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• C. Ephrem Stuyvesant

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7: Structure of Groups

7.1: Isomorphism Theorems; Automorphisms

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Isomorphism Theorems; Automorphisms 1. In G=ℤ×32 find cyclic subgroups H of order 2 and K of order 8 with HK = G and H ∩ K={e }. Conclude that ℤ×32≃ℤ 2×ℤ8 . First, ℤ×32={1, 3,5, 7, 9,11, 13,15, 17,19, 21, 23, 25, 27, 29, 31}. We see that o (3)=8 and o (15)=2, so 〈3〉={1, 3, 9,11, 17,19, 25, 27}≃ℤ8 and 〈15〉={1, 15}≃ℤ2 We see that 〈3〉∩〈15〉={1} and 〈3〉 〈15〉={1, 3,5, 7, 9,11, 13, 15,17, 19, 21, 25, 27, 29,31}. Thus, by Theorem 7.1.3, ℤ×32≃ℤ 2×ℤ8 . 2. Prove that D 6 ≃S 3×ℤ2 . Consider the sets {e , a b} and {e , a 2 , a 4 , b , a 2 b , a 4 b}. Both are subgroups: the first is cyclic of order 2 and the second has the same structure as S 3 . We also see that the intersection of the subgroups is the identity and that their product yields all of D 6 . Thus, by Theorem 7.1.3, D6 ≃S 3×ℤ2 . 3. Determine Aut (ℤ2×ℤ 2). First, ℤ2 ×ℤ2={(0,0),(0,1),(1,0),(1,1)}. Now, since ∣ℤ2×ℤ2∣=4, the only possible proper, nontrivial subgroups are of order 2. (This is by Lagrange's Theorem. Since it is such a fundamental result, I will no longer cite it when it is used.) There are clearly 3 subgroups of this type: {(0,0),(0,1)}, {(0,0),(1,0)}, and {(0,0) ,(1,1)}. Since an isomorphism must preserve group structure, these subgroups must be mapped among each other. In other words, Aut (ℤ2×ℤ 2)≃S 3 . 4. Let G be a finite abelian group of order n, and let m be a positive integer with (n, m) = 1. Show that ϕ :G →G defined by ϕ ( g )=g m for all g ∈G belongs to Aut(G). (1-1) We know that ker ϕ={g ∈G : g m=e }. Thus, if g ∈ker ϕ , we know g m=e , so o ( g )∣m. As always, o ( g )∣n as well. But since (n, m)=1, we must have g = e. Thus, ker ϕ={e}. (Onto) Any finite map that is 1-1 is also onto. (Preservation) ϕ (ab)=(ab) m=a m b m=ϕ (a ) ϕ(b). Thus, the map is an automorphism. 5. Let ϕ :G →G be the function defined by ϕ (g )=g −1 for all g ∈G. Find conditions on G such that ϕ is a homomorphism. The map is 1-1 by uniqueness of inverses. It is also onto since inverses must exist in a group. However, ϕ (ab)=( ab)−1=b−1 a−1 . The only way ϕ (ab)=a−1 b−1 is if the elements commute. Thus, G must be abelian for the map to be an automorphism.

7: Structure of Groups

7.1: Isomorphism Theorems; Automorphisms

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6. Show that for G=S 3 , Inn (G)≃G . Examine Z ( G). Since S 3≃ D3 , which is defined by two elements a and b which do not commute, we know Z ( G)={e}. Thus, by Theorem 7.1.8, Inn (S 3)≃S 3 . 7. Determine Aut (S 3). We have already shown (Q2) that S 3 has four subgroups: one isomorphic to ℤ3 and three isomorphic to ℤ2 . Now, the subgroup of order 3 must be mapped to itself, leaving the three isomorphic copies of ℤ2 to be permuted with each other. Thus, Inn (S 3)≃S 3 . 8. For groups G1 and G 2 , determine the center of G1 ×G 2 . Let ( a , b)∈Z ( G1×G 2). Then, ( a , b)( g 1 , g 2 )=( g 1 , g 2)( a , b) for all (g 1 , g 2 )∈G1×G 2 . This implies that ( ag 1 , bg 2 )=( g 1 a , g 2 b). Thus, (a , b)∈Z (G1)×Z (G 2). Conversely, if (a , b)∈ Z (G 1)×Z (G 2), then (ag 1 , bg 2 )=( g 1 a , g 2 b) implying that (a , b)( g 1 , g 2 )=(g 1 , g 2)(a , b). Thus, (a , b)∈ Z ( G1×G 2). Therefor, Z ( G1×G 2)=Z (G1 )×Z (G 2). 9. Show that G /Z (G) cannot be a nontrivial cyclic group. Suppose G /Z (G)={Z (G) , a Z (G), a 2 Z (G),...}. Let b , c ∈G. Without loss of generality, assume j k b∈ a j Z (G) and c ∈a k Z (G) . In other words, b=a z 1 and c=a z 2 . Then, j k j k k j k j bc=a z 1 a z 2=a (a z 2 ) z 1=a a z 2 z 1=a z 2 a z 1=cb . Thus, c ,b ∈Z (G) and Z (G)=G. 10. Describe the centers Z ( D n ) of the dihedral groups D n , for all integers n≥3. If for some m, a m =a n−m , then a m ∈ Z ( D n ). These elements and the identity can be the only elements that commute since the group is generated by 〈a , b : a m=b 2 , a m b=ba m−n 〉. 11. In the group GL 2 (ℂ) of all invertible 2×2 matrices with complex entries, let Q be the following 1 0 ,± i 0 ,± 0 1 ,± 0 i . set of matrices: ± 0 1 0 −i −1 0 i 0 D . (a) Show that Q is not isomorphic to 4 (b) Find the center Z (Q)of Q .

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1 0 (to name two) commute with everything, but in D 4 , only the 0 1 element a 2 commutes with everything. Thus, Q ≄ D 4 . (a) In Q, the elements ±

(b) From p. 122, it is clear that only

commutes with every member of Q.

7: Structure of Groups

7.1: Isomorphism Theorems; Automorphisms

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m n , 12. Let F 20 be the subgroup of GL 2 ( ℤ5 ) consisting of all matrices of the form such that 0 1 m, n∈ℤ5 and m≠0, as defined in Exercise 23 of Section 3.8. This group will be called the Frobenius group of degree 5. Find the center of F 20 .

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m0 n 0 m 0 n0 m n ∈Z ( F 20 ). Then a b = 0 0 a b . Thus, 0 1 0 1 0 1 0 1 0 1 a m0 a n0+b m a m0 b+n0 = 0 . So, we need a n 0+b=m0 b+n0 for all a and b in ℤ5 . In 0 1 0 1 m 0 particular, when a = b = 0, we must have n0=0. Thus, Z ( F 20 )= 0 where the first entry is 1, 2, 0 1 3, or 4. Suppose

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13. Show that the Frobenius group F 20 defined in Exercise 12 can be defined by generators and 1 1 and b= 2 0 . relations as follows. Let a= 0 1 0 1 2 (a) Show that o (a)=5 , o(b)=4 , and b a=a b . (b) Show that each element of F 20 can be expressed in the form a i b j for 0≤i≤4 and 0≤ j≤3.

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(a) 〈 a 〉=

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1 0 1 1 1 2 1 3 1 4 , , , , . 〈b〉= 1 0 , 2 0 , 4 0 , 3 0 . 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1

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2 2 2 =a b. 0 1 m n m n = 1 n m 0 =a n bk . ∈F 20 . Then, (b) Let 0 1 0 1 0 1 0 1 ba=

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14. Let G be the subgroup of GL 2 (ℝ ) consisting of all matrices

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a 11 a 12 such that a 21 =0 and a 21 a 22

a 22=1. (a) Let N be the set of matrices in G with a 11=1. Show that N is a normal subgroup of G. 1 1 and b= 2 0 . (b) Let a= Show that if H =〈a 〉 , then b H b−1 is a proper subset of H. 0 1 0 1 Conclude that H is not normal in any group that contains b.

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x 0

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y r s −1 x / r y−sx /r and B= . Suppose AB = ∈N. This implies that x =r 1 0 1 0 1 −1 1 1/ x ( s− y) ∈ N . since the first entry must be 1. Then, A B= Conversely, if we evaluate 0 1 A−1 B= r / x 1/ x (s− y) first, we again see r = x. Thus, N ⊴G . 0 1 (a) Let A=

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7: Structure of Groups

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7.1: Isomorphism Theorems; Automorphisms

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m −1 1 2 m =a 2 m ∈ H. (b) ba b = Since we get only even multiples, we know b H b−1 ⊊ H . This 0 1 implies that b H ≠ H b so H ⋬ G.

15. Give another proof of Theorem 7.1.1 by constructing an isomorphism from ( HN )/ N onto H /( H ∩N ). Define f : HN → G/( H ∩N ) via f ( hn)=h( H ∩N ). Clearly ker f =N and f ( HN )=H /( H ∩N ). Thus, HN ≃H /( H ∩N ) .

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