(5.11) (Second Isomorphism Theorem) If K G and N G, then K/(N K) = NK/N. PF: Verify N HK. Find a homomorphism f : K HK/N with ker(f) = (N K).


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1 Lecture Note of Week 3 6. Normality, Quotients and Homomorphisms (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. (5.8) 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 (5.1). (ii) of (5.8) = (v) of (5.1). (5.9) For any H G/N, N π 1 ( H) G. Proof It suffices to show that π 1 ( H) G. Use (i) of (5.8). (5.10) (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 Isomorphism Theorem) f is an isomorphism iff f is an epimorphism and N = ker(f). (5.11) (Second Isomorphism Theorem) If K G and N G, then K/(N K) = NK/N. PF: Verify N HK. Find a homomorphism f : K HK/N with ker(f) = (N K). (5.12) (Third Isomorphism Theorem) If K G and N G and if K < H then H/K G/K and (G/K)/(H/K) = G/H. PF: Verify H/K G/K. Find a homomorphism f : G/K G/H with ker(f) = H/K. (5.13) (Thm 5.11) Let f : G H be an epimorphism (onto homomorphism) of groups. Then the assignment 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}. 1
2 (5.14) Z(G) G. (5.14a) If G/Z(G) is cyclic, then G is abelian. (5.15) A group G is simple if G > 1 and if H {< 1 >, G} whenever H G. (5.15a) The only simple abelian groups are Z p, for prime p s. Proof (5.1). (5.16) 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. 6. Symmetric, Alternating, and Dihedral Groups (6.1) Review of symmetric group on n elements. Any permutation φ S n is a product of cycles (called the cycle decomposition of φ). A 2cycle is a transposition. Any permutation can be written as 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 ). (6.2) For each n 3, the dihedral group D n is a subgroup permutations of order 2n generated by Any group with two generator a, b with the relations a n = e, b 2 = e and ba = a 1 b is isomorphic to D n, and is also called the dihedral group of order 2n. Example: D 4. (6.3) 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) (6.2) holds for φ = e. 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. 2
3 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 further 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 transposition 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. (6.4) (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 Group of degree n. Proof Use (2.3) to show A n S n. 3
4 (6.5) Let a = (123 n) and (1n)(2(n 1)) ( n 2 b = ( n 2 + 1)) (1n)(2(n 1)) ( n 1 n if n is even, n+1 )( 2 ) if n is odd. The subgroup D 2n =< a, b > is called the dihedral group of order 2n. The presentation of D 2n is a n = b 2 = 1, and ba = a 1 b. We denote that D 2n =< a, b a n = b 2 = 1, and ba = a 1 b >. 8. Direct Products and Direct Sums (8.1) The direct product (also refereed as complete direct sum) of a collection of groups G i, i I consists of the Cartesian product (of sets) G i = {f : I i I G i such that for each i I, f(i) G i } i I in which the binary operation is defined componentwise. That is, if f, g i I G i, then for each i I, fg(i) = f(i)g(i) with the multiplication taking place in G i. One can routinely verify that this is a group. Elements in i I G i are also commonly written in a vector form a = {a i }, where a i = a(i). In this case, the product (sum, if in additive notation) of {a i } and {b i } will be {a i b i } (or respectively, {a i + b i }, in additive notation.) (8.2) For a fixed i I, let J = I {i}. Define a map ι i : G i j I G j by defining ι i (g) to be the element in j I G j such that for any j I, g ι i (g)(j) = e Gj if j = i otherwise Then ι i is an injective homomorphism, (referred as the inclusion map, or the canonical injection), which is an isomorphism between G i and a subgroup ι i (G i ). (8.2A) Let φ i : j I G j j J G j by φ(f)(j) = f(j) if j i and φ(f)(j) = e Gi if j = i. Then φ i is an onto homomorphism with kernel ι i (G i ). Therefore, j I G j/ι i (G i ) = j J G j. (8.3) Fix i I. Define π i : j J G j G i by π i (f)(j) = f(i) if j = i and π i (f)(j) = e Gj if 4
5 j i. Then π i is also a homomorphism, (referred as the canonical projection map onto the ith component). (8.4) Let w G i = {f G i : for all but finitely many i I, f(i) = eg i }. i I i I Then w i I G i is also a group, called the (external) weak direct product (also referred as the (external) direct sum) of the G i s. (8.5) Let {N i i I} be a collection of normal subgroups of a group G. If each of the following holds: (i) G = i I N i. (ii) for each k I, N k i I {k} N i = {e G }. Then G = w i I N i. Proof For each f w i I N i, there is a finite set I f I such that if j I I f, then f(j) = e Nj. Define a map φ : w i I N i G by φ(f) = i I f f(i). As I f <, φ(f) is a welldefined product in G. By (ii), when i j, a i N i and a j N j, a i a j = a j a i. This, together with (i), implies that φ is an onto homomorphism. Note Ker φ = {f w i I N i i I f f(i) = e G }. Apply (ii) again to see that Ker φ = {e}, where e(i) = e Ni, i I. (8.6) Let G and the N i s satisfy the hypotheses (i) and (ii) in (1.4). Then G is called the internal weak direct product (also referred as the internal direct sum) of the N i s. (8.7) Let {f i : G i H i } be a collection of group homomorphisms, and let f = i I f i denote the map f : i I G i i I H i such that for every {a i } i I G i, f({a i }) = {f i (a i )} i I H i. Then f is a homomorphism. such that f( w i I G i w i I H i, Ker f = i I Ker f i, and Im f = i I Im f i. Proof Routine verification. (8.8) Let {G i i I} and {N i i I} be collections of groups such that for each i I, N i G i. Then each of the following holds. (i) i I N i i I G i, and i I G i/ i I N i = i I G i/n i. 5
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