ABSTRACT ALGEBRA I: HOMEWORK 4 SOLUTIONS ANDY EISENBERG 1 a b Problem . R (pgs. 111-114, 9) Let G = GL3( ). Show that H = 0 1 c is 0 0 1 a subgroup of G. Proof. Since the determinant of any matrix in H is 1, they are all invertible. It is 1 a b easy to verify that if A = 0 1 c ∈ H is an arbitrary element, then 0 0 1 1 a b 1 −a ac − b 1 −a ac − b 1 a b 1 0 0 0 1 c 0 1 −c = 0 1 −c 0 1 c = 0 1 0 . 0 0 1 00 1 00 1 0 0 1 0 0 1 1 d e Thus the inverse is also in H. Now let B = 0 1 f ∈ H be another arbitrary 0 0 1 element. Then 1 a b 1 d e 1 a + d af + b + e 0 1 c 0 1 f = 0 1 c + f . 0 0 1 0 0 1 00 1 Thus for any A, B ∈ H, we have A−1 ∈ H and AB ∈ H. Since H is closed under the group operation and under taking inverses, it is a subgroup. Problem (pgs. 111-114, 14). Let G be an abelian group. Show that the set of all elements of G of finite order forms a subgroup of G. Proof. Let H be the set of all elements of G of finite order. Let a,b ∈ H be arbitrary elements. Since a is of finite order in G, we have an = e for some n. Then (a−1)n = a−n = (an)−1 = e−1 = e, so a−1 ∈ H. Since b is also of finite order, there is some m so that bm = e. Now (ab)nm = anmbnm = emen = e, and ab ∈ H. Since H is closed under the group operation and under taking inverses, it is a subgroup. Problem (pgs. 111-114, 15). Prove that any cyclic group is abelian. Proof. Let G = hai be a cyclic group generated by a. Take x, y ∈ G. Then there exist n,m ∈ Z such that x = an, y = am. Hence xy = anam = an+m = aman = yx, so G is abelian. Problem (pgs. 111-114, 17). Prove that the intersection of any collection of sub- groups of a group is again a subgroup. Date: Fall 2009. 1 2 ANDY EISENBERG Proof. Let G be a group, and let {Hα}α∈A be a collection of subgroups of G indexed by an arbitrary set A. Define H = α∈A Hα. Take arbitrary elements a,b ∈ H. T −1 Since a ∈ H, a ∈ Hα for all α ∈ A. For each α, since Hα is a subgroup, a ∈ Hα. −1 −1 Since a ∈ Hα for all α ∈ A, a ∈ α Hα = H. Similarly, since a,b ∈ H, we have a,b ∈ Hα for all α. Since each HTα is a subgroup, we have ab ∈ Hα. Since ab ∈ Hα for all α ∈ A, we have ab ∈ α Hα = H. Since H is closed under the group operation and under taking inverses,T it is a subgroup. Problem (pgs. 111-114, 19). Let G be a group, and let a ∈ G. The set C(a) = {x ∈ G | xa = ax} of all elements of G that commute with a is called the centralizer of a. (a) Show that C(a) is a subgroup of G. (b) Show that hai⊂ C(a). Proof. (a) Let g,h ∈ C(a). Then ga = ag. Multiplying on the left by g−1, we have a = g−1ga = g−1ag. Multiplying on the right by g−1, we have ag−1 = g−1agg−1 = g−1a. Since ag−1 = g−1a, we have g−1 ∈ C(a). We also have ha = ah. So (gh)a = g(ha)= g(ah) = (ga)h = (ag)h = a(gh), so gh ∈ C(a). Since C(a) is closed under the group operation and under taking inverses, it is a subgroup. (b) A typical element of hai is an for some n ∈ Z. Recall that anam = an+m, so ana = an+1 = aan, and an ∈ C(a). This shows hai⊂ C(a). Problem (pgs. 111-114, 21). Let G be a group. The set Z(G) = {x ∈ G | xg = gx, ∀ g ∈ G} of all elements that commute with every other element of G is called the center of G. (a) Show that Z(G) is a subgroup of G. (b) Show that Z(G)= a∈G C(a). T (c) Compute the center of S3. Proof. (a) Suppose a,b ∈ Z(G). Then for any x ∈ G, we have abx = axb = xab, so ab ∈ Z(G). Also, we have ax = xa. Multiplying both sides on the right by a−1, we have axa−1 = x. Multiplying on the left by a−1, we have xa−1 = a−1x, hence a−1 ∈ Z(G). Since Z(G) is closed under the group operation and under taking inverses, it is a subgroup. (b) Suppose g ∈ Z(G). Then ga = ag for every a ∈ G, so g ∈ C(a) for every a ∈ G, hence g ∈ C(a). That is, Z(G) ⊂ C(a). Conversely, suppose g ∈ C(a). Then ga =Tag for every a ∈ G. By definition,T g ∈ Z(G), hence Z(G)= T C(a). T (c) The elements of S3 besides the identity are either transpositions or 3-cycles. Consider the transposition (1 2). We have (1 2)(1 3) = (3 2 1) 6= (2 3 1) = (1 3)(1 2). This shows that (1 2) ∈/ Z(S3). We can similarly demonstrate that no transpo- sition can be in Z(S3). Now consider (1 2 3). We have (1 2 3)(1 2) = (1 3) 6= (3 2) = (1 2)(1 2 3). By a similar calculation, no 3-cycle is in Z(S3). This shows that Z(S3)= {e}. ABSTRACT ALGEBRA I: HOMEWORK 4 SOLUTIONS 3 Problem (pgs. 111-114, 26). Let G be a group with a,b ∈ G Assume that o(a) and o(b) are finite and relatively prime, and that ab = ba. Show that o(ab)= o(a)o(b). Proof. Since o(ab) = lcm(o(a), o(b)) and gcd(n,m) = 1, we have o(ab) = o(a)o(b). .