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MATH 3005 Homework Solution Han-Bom MoonHomework 9 SolutionChapter 8. Prove or disprove thatZ⊕Zis a cyclic group. IfZ⊕Zis a cyclic group, then all elements are multiple of a generator(a,b)∈ Z⊕Z. In particular, there is an integermsuch that(1,0) =m·(a,b). Sob= 0. Also there isn∈Zsuch that(0,1) =m·(a,...

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MATH 3005 Homework Solution

Han-Bom Moon

Homework 9 Solution Chapter 8.

5. Prove or disprove that Z ⊕ Z is a cyclic group. If Z ⊕ Z is a cyclic group, then all elements are multiple of a generator (a, b) ∈ Z ⊕ Z. In particular, there is an integer m such that (1, 0) = m · (a, b). So b = 0. Also there is n ∈ Z such that (0, 1) = m · (a, b) and a = 0. Therefore (a, b) = (0, 0) but it is not a generator of Z ⊕ Z. Thus Z ⊕ Z is not a cyclic group. 6. Prove, by comparing orders of elements, that Z8 ⊕Z2 is not isomorphic to Z4 ⊕ Z4 . In Z8 ⊕ Z2 , (1, 0) has order lcm(8, 1) = 8. But because any element in Z4 has order 1, 2, or 4, in Z4 ⊕ Z4 , the order of an element is 1, 2, or 4. Therefore Z8 ⊕ Z2 is not isomorphic to Z4 ⊕ Z4 . 7. Prove that G1 ⊕ G2 is isomorphic to G2 ⊕ G1 . State the general case. Let φ : G1 ⊕ G2 → G2 ⊕ G1 be a map defined by φ(a, b) = (b, a). Step 1. φ is one-to-one. φ(a1 , b1 ) = φ(a2 , b2 ) ⇒ (b1 , a1 ) = (b2 , a2 ) ⇒ b1 = b2 , a1 = a2 ⇒ (a1 , b1 ) = (a2 , b2 ) Step 2. φ is onto. For any (b, a) ∈ G2 ⊕ G1 , φ(a, b) = (b, a). Step 3. φ has the operation preserving property. φ(a1 , b1 )·φ(a2 , b2 ) = (b1 , a1 )·(b2 , a2 ) = (b1 b2 , a1 a2 ) = φ(a1 a2 , b1 b2 ) = φ((a1 , b1 )·(a2 , b2 )) Therefore φ is an isomorphism. In general, for any n groups G1 , G2 , · · · , Gn and a permutation σ ∈ Sn , G1 ⊕ G2 ⊕ · · · ⊕ Gn ≈ Gσ(1) ⊕ Gσ(2) ⊕ · · · ⊕ Gσ(n) . 12. Give examples of four groups of order 12, no two of which are isomorphic. Give reasons why no two are isomorphic. Z12 , Z6 ⊕ Z2 , D6 , A4 The first two groups are Abelian so they are not isomorphic to the latter two groups. Z12 has an order 12 element, but on Z6 ⊕ Z2 , the maximum order of an element is lcm(6, 2) = 6. So Z12 6≈ Z6 ⊕ Z2 . On the other hand, D6 has an order 6 rotation. But every non-identity element in A4 is of the form (abc) or (ab)(cd ), so the order of an element in A4 is 1, 2, or 3. Therefore D6 6≈ A4 . 1

MATH 3005 Homework Solution

Han-Bom Moon

16. Suppose that G1 ≈ G2 and H1 ≈ H2 . Prove that G1 ⊕ H1 ≈ G2 ⊕ H2 . State the general case. Let φ : G1 → G2 and ψ : H1 → H2 be two isomorphisms. Define a map f : G1 ⊕ H1 → G2 ⊕ H2 as f (g, h) = (φ(g), ψ(h)). We claim that f is an isomorphism. Step 1. f is one-to-one. f (g1 , h1 ) = f (g2 , h2 ) ⇒ (φ(g1 ), ψ (h1 )) = (φ(g2 ), ψ (h2 )) ⇒ φ(g1 ) = φ(g2 ), ψ(h1 ) = ψ(h2 ) Because φ and ψ are isomorphisms, they are one-to-one. So we have g1 = g2 and h1 = h2 , and these imply (g1 , h1 ) = (g2 , h2 ). Step 2. f is onto. For any element (x, y) ∈ G2 ⊕ H2 , there is a ∈ G1 such that φ(a) = x because φ is onto. Similarly, there is b ∈ H1 such that ψ(b) = y. Then f (a, b) = (φ(a), ψ(b)) = (x, y). Step 3. f has the operation preserving property. f (g1 , h1 ) · f (g2 , h2 ) = (φ(g1 ), ψ(h1 )) · (φ(g2 ), ψ (h2 )) = (φ(g1 )φ(g2 ), ψ(h1 )ψ(h2 )) = (φ(g1 g2 ), ψ(h1 h2 )) = f (g1 g2 , h1 h2 ) = f ((g1 , h1 ) · (g2 , h2 )) Therefore f is an isomorphism. In general, for i = 1, 2, · · · , n, if Gi ≈ Hi , then G1 ⊕ G2 ⊕ · · · ⊕ Gn ≈ H1 ⊕ H2 ⊕ · · · ⊕ Hn . 18. In Z40 ⊕ Z30 , find two subgroups of order 12. In Z40 , |10| = 40/10 = 4. In Z30 |10| = 30/10 = 3. So in Z40 ⊕ Z30 , |(10, 10)| = lcm(4, 3) = 12. Therefore h(10, 10)i is a subgroup of order 12. Similarly, in Z30 , |5| = 30/5 = 6. Therefore in Z40 ⊕ Z30 , |(10, 5)| = lcm(4, 6) = 12. So h(10, 5)i is a subgroup of order 12. We need to show that these two groups are not equal. Indeed, (10, 5) ∈ h(10, 10)i because for any elements in h(10, 10)i, the second coordinate is one of 0, 10, 20. So h(10, 5)i 6= h(10, 10)i. 30. Find all subgroups of order 4 in Z4 ⊕ Z4 . Case 1. Cyclic subgroups. Let (a, b) ∈ Z4 ⊕ Z4 be an order 4 element. Then lcm(|a|, |b|) = 4. Because |a|, |b| can be 1, 2, or 4, one of |a|, |b| must be order 4. So (1, b), (3, b), (a, 1), or (a, 3) are order 4. Now h(1, 0)i = h(3, 0)i, h(1, 1)i = h(3, 3)i, h(1, 2)i = h(3, 2)i, h(1, 3)i = h(3, 1)i, h(0, 1)i = h(0, 3)i, h(2, 1)i = h(2, 3)i are order 4 cyclic subgroups. Case 2. Subgroups isomorphic to Z2 ⊕ Z2 . 2

MATH 3005 Homework Solution

Han-Bom Moon

Any non-identity element in Z2 ⊕ Z2 has order 2. There are only 3 elements of order 2 in Z4 ⊕ Z4 : (2, 0), (0, 2), and (2, 2). It is straightforward to check that h2i ⊕ h2i = {(0, 0), (2, 0), (0, 2), (2, 2)} is a subgroup isomorphic to Z2 ⊕ Z2 . In summary, there are 7 order 4 subgroups in Z4 ⊕ Z4 . 36. Find a subgroup of Z12 ⊕ Z4 ⊕ Z15 that has order 9. Consider 4 ∈ Z12 and 5 ∈ Z15 . Then h4i is a cyclic subgroup of order 3 and h5i is a cyclic subgroup of order 3. Consider h4i ⊕ h0i ⊕ h5i ≤ Z12 ⊕ Z4 ⊕ Z15 . Then h4i ⊕ h0i ⊕ h5i ≈ Z3 ⊕ h0i ⊕ Z3 ≈ Z3 ⊕ Z3 . Therefore h4i ⊕ h0i ⊕ h5i is a subgroup of order 9. For three positive integers m, n, k, if lcm(m, n, k) = 9, then all m, n, k are one of 1 ,3, and 9, and at least one of m, n, k is 9. Because 9 is not a divisor of 12, 4, nor 15, there is no element of order 9 in Z12 ⊕ Z4 ⊕ Z15 . Therefore there is no cyclic subgroup of order 9 in Z12 ⊕ Z4 ⊕ Z15 . 37. Prove that R∗ ⊕ R∗ is not isomorphic to C∗ . In C∗ , because i2 = −1, i3 = −i, i4 = 1, |i| = 4. We claim that there is no order 4 element in R∗ ⊕ R∗ . Indeed, for (a, b) ∈ R∗ ⊕ R∗ , if (a, b)4 = (a4 , b4 ) = (1, 1) = e, then a4 = 1 and b4 = 1. But in R∗ there are only two solutions of x4 = 1, x = 1 and x = −1. So (a, b) is one of (1, 1), (1, −1), (−1, 1), (−1, −1). It is easy to see that (a, b)2 = (1, 1) so |(a, b)| ≤ 2. 63. Let p be a prime. Prove that Zp ⊕ Zp has exactly p + 1 subgroups of order p. In Zp ⊕ Zp , all elements have order p or 1. The only order 1 element is the identity, there are |Zp ⊕ Zp | − 1 = p2 − 1 order p elements. If G is a subgroup of order p, then it is cyclic so there are p − 1 elements of order p in G. Furthermore, if G and H are two distinct subgroups of order p, then |G ∩ H|||G| = p and |G ∩ H| = 6 p. So G ∩ H = {e}. Therefore if we denote the number of order p subgroups by k , then the number of order p elements in G is (p − 1)k. Since (p − 1)k = p2 − 1 = (p − 1)(p + 1), k = p + 1.

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