Abstract Algebra

Abstract Algebra

1 Groups Definition 1.1 (Group). A Group G is a set with a binary operation m : G × G ! G (m(g; h) := gh) such that 1. 8a; b; c 2 G, (ab)c = a(bc) 2. 9e 2 G such that 8a 2 G, ea = ae = a 3. 8a 2 G, 9a−1 2 G such that aa−1 = a−1a = e If only 1 is satisfied, then G is a semigroup. If only 1 and 2 are satisfied, then G is a monoid. If a structure has 8a; b 2 G, ab = ba, then it is abelian. Definition 1.2 (Group Homomorphism). A map f : G ! H between groups is a homomorphism if f(ab) = f(a)f(b) If the homomorphism is injective, it is a monomorphism. If the homomorphism is surjective, it is an epimorphism. If the homomorphism is bijective, it is an isomorphism. Lemma 1.1. Let ' : G ! H be a group homomorphism. Then '(eG) = eH and '(a−1) = '(a)−1 Proof. '(a) = '(aeG) = '(a)'(eG) −1 −1 '(a) '(a) = '(a) '(a)'(eG) = eH = eH ')eG) = '(eG) The other part is similar. Definition 1.3 (Kernel of a Homomorphism). The kernel of a homomorphism f : G ! H is the set fa 2 G : f(a) = eH g and is denoted ker f Definition 1.4 (Subgroup). If G is a group and H ⊆ G is itself a group under G's multiplication, then H is a subgroup of G, denoted H < G Trivially, ker f < G Lemma 1.2. A nonempty subset H ⊆ G is a subgroup iff 8a; b 2 H, ab−1 2 G Definition 1.5 (hom(G; H)). If G; H are groups, then hom(G; H) is the set of homomorphisms from G to H. Definition 1.6 (Cosets). Let H < G. The the left coset of H containing a is aH = fah : h 2 Hg. Right cosets are similarly defined. Lemma 1.3. If jHj < 1 then jaHj = jHj = jHaj Proof. Let h : H ! aH : h 7! ah. Then ah = ah0 ) h = h0, so f is bijective. Lemma 1.4. Let H < G. Then for a; b 2 G, either aH = bH or aH \ bH = ;. Also, aH = bH iff a−1b 2 H 1 Proof. Suppose aH \ bH 6= ;. Then 9h; h0 2 H such that ah = bh0. So a = bh0h−1 thus a 2 bH. Thus aH ⊂ bH. This argument is symmetric, thus aH = bH. Then, if aH = bH we have a−1b = hh0−1 2 H If a−1b = h 2 H, then b = ah so b 2 aH and aH \ bH 6= ;, so aH = bH. Definition 1.7 (Order of g). The order of g 2 G is the smallest positive n such that gn = e, or 1 is no such n exists. Theorem 1.5 (Lagrange's Theorem). If G is a finite group, then the order of any element divides jGj. Proof. We first check that each element has finite order. Since jGj < 1, there must be m; n > 0 such that gn = gm, where n < m So gm−n = e. Suppose that g has order n. Then consider fe; g; g2; : : : ; gn−1g = H The left cosets of H partition G, and each is size n, so jGj = nk, where k is the number of left cosets of G. Definition 1.8 (Direct Product Group). If G; H are groups, then G × H has elements f(g; h): g 2 G; h 2 Hg and multiplication (g; h)(g0; h0) = (gg0; hh0) Definition 1.9 (Normal Subgroup). A subgroup H of G is said to be normal if 8a 2 G; aH = Ha. We write H E G −1 −1 Lemma 1.6. N E G iff aNa = H 8a 2 G iff aNa ⊆ N 8a 2 G. Proof. If N is normal, then aN = Na, so aNa−1 ⊆ N. Also, 8n0 2 N, 9n 2 N such that an = n0a so ana−1 = n0 2 N, thus n0 2 aNa−1. Thus aNa−1 = N If aNa−1 = N, then 8n 2 N, 9n0 2 N such that an0a−1 = n so an0 = na and thus Na ⊂ aN. This argument is symmetric, so N E G. Definition 1.10 (Index). If H < G then the index of H in G, written [G : H] is the number of left cosets of H. This may be infinite. Theorem 1.7. If N E G then the set faN : a 2 Gg is a group of order [G : N] with operation (aN)(bN) = (ab)N Proof. If aN = a0N, bN = b0N then abN = a0b0N. as aN = a0N; a−1a0 2 N, similarly b−1b0 2 N. (ab)−1(a0b0) = b−1a−1a0b0 = b−1nb for some n 2 N. Then, there is an n0 such that b−1nb0 = b−1b0n0 = n00 2 N Thus, abN = a0b0N This multiplication is associative, because G is a group, and the other group properties follow similarly. Definition 1.11 (Quotient Group). The group in the theorem above is called G=N. 2 Note: f : G ! H a homomorphism, then ker f E G. Theorem 1.8. If f : G ! H is a homomorphism and N E G, N ⊆ ker f then 9!f~ : G=N ! H such that f~(aN) = f(a) 8a 2 G. Then Im f~ = Im(f) and ker f~ = ker f=N and f~ is an isomorphism iff f is surjective and N = ker f. Proof. See Hungerford. Corollary 1.9 (First Isomorphism Theorem). If f : G ! H is a group homo- morphism, then G= ker f ' Im f Definition 1.12 (HK). If H and K are subsets of a group, then HK = fhk : h 2 H; k 2 Kg Definition 1.13 (Join). If H; K < G then H _ K is the smallest subgroup of G containing both H and K. Whenever the smallest subgroup of G containing some set S is mentioned, it is the subgroup \Hi, where Hi < G and S ⊆ Hi. Lemma 1.10. If N E G; K < G, then 1. N \ K E K 2. N E N _ K 3. NK = N _ K = KN Proof. 1. If n 2 N\K, a 2 K, then ana−1 2 N\K, so a(N\K)a−1 ⊂ N\K, so N \ K E K 2. N E G and N < N _ K < G, so N E N _ K. 3. Since N _K is closed under multiplication, NK ⊆ N _K. Let a 2 N _K, write a = n1k1 : : : ; nrkr 0 0 Since N is normal, 8n 2 N, we can write kjn = n kj for some n 2 N. So a = nk1 : : : kr = nk with n 2 N; k 2 K. So a 2 NK. Theorem 1.11 (Second Isomorphism Theorem). If K; N < G and N E G then K=N \ K ' NK=N Proof. Define a homomorphism f : K ! NK=N by f(k) = kN. f ... K ............................................................................................................................................................................................................................................... NK=N .... ..... ....... ..... ...... ..... ...... ..... ..... ..... ..... ..... ..... ..... ..... ..... ι ..... ..... π ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ...... ..... ........ ..... N _ K ker f = N \ K. By the first isomorphism theorem, K=N \ K ' Im f 3 Thus, it remains to show that Im f = NK=N. Consider nkN 2 NK=N. Since N is normal, 9n0 such that nkN = kn0N = kN = f(k). Theorem 1.12 (Third Isomorphism Theorem). Let K E G and H E G, with K < H. Then H=K E G=K and (G=K)=(H=K) E G=H Proof. Define a homomorphism f : G=K ! G=H by f(aK) = aH. f is a homomorphism, as K ⊂ ker(G ! G=H) ker f = H=K, and f is surjective by definition, so by the first isomorphism theorem, the result follows. Definition 1.14 (Simple Group). A group G is a simple group if it has no proper nontrivial normal subgroups. Now we will determine solutions to the problem of how can a group be described. 1. Listing the elements and making a table 2. Give the generators for G as a subgroups of Sn, finite groups only. 3. Give it as Aut X for some structure X. 4. Build up from simpler groups 5. Give generators for G as a subgroup of GLn (giving a homomorphism f : G ! GLn is the beginning of representation theory) 6. Generators and Relations. Theorem 1.13 (Cayley). Any finite group G is isomorphic to a subgroup of Sn for some n. Proof. Let n = jGj, and fix a bijection f1; : : : ; ng ! G such that G = fg1; : : : ; gng. Given g 2 G, 8i9j such that ggi = gj. Note that if ggi = ggk then gi = gk. Define σg : f1; : : : ; ng ! f1; : : : ; ng by σg(i) = j if ggi = gj This is an injection from a finite set into itself, and so a bijection. We will define ' : G ! Sn : g 7! σg Check that σgh = σgσh. ' is a homomorphism. Also, if ggi = gi then g = e, so σg 6= 1 for g 6= e. So ' is an injection, this '(G) ' G and is a subgroup of Sn. Corollary 1.14. Every finite group is isomorphic to a subgroup of GLn(R) for n = jGj. Proof. By Cayley's Theorem, it suffices to show that Sn is isomorphic to a subgroup of GLn 1 if σ(j) = i Define ' : S ! GL ( ): σ 7! A written A = n n R ij 0 else 4 Definition 1.15 (F (X)). Given a set X, we choose a set X−1 which is disjoint from X and has the same cardinality, with a bijection X ! X−1 : x 7! x−1, and we define an element 1 2 X [ X−1.

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