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2 Chapter 1 Groups 1.1 Isomorphism theorems Throughout the chapter, we ll be studying the category of groups. Let G, H be groups. Recall that a homomorphism f : G H means a function such that f(g 1 g 2 ) = f(g 1 )f(g 2 ) for all g 1, g 2 G. A homomorphism is called a monomorphism if it is injective, i.e. there exists a set function f : H G such that f f = id G. It s an epimorphism if its surjective, i.e. there exists a set function f : H G such that f f = id H. It s an isomorphism if it s bijective, i.e. both injective and surjective. Note this definition of isomorphism of groups is equivalent to the definition of isomorphism given earlier for an arbitrary category: luckily for groups a morphism is an isomorphism if and only if it is a bijection. We also talk about endomorphisms (homomorphisms from G to itself) and automorphisms (isomorphisms from G to itself). Given a homomorphism f : G H, we set im f = f(g) = {f(g) g G}, ker f = f 1 (1 H ) = {g G f(g) = 1 H }, its image and kernel. A subgroup of a group G is a non-empty subset K of G such that k 1 k 1 2 K for every k 1, k 2 K. This condition implies that K is itself a group in its own right, the operation being the restriction of the operation on G to K. We usually write K G to indicate that K is a subgroup of G. For example, im f H (and f is an epimorphism if and only if im f = H); ker f G (and f is a monomorphism if and only if ker f = {1 G }). Actually, ker f is more than just a subgroup of G: it is a normal subgroup. By definition, a subgroup N G is normal if any of the following equivalent conditions hold: (N1) gng 1 = {gng 1 n N} = N for all g G; (N2) gng 1 = {gng 1 n N} N for all g G; (N3) gng 1 N for all g G, n N; (N4) Ng = gn for all g G. We usually write N G to indicate that N is a normal subgroup of G. The sets Ng = {ng n N} and gn = {gn n N} appearing in (N4) are the right cosets and left cosets of N in G. Let s make this notion more precise. Let K be any subgroup of G. Define an equivalence relation on G called left congruence modulo K by g 1 g 2 if g1 1 g 2 K. The equivalence classes are called the left cosets of K in G: each equivalence class has the form gk for some (any) representative g G of the equivalence class. The set of all left cosets of K in G is denoted G/K. One also defines right cosets to be the equivalence classes of a relation called right congruence modulo K, namely, g 1 g 2 if g 1 g2 1 K. Right cosets look like Kg for g G, and the set of all 11

3 12 CHAPTER 1. GROUPS right cosets is denoted K\G. Note the map g g 1 induces a bijection between G/K and K\G, so it usually doesn t matter whether you choose work with left or right cosets. In case G is a finite group and K G, we let [G : K] = G/K denote the number of left cosets of K in G, the index of K in G. Now, all the left cosets have the same size, namely, K. Since equivalence classes partition G into disjoint subsets, we deduce: Lagrange s theorem. G = K [G : K]. Lagrange s theorem shows in particular that the order of any subgroup of a finite group G divides the order of the group. Now return to the case that N is a normal subgroup of G. Then, (N4) says that G/N = N\G, i.e. left cosets are the same as right cosets so we can just call them simply cosets. Define a multiplication on the set of cosets G/N by (g 1 N)(g 2 N) = (g 1 g 2 )N. The fact that this is well-defined depends on N being a normal subgroup. This multiplication gives G/N the structure of a group in its own right, called the quotient group of G by the normal subgroup N. Note that there is a canonical homomorphism π : G G/N, g gn, which is an epimorphism of G onto G/N with kernel exactly N. The group G/N together with the map π : G G/N has the following universal property: Universal property of quotients. Let N G and π : G G/N be the canonical epimorphism. Given any homomorphism f : G H with N ker f, there exists a unique homomorphism f : G/N H such that f = f π. Exercise. Let G be a group and H, K be subgroups. We can consider the subset HK = {hk h H, k K} of G. (1) HK is a subgroup of G if and only if HK = KH as subsets of G. (2) If either of H or K is a normal subgroup of G, then HK = KH, hence is a subgroup of G. G. (3) If in fact both H and K are normal subgroups of G, then HK is a normal subgroup of G. Now let me state the isomorphism theorems for groups. These are all really consequences of the universal property of quotients: you should be able to prove them for yourselves. First isomorphism theorem. Let f : G H and N = ker f. Then, N G and f factors through the quotient G/N to induce an isomorphism f : G/N im f. Second isomorphism theorem. Let K G, N G. Then, N KN, K N K and K/K N = KN/N. Third isomorphism theorem. Let K N G with K G, N G. Then, N/K G/K and G/N = (G/K)/(N/K). There is one other important result traditionally included with the isomorphism theorems: the lattice isomorphism theorem. First, recall that a relation on a set X is called a partial ordering (and X is called a partially ordered set) if for all x, y, z X (R1) x x; (R2) x y and y x implies y = x; (R3) x y and y z implies x z.

4 1.2. CYCLIC AND DIHEDRAL GROUPS 13 If in addition we have that for all x, y X one of x < y, x = y, x > y holds, then < is called a total order or a linear order (and X is called a totally ordered set). Now let X be a partially ordered set and A X. An element a X is called the join or least upper bound of A if a is the unique minimal element of {x X a x a A}. The join A may or may not exist: for instance there could be many such minimal elements or no minimal element at all. Similarly, an element a X is called the meet or greatest lower bound of A if a is the unique maximal element of {x X a x a A}. A partially ordered set is called a lattice if every pair of elements of X has both a join and a meet in X. A partially ordered set is called a complete lattice if every non-empty subset of X has both a join and a meet in X. Let G be any group and X be the set of all subgroups of G, partially ordered by inclusion. Then, X is a complete lattice. The meet of a set of subgroups of G is simply their intersection, also a subgroup of G. The join of a set of subgroups of G is the subgroup generated by the subgroups, that is, the intersection of all subgroups of G that contain all the given subgroups. More generally, given any subset A of G, we write A for the subgroup of G generated by A, namely, the intersection of all subgroups of G that contain A. In the special case A = {a} for some a G, we write simply a for the subgroup of G generated by the element a. As a variation, let G be any group and X be the set of all normal subgroups of G, partially ordered by inclusion. One checks that the intersection of a family normal subgroups is normal, and that the group generated by a family of normal subgroups is normal. Hence, X in this case is again a complete lattice, with join and meet being as for subgroups. Lattice isomorphism theorem. Let f : G G be an epimorphism with kernel K. Then, the map H f(h) gives an isomorphism between the lattice of subgroups (resp. normal subgroups) of G containing K and the lattice of subgroups (resp. normal subgroups) of G. (The inverse map sends a subgroup H of G to its preimage f 1 (H ) in G.) 1.2 Cyclic and dihedral groups It s nearly time for some examples of groups. But first, let G be any group and g G an element. Recall g denotes the subgroup of G generated by the element g. So, g = {1 G, g ±1, g ±2, g ±3,... }, and is called the cyclic subgroup generated by g. It may or may not be a finite group; if it is finite, its order is denoted o(g), the order of the element g. Note Lagrange s theorem implies that in a finite group, the order of every element divides the order of the group. A group G is called cyclic if G = g for some element g G. We then say that g is a generator of the cyclic group G. For example, the group (Z, +) of integers under addition is an infinite cyclic group generated by the integer 1 (the identity element of Z is the integer 0 of course we use the additive notation in this Abelian group!). Actually, any infinite cyclic group is isomorphic to the group (Z, +). We can classify all subgroups of the infinite cyclic group: Lemma. If H (Z, +), then H = n for some integer n 0. Proof. Take n > 0 minimal subject to the condition that n H (if no such n exists, then H = {0} and there is nothing to prove). Then, n H. Now take any m H, and write m = qn + r for integers q, r with 0 r < n. Then, r = m qn H, so by minimality of n we have that r = 0. Hence, m = qn, so that m n. This shows that H = n. Note that n m if and only if m n. It follows that the lattice of subgroups of Z is isomorphic to the opposite of the lattice of non-negative integers partially ordered by divides. The non-negative integer n corresponds under the isomorphism to the subgroup n. Given n N, we define the group (Z n, +), the group of integers modulo n under addition, to be the quotient group Z/ n. Thus, the elements of the group Z n are the cosets of n in Z, namely, {[0], [1],..., [n 1]} where [i] = {i + jn j Z}. We have that [i] + [j] = [i + j]. The group Z n is the cyclic group of order n: it is generated by the element [1] which has order n in Z n. Sometimes

5 14 CHAPTER 1. GROUPS we denote the cyclic group of order n instead by C n, to indicate that we re writing the operation multiplicatively instead. So C n = {1, x, x 2,..., x n 1 } where x is any element of C n of order n. By the lattice isomorphism theorem, the lattice of subgroups of Z n is isomorphic to the opposite of the lattice of divisors of n. The divisor d of n corresponds under the isomorphism to the cyclic subgroup of Z n generated by [d], which is a subgroup of order n/d. In other words: Lemma. The group Z n has a unique subgroup of order d for each divisor d of n, namely, the cyclic subgroup generated by [n/d]. Given groups G 1, G 2, their (external) direct product is the set G 1 G 2 (Cartesian product) with coordinatewise multiplication. Actually, G 1 G 2, with the obvious projections π i : G 1 G 2 G i is the product of G 1 and G 2 in the categorical sense. More generally given a family G i (i I) of groups, their product i I G i is simply their Cartesian product as sets with coordinatewise multiplication. In other words, the category of groups possesses arbitrary products (see (0.3.2)) Lemma. If n = st with (s, t) = 1, then Z n = Zs Z t. Proof. The element ([1] s, [1] t ) Z s Z t has order st as (s, t) = 1. Hence it generates a cyclic subgroup of Z s Z t of order st. Hence, it generates all of Z s Z t, which is therefore isomorphic to the cyclic group of order n. The Euler φ function is defined by φ(n) = #{x Z n o(x) = n} = #{1 k < n (k, n) = 1}. By the lemma, if n = st with (s, t) = 1, then φ(n) = φ(s)φ(t). It follows immediately that to compute φ(n) it suffices to know φ(p n ) for each prime power p n. In this special case, it is an exercise to show that ( φ(p n ) = p n 1 1 ). p Here s an important number theoretic fact about the Euler φ function: Lemma. For n N, n = d n φ(d). Proof. We have from Lagrange s theorem that n = Z n = d n (the number of elements of Z n of order d). Now if x Z n has order d, it generates a subgroup of order d, isomorphic to Z d. By Lemma 1.2.2, Z n has a unique such subgroup of order d. Hence, the number of elements of Z n of order d equals the number of elements of Z d of order d, namely φ(d). I end the section by introducing one other basic example of a finite group: the dihedral group D n of order 2n. This is probably easiest left as an exercise. So, start with the group O 2 of orthogonal linear transformations of the plane R 2. So O 2 consists of all rotations (det = +1) around the origin and all reflections (det = 1) in axes passing through the origin. Consider the elements f, g O 2 where g is reflection in the x-axis and f is counterclockwise rotation through the angle 2π/n. Obviously, f n = 1, g 2 = 1. Let D n be the subgroup of O 2 generated by f and g.

6 1.3. THE SYMMETRIC GROUP 15 (1) Show that the product f i g is reflection in the line at angle iπ/n to the x-axis. Hence, the transformations 1, f, f 2,..., f n 1, g, fg,..., f n 1 g are all distinct. (2) Now show that the elements {1, f, f 2,..., f n 1, g, fg,..., f n 1 g} form a subgroup of O 2. (3) Deduce that D n = {1, f, f 2,..., f n 1, g, fg,..., f n 1 g} is a group of order 2n. Picture: D n is the group of symmetries of a regular n-gon. It contains the cyclic group C n as a subgroup, namely, the subgroup {1, f, f 2,..., f n 1 } consisting of all rotations. (4) Find all normal subgroups of D n. (5) For any group G, its centre is defined as Determine Z(D n ). 1.3 The symmetric group Z(G) = {x G xy = yx for all y G}. We come now to one of the most important examples of a group. Let X be any set. Denote by A(X) the set of all bijective ( invertible ) functions from X to itself. Then, A(X) is a group with multiplication being composition of functions. The identity element of the group A(X) is the identity function id X : X X, x x. The group A(X) is called the symmetric group on X. In the special case that X is the set {1, 2,..., n}, we denote the group A(X) instead by S n and call it the symmetric group on n letters (even though they re numbers!). Obviously, S n = n! so S n is a finite group. Now let G be any group. Then, A(G) is also a group. Define λ : G A(G), g λ g where λ g A(G) is the function x gx (i.e. λ g is left multiplication by g ). Note λ is a group homomorphism. Moreover, it s injective, for if λ g = id G, then λ g (1 G ) = 1 G = g so ker λ = {1 G }. Hence, λ defines an isomorphism between G and im λ A(G). We ve proved: Cayley s theorem. Every group is isomorphic to a subgroup of the symmetric group A(X) for some set X. Every finite group, is isomorphic to a subgroup of S n for some n. By the way, you can also define ρ : G A(G), g ρ g, where ρ g is the function right multiplication by g. But ρ satisfies ρ(gh) = ρ(h)ρ(g), the wrong way round to be a homorphism. In fact, ρ is an antihomomorphism from G to A(G). Now let s discuss the finite group S n in more detail. So it s the group of all permutations of {1,..., n}. Let x 1,..., x a be distinct elements of {1,..., n}. Denote the permutation which maps x a x 1, x i x i+1 for each i = 1,..., a 1 and fixes all other points, by (x 1 x 2... x a ). This is called an a-cycle. A 2-cycle is called a transposition. Disjoint cycle notation. Every permutation f S n can be written as a product of disjoint cycles f = (x 1 x 2... x a )(y 1 y 2... y b ).... (z 1 z 2... z a ). Moreover, this representation of the permutation f is unique up to deleting 1-cycles and reordering the product (disjoint cycles commute!). (You should be quite familiar with using the disjoint cycle notation.) Now define a function sgn : S n {±1} as follows. Take g S n. Write g = c 1... c a as a product of disjoint cycles, where c i is an o i -cycle. Set a sgn(g) = ( 1) oi 1. So for example, sgn(1) = 1, sgn(t) = 1 for t a transposition,... i=1

7 16 CHAPTER 1. GROUPS Lemma. Every element g S n can be expressed as a product of transpositions. Proof. Using the disjoint cycle decomposition, it suffices to show that any a-cycle can be written as a product of transpositions. Then, for example, the cycle (12... n) equals (12)(23)(34)... (n 1 n) Theorem. sgn is a group homomorphism. Proof. We first prove: Claim. If s is a transposition and w S n is arbitrary, then sgn(sw) = sgn(w). To see this, say s = (a b) and w = c 1... c m as a product of disjoint cycles. If s and w are disjoint, the conclusion follows by definition. Else, without loss of generality, we may assume a appears in the cycle c 1. If b appears in c 1 too, then c 1 = (a x 1... x k b y 1... y l ) and (a b)c 1 = (b y 1... y l )(a x 1... x k ) which is a product of disjoint cycles. In then follows by definition of sgn that sgn((a b)w) = sgn(w). The other possibility is if b appears in another of the cycles, say c 2 (allowing 1-cycles). Now another similar explicit calculation with cycle notation expresses (a b)c 1 c 2 as a product of disjoint cycles, and the conclusion again follows using the definition of sgn. Now we can prove the theorem. We need to show that sgn(xw) = sgn(x) sgn(w) for any x, w S n. Write x = s 1... s m as a product of transpositions, applying Lemma 1.3.1, and proceed by induction on m, the case m = 1 being the claim. For m > 1, set y = s 1 x = s 2... s m. Then, by the claim, sgn(x) = sgn(y); by the induction hypothesis, sgn(yw) = sgn(y) sgn(w) = sgn(x) sgn(w). Hence, using the claim once more, sgn(xw) = sgn(s 1 (yw)) = sgn(yw) = sgn(x) sgn(w) Corollary. If w S n is written as a product of m transpositions, then sgn(w) = ( 1) m. We define the alternating group A n to be the kernel of the homomorphism sgn : S n {±1}. Providing n > 1, A n is a normal subgroup of S n of index 2. The elements of A n are called even permutations. A group G is called simple if it has no proper normal subgroups. For instance, the cyclic group C p where p is a prime is a simple group for Lagrange s theorem implies that C p has no proper subgroups at all. The goal in the remainder of the section is to prove that A n is simple for n 5. (On the other hand, A 4 is not simple, for it contains the Klein 4 group V 4 = {1, (12)(34), (13)(24), (14)(23)} as a normal subgroup of index 3.) As a first step to the goal, we need to understand the conjugacy classes of the group S n. In any group G, the conjugate of an element x G by g G is defined to be g x := gxg 1. The conjugacy class of x is the set G x := { g x g G}. The conjugacy classes of G partition it into disjoint subsets (because conjugacy classes are the equivalence classes of the equivalence relation is conjugate to ). Observe that a subgroup N G is a normal subgroup if and only if it is a union of conjugacy classes of G: so once we understand conjugacy classes we can quite easily test subgroups for normality Lemma. Take g, x S n and write as a product of disjoint cycles. Then, x = (a 1... a s )(b 1... b t )... gxg 1 = (ga 1... ga s )(gb 1... gb s )....

8 1.3. THE SYMMETRIC GROUP 17 Proof. Calculate what each side does to an integer i. Define the cycle-type of x S n to be the ordered tuple (o 1, o 2,..., o l ) consisting of the orders of the disjoint cycles of x in cycle notation, arranged so that o 1 o 2 o l > 0. You should include all the trivial 1-cycles so that in addition l i=1 o i = n. For instance (1 2) S 5 has cycle-type (2, 1, 1, 1). The lemma immediately implies the description of the conjugacy classes of S n : Theorem. The conjugacy classes of S n are precisely the {x S n x has cycle-type λ} as λ runs over all conceivable cycle-types. Now we can prove that A n is simple for n 5. We proceed with a series of lemmas Lemma. The group A 5 is simple. Proof. Let us list the conjugacy classes in S 5. Cycle-type Size sgn Splits in A 5? (5) 24 + yes (4, 1) 30 (3, 2) 20 (3, 1, 1) 20 + no (2, 2, 1) 15 + no (2, 1, 1, 1) 10 (1, 1, 1, 1, 1) 1 + no Now an S n -conjugacy class that is contained in A n can either be equal to a single A n -conjugacy class, or else it can be a union of two A n -conjugacy classes of the same size. We have listed in the table which of the S 5 -classes split as A 5 -classes in this way. Therefore, the conjugacy classes of A 5 have sizes 12, 12, 20, 15, 1. Now, a normal subgroup of A 5 must be a union of these conjugacy classes, must contain 1 and must be of order dividing 60. There s no way to do this by elementary arithmetic based on the orders of the classes in A 5! Lemma. If H A n for n 5 and H contains a 3-cycle, then H = A n. Proof. Show that A n is generated by 3-cycles and all 3-cycles are conjugate in A n Lemma. The group A 6 is simple. Proof. Let {1} H A 6. Suppose 1 g H has a fixed point. Without loss of generality, suppose g6 = 6. Then, g lies in the naturally embedded subgroup A 5 < A 6. So, {1} H A 5 A 5, so H A 5 = A 5 by simplicity of A 5. Hence, H contains a 3-cycle, so H = A 6 by the previous lemma. This reduces to the case that all elements 1 g H move all 6 points. Now consider the two possible cycle types one by one and get a contradiction. For instance, if (12)(3456) H, then its square is a non-identity element with a fixed point. Otherwise, (123)(456) H (or an element of the same cycle-type). In this case, conjugating gives that But this is a non-identity element that fixes 1. (123)(456)(234)(123)(456)(243) H Theorem. For n 5, A n is simple.

9 18 CHAPTER 1. GROUPS Proof. We may assume that n 7. Take {1} = H A n. Using induction on n, the argument in the proof of the previous lemma reduces to the case when all 1 g H move all of 1,..., n. Take such a g such that, without loss of generality, g1 = 2. Consider (234)g(243)g 1 H. Its a non-trivial element since it sends 2 to 3. But its a product of two 3 cycles, so it can move at most 6 points, so as n 7 it must have a fixed point. 1.4 Free groups Let X be a set. A group G is said to be free on X if (1) X G; (2) for every set function f : X H to a group H, there exists a unique group homomorphism f : G H whose restriction to X G equals f. (It is sometimes convenient to weaken condition (1) to say just that there is a given distinguished injective function i : X G. In this formulation, (2) would say that there exists a unique f : G H such that f = f i.) This definition is our first detailed example of a definition by a universal property. It will always be the case with such definitions that if such an object exists, it is unique up to canonical isomorphism. The proof always follows the same pattern, and the one below is the only one I ll ever give: Lemma. If G and G are both free on X then there is a unique isomorphism j : G G such that j(x) = x for all x X. Proof. Consider the inclusion maps f : X G and g : X G. Since G is free on X, there exists a unique f : G G which is the identity on X. Since G is free on X, there exists a unique ḡ : G G which is the identity on X. Now, f ḡ : G G is a group homomorphism that is the identity on X. But since G is free on X, there exists a unique such homomorphism, namely, the identity on G. Hence, f ḡ = id G. Similarly, ḡ f = id G. In view of the lemma, we will henceforth abuse notation and call G the free group on X if it satisfies the properties (1) and (2) above. There is now a problem: it is not at all clear whether such a group even exists at all! This is always the case with univeral property definitions: we now need to give an explicit construction to prove existence Theorem. For every set X, there exists a group G that is free on X. Proof. We need a construction. Let X 1 = {x 1 x X} be a set of distinct elements disjoint from X. By convention, we set (x 1 ) 1 = x for x X. Let 1 be an element disjoint from X X 1. A word means an ordered tuple (a 1, a 2,..., a n ) for n 1 with each a i {1} X X 1. A word is called reduced if (1) a 2,..., a n 1, and a 1 1 in case n > 1; (2) a i a 1 i+1 for i = 1,..., n 1. Given two reduced words, we define their product by concatenation followed by replacement of xx 1 s by 1 s and removal of 1 s whenever possible. Then we would like to check that the set of reduced words on X, with this operation, forms a group G. But this is hard: the problem is proving associativity. So instead, we proceed indirectly via a trick. Let W be the set of reduced words. For each x X, define functions x and x 1 from W to W by { x ε x (x ε xεn n ) = ε x ε xεn n if x ε x ε1 1, x ε xεn n if x ε = x ε1 1.

10 1.4. FREE GROUPS 19 Note x is the inverse function to x 1. Hence, each x is an element of the symmetric group A(W ). Now define G to be the subgroup of A(W ) generated by the elements { x x X}. We claim that G, together with the function i : X G, x x, is free on X. We first need to observe that i is injective. To see this, note any element of G can be expressed as a reduced word x ε xεn n in the x, x 1. Moreover this reduced representation is unique, for applying this function to 1 yields x ε xεn n, a reduced word in W, and words in W have unique reduced spelling by definition. In particular, since each x for x X is reduced, we get from this that i is injective. Now we verify that G, together with the map i : X G, really is free on X. Take f : X H any function to a group H. Define f : G H on a reduced word x ε xεn n by f( x ε xεn n ) = f(x 1 ) ε1... f(x n ) εn. This is well-defined since we showed that reduced words have unique spelling. It just remains to check that f is a group homomorphism. But this follows since cancellation in G implies cancellation in H Corollary (of proof). If G is free on X, then G is generated by X. Proof. in X. We showed in the construction that every element of G can be written as a reduced word Now suppose that X is a set and R is a set of reduced words in X. Define the group X R, the group given by generators X and relations R, to be the quotient group F/N, where F is the free group on X and N is the normal subgroup of F generated by R (the intersection of all normal subgroups of F containing R). Informally, X R is the largest group that can be generated by the elements X in which the relations R hold. Formally: Theorem. Given any group G generated by (not necessarily distinct) elements {t x x X} satisfying the relations R, there is a unique epimorphism f : X R G such that f(x) = t x for each x X. Proof. Let F be the free group on X, and π : F X R be the canonical quotient map. By the universal property of free groups, there exists a unique homomorphism h : F G such that h(x) = t x for each x X. Since the relations are satisfied by the t x, R ker h. Hence, the normal subgroup of F generated by R lies in ker h. Hence, by the universal property of quotients, h factors through the quotient X R of F to induce a unique f : X R G such that f(x) = t x for each x X. Now for some examples of groups given by generators and relations (which can often be an entirely intractable way of defining a group!). Consider G = a, b a n = b 2 = 1, ab = ba n 1. Using the relations, one easily shows that every element of G can be reduced to one of the words a i b j, i = 0, 1,..., n 1, j = 0, 1. Hence, G 2n. To prove that G = 2n is hard from the point of view of generators and relations. However, we have already seen the dihedral group D n which is generated by a rotation f of order n and a reflection g, and these satisfy the relation fg = gf n 1. Hence, applying Theorem 1.4.4, the map a f, b g extends to a unique epimorphism G D n. This shows that G D n = 2n. Hence, G = 2n and G = D n. Here s a harder example giving generators and relations for the symmetric group S n. Let G n be the group with generators {s 1, s 2,..., s n 1 } subject to the relations s 2 i = 1, s i s j = s j s i for i j > 1 and s i s i+1 s i = s i+1 s i s i+1. Let S n denote the symmetric group, and t i denote the basic transposition (i i + 1) in S n. (1) Prove that the t i satisfy the same relations as the s i. (2) Embed S n 1 into S n as the subgroup consisting of all permutations fixing n. Prove that {1, t n 1, t n 2 t n 1,..., t 1 t 2... t n 1 } is a set of S n /S n 1 -coset representatives. (3) By considering the subgroup G n 1 of G n generated by s 1,..., s n 2 only and using induction, prove that G n = Sn.

11 20 CHAPTER 1. GROUPS 1.5 Group actions We say a group G acts (on the left) on a set X if there is a given map G X X, (g, x) gx such that (A1) 1x = x for all x X; (A2) (gh)x = g(hx) for all x X, g, h G. We call X a (left) G-set if G acts on X. Equivalently, X is a G-set if the map α : G A(X), g α g, where α g (x) = gx for x X, is a group homomorphism. Then, the G-set X is called faithful if this map α is injective. So the action is faithful if and only if gx = x for all x X implies g = 1. The action of G on X is called transitive if for every x, y X, there exists at least one g G such that gx = y. More generally, let X be a G-set and define an equivalence relation on X by x y if there exists a g G such that gx = y. The equivalence classes are called the orbits of G on X; the orbit of x X is denoted Gx = {gx g G}. So the action is transitive if and only if there exists just one orbit. Elements x X with gx = x for all g G (i.e. the orbit of x is just {x}) are called fixed points, and we write F ix(x) for the set of all fixed points. Given x X, its stabilizer G x, or the isotropy group of x, is defined to be G x = {g G gx = x}. Note G x is a subgroup of G. So, x is a fixed point if and only if G x = G. Given x X, the resulting orbit map is the map f : G X, g gx. Clearly, f(g) = f(g ) if and only if g 1 g G x, i.e. gg x = g G x, i.e. g and g lie in the same left G x -coset. Hence, the orbit map induces a well-defined map G/G x Gx, gg x gx. This map is clearly surjective, and we checked above that it is injective. In other words, the orbit map induces a bijection between the cosets G/G x of G x in G and the orbit Gx of x. In particular, if G is finite, we get that Gx = [G : G x ], so the size of an orbit divides the order of G Lemma. If x and y lie in the same orbit, then G x and G y are conjugate in G, i.e. there exists g G such that gg x g 1 = G y. Proof. Pick g G such that gx = y, i.e. x = g 1 y. Then, for h G x, we have that ghg 1 y = ghx = gx = y, so gg x g 1 G y. Similarly, g 1 G y g G x. Hence, G y gg x g 1. Now we have the language, we are ready for examples Given any G-set X, a subset Y X is called G-stable if gy Y for each g G, y Y. So, Y is G-stable if and only if it is a union of orbits. In that case, Y is itself a G-set in its own right via the restriction of the action of G on X to Y If H < G is any subgroup, then H acts on G by left multiplication, i,e, (h, g) hg. The orbits are the right cosets H\G If H < G is any subgroup, then H acts on G by (h, g) gh 1. The orbits are the left cosets G/H If H, K < G, then H K acts on G by ((h, k), g) hgk 1. The orbits are the double cosets K\G/H, subsets of the form HgK = {hgk h H, k K}.

12 1.5. GROUP ACTIONS G acts on itself by conjugation, (g, x) gxg 1. The orbits are just the conjugacy classes of G. In this case, given x G, the stabilizer G x = {g G gxg 1 = x} is called the centralizer of x in G. We often write C G (x) for the centralizer of x in G. The order of the conjugacy class G x is the index of C G (x) in G: G x = [G : C G (x)]. The fixed points for this action are precisely the elements in the center Z(G). An element x G lies in Z(G) if and only if C G (x) = G More generally, G acts on the set of subsets of G by conjugation. So for a subset S G, the action is by (g, S) gsg 1 = {gsg 1 s S}. The stablizer G S of S under this action is usually called the normalizer of S in G, denoted N G (S). So, N G (S) = {g G gsg 1 = S}. In particular, if H is a subgroup of G, we have its normalizer N G (H), and H N G (H). In particular, H G if and only if G = N G (H). For an arbitrary subgroup H G, the number of distinct conjugates of H in G is given by the formula G H = [G : N G (H)] The symmetric group S n acts faithfully on the numbers {1,..., n} by the very definition of S n! This action is transitive so that there is a single orbit, and each point stabilizer is isomorphic to the symmetric group S n The dihedral group D 4 of symmetries of the square acts naturally on the set of vertices (labelled {1, 2, 3, 4} say) of the square. This yields a homomorphism α : D 4 S 4 which is in fact injective, hence D 4 embeds as a subgroup of S 4. In other words, the action of D 4 on the vertices of the square is faithful. Given a finite G-set X, we have the class equation X = F ix(x) + x [G : G x ] where the sum is over a set of representatives x of the non-trivial orbits. Applying this in particular to the conjugation action of G on itself, we obtain G = Z(G) + x [G : G x ] summing over a set of representatives for the conjugacy classes of non-central elements. Let p be a prime. We call a finite group G a p-group if the order of G has order a power of p. (More generally, a possibly infinite group G is called a p-group if every element of G has order a power of p). Now if X is a G-set and x X is not a fixed point, then G x is a proper subgroup of G, so [G : G x ] is a divisor of G strictly greater than 1. So if G is a finite p-group, [G : G x ] is a proper power of p, so [G : G x ] 0 (mod p). Taking both sides of the class equation above modulo p, we deduce: Lemma. If G is a finite p-group and X is a finite G-set, then F ix(x) X (mod p). Applying this to the conjugation action of G on itself, this shows in particular: Corollary. If G is a finite p-group, then Z(G) {1}. Proof. By the lemma, Z(G) G 0 (mod p). But 1 Z(G), so Z(G) 1. But 1 0 (mod p) so we cannot have Z(G) = 1. You should be able to prove now that if G = p 2 for a prime p, then G is Abelian.

13 22 CHAPTER 1. GROUPS Lemma. Let G be a finite Abelian group and p be a prime dividing the order of G. Then, there exists x G with o(x) = p. Proof. Say G = pm. If m = 1, then G = C p and the result is obvious. If m > 1, take 1 x G. If o(x) = pk for some k, then x k has order p and we re done. So we may assume that p o(x). Then, p G/ x, so by induction there exists an element y G such that y / x but y p x. Then, y ph = 1 for some h coprime to p (indeed, h divides the order of x). So the order of y h divides p, so either equals p, as required, or 1. But in the latter case, we have that y h = 1. Write 1 = ah + bp for integers a, b. Then, y = y ah+bp = y bp which lies in x as y p x. Hence, y x, a contradiction. Cauchy s theorem. If G is a finite group and p is a prime dividing the order of G, then there exists an element x G with o(x) = p. Proof. Let G = pm. If m = 1, the conclusion is obvious. Otherwise, we have the class equation: G = Z(G) + x [G : C G (x)]. For each x in the sum, C G (x) < G, so if p C G (x) for any such x we get the conclusion by induction. Hence, p C G (x) for any such x, in other words, p divides each [G : C G (x)]. Since p also divides G, we deduce that p divides Z(G). Then Z(G) contains an element of order p by the lemma. 1.6 The Sylow theorems Now suppose that G is a finite group of order p n k for a prime p with (p, k) = 1. A subgroup H G with H = p n is called a Sylow p-subgroup. Equivalently, H is a Sylow p-subgroup of G if H is a p-group and p [G : H] Lemma. If H is a p-subgroup of G, then [N G (H) : H] [G : H] (mod p). In particular, if p [G : H] (i.e. H is not a Sylow p-subgroup of G) then N G (H) H. Proof. Let H act on G/H by left multiplication. Then, ah F ix(g/h) if and only if HaH = ah which is if and only if a 1 Ha = H which is if and only if a N G (H). Hence, [N G (H) : H] = F ix(g/h). Now apply Lemma First Sylow theorem. Let H < G be a p-subgroup of G that is not a Sylow p-subgroup. Then, there exists a p-subgroup K G with H K and K = p H. In particular, H can be embedded into a Sylow p-subgroup of G. Proof. Since p [G : H], we have by the preceeding lemma that H N G (H), H N G (H) and N G (H)/H is a group of order divisible by p. Pick any x N G (H)/H of order p. Then, the pre-image in N G (H) of x is a subgroup of order p H which contains H as a normal subgroup. Second Sylow theorem. All Sylow p-subgroups of G are conjugate. Proof. Let H, K be two Sylow p-subgroups of G. Let H act on G/K by left multiplication. Then, ak F ix(g/k) if and only if HaK = ak which is if and only if a 1 Ha K, i.e. a 1 Ha = K since they have the same order. So to prove that H and K are conjugate in G, it suffices to show that F ix(g/k) 0. But F ix(g/k) G/K 0 (mod p) by Lemma

14 1.6. THE SYLOW THEOREMS 23 Third Sylow theorem. Let m be the number of distinct Sylow p-subgroups of G. Then, m 1 (mod p) and m G. Proof. Let H be a Sylow p-subgroup (exists by the first Sylow theorem). Applying the second Sylow theorem, m is the number of conjugates of H, which is [G : N G (H)]. This gives that m G. Now let X be the set of all Sylow p-subgroups of G and let H act on X by conjugation. Then, F ix(x) m (mod p). It therefore suffices to show that F ix(x) = 1. Now, K F ix(x) if and only if hkh 1 = K for all h H, i.e. H N G (K). But H N G (K) if and only if H is a Sylow p-subgroup of N G (K). By the second Sylow theorem, K is the unique Sylow p-subgroup of N G (K), so this is if and only if H = K. Hence, F ix(x) = {H}. We give some examples to illustrate applying Sylow theorems Let G = pq with p > q prime and G not Abelian. Then, q p 1 and G = a, b a q = b p = 1, aba 1 = b r for some 1 < r < p with r q 1 (mod p). Proof. Let K = b be a Sylow p-subgroup of G. We have that n p, the number of Sylow p- subgroups, divides pq and is congruent to 1 mod p. Since p > q, this means that n p = 1. Hence, K must be normal in G. Now let H = a be a Sylow q-subgroup of G. Since G is not Abelian, we cannot have that H G. (If it was then for any h H, k K we would have that hkh 1 k 1 H K = {1}. Hence hk = kh for all h H, k K which implies that G = H K is Abelian.) One deduces that the number of Sylow q-subgroups of G must be p, and that q p 1 by the third Sylow theorem. Now, since K G, aba 1 K so equals b r for some 1 < r < p (cannot have r = 1 since G is not Abelian). Next, a q ba q = b rq = b implies that r q 1 (mod p). We ve now shown that G satisfies the given relations. Hence, there exists an epimorphism from the group with the given relations to G, which is injective since you easily check that any group satisfying the given relations has order at most pq As a special case of the previous example, we get that a group of order 2p for p an odd prime is either isomorphic to C 2p or to D p There is no simple group of order 12. Proof. Let G be a simple group of order 12. Consider n 2, the number of Sylow 2-subgroups. It is either 1 or 3. In the former case, G has a normal Sylow 2-subgroup so is not simple. Hence, n 2 = 3. Now, G acts on the Sylow 2-subgroups by conjugation, giving a non-constant homomorphism G S 3. The kernel is a proper normal subgroup of G since G = 12, S 3 = 6. Here is a table listing all groups of order 12: Order G 1 {1} 2 C 2 3 C 3 4 V 4 = C2 C 2, C 4 5 C 5 6 C 2 C 3 = C6, D 3 = S3 7 C 7 8 C 8, C 4 C 2, C 2 C 2 C 2, D 4, Q 8 9 C 9, C 3 C 3 10 C 2 C 5, D 5 11 C C 12, C 4 C 3, C 2 C 2 C 3, D 6, A 4, T For the definitions of the groups T of order 12 and the quaternion group Q 8 of order 8, see Hungerford (or section 2.1 for Q 8 ).

15 24 CHAPTER 1. GROUPS 1.7 Semidirect products In (1.6.2), we showed that if G is a non-abelian group of order G = pq with p > q prime, then q p 1 and G = a, b a q = b p = 1, aba 1 = b r for some 1 < r < p with r q 1 (mod p). We stopped short of showing that there actually exists such a group G for all possible choices of p, q, r. However, we showed in the proof that G contained a normal subgroup K = b = C p. The quotient group G/K is isomorphic to C q. So you can think of G as being built by glueing the group C q on top of the group C p. In other words, G is an extension of C q by C p. For the general definition, let H, K be two groups. We call a group G an extension of H by K if G contains a copy of K as a normal subgroup and G/K = H. We can represent this by the following exact sequence : 1 K G H 1 (the precise meaning of this will be given later when studying modules.) It is important to change the point of view and ask: given H and K what possible extensions G of H by K can we form? For instance, G = H K is an example of an extension of H by K: the kernel of the canonical projection G H is isomorphic to K. (Equally well, you could call G = H K an extension of K by H.) But this extension H K is not really very interesting! How can we build more complicated extensions? We discuss here one important class of extensions which are not too complicated to understand, known as semidirect products. This is really a generalization of the construction of the direct product there are in general other extensions which are not semidirect products. So now let H and K be given groups and suppose we also have a group homomorphism θ : H Aut K, h θ h (where Aut K is the group of all automorphisms of the group K). Define G = K θ H, the external semidirect product, to be the group equal to the Cartesian product K H as a set, with multiplication defined by (k, h)(k, h ) = (kθ h (k ), hh ). You need to check that this really is a multiplication making K H into a group! For instance, (k, h) 1 = (θ h 1(k 1 ), h 1 ). Now, (k, 1)(k, 1) = (kk, 1), so the set K = {(k, 1) k K} is a subgroup of G isomorphic to K. The map G H determined by projection onto the second coordinate is a surjective group homomorphism, with kernel K. This verifies that G has a copy of K as a normal subgroup and the quotient group is isomorphic to H. Hence: the semidirect product G is an extension of H by K. Observe moreover that: (1, h)(k, 1)(1, h 1 ) = (θ h (k), h)(1, h 1 ) = (θ h (k), 1). This shows that the homomorphism θ : H Aut K can be recovered from the semidirect product: θ h is precisely the homomorphism determined by conjugating by the element (1, h) of G. Actually, K θ H is a very special sort of extension: the map π : G H determined by projection onto the second coordinate is split. This means that there is a group homomorphism σ : H G called a splitting such that π σ = id H, namely, the map h (1, h). In fact an extension of H by K is a semidirect product if and only if such a splitting map exists; more general extensions need not have a splitting map (i.e. need not be semidirect products). This last observation is the key to recognizing whether or not a given group G is isomorphic to a semidirect product of groups H and K. Indeed, given a group G and subgroups H and K, we say that G is the internal semidirect product of H and K if (1) K G; (2) H K = {1}; (3) G = KH = {kh k K, h H}.

16 1.8. SOLVABLE AND NILPOTENT GROUPS 25 If G is the internal semidirect product of H and K, we define a map θ : H Aut K by setting θ h (k) = hkh 1 for all k K. Then, G is isomorphic to the external semidirect product K θ H. Conversely, any external semidirect product K θ H is the internal semidirect product of {1} H and K {1}. Take the very special case of external semidirect product where θ h = id K for each h H. Then, K θ H is exactly the same as the definition of the direct product K H: so semidirect products are a generalization of direct products. The previous paragraph gives in this special case that a group G is isomorphic to the direct product of subgroups H, K G if H G, K G, H K = {1} and G = HK. We return to the example (1.6.2). So we have primes p > q with q p 1 and 1 < r < p with r q 1 (mod p). We wish to prove that there exists a non-abelian group G of order pq. Let K = Z p, H = Z q (both written additively) and define a map θ : H Aut K as follows. Define θ 1 : Z q Z q, [i] [i r ]. Note (θ 1 ) q ([i]) = [i rq ] = [i 1 ] = [i]. This shows in particular that θ 1 is an automorphism of Z q. Now we obtain a well-defined map θ : H Aut K, [n] (θ 1 ) n. Finally, define G = K θ H. It s an extension of H by K, you just need to check that it is non-abelian to complete the proof of the existence of such a group. 1.8 Solvable and nilpotent groups In the previous section, we started thinking of extensions of groups. Now we take the idea further. Define a subnormal series of a group G to be a chain of subgroups G = G 0 G 1 G n G n = {1} such that G i G i 1 for each i = 1,..., n. The factors of the subnormal series are the groups G i 1 /G i for i = 1,..., n. The length of the series is the number of non-trivial factors. A normal series is a subnormal series in which in addition each G i is a normal subgroup of G. The basic idea to keep in mind is that the structure of the factors in a subnormal series should tell you something about the structure of the group G. In this section, we focus on two special classes of group defined in terms of certain subnormal series: solvable and nilpotent groups. So now let G be a group and H, K be subgroups. Define [H, K] = [h, k] h H, k K where [h, k] denotes the commutator hkh 1 k 1. Note that [H, K] = [K, H], and that [H, K] = {1} if and only if every element of H commutes with every element of K Lemma. If H G then [H, G] G, [H, G] H and H/[H, G] is Abelian. Proof. Exercise. The subgroup G := [G, G] of G is called the commutator subgroup of G. By the lemma, it is a normal subgroup of G and G/G is Abelian. Indeed, given any normal subgroup N G such that G/N is Abelian, we have for any g 1, g 2 G that [g 1 N, g 2 N] = N, hence [g 1, g 2 ] N, hence G N. This shows that G is the unique smallest normal subgroup of G whose associated quotient group is Abelian. The group G/G is sometimes called the Abelianization of G ( Abelianization is a functor!). Now, define G (0) = G and inductively set G (i) = [G (i 1), G (i 1) ], the commutator subgroup of G (i 1). So, G (i) is the unique smallest normal subgroup of G (i 1) with Abelian quotient. It follows by induction on i that in fact G (i) is a normal subgroup even of G. Proof: by induction, G (i 1) is normal in G and so gg (i) g 1 is the unique smallest normal subgroup of gg (i 1) g 1 = G (i 1) with Abelian quotient hence coincides with G (i).

17 26 CHAPTER 1. GROUPS We therefore obtain a descending chain of normal subgroups of G: G = G (0) G (1) G (2)... called the derived series of G. The group G is called solvable if for some n >> 1 we have that G (n) = {1}. Then, the derived series is in fact a normal series of G. The basic properties of solvable groups are given in the following lemma: Properties of solvable groups. (1) If G is solvable, then so is every subgroup. (2) Given N G, G is solvable if and only if both N and G/N are solvable. (3) G is solvable if and only if G has a subnormal series with Abelian factors. (4) Finite direct products of solvable groups are solvable. (5) A semidirect product of two solvable groups is solvable. Proof. (1) Let H G with G solvable. Show by induction on n that H (n) G (n). Since G (n) = {1} for some n we get that H (n) = {1} for some n, so H is solvable. (2) Induction shows that (G/N) (n) = G (n) N/N. It follows easily using (1) too that if G is solvable, so are both N and G/N. Conversely, if N and G/N are solvable, there exists n such that G (n) N and m such that N (m) = {1}. Then, G (n+m) N (m) = {1} so G is solvable too. (3) If G is solvable, the derived series is a subnormal series with Abelian quotients. Conversely, suppose G has a subnormal series with Abelian quotients; we prove G is solvable by induction on the length of the subnormal series. If the series has length 1, the result is clear. Else, consider the first subgroup G 1 G in the subnormal series different from G. It has a subnormal series with Abelian quotients of length one less than that of G, so is solvable by induction. Also, G/G 1 is Abelian so solvable. Hence by (2), G is solvable. (4) Apply (2) and induction on the number of direct factors. (5) Apply (2). For example, the dihedral group D n is an internal semidirect product of its rotation subgroup C n (which is normal of index two) and any subgroup generated by a reflection. So (5) implies that D n is solvable. You should also easily be able to prove directly that the symmetric group S 4 is solvable. On the other hand, no non-abelian simple group is solvable, so no A n (n 5) is solvable. In particular, this shows that S n for n 5 is not solvable. We turn now to nilpotent groups. The definition is very similar to the definition of a solvable group, but involves a different descending chain of subgroups : set G 0 = G and inductively define G i = [G, G i 1 ]. By Lemma 1.8.1, G i is a normal subgroup of G and G i 1 /G i is Abelian. We obtain a descending chain of normal subgroups of G G = G 0 G 1... called the descending central series. The group G is nilpotent if G n = {1} for some n >> 1, in which case the descending central series is in fact a normal series of G with Abelian factors. By property (3) of solvable groups, we have at once that nilpotent groups are solvable. Properties of nilpotent groups. (1) If G is nilpotent, so is every subgroup. (2) If G is nilpotent and N G, then G/N is nilpotent. (3) A non-trivial group G is nilpotent if and only if Z(G) {1} and G/Z(G) is nilpotent. (4) Direct products of finitely many nilpotent groups are nilpotent. (5) If G is nilpotent and H < G is a proper subgroup, then H N G (H). Proof. (1) Show by induction on n that H n G n for a subgroup H G. (2) Show by induction on n that (G/N) n = G n N/N. (3) If G is nilpotent, consider the last non-trivial subgroup C in the descending central series of G. Then, [G, C] = {1} so that {1} < C Z(G), giving that Z(G) {1}. Then by (2), G/Z(G)

18 1.9. JORDAN-HÖLDER THEOREM 27 is nilpotent. Conversely, suppose that Z(G) {1} and that G/Z(G) is nilpotent. Then, using the result proved in (2), for some n we have that G n Z(G). Then, G n+1 [G, Z(G)] = {1} so G is nilpotent. (4) Let me prove this just for the direct product of two nilpotent groups (the general case following by induction from this). Consider G H with G, H nilpotent. Then, we can find some n >> 1 such that both G n = {1} and H n = {1} Then, (G H) n = G n H n = {1} too. (5) Choose i so that G i+1 H but G i H. Then, [G i, H] [G i, G] = G i+1 H. In other words, G i N G (H). Since G i H, this shows that N G (H) is strictly larger than H. Now we can give an example of a solvable group that is not nilpotent: the dihedral group D n. You should be able to prove that Z(D n ) = {1} if n is odd or {1, t} if n is even, where t is the 180 degree rotation. So if n is odd, D n is certainly not nilpotent by (3), but D n is always solvable as its a semidirect product of two cyclic groups. If n is even, D n /Z(D n ) = D n/2, and using this and (3) you show that in fact D n is nilpotent if and only if n is a power of 2. We end with an important characterization of finite nilpotent groups in terms of Sylow p- subgroups. (There is a similar, but more complicated, such characterization of finite solvable groups called P. Hall s theorem, which is really a generalization of the Sylow theorems for solvable groups.) Characterization of finite nilpotent groups. Let G be a finite group. Then, G is nilpotent if and only if G is isomorphic to the direct product of its Sylow p-subgroups for all p. Proof. Suppose G is the direct product of its Sylow p-subgroups for all p. Using property (4), to show that G is nilpotent, we just need to show that every finite p-group is nilpotent. This follows by induction on the order using property (3) and Corollary Conversely, suppose that G is nilpotent and take a prime p dividing the order of G. Let P be a Sylow p-subgroup. If P = G, we are done: there is nothing to prove. We claim that N G (N G (P )) = N G (P ). Indeed, P is the unique Sylow p-subgroup of N G (P ). Take any x N G (N G (P )). Then, xp x 1 is the unique Sylow p-subgroup of xn G (P )x 1 = N G (P ). Hence, xp x 1 = P. Hence, x N G (P ). This proves the claim. It follows from property (5) that in fact, N G (P ) = G, so that P is normal in G. This means that G has a unique Sylow p-subgroup for each prime p dividing its order. Now let p 1,..., p n be the primes dividing G and let P i be the unique Sylow p i -subgroup. Define a map µ : P 1 P n G, (g 1,..., g n ) g 1... g n. To see that this is a group homomorphism, it suffices to see that [P i, P j ] = {1} for i j, so that elements of P i and P j pairwise commute. But [P i, P j ] is a subgroup of both P i and P j, hence of P i P j which is trivial by Lagrange s theorem. Now suppose that we have (g 1,..., g n ) ker µ. Then, we have that g 1 = g2 1 g gn 1 in G. But g 1 has order a power of p 1, whereas the order of the element on the right hand side is coprime to p 1. This shows that g 1 = 1. Similarly, each g i = 1. Hence, µ is injective. Finally, µ is a surjection by considering orders. Hence, G is isomorphic to the direct product of its Sylow p-subgroups for all p. 1.9 Jordan-Hölder theorem Recall the definition of a subnormal series of a group G from the previous section. Let G = G 0 G 1 G n = {1} be a subnormal series. A one-step refinement means a subnormal series G = G 0 G 1 G i H G i+1 G n = {1} for some subgroup H G i. A refinement of a subnormal series means a subnormal series obtained by finitely many one-step refinements.

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