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Homework 10 SolutionChapter 9. LetH={(1),(12)}. IsHnormal inS 3? (13)H={(13)(1),(13)(12)}={(13),(123)}H(13) ={(1)(13),(12)(13)}={(13),(132)} Because(13)H 6 =H(13),His not a normal subgroup ofS 3. Viewing〈 3 〉and〈 12 〉as subgroups ofZ, prove that〈 3 〉/〈 12 〉is isomorphic toZ 4. Similarly, prove that〈...

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MATH 3005 Homework Solution

Han-Bom Moon

Homework 10 Solution Chapter 9.

1. Let H = {(1), (12)}. Is H normal in S3 ? (13)H = {(13)(1), (13)(12)} = {(13), (123)} H(13) = {(1)(13), (12)(13)} = {(13), (132)} Because (13)H 6= H (13), H is not a normal subgroup of S3 . 8. Viewing h3i and h12i as subgroups of Z, prove that h3i/h12i is isomorphic to Z4 . Similarly, prove that h8i/h48i is isomorphic to Z6 . Generalize to arbitrary integers k and n. I will prove the general formula: For any positive integers k and n, two groups hki/hkni and Zn are isomorphic. Sol 1. Because hki is cyclic, all elements in hki is of the form mk for m ∈ Z. So all elements in hki/hkni is of the form mk + hkni = m(k + hkni). Therefore hki/hkni is cyclic and it is generated by k +hkni. So it suffices to check the order of k +hkni. Note that n(k + hkni) = nk + hkni = hkni and for 0 < m < n, m(k + hkni) 6= hkni. Therefore |k + hkni| = n and hki/hkni ∼ = Zn . Sol 2. Note that an element in hki/hkni is of the form mk + hkni for some m ∈ Z. Define a map φ : hki/hkni → Zn as φ(mk + hkni) = m mod n. Step 0. φ is well-defined. If m1 k + hkni = m2 k + hkni, then −m2 k + m1 k + hkni = hkni. So (m1 − m2 )k = −m2 k + m1 k ∈ hkni. Therefore m1 − m2 is a multiple of n so m1 mod n = m2 mod n. This implies that φ(m1 k + hkni) = φ(m2 k + hkni). Step 1. φ is one-to-one. If φ(m1 k + hkni) = φ(m2 k + hkni), then m1 mod n = m2 mod n. Therefore n|m1 − m2 and nk |m1 k −m2 k. So m1 k −m2 k ∈ hkni, and m1 k − m2 k +hnk i = hnk i. So m1 k + hnk i = m2 k + hnki. Step 2. φ is onto. Obviously, for m ∈ Zn , φ(mk + hkni) = m mod n = m. Step 3. φ has the operation preserving property. φ(m1 k + hkni)φ(m2 k + hkni) = m1 mod n + m2 mod n = m1 + m2 mod n = φ((m1 + m2 )k + hkni) = φ((m1 k + hkni) + (m2 k + hkni)). Therefore φ is an isomorphism. 1

MATH 3005 Homework Solution

Han-Bom Moon

9. Prove that if H has index 2 in G, then H is normal in G. Because H has index 2, there are exactly two left cosets (say {H, aH }) and two right cosets ({H, H b}). Note that the disjoint union of H and aH is G. Also the disjoint union of H and H b is G. So aH = G − H = H b. If x ∈ H, then xH = H = H x. If x ∈ / H, then xH = G − H = H x. So in any cases, the left coset is equal to the right coset. Therefore H is normal. 11. Let G = Z4 ⊕ U (4), H = h(2, 3)i, and K = h(2, 1)i. Show that G/H is not isomorphic to G/K. (This shows that H ≈ K does not imply that G/H ≈ G/K .) In G/H , ((1, 3)H)2 = (1 + 1, 32 )H = (2, 1)H, ((1, 3)H)3 = (1 + 1 + 1, 33 )H = (3, 3)H, ((1, 3)H)4 = (1 + 1 + 1 + 1, 34 )H = (0, 1)H = H (Note that (0, 1) is the identity of G.). So |(1, 3)H| = 4 in G/H . On the other hand, in G/K, for any element (a, b)H, ((a, b)H)2 = (a + a, b2 )H = (2a, b2 )H. But in U (4) = {1, 3}, all squares are 1, so (2a, b2 )H = (2a, 1)H = (2, 1)aH = H because (2, 1)a ∈ h(2, 1)i = H. Therefore |(a, b)H| ≤ 2. So G/H 6≈ G/K . 12. Prove that a factor group of a cyclic group is cyclic. Let G = hai and H ⊳ G. Then any element in G/H is of the form ak H, which is (aH )k for some k ∈ Z. Therefore G/H = haH i and it is cyclic. 14. What is the order of the element 14 + h8i in the factor group Z24 /h8i? 14 + h8i = 6 + h8i 2 · (6 + h8i) = 12 + h8i = 4 + h8i 3 · (6 + h8i) = 18 + h8i = 2 + h8i 4 · (6 + h8i) = 24 + h8i = h8i So |14 + h8i| = 4. Can you generalize it? Answer: In Zn , |a + hbi| = lcm(a, gcd(n, b))/a. 21. Prove that an Abelian group of order 33 is cyclic. Let G be an Abelian group of order 33. There is a ∈ G with |a| = 11 and b ∈ G with |b| = 3. Sol 1. Because G is Abelian, (ab)i = ai bi . The order |ab| is one of 1, 3, 11, 33. If |ab| = 1, then ab = e and b = a−1 . So we obtain a contradiction |a| = |b|. If |ab| = 3, then e = (ab)3 = a3 b3 = a3 . This is impossible because |a| = 11. If |ab| = 11, then e = (ab)11 = a11 b11 = b11 . From b9 = e, we have b2 = e, which is impossible too. Therefore |ab| = 33 and G = habi. Sol 2. Let H = hai and K = hbi. H ∩K = {e} because |H ∩K| is a common divisor of |H| = 11 and |K| = 3. Because |HK | = |H||K|/|H ∩ K| = 11 · 3/1 = 33 = |G|, G = H K. Since G is Abelian, both H and K are normal subgroups. Therefore G = H × K ≈ H ⊕ K ≈ Z11 ⊕ Z3 ≈ Z33 . 2

MATH 3005 Homework Solution

Han-Bom Moon

22. Determine the order of (Z ⊕ Z)/h(2, 2)i. Is the group cyclic? For (1, 0) ∈ Z ⊕ Z, m(1, 0) = (m, 0) ∈ / h(2, 2)i for all m > 0. This implies that m((1, 0) + h(2, 2)i) 6= h(2, 2)i for any m > 0. So |(1, 0) + h(2, 2)i| = ∞, and |(Z ⊕ Z)/h(2, 2)i| = ∞. 2((1, 1) + h(2, 2)i) = (2, 2) + h(2, 2)i = h(2, 2)i So |(1, 1) + h(2, 2)i| = 2. On the infinite cyclic group Z, except the identity, there is no element with finite order. Therefore (Z ⊕ Z)/h(2, 2)i is not cyclic. 24. The group (Z4 ⊕ Z12 )/h(2, 2)i is isomorphic to one of Z8 , Z4 ⊕ Z2 , or Z2 ⊕ Z2 ⊕ Z2 . Determine which one by elimination. Note that h(2, 2)i = {(0, 0), (2, 2), (0, 4), (2, 6), (0, 8), (2, 10)}. For (0, 1) ∈ Z4 ⊕ Z12 , k(0, 1) = (0, k) ∈ h(2, 2)i only if 4|k. So |(0, 1)+h(2, 2)i| = 4. So it is not isomorphic to Z2 ⊕ Z2 ⊕ Z2 where all nonidentity elements have order 2. Furthermore, an element (a, b) ∈ Z4 ⊕ Z12 has order lcm(|a|, |b|). Because both |a| and |b| are divisors of 12, |(a, b)| = lcm(|a|, |b|) is a divisor of 12. Note that 12((a, b) + Z4 ⊕ Z12 ) = 12(a, b) + Z4 ⊕ Z12 = Z4 ⊕ Z12 . So |(a, b) + Z4 ⊕ Z12 | is a divisor of 12 and there is no order 8 element. Therefore given group is not isomorphic to Z8 . So it is isomorphic to Z4 ⊕ Z2 . 32. Prove that D4 cannot be expressed as an internal direct product of two proper subgroups. If D4 = H × K, then because |D4 | = 8, |H| = 4 and |K| = 2 (or vice versa.). Then K ≈ Z2 , and H ≈ Z4 or Z2 ⊕ Z2 . In particular, both K and H are Abelian groups. Because D4 = H × K ≈ H ⊕ K, D4 must be Abelian, too. But D4 is not, so it is not an internal direct product. 34. In Z, let H = h5i and K = h7i. Prove that Z = H K. Does Z = H × K ? Note that Z is an additive group. So H K = {a + b | a ∈ H, b ∈ K}. Because gcd(5, 7) = 1, there are two integers x and y such that 5x + 7y = 1. So for any m ∈ Z, m = 5mx + 7my ∈ h5ih7i = H K. Therefore Z = H K . But H ∩ K = hlcm(5, 7)i = h35i. So Z 6= H × K . 39. If H is a normal subgroup of a group G, prove that C(H), the centralizer of H in G, is a normal subgroup of G. We will show that xC(H)x−1 ⊂ C(H) for all x ∈ G. Let a ∈ xC(H)x−1 . Then a = xyx−1 for some y ∈ C(H). We need to show that ah = ha, or aha−1 = h for all h ∈ H . aha−1 = (xyx−1 )h(xyx−1 )−1 = xyx−1 hxy −1 x−1 . Because H ⊳ G, x−1 hx ∈ x−1 H x ⊂ H. So yx−1 hx = x−1 hxy. Thus we have xyx−1 hxy−1 x = xx−1 hxyy−1 x−1 = h. Therefore aha−1 = h and a ∈ C(H). 3

MATH 3005 Homework Solution

Han-Bom Moon

44. Observe from the table for A4 given in Table 5.1 on page 111 that the subgroup given in Example 9 of this chapter is the only subgroup of A4 of order 4. Why does this imply that this subgroup must be normal in A4 ? Generalize this to arbitrary finite groups. Generally, if a finite group G has only one subgroup of H of fixed order k, then H ⊳ G. Indeed, for any x ∈ G, xH x−1 ≤ G and |xH x−1 | = |H| = k. From the assumption, xH x−1 = H. This implies that H ⊳ G. 51. Let N be a normal subgroup of G and let H be a subgroup of G. If N is a subgroup of H, prove that H/N is a normal subgroup of G/N if and only if H is a normal subgroup of G. Suppose that H is a normal subgroup of G. Then for any a ∈ G, aH a−1 ⊂ H . So for any hN ∈ H/N , aN · hN · (aN )−1 = aha−1 N ∈ H/N . In other words, aN (H/N )(aN )−1 ⊂ H/N for any aN ∈ G/N . Therefore H/N ⊳ G/N . Conversely, suppose that H/N ⊳G/N . Then for any aN ∈ G/N , aN (H/N )(aN )−1 ⊂ H/N . So aha−1 N = aN · hN · (aN )−1 ∈ H/N for all h ∈ H. Therefore there exists h′ ∈ H such that aha−1 N = h′ N . So aha−1 = h′ n for some n ∈ N . Then aha−1 = h′ n ∈ H, because N ≤ H. This implies that aH a−1 ⊂ H. So H ⊳ G. 58. If N and M are normal subgroups of G, prove that N M is also a normal subgroup of G. In general, N M is not a subgroup! But if (at least one of) N and M are normal, we can prove that N M is also a subgroup. e = ee ∈ N M , so N M is nonempty. −1 Take n1 m1 , n2 m2 ∈ N M . Then (n1 m1 )(n2 m2 )−1 = n1 m1 m−1 2 n2 . Because M −1 −1 −1 is normal, there is m3 ∈ M such that m1 m2 n2 = n2 m3 . So n1 m1 m2−1n−1 2 = m ∈ N M and N M ≤ G. n1 n−1 3 2 Because N and M are normal subgroups, aN a−1 ⊂ N and aM a−1 ⊂ M for any a ∈ G. Now aN M a−1 = aN a−1 aM a−1 ⊂ N M . Therefore N M ⊳ G also.

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