Lecture Note of Week 2

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1 Lecture Note of Week 2 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. (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 ). 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 1

2 (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 2

3 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)(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. 3

4 4. Cosets and Counting (4.1) (Thm 4.2) Let G be a group 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. (4.2) (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. (4.3) (Thm 4.2) g G, gh = H = Hg. PF: define a bijection. (4.4) Let H < G. The index of H in G, denoted [G : H], is the cardinal number of the set of distinct left cosets of H in G. (4.5) If K < H < G, then [G : K] = [G : H][H : K]. PF: Use (4.2). Show that K has [G : H][H : K] cosets in G. (4.6) (Cor. 4.6: Lagrange) If H < G, then G = [G : H] H. (4.7) Let H and K be finite subgroup of G, then HK = H K / H K. 4

5 5. Normality, Quotients and Homomorphisms (5.1) Let φ : G H be a group homomorphism. (i) φ(e G ) = e 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) = e H } G. (v) The image of G under φ, imφ = {h H φ(g) = h, for some g G} H. Proof (i): Use φ(e G e 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). (5.2) 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. (5.3) 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 5

6 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). (5.4) (5.1) 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). (5.4a) Examples: φ : Z nz, for any fixed n 1 and n Z. Projections in R 2. φ : S 3 Z 3. (5.5) Let N G. Then un = vn u 1 v N. Proof un = vn = u vn = u 1 v N = v un = un = vn. (5.6) (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, (eg 1 )N = (ng 1 )N, and so by (5.5), 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, 6

7 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 (5.2). (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. (5.7) A subgroup N satisfying any one properties of (5.6) is called a normal subgroup of G. Denote this fact by N G. The homomorphism π in the proof of (v) = (vi) in (5.6) is called the natural projection or canonical homomorphism of G onto G/N. 7

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