e G, thus H is a conjugate of itself). (iii) If S = G where G acts by conjugation. The conjugacy class of x is [x] = {gxg 1 : g G}. Instead of subgroups, then, we ask, what does it mean for [x] = 1? It means that gxg 1 = x for all g G, thus x C(G), the center. With all of these special cases, it is natural to ask if there is anything we can say about group actions on a set S and the conjugacy classes in general. With this in mind, we state the following theorem. Theorem [s] = [G : G s ]. Proof. There exists a natural bijection between the set [s] and the set G/G s of left cosets. Define f : G/H [s] by f(gh) = gs. We must check this is well-defined, but this is clear since if g 1 = g 2 h for some h H, then g 1 s = g 2 h(s) = g 2 (hs) = g 2 (s). We can verify that it is onto; any element of [s] is gs for some g G, and hence it gets mapped to by gh. Similarly, it is one-to-one as suppose g 1 s = g 2 s, then g 1 H g 1 s = g 2 s and g 2 H g 2 s = g 1 s, thus g1 1 g 2 s = g1 1 (g 2 s) = g1 1 (g 1 (s)) = s, thus g1 1 g 2 G s = H, which implies g 2 g 1 G s = g 1 H. Therefore, g 2 H = g 1 H. Corollary [s] divides G. Corollary If G is a group and H is a subgroup, then the number of conjugate subgroups of G equals [G : N G (H)]. Corollary If G is a group, x G, then the number of conjugates of x in G equals [G : C G (x)]. 25

27 CHAPTER 1. GROUP THEORY 1.10 Applications of Group Actions Let G be a group and suppose it acts on a set S. Then observe that which means S = different orbits S = orbits, orbit = [G : G s ]. Lets see an application of this idea. Suppose S = G and G acts on S by conjugation. In this case, x is a fixed point if and only if x C(G). Thus, we can write the above formula as G = C(G) + [G : G x ]. Theorem Suppose G is a finite p-group. Then the center is nontrivial. Proof. Suppose the order of G is p r. Then p r = Z(G) + [G : G x ], where each [G : G x ] > 1 and [G : G x ] divides p r. Thus, p divides [G : G x ]. Since the right-hand side of the equation is divisible by p, we must have that p divides Z(G). Theorem If G/Z(G) is cyclic, then G is abelian. Proof. Let Z = C(G) and suppose G/Z = az. Let x, y G. Then xz = (az) i = a i Z, thus x = a m z 1 for some z 1 Z. Likewise, we know y = a n z 2. Then xy = a m z 1 a n z 2 = a n z 2 a m z 1 = yx, therefore, G is abelian. Corollary Any group of order p 2 is abelian. Proof. By the previous theorems, the center of G is nontrivial, hence C(G) = p or C(G) = p 2. If it s the latter case, we re done. If not, then G/C(G) is cyclic since it is a group of prime order, thus G is abelian, which implies the center has order p 2, a contradiction since we assumed Z(G) p 2. Hence, the center has order p 2 and we re done. Proposition Let G be a finite group. Let H be a subgroup of index p where p is the smallest prime dividing G. Then H G. Proof. Let N = N G (H). Then H N G, hence N = H or N = G. In the latter case, we are done, so assume N = H. Let S be the set of all conjugates of H, let G act on S by conjugation. Then there exists a homomorphism ϕ : G Perm(S) given by this action. Observe that S = [G : N G (H)] = [G : H]. So Perm(S) = S p, so this gives a homomorphism ϕ : G S p. Let K = ker(ϕ ). Then G/K S p, so G/K divides S p = p!, that is, [G : K] divides p!. But K N G (H) = H, therefore [G : K] = [G : H][H : K] = p[h : K], thus p[h : K] divides p!, i.e., [H : K] divides (p 1)!. But also, H = K [H : K], thus [H : K] divides the order of G by Lagrange s Theorem. This mean [H : K] = 1, which is true if and only if H = K, but then H is the kernel of a homomorphism, thus it is normal in G, a contradiction to our assumption that H = N G (H). Therefore, we have H G. 26

28 CHAPTER 1. GROUP THEORY For an alternate proof of the above theorem, see homework set 5. Let J n = {1, 2,..., n} and let S n = Perm(J n ). Definition A cycle (i 1 i 2... i s ) is short-hand notation to say i 1 i 2 i s i 1. Note that the action of σ S n on any given orbit is a cycle and that cycle fixes anything not in the orbit. Definition A transposition is a 2-cycle written (ij). Proposition Any permutation can be written as a product of (not necessarily disjoint) transpositions. Proof. Let (i 1 i 2... i n ) be a cycle. Then (i 1 i 2... i n ) = (i 1 i n )(i 1 i n 1 )... (i 1 i 2 ). Definition A permutation is called an even permutation if it can be written as a product of an even number of transpositions. The definition for an odd permutation is analogous. Theorem No permutation is both even and odd. Proof. Let A M n n (R) and let σ S n. Define σ(a) to be the new matrix achieved by rearranging the rows of A according to σ. Note that τ(σ(a)) = (τσ)(a). Define a map ε : S n { 1, 1} by ε(σ) = det(σ(i n )). Check that ϕ is a homomorphism. We claim that if τ is a transposition, then ε(τ) = 1. Thus, we have that ε : S n { 1, 1} is a homomorphism that sends transpositions to -1, thus odd permutations are sent to -1, which means even permutations are sent to 1, and nothing is sent to both 1 and -1. Then ker(ε) = A n, the set of all even permutations. Theorem S n is not solvable if n 5. S n is solvable if 1 n 4. Lemma Let G = S n and H, N be two subgroups with N H. Then if H contains all 3-cycles, so does N. We will motivate the lemma, then prove the lemma. Proof of Theorem. Assume the lemma, and let us prove the theorem. Suppose S n with n 5 is solvable. Then G 0 = S n G 1 G s = {e} where each G i /G i+1 is abelian. We know G = G 0 contains all 3 cycles and G i /G i+1 is abelian, so G 1 must contain all 3-cycles. Continuing in this manner, we see that G s must contain all 3-cycles, a contradiction since we assumed G s = {e}. 27

29 CHAPTER 1. GROUP THEORY Proof of Lemma. Assume N H, G/N abelian, and H contains all 3-cycles. Let (rki) be an arbitrary 3-cycle; we will show (rki) N. Since n 5, there exist at least 2 elements other than r, k, and i, say j and s. Let σ = (ijk) H and τ = (krs) H. Compute στσ 1 τ 1 = (rki), thus (rki) G C, and since G/N is abelian, N contains the commutator subgroup. Hence, N contains all 3-cycles. Claim A n is generated by 3-cycles. Proof. It suffices to show that any even permutations can be written as a product of 3-cycles since A n clearly contains 3-cycles. Thus, it suffices to show any product of two transpositions can be written by the product of 3-cycles. This leaves us with 3 cases: (i) the two transpositions have two elements in common, in which case they are inverses of each, hence the product is the identity, an even permutation; (ii) the two transpositions have one element in common, thus their product is a 3-cycle; (iii) the two transpositions do not share any elements, thus (ab)(cd) = (abc)(bcd), and we have their product can be written as a product of 3-cycles. This completes the proof. Claim All 3-cycles are conjugate to each other in S n. Proof. See homework set 6 (we actually prove something more general). Claim All 3-cycles in S n are conjugate to each other in A n for n 5. Proof. Let (ijk) and (i j k ) be 3-cycles. Then, by the second claim above, we have (i j k ) = γ(ijk)γ 1 for some γ S n. If γ A n, we re done. If γ / A n, it must be odd. So, choose r and s distinct from i, j, and k (which we can do since there are at least five elements). Replace γ by γ(rs), then γ(rs) A n and the distinctness of r and s from i, j, k allows them to commute, thus γ(rs)(ijk)(rs)γ 1 = γ(ijk)γ 1. Theorem A n is simple (and therefore not solvable) if n 5. Proof. Let N A n be a nontrivial normal subgroup. We will show N contains a 3-cycle. Then, by the third claim above, since N A n, N contains all 3-cycles, thus N = A n. Let id σ N with the maximal number of fixed points acting on {1, 2,..., n}. We will be done if we show σ is a 3-cycle. We split this into cases. 28