Introduction to Groups


 Cornelia May
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1 Introduction to Groups HongJian Lai August Basic Concepts and Facts (1.1) A semigroup is an ordered pair (G, ) where G is a nonempty set and is a binary operation on G satisfying: (G1) a (b c) = (a b) c, a, b, c G. A semigroup G is a monoid if it also satisfies: (G2) G has an element e (sometimes denoted by 1 G, called the identity of G), such that e a = a e = a, a G. A monoid is a group if (G3) below is satisfied. (G3) For each a G, there is an a 1 G such that a a 1 = a 1 a = e. (a 1 is called the inverse of a). (1.1a) Z, Q, R, Z n and vector spaces with addition, and the corresponding multiplicative groups. Groups obtained from taking products. (1.1b) Permutations, S n, symmetric group on n letters, and D n, the dihedral group. (Display n = 3, 4). (1.1c) Let F be a field. For each integer n 1, let GL n (F ) = {A A is an n n matrix over F with det(a) 0}. Then GL n (F ) with matrix multiplication forms a group, called the general linear group of degree n, whose identity is the identity matrix. (1.2) (Thm 1.2) If G is a monoid, then identity of G is unique. If G is a group, then (i) c G and cc = c = c = e (ii) a, b, c G, ab = ac = b = c and ba = ca = b = c. (Cancellation Laws). (iii) a G, a 1 is unique. (iv) (a 1 ) 1 = a. 1
2 (v) (a b) 1 = b 1 a 1. (vi) a, b G, ax = b and ya = b have unique solutions in G : x = a 1 b and y = ba 1. (vii) (Generalized Associative Law, Thm 1.6) For any a 1, a 2,, a n G, the value of a 1 a 2 a n is independent of how the expression is bracketed. Proof (1)  (vi) by definitions. For (vii), use induction on n to show every such expression is equal to a 1 (a 2 ( a n ) ). (G1) implies the case when n=3. (1.3) Let G be a semigroup. The G is a group iff each holds: (i) e G such that a G, ea = a (left identity). (ii) For each a G, a 1 G such that a 1 a = e (left inverse). (1.3A) Let G be a semigroup. The G is a group for all a, b G, the equations ax = b and ya = b have solutions. PF: Apply (1.2) in (1.3), and apply (1.3) to (1.3A). (1.4) (Thm 1.5) Let R be an equivalence relation on a monoid G such that a 1 Ra 2 and b 1 Rb 2 imply (a 1 b 1 )R(a 2 b 2 ). Then the set G/R of all equivalence classes of G under the binary operation ā b = ab is also a monoid. (1.4a) Z n = Z/nZ. (1.4b) Q/Z. (1.4c) For a group G, an element x G has order n if x n = 1 and n is the smallest such positive integer, (n can be finite or infinite). The order of x is x. (1.4d) In Z n, 1 = n; in S 3, (12) = 2. For any n Z {0}, n =. (1.5) A group G is abelian if for any pair a, b G, ab = ba. (Commutative Law holds). For abelian groups, we usually use addition to denote the binary operation. (1.5a) ((Thm 1.7) Generalized Commutative Law) If G is a commutative semigroup, and if a 1,..., a n G, then for any permutation i 1,..., i n of 1, 2,..., n, a i1 a i2 a in = a 1 a 2 a n. PF: Induction on n 2. (1.6) Notation: (1.6a) (multiplication notation) a 1 = a and a n = a a n 1, for n 2, and a 0 = 1 G. If g G and H G, then gh = {gh h H} and Hg = {hg h H}. 2
3 If H, K G, then HK = {hk h H k K}. (1.6a) (addition notation) 1 a = a and na = a(n 1)a, for n 2, and 0 a = 0 G. If g G and H G, then g + H = {g + h h H}. If H, K G, then H + K = {h + k h H k K}. (1.7) (Thm 1.9) Let G be a group. For any a G and m, n Z, (a n ) m = a nm and a m a n = a m+n. (Same result for addition notation). (1.8) (Direct Products) Let G and H be two groups. Let G H = {(g, h) : g G and h H}. Then the operation (g 1, h 1 )(g 2, h 2 ) = (g 1 g 2, h 1 h 2 ) is a binary operation on G H. One can verify that the set G H with this binary operation will form a group. (When both operations in G and in H are additive, we often use G H for G H). (1.8a) If both G and H are abilian, then G H is also abelian. 2. Homomorphisms and Subgroups (2.1) Let G and H be groups. A map f : G H is a homomorphism if for all x, y G, f(xy) = f(x)f(y). f is an isomorphism if it is bijective. If f : G G is a homomorphism (isomorphism, resp.), then f is also called an endomorphism (automorphism), resp.) of G. (2.1a) Linear transformations of vector spaces are examples of homomorphisms; Z 2 Z 3 = Z 6 ; GL 2 (Z 2 ) = S 3. (2.1b) Let f : G H is a group homomorphism. The kernel of f is ker(f) = {a G : f(a) = e in H}. For A G, f(a) = {f(a) : a A} is the image of A, and we denote Im(f) = f(g), called the image of f. If B H, then f 1 (B) = {a G : f(a) B} is the inverse image of B. (2.1c) (Thm 2.3) Let f : G H be a group homomorphism, and let e G and e H denote the identities of G and H, respectively. Let 1 G and 1 H denote the identity maps in G and in H, respectively. Then (i) f(e G ) = e H. (ii) For any a G, f(a 1 ) = [(f(a)] 1. 3
4 (iii) f is injective (called a monomorphism) iff ker(f) = {e}. (iv) f is onto (called an epimorphism) iff Im(f) = H. (v) f is an isomorphism iff there exists a homomorphism f 1 : H G such that ff 1 = 1 H and f 1 f = 1 G. (2.1d) Let G be a group, and let Aut(G) denote the set of all automorphisms of G. Then Aut(G) with the map composition forms a group itself, called the automorphism group of G. (2.1e) AutZ = Z 2 = Aut(Z6 ). (2.2) Let G be a group and let H G. If H is also a group, then H is a subgroup of G, denoted by H < G. (2.3) Let G be a group and let H G be a nonempty subset of G. Then TFAE: (i) H is a subgroup of G. (ii) a, b H, ab 1 H. (iii) a, b H, a 1 H and ab H. Proof (i) = (ii) = (iii) = (i). (Show 1 H first). (2.4) If H 1, H 2 are subgroups of G, then H 1 H 2 is also a subgroup of G. Proof Use (2.3). (2.5) Let X G. Denote < X >= X H H G Then < X > G. (< X > is the smallest subgroup of G with X < X >). Call < X > the subgroup generated by X, and elements in X are the generators of < X >. Proof Use (2.3). (2.5a) Let A G, and let Then < A >= Ā. Ā = {a ɛ 1 1 a ɛ 2 2 a ɛn n n Z, n 0, ɛ i = ±1 for each i}. Proof: Let H be a subgroup of G such that A H. Since elements in a subgroup are closed under multiplication, Ā H. In particular, Ā < A >. On the other hand, it can be checked by using (2.3) that Ā itself is a subgroup containing A, and so < A > Ā. (2.5b) Let A G such that a 1, a 2 A, a 1 a 2 = a 2 a 1. Then < A > is abelian. (2.5c) Let H G. If there is an a H such that H =< a >, then H is a cyclic (sub)group H. 4
5 and a is a generator of H. (2.5d) (Z, +) =< 1 >, (Z n, +) =< 1 >. Q and R are not cyclic. (2.6) Let σ = (123 n) and (1n)(2(n 1)) ( n 2 τ = ( n 2 + 1)) (1n)(2(n 1)) ( n 1 n if n is even, n+1 )( 2 ) if n is odd. Let X = {σ, τ} S n. The D 2n =< X > is called the dihedral group of order 2n. The presentation of D 2n is D 2n = {σ, τ σ n = τ 2 = 1, στ = τσ 1 }. (2.6a) Group of rigid motions in R 2 and in R 3. (2.6b) < σ > is a cyclic subgroup of D 2n. So is < τ >. (2.7) Let G be a group and A G. The centralizer of A is C G (A) = {g G gag 1 = a, a A}. The normalizer of A is N G (A) = {g G gag 1 = A}, where gag 1 = {gag 1 a A}. The center of G is Z(G) = C G (G). (2.8) C G (A) G and N G (A) G. Proof Use (2.3). (2.8a) N D8 (< σ >) = D 8, C D8 (< σ >) =< σ >, and Z(D 8 ) =< τ >. 3. Cyclic Groups (3.1) Recall that the order of x is x. If x = n and if x m = e, then n m. Proof by Long Division, m = qn + r, where 0 r < n. (x n = 1) (x m = e) = x r = e. Hence r = 0. (3.2) (Thm 3.1) Let H be a subgroup of the additive group Z. (i) Either H =< 0 >, or 5
6 (ii) for some m Z {0}, H =< m >, and H =. (3.3) (Thm 3.2) Let H =< x > be a cyclic group. (i) If H = n <, then H = {x i 0 i n} and the order of x is n. Moreover, H = Z n. (ii) If H =, then H = {x i i Z} and no element of H {1} has a finite order. Moreover, H = Z. Proof Use definition of order. (3.3A) Any two cyclic group with the same order are isomorphic. (finite or infinite) Proof They are iso to either Z n of or to Z. (3.4) Let x G, and let n 0 be an integer. (i) If x =, then x n =. (ii) If x = m <, then x n = m (m, n) = l. (iii) If x = m < and d m, then x d = m/d. Proof: (i) follows by (ii) of (3.2). (ii) Let y = x n and d = y. First y l = e and so by (3.1), d l. Since 1 = (x n ) d = x dn, by (3.1), m (dn) = m/(n, m) dn/(n, m). Since (m/(n, m), n/(n, m)) = 1, l d. (iii) follows by (ii). (3.5) Let H =< x >. (i) Assume x =. Then H =< x m > m = ±1. (ii) Assume x = n <. Then H =< x m > (n, m) = 1. Proof Use (3.3) and then (3.4). (3.6) (Structure of Subgroups of a Cyclic Group) Let H =< x >. (i) If K H, then either K = {e}, or K =< x d >, where d is the smallest positive integer such that x d K. (ii) If H =, then for any distinct nonnegative integer n and m, < x n > < x m >. Furthermore, m Z, < x m >=< x m >. (Thus the number of distinct subgroups of H is the same as the cardinality of Z.) (iii) If H = n <, then for each positive integer m n, there is a unique subgroup < x d > H such that < x d > = m, where d = n/m. Furthermore, < x m >=< x (n,m) >. Proof (i) Assume K {e}. Let P = {(n Z) (n > 0) x n K}. Let d = min P. Then 6
7 x d K. Use long division to show K < x d >. (ii) < x n >=< x m >, then n m and m n and so n = m. (iii) By (3.4)(3.4)(iii), < x d > = n/d = m. Let K H be such that K = m. By (3.6)(i), K =< x l >, where l is the smallest non negative integer such that x l K. To prove the uniqueness, write n = ql + r, with 0 r < l. As x r = (x n )(x ql ) 1 e(x ql ) 1 = (x ql ) 1 K, and by the minimality of l, we have r = 0 and so l n. By (3.4)(ii), and so l = [d and K =< x d >. n l = n n, l = xl = K = m = n d, (3.7) More examples of groups: (3.7a) (Define direct product G H) V 2 = Z 2 Z 2, a group each of whose proper subgroups are cyclic, but the group is not cyclic. (3.7b) Q 8, the quaternion group, is defined by Q 8 = {1, 1, i, i, j, j, k, k}, whose identity is 1 and whose multiplication is defined as follows: ( 1) 2 = 1, ( 1)a = a, a Q 8, b 2 = 1, b Q 8 {1, 1}, and ij = k, jk = i, ki = j, ji = k, kj = i, ik = j. Each of the proper subgroup of Q 8 is cyclic, but Q 8 is not abelian. 4. Alternating Groups (4.1) Recall permutations, cycles, and transpositions (2cycles). Every φ S n is a product of transpositions. Proof Every cycle is a product of transpositions: (i 1 i 2 i k ) = (i 1 i k )(i 1 i k 1 ) (i 1 i 3 )(i 1 i 2 ). Every φ S n is a product of cycles (called the cycle decomposition of φ). (4.2) No φ S n (n 2) can be expressed both as a product of an even number of transpositions and as a product of an odd number of transpositions. Proof We use e to denote the identity of S n. (Step 1) (4.2) holds for φ = e. 7
8 Suppose that S n is the set of all permutations on the set {1, 2,, n}, and that e = τ k τ 1, where each τ i is a transposition. Let X = {x : 1 x n and x is involved in some of the τ s }, and s = X. Argue by induction on s. If s = 2, then we may assume that the involved letters are 1 and 2, and e = τ k τ 1, where each τ i = (1, 2). Since e = (1, 2)(1, 2), k must be even. Assume that s 3 and that (Step 1) holds for smaller values of s. Suppose that e = τ k τ 1, where each τ i is a transposition, and where the involved letters are in {1, 2,..., s}. We further argue by induction on k. (Step 1) holds trivially if k = 2, and so we assume futher that (Step 1) holds for smaller values of k. Pick m X. Let τ j be the 1st transposition (from R to L) that contains m. Then τ j+1 τ j must be one in the left side of (x, m)(x, m) = e (m, y)(m, x) = (m, x)(x, y) (y, z)(m, x) = (m, x)(y, z) (x, y)(x, m) = (m, y)(x, y) Hence the substitution of the left by the right either reduces the number of transpositions by 2; whence by induction on k, (Step 1) holds; or moves the 1st transposition containing m to the left by one step. Repeat this process (assuming that k remains unchanged) until the first τ j containing m is τ k 1. Then τ t au k 1 must be one of the four cases listed above. In this case, only the case τ k = τ k 1 = (x, m) will occur, as otherwise, after the process of pushing m to the left, the right most transposition of a factoring of e is the only transpotition in the factorization of e contains the element m, and so m must be moved, contrary to the fact that e S n is the identity permutation. Therefore, such a process can eliminate the element m, without introducing any new elements involved in the factorization, and without changing the parity of k. becomes X 1, and so by induction on X, (Step 1) holds also for all values of k. (Step 2) General Case: Suppose φ S n has two factorizations: φ = τ 1 τ 2 τ r = τ 1τ 2 τ t, Now X where τ i s and τ j s are transpositions. Then φ 1 = τr 1 τ1 1 and so e = φφ 1 = τr 1 τ1 1 τ 1 τ 2 τ t. Hence r + t must be even, and so r and t must have the same parity. (4.3) (Even and Odd Permutations) A permutation in S n is even (or odd) if it can be expressed as a product of an even (or odd) number of transpositions. The set of all even permutations in S n is denoted by A n. A n is a subgroup of S n, called the Alternating 8
9 Group of degree n. Proof Use (2.3) to show A n S n. 9
10 5. Cosets and Counting (5.1) (Thm 4.2) Let G be agroup and let H < G. For any a, b G, define a l b (mod H) iff a 1 b H (a r b (mod H) iff ab 1 H, resp.). Then both l and r are equivalence relations. (5.2) (Them 4.2) Each equivalence class of l has the form gh, where g G, and is called a left coset of H in G. Each equivalence class of r has the form Hg, where g G, and is called a right coset of H in G. Any element in a coset if a representative of the coset. (Every statement below about left cosets can also have a right coset version.) PF: Show that a and b are in the same class iff ah = bh. (5.3) (Thm 4.2) g G, gh = H = Hg. PF: define a bijection. (5.4) Let H < G. The index of H in Gdenoted [G : H], is the cardinal number of the set of distinct left cosets of H in G. (5.5) If K < H < G, then [G : K] = [G : H][H : K]. PF: Use (5.2). Show that K has [G : H][H : K] cosets in G. (5.6) (Cor. 4.6: Lagrange) If H < G, then G = [G : H] H. (5.7) Let H and K be finite subgroup of G, then HK = H K / H K. 10
11 6. Normality, Quotients and Homomorphisms (6.1) Let φ : G H be a group homomorphism. (i) φ(1 G ) = 1 H. (ii) φ(g 1 ) = (φ(g)) 1, g G. (iii) φ(g n ) = (φ(g)) n, n Z. (iv) The kernel of φ, kerφ = {g G φ(g) = 1 H } G. (v) The image of G under φ, imφ = {h H φ(g) = h, for some g G} H. Proof (i): Use φ(1 G 1 G ) and Cancellation Law. (ii): Use uniqueness of inverse. (iii): Induction on n for n > 0, and use (ii) for negative n s. (iv) and (v): (Check ab 1 kerφ). (6.2) For a map φ : X Y and for each y Y, the subset φ 1 (y) = {x X φ(x) = y} is called a fiber of φ. Given a group homomorphism φ : G H with K = kerφ, G/K denotes the set of all fibers of φ. Define a binary operation on G/K by φ 1 (a) φ 1 (b) = φ 1 (ab). Then (i) is well defined. (φ 1 (ab) is independent of the choices of a and b). (ii) (G/K, ) is a group, called the quotient group of factor group. The identity of G/K is K and the inverse of gk is g 1 K. (iii) φ 1 (a) = ak = {ak k K} = Ka = {ka k K}. Proof: (i) Suppose that a φ 1 (a) and b φ 1 (b). The φ(a ) = φ(a) and φ(b ) = φ(b). Thus φ(a b ) = φ(a )φ(b ) = φ(a)φ(b) = φ(ab). (ii) Verify the group axioms. The inverse and the identity conclusions follow from the definition of the binary operation and the uniqueness of identity and inverse. (iii) Since φ(ak) = a, ak φ 1 (a). x φ 1 (a), we can write x = ay (y = a 1 x). Thus φ(a) = φ(x) = φ(a)φ(y) and so y K. (6.3) For any N G and g G, gn and Ng are called the left coset and the right coset of N in G. Any element in a coset if a representative of the coset. (Every theorem below about left cosets can also have a right coset version.) If G is finite, then (i) g G, gn = N, and (ii) G is the disjoint union of distinct left (or right) cosets of N. (Valid even when G =.) (iii) If φ : G H is a homomorphism with ker(φ) = K, then every fiber of φ has the 11
12 same cardinality. (iv) A homomorphism φ is injective iff ker(φ) = {1}. Proof (i) It suffices to show that if n 1 n 2 and n 1, n 2 N, then gn 1 gn 2, which is assured by Cancellation Laws. (ii) Since G = {g G} g G gn, it suffices to show that if g 1 N g 2 N, then g 1 N g 2 N =. In fact, if g 1 n 1 = g 2 n 2 for some n 1, n 2 N, then g 1 = g 2 n 2 n 1 1 g 2 N, and so g 1 N g 2 N. Similarly, g 2 N g 1 N. (iii) follows from (i) and (iv) follows from (iii). (6.4) (6.1) can be restated in terms of left and right cosets. Let G be a group and let K G be the kernel of some homomorphism from G. Then the set of all left (or all right) cosets of K with the operation defined by uk vk = (uv)k (or Ku Kv = K(uv)) is a group, denoted by G/K. The operation is well defined (independent of the choices of the representatives). (6.4a) Examples: φ : Z nz, for any fixed n 1 and n Z. Projections in R 2. φ : S 3 Z 3. (6.5) (This is another way to state (5.2)) Let N G. Then un = vn u 1 v N. Proof un = vn = u vn = u 1 v N = v un = un = vn. (6.6) (Thm 5.1 and Thm 5.5) Let N G. TFAE: (i) The operation on the left cosets of N by un vn = (uv)n is well defined. (ii) g G, and n N, gng 1 N. (iii) g G, gng 1 N. (iv) g G, gn = Ng. (v) N G (N) = G or equivalently g G, gng 1 = N. (vi) N is the kernel of some homomorphism from G. Proof (i) = (ii). Suppose that is well defined. g G and n N, (1g 1 )N = (ng 1 )N, and so by (6.5), gng 1 N. (ii) = (i). Suppose that u un and v vn. Want to show (u v )N = (uv)n. Since u = un and v = vn, for some n, n N, u v = unvn = uv(v 1 nv)n = uvn (uv)n, 12
13 where n = (v 1 nv)n. (ii) (iii). Definition. (iii) = (iv). By (iii), we have gng 1 N, and so gn Ng. Replace g by g 1 to get Ng gn. (iv) (v). N G (N) = {g G gng 1 = N} = G. (vi) = (i). (i) of (6.2). (v) = (vi). Let P denote all the left cosets of N in G. By (i), (P, ) is a group. Define a map π : G P by π(g) = gn, g G. Then π(gg ) = (gg )N = g(g Ng 1 )gn = gng N = π(g)π(g ), and so a homomorphism. The kernel of π, by (ii) of (6.2), is ker(π) = {g G φ(g) = N} = {g G gn = 1N} = by (6.5) {g G g N} = N. (6.7) A subgroup N satisfying any one properties of (4.6) is called a normal subgroup of G. Denote this fact by N G. The homomorphism π in (vi) of (6.6) is called the natural projection of G onto G/N. (6.8) Let φ : G H be a homomorphism. (i) if H H, then φ 1 (H ) G. (ii) If G G, then φ(g ) H. Proof (i) Use (2.3) and (ii) of (4.1). (ii) of (4.8) = (v) of (4.1). (6.9) For any H G/N, N π 1 ( H) G. Proof It suffices to show that π 1 ( H) G. Use (i) of (6.8). (6.10) (Thm 5.6) Let f : G H be a group homomorphism and N G such that N < ker(f). Then there exists a unique homomorphism f : G/N H such that (i) f(an) = f(a), a G. (ii) Im(f) = Im( f) and ker( f) = ker(f)/n. Moreover, (First Isomorphisn Theorem) f is an isomorphism iff f is an epimorphism and N = ker(f). (6.11) (Second Isomorphisn Theorem) If K G and N G, then K/(N K) = NK/N. PF: Justfy N HK. Find a homomorphism f : K HK/N with ker(f) = (N K). 13
14 (6.12) (Third Isomorphisn Theorem) If K G and N G and if K < H then H/K G/K and (G/K)/(H/K) = G/H. PF: Justfy H/K G/K. Find a homomorphism f : G/K G/H with ker(f) = H/K. (6.13) (Thm 5.11) Let f : G H be an epimorphism of groups. Then the assignement K f(k) is (i) a bijection between the set S f (G) = {K G : ker(f) K G} and the set S(H) = {N H}; and (ii) a bijection between the set S f (G) = {K G : ker(f) K G} and the set S(H) = {N H}. (6.14) Z(G) G. (6.14a) If G/Z(G) is cyclic, then G is abelian. (6.15) A group G is simple if G > 1 and if H {< 1 >, G} whenever H G. (6.15a) The only simple abelian groups are Z p, for prime p s. Proof (5.1). (6.16) An example: being normal is not a transitive relation. Let G = D 8 =< r, s r 4 = 1, s 2 = 1, rs = sr 1 >. Let H = {1, r 2, s, sr 2 }, and K = {1, s}. Then since G : H = 2, and H : K = 2, both K H and H G. However, rsr 1 = r 2 s K and so K G. 7. Some Counting Lemmas (7.1) If G <, then g G, g divides G. Proof H =< g >. (7.2) If G = p is a prime, then G = Z p. Proof Pick g G {1} and set H =< g >. (7.3) If G : H = 2, then H G. Proof By (5.6), G = H gh = H Hg, for any g H. Therefore, g G H, g 1 G. Hence gh = Hg and so by (iv) of (6.6), H G. 14
15 (7.4) If H G and K G with max{ H, K } <, then HK = H K H K. Proof Note that HK = h H hk = m i=1h i K, where h 1 K, h 2 K,, h m K are the distinct cosets of the form hk with h H. (How many distinct left cosets?) Note that for h, h H, by (4.5) hk = h K h 1 h H K h(h K) = h (H K). Hence m = number of distinct left cosets of (H K) in H, and so by (5.6), m = H / H K. Then use (i) of (4.3). (7.5) Suppose K G and H G. Then HK G HK = KH. Proof Assume HK G. Then H, K HK and so KH HK. To see the reverse containment, by HK G, (hk) 1 = h k HK, and so hk = (h k ) 1 = k 1 h 1 KH = HK KH. Conversely, assume HK = KH. Use (2.3). (7.6) If H G, K G, and if H N G (K), then HK G. Proof H N G (K) = HK = KH. Use (7.5). (7.7) If H G, K G, and if K G, then HK G. Proof K G implies that H N G (K). 15
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