II. Products of Groups

II. Products of Groups Hong-Jian Lai October 2002 1. Direct Products (1.1) 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.) (1.2) 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 ). (1.2A) 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. (1.3) 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 j i. Then π i is also a homomorphism, (referred as the canonical projection map onto 1

the ith component). (1.4) 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. (1.5) 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 well-defined 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. (1.6) 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. (1.7) 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. (1.8) 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. ii w i I N i w i I G i, and w i I G i / w i I N i = wi I G i /N i. 2

Proof Let π i : G i G i /N i be canonical epimorphism. Then i I π i : i I G i i I G i/n i is also an epimorphism with kernel i I N i. Then apply the 1st isomorphism theorem. 2. Free Abelian Groups (2.1) All groups will be in additive notation in this section, and with 0 as the identity. Therefore, if G is an abelian group and a G, then for any integer n > 0, na = a+a+ +a, ( n)a = ( a) + ( a) + + ( a) with n summands in either case. (2.1A) The cyclic group a = {na n Z}. More generally, an abelian group generated by a set X has the form X = {n 1 a 1 + n 2 a 2 + + n k a k with k > 0, and n 1, n 2,, n k, k Z, a 1, a 2,, a k X}. An expression like n 1 a 1 +n 2 a 2 + +n k a k will be called a linear combination of elements in X. (2.1B) For an abelian group G, a basis of G is a subset X G satisfying the following two conditions: (i) G = X. (ii) If n 1 a 1 +n 2 a 2 + +n k a k = 0, for some a i X and n i Z, then for every i, n i = 0 in Z. (2.2) Let F be an abelian group. The following are equivalent for F (any abelian group satisfying any of the following will then be called a free abelian group over the set X): (i) F has a nonempty basis. (ii) F is the (internal) direct sum of a family of infinite cyclic subgroups. (iii) F is isomorphic to i I Z. (iv) There exists a nonempty set X and a map ι : X F such that given any abelian group G and a map f : X G, there exists a unique homomorphism of groups f : F G such that fι = f. Proof (i) = (ii). By (2.1B)(ii), a X, N a = a = Z. Then verify that G = a X N a and N a ( x X {a} N x ) = {0}. (ii) = (iii). Exercise. (iii) = (i). Assume F = i I Z. It suffices to show that i I Z has a basis. For each 3

i I, let u i i I Z be such that u i (j) = 1 if j = i 0 otherwise, and let X = {u i i I}. Show that X is a basis of i I Z. (i) = (iv). Let X be a basis of F and ι : X F be the inclusion map. Let f : X G be a map. Linearly expand f to a homomorphism f : F G. Verify that fι = f. For uniqueness, show that if gι = f, then we must have g = f, by the linearity. (iv) = (i). Consider G = i X Z and take the standard basis u i as in (iii) = (i). Then verify that ι(x) is a basis of F. (2.3) Any two basis of a free abelian group F have the same cardinality, (called the rank of F ). Proof Let X be a fixed basis of F, and let Y be another basis of F. If X = n <, then F = n i=1 Z. It follows that F/2F = (Z 2 ) n, and so F/2F = 2 n. If Y = r, then similarly 2 n = F/2F = 2 r, and so r < and r = n. Now assume that X =. Then Y = as well. We shall show that X = F. Clearly X F. Let X n = n i=1 X, and S = n Z,n>0 X n. Verify that X = S. For each s = (x 1, x 2,, x n ) X n, let G s = x 1, x 2,, x n. Then G s = Z. As F = s S G s, we have F Z S = S. But X = S. (2.4) If F is a free abelian group of finite rank n, and G is a nonzero subgroup of F. Then there exist a basis {x 1, x 2,, x n } of F, an integer r (with 0 < r n), and positive integers d 1, d 2,, d r with d 1 d 2 d r, and G is free abelian with basis {d 1 x 1,, d r x r }. (2.4A) Lemma: Suppose that F is a finite rank free abelian group and {x 1, x 2,..., x n } is a basis of F. Let {y 1, y 2,..., y n } F be a set of elements and A be an integral matrix such that (y 1, y 2,..., y n ) = (x 1, x 2,..., x n )A. Then {y 1, y 2,..., y n } is a basis of F if det A = 1. Proof: If det A = 1, then A 1 is also an integral matrix and so (x 1, x 2,..., x n ) = (y 1, y 2,..., y n )A 1. It follows that {y 1, y 2,..., y n } is also a basis. Proof for (2.4) We argue by induction on n. As (2.4) holds trivially for n = 1, we assume that n > 1, and (2.4) holds for free abelian group with rank smaller than n. Let S = {s Z {0} : such that there exists a basis {y 1, y 2,..., y n } of F and for some k 2,..., k n Z, sy 1 + k 2 y 2 + + k n y n G}. As G {0}, we have S. Fact 1 If s S, then s S; if sy 1 + k 2 y 2 + + k n y n G, then for each i 2, k i S. (Reason: G is a group; and {y i, y 1,..., y i 1, y i+1,..., y n } is also a basis of F ). 4

Therefore, d 1 = min{ s : s S} > 0. Pick g 1 = sy 1 + k 2 y 2 + + k n y n G. By division, q i, r i Z with k i = q i d 1 + r i for each 2 i n such that 0 r i < d 1. Let x 1 = y 1 + q 2 y 2 + + q n y n F. Fact 2 By Lemma 2.4A, {x 1, y 2,..., y n } is also a basis of F. Therefore, with H = y 2, y 3,..., y n, F = x 1 H. Note that the element g 1 = d 1 x 1 + r 2 y 2 + + r n y n G. As {x 1, y 2,..., y n } in any order is also a basis of F, each r i S if r i 0. By the minimality of d 1, each r i = 0 and so g 1 = d 1 x 1. Fact 3 G = d 1 x 1 (G )H. (Reason: By Fact 2, d 1 x 1 (G )H = {0}. For any g = t 1 x 1 + t 2 y 2 + + t n y n G, t 1 = d 1 l 1 and so g d 1 x 1 (G )H. On the other hand, d 1 x 1 G and so d 1 x 1 (G )H G.) By Fact 3, if G H = {0}, then G = d 1 x 1. Otherwise, by induction of n, H has a basis {x 2, x 3,..., x n } and there exist some integers r, d 2,..., d r such that G H is free abelian with basis {d 2 x 2, d 3 x 3,..., d r x r }, and such that d 2 d 3 d r. To prove d 1 d 2, we write d 2 = qd 1 + r with 0 r < d 1, and we want to show r = 0. Note that d 1 x 1, d 2 x 2 G. Then d 1 x 1 + d 2 x 2 = d 1 (x 1 + qx 2 )rx 2 G. As 0 r < d 1, by the minimality of d 1, it suffices to show that {x 2, x 1 + qx 2, x 3,, x n } is also a basis of F. (2.5) A group G is finitely generated if there is a finite subset X G such that G =< X >. (2.6) The Fundamental Theorem of Finitely Generated Abelian Groups Let G be a finitely generate abelian group. G = Z n Z m1 Z m2 Z mt, for some integer n 0 (the free rank), integer t 0, and (when t > 0) integers m 1, m 2,, m t (the invariant factors) such that m 1 > 1 and m i m i+1, (1 i t 1). (Notation: When n = 0, we say that the finite abelian group G is of type (m 1, m 2,, m t ).) Proof Let X be a set with X = n such that G = X. By (2.2)(iv), there is a free abelian group F with rank n and an epimorphism π : F G. Let N = Ker π. By the 1st Isomorphism Theorem, G = F/N. Therefore, we may assume that N is nontrivial, and so there exist a basis {x 1, x 2,, x n } of F, an integer r (with 0 < r n), and positive integers d 1, d 2,, d r with d 1 d 2 d r, and N is free abelian with basis {d 1 x 1,, d r x r }. Let d r+1 = = d n = 0. Then F = n i=1 x i and 5

N = n i=1 d i x i. Thus (m i will be those d j s such that d j > 1) G = F/N n n = x i / d i x i = Z di. i=1 i=1 (2.7) Let G be an abelian group, m, p are integers such that p is a prime. Each of the following holds. (i) Each of the following is a subgroup of G: mg = {ma : a G}, G[m] = {a G : ma = 0}, G(p) = {a G : a = p n for some integer n 0}, and G t = {a G : a < } (called the torsion subgroup of G). (ii) There are isomorphisms Z p n[p] = Z p (n 1) and p m Z p n = Z n m p (m < n). (iii) Let H and G i (i I) be abelian groups. If φ : G i I G i is an isomorphism, then the restrictions of φ to mg and G[m] respectively are isomorphisms mg = i I mg i and G[m] = i I G i[m]. (iv) If φ : G H is an isomorphism, then the restrictions of φ to G t and G(p) respectively are isomorphisms G t = Ht and G(p) = H(p). Proof (i). Use the subgroup criterion. (Exercise: Find a counterexample to (i) if G is not abelian). (ii). Z p n[p] = p n 1. Also, pa 0 (mod p n ) iff a 0 (mod p n 1 ). (iv). p n φ(a) = φ(p n a) = φ(g(p)) H(p). Now consider φ 1, and the restrictions of φφ 1 in H(p) and φ 1 φ in G(p). (2.8) Let G be a finitely generated abelian group. Each of the following holds. (i) There is a unique nonnegative integer s such that the number of infinite cyclic summands in any decomposition of G as a direct sum of cyclic groups is precisely s. (ii) Either G is free abelian, or there is a unique list of (not necessarily distinct) positive integers m 1, m 2,..., m t such that m 1 > 1, m 1 m 2 m t and for a free abelian group F, G = Z m1 Z m2 Z mt F. (iii) Either G is free abelian, or there is a list of prime powers p n 1 1, pn 2 2,..., pn k k, where p i s are not necessarily distinct primes, and where n j s are not necessarily distinct positive integers, such that for a free abelian group F, G = Z p n 1 1 Z p n 2 2 Z p n k k F. 6

3. Semidirect Products (3.1) Let H and K be groups and let φ : K Aut(H) be a homomorphism. (Notation: φ(k)(h) = k h.) Let G = {(h, k) h H and k K} with binary operation (h 1, k 1 )(h 2, k 2 ) = (h 1 (k 1 h 2 ), k 1 k 2 ). Then each of the following holds: (i) G with this operation forms a group, denoted by G = H φ K, called the semidirect product of H and K with respect to φ. (ii) The sets H = {(h, 1) h H} and K = {(1, k) k K} are subgroups of G, isomorphic to H and K, respectively. (iii) H G. (iv) H K = {1}. (v) h H and k K, khk 1 = k h = φ(k)(h). Proof Check definitions. (3.1a) When φ is the identity homomorphism, then H φ K H K. (3.2) If K, H G with G = HK and H K = 1, then H is a complement of K in G. If each of the following holds: (i) H is a complement of K in G, (ii) H G, (iii) there is a homomorphism φ : K Aut(H) such that φ(k)(h) = khk 1, h H, k K, then G = H φ K. Proof The map hk (h, k) is an isomorphism G H φ K. (3.2A) Groups of order pq, p and q are primes with p > q. Cases: q (p 1), and q (p 1). (3.2B) Groups of order 96. Pick H, K Syl 2 (G). Estimate HK and N G (H K). (3.2C) Groups of order p 3, where p > 2 is a prime. 7

4. The Krull-Schmidt Theorem (4.1) A group G is indecomposable if G > 1 and G is not the internal direct product of two of its proper subgroups. (Examples include Z, Z p n and simple groups). (4.2) A group G satisfies the ascending chain condition (ACC) on (normal) subgroups if for every chain G 1 < G 2 < of (normal) subgroups of G, there is an integer n such that G i = G n whenever i n; G satisfies the descending chain condition (DCC) on (normal) subgroups if for every chain G 1 > G 2 > of (normal) subgroups of G, there is an integer n such that G i = G n whenever i n. (4.2A) Z satisfies ACC but not DCC. (4.2B) Z(p ) satisfies DCC but not ACC. (4.3) If G satisfies ACC or DCC on normal subgroups, then G is the direct product of a finite number of indecomposable subgroups. (4.4) An endomorphism f of G is a normal endomorphism if a, b G, af(b)a 1 = f(aba 1 ). Suppose that G satisfies ACC (resp. DCC) on normal subgroups, and f is an endomorphism (resp. normal endomorphism) of G. Then f is an automorphism iff f is an epimorphism (resp. monomorphism). (4.5) (Fitting s Lemma) If G satisfies both ACC and DCC on normal subgroups, and f is a normal epimorphism of G. Then for some n 1, G = Ker f n Im f n. (4.6) An epimorphism of G is nilpotent if for some n > 0, f n (g) = e G for all g G. If G is an indecomposable group that satisfies both ACC and DCC on normal subgroups, and f is a normal epimorphism of G, then either f is nilpotent, or f is an automorphism. (4.7) Let G be a nontrivial indecomposable group that satisfies both ACC and DCC on normal subgroups. If f 1, f 2,..., f n are normal nilpotent epimorphisms of G such that every f i1 + f i2 + + f ir (1 i 1 i 2 i r n) is an endomorphism, then f 1 + f 2 + + f n is nilpotent. (4.8) Let G be a nontrivial indecomposable group that satisfies both ACC and DCC on normal subgroups. If G = s i=1 G i and G = t j=1 H j with each G i and H j indecomposable, 8

then s = t and after reindexing, G i = Hi for every i and for each r < t, G = G 1 G r H r+1 H t. 9