Abstract Algebra Exam 3

Find all subgroups of order 3 in Z9+Z3.

1. A subgroup of order 3 has prime order and is thus cyclic. Consider all elements of order 3 in Z9+Z3. Elements of order 3 in Z9 are 3 and 6. Elements of order 3 in Z3 are 1 and 2. Elements of order 1 in Z9 is 0 and elements of order 1 in Z3 is 0. Through combining these elements in pairs such that the least common multiple of their orders is 3, we get 8 elements of order 3: (3,0), (6,0), (0,1), (0,2), (3,1), (3,2), (6,1), and (6,2). 2. Each subgroup of order 3 has 2 elements of order 3 and the identity element (0,0). Thus, there are 8/2 = 4 distinct cyclic subgroups of order 3. 3. These are <(3,0)>, <(3,1)>, <(3,2)>, and <(0,1)>.

Prove that A5 cannot have a normal subgroup of order 2.

1. ABWOC that A5 has a normal subgroup H of order 2. 2. If H has order 2, it contains the identity element and an element of order 2 that commutes with every element of A5. 3. Elements of A5 with order 2 are of the form (ab)(cd), but it is clear that (abc)(ab)(cd) does not equal (ab)(cd)(abc). (--><--) 4. Hence, A5 cannot have a normal subgroup of order 2.

If G is non-Abelian, show that Aut(G) is not cyclic.

1. ABWOC that Aut(G) is cyclic. 2. Since Inn(G) is an automorphism of G, it follows that Inn(G) is also cyclic. 3. By Thm 9.4 (Inn(G) is isomorphic to G/Z(G)), it follows that G/Z(G) is cyclic. 4. By Thm 9.3, G is Abelian since G/Z(G) is cyclic (--><--). 5. Hence, Aut(G) is not cyclic.

Prove that if G is a finite group, the index of Z(G) cannot be prime.

1. ABWOC: Suppose that |G|/|Z(G)| = p, where p is prime. 2 Let a be in G, but a not be in Z(G). Since C(a) contains both a and Z(G), we know that Z(G) is a proper subgroup of C(a). 3. Since p = (|G|/|C(a)|)(|C(a)|/|Z(G)|), we have |G|/|C(a)| = 1 and therefore C(a) = G and a is an element of Z(G).

Without doing any calculations in Aut(Z105), determine how many elements of Aut(Z105) have order 6.

1. Aut(Z105) = U(105) = U(3)+U(5)+U(7) = Z2+Z4+Z6. 2. Let (a,b,c) be in Z2+Z4+Z6. Possible orders: |a|={1,2}, |b|={1,2,4}, and |c|={1,2,3,6}. 3. Solve 6 = |(a,b,c)| = lcm(|a|,|b|,|c|). 4. Possible solutions: (1,1,6), (1,2,6), (2,1,6), (2,2,6), (1,2,3), (2,1,3), (2,2,3). 5. Using the Euler Totient Function, there are 14 elements of order 6 in Aut(Z105)

Let |G| = 8. Show that G must have an element of order 2.

1. By Cor. 2 to Lagrange's Thm, for any element g in G, |g| divides |G|. 2. Divisors of |G| are 1, 2, 4, and 8. 3. Consider g in G that is not the identity element. 4. If |g| = 8, then |g^4| = 2. 5. If |g| = 4, then |g^2| = 2. 6. The |g| = 2 case is trivial.

Let G be a group and |G|=21. If g E G and g^14 = e, what are the possibilities for |g|?

1. By Cor. 2 to Lagrange's Thm, |g| divides |G|. 2. By Cor. 2 to Thm. 4.1, g^k = e implies |g| divides k. 3. Therefore, |g| must divide both 21 and 14. 4. |g| can be either 1 or 7.

What is the smallest positive integer k such that x^k = e for all x in U(7*17)?

1. By Cor. to Thm 8.3, U(7*17) is isomorphic to U(7)xU(17). 2. By Gauss, U(7)xU(17) is isomorphic to Z6xZ16. 3. Using Cor. 4 of Lagrange's Thm, consider x^k = x^(6m) = x^(16n) and note lcm(6,16) = 48. Thus, k=48. 4. In general, for U(pq) = U(p)xU(q) = Z(p-1)xZ(q-1), k = lcm(p-1,q-1).

Compute 5^15mod7 and 7^13mod11.

1. By Fermat's Little Theorem, 5^(7-1)=1mod7 and 7^(11-1)=1mod13. 2. Thus, 5^15mod7 = 5^6*5^6*5^3mod7 = 5^3mod7 = 125mod7 = 6. 3. Also, 7^13mod11 = 7^10*7^3mod11 = 7^3mod11 = 343mod11 = 2.

Let |a| = 30. How many left cosets of <a^4> in <a> are there? List them.

1. By Lagrange's Thm, we know that the number of left cosets of <a^4> in <a> is |<a>|/|<a^4>|. 2. Consider |<a^4>| = |<a>|/gcd(4,|<a>|) = 30/gcd(4,30) = 15. 3. Thus, |<a>|/|<a^4>| = 30/15 = 2 (i.e., there are two left cosets). 4. Note that <a^4> generates the set of all a^k where k is even. 5. Note that a<a^4> generates the set of all a^k where k is odd. 6. Thus, the two cosets of <a^4> in <a> are <a^4> and a<a^4>.

Suppose that G is a non-Abelian group of order p^3, where p is a prime and Z(G) does not equal {e}. Prove that |Z(G)| = p.

1. By Lagrange's Thm, |Z(G)| = {1,p,p^2, or p^3}. However, it is given that Z is not {e} and that G is non-Abelian, so |Z(G)| is not 1 or p^3. 2. Suppose |Z(G)| = p^2. Since Z(G) is normal in G, consider the factor group G/Z(G). Observe |G/Z(G)| = |G|/|Z(G)| = p^3/p^2 = p. By Cor. to Lagrange's, we know that groups of prime order are cyclic. However, by Thm 9.3, G/Z(G) is not cyclic. (--><--). 3. Hence, |Z(G)| = p.

Prove that (1,1) is an element of largest order In Zn1 + Zn2. State the general case.

1. By Thm 8.1, |(1,1)| = lcm(|1|,|1|) = lcm(n1,n2) (because order of 1 in any Zn is n). Then lcm(n1,n2) = n1n2 = |Zn1+Zn2|. 2. I.e., |(1,1)| = |Zn1+Zn2|. By Lagrange's Theorem, no element of a group has an order that isn't a divisor of the order of the group, so |(1,1)| has the maximum possible order. 3. In general, (1,1,...,1) is an element of largest order in Zn1+...+Zn2. Note that because the order of 1 in each component is the order of the group in that component, |(1,1,...,1)| = lcm(n1,n2,...,nt) and the order of every element in the product must divide lcm(n1,n1,....,nt).

Prove that an Abelian group of order 33 is cyclic. Does your proof hold when 33 is replaced by pq where p and q are distinct primes?

1. By Thm 9.5, the group has an element a of order 3 and an element b of order 11. 2. Because (ab)^33 = (a^3)^11(b^11)^3 = ee = e, we know that |ab| divides 33. 3. Note that |ab| is not 1, for otherwise |a| = |b^-1| = |b|. 4. Note that |ab| is not 3, for otherwise e = (ab)^3 = a^3b^3 = b^3, which is false. 5. Note that |ab| is not 11, for otherwise e = (ab)^11 = a^11b^11 = a^2, which is false. 6. So, |ab|=33 and <ab> = G.

Suppose that G is a group with more than one element and G has no proper, nontrivial subgroups. Prove that |G| is prime. Do not assume that G is finite.

1. Case 1: Suppose G is infinite. Let a nontrivial x be in G. Then G=<x>. But <x^2> contradicts the hypothesis. (--><--). 2. Case 2: Suppose G is finite. let a nontrivial x be in G. Then G=<x> (otherwise <x> contradicts the hypothesis). 3. By Thm 4.3, G has a subgroup of each divisor of |G|. The only subgroups of G have order |G| and 1, so the only divisors of |G| are |G| and 1 so |G| is prime.

How many elements of order 4 does Z4+Z4 have? Do not do this by examining each element. Explain why Z4+Z4 has the same number of elements of order 4 as does Z80000000+Z400000. Generalize to the case Zm+Zn.

1. Consider (a,b) in Z4+Z4. By Lagrange's Thm, possible orders for both a and b are 1,2,4. By Thm 8.1, solve 4 = |(a,b)| = lcm(|a|,|b|) = lcm(1,4) = lcm(2,4) = lcm(4,4) = lcm(4,2) = lcm(4,1). Using the Euler Totient Function, there are 12 elements of order 4. 2. In Z80000000+Z400000, an element of order 4 must again satisfy Thm 8.1 so by the same process there are 12 elements of order 4. 3. In general, for a group Zm+Zn with order k, if d divides k then the number of elements of order d depends solely on d (Thm 4.4).

Give an example of a group of order 12 that has more than one subgroup of order 6.

1. Consider (a,b) in Z6+Z2. We will find the number of cyclic subgroups of Z6+Z2. 2. Note that possible orders for a are 1,2,3,6 and possible orders for b are 1,2. By Thm 8.1, solve 6 = |(a,b)| = lcm(|a|,|b|) = lcm(6,1) = lcm(6,2) = lcm(3,2). 3. Using the Euler Totient Function, there are 2 + 2 + 2 = 6 elements of order 6 in Z6+Z2. 4. A subgroup of order 6 contains 2 elements of order 6. No two cyclic subgroups can have an element of order 2 in common. Thus, there must be 6/2 = 3 subgroups of order 6 in Z6+Z2.

Prove, by comparing orders of elements, that Z8+Z2 is not isomorphic to Z4+Z4.

1. Consider (a,b) in Z8+Z2. By Lagrange's Thm, possible orders of a are 1,2,4,8 and possible orders of b are 1,2. Thus, when considering the largest possible order of any element of Z8+Z2, we have |(a,b)| = lcm(8,2) = lcm(8,1) = 8. 2. Consider (a,b) in Z4+Z4. By Lagrange's Thm, possible orders of both a and b are 1,2,4. Thus, when considering the largest possible order of any element of Z4+Z4, we have |(a,b)| = lcm(4,4) = lcm(4,2) = lcm(4,1) = lcm(1,4) = lcm(2,4) = 4. 3. Isomorphisms preserve orders, but the largest possible order of any element in Z8+Z2 is not equal to the largest possible order of any element in Z4+Z4. Thus, the two groups are not isomorphic.

What is the order of the element 14+<8> in the factor group Z24/<8>?

1. Consider <8>, a subgroup of Z24. Note that <8> = {0, 8, 16} and thus |<8>| = 3. 2. It follows that |Z24/<8>| = |Z24|/|<8>| = 24/3 = 8, so Z24/<8> is a cyclic group of order 8, generated by 1+<8>. 3. To find |14+<8>|, consider (14+<8>)^1 = 14+<8>, which does not equal <8>. 4. Consider (14+<8>)^2 = 28+<8> = 4+<8>, which does not equal <8>. 5. Consider (14+<8>)^3 = 42+<8> = 18+<8>, which does not equal <8>. 6. Consider (14+<8>)^4 = 56+<8> = 8+<8> = <8>.

Let p be a prime. Show that if H is a subgroup of a group of order 2p that is not normal, then H has order 2.

1. Consider G with subgroup H, where |G|=2p. By Lagrange's Thm, |H| = {1,2,p,2p}. 2. (Case 1) If |H| = 1, then H = <e>, which is normal. 3. (Case 2) If |H| = 2p, then H = G, which is normal. 4. (Case 3) If |H| = p, then |G:H| = 2. One coset is H and the other (for some x in G) is xH=Hx, which is normal. 5. Hence, the order of H must be 2.

Prove or disprove: U(200) is isomorphic to U(50)+U(4)

1. Consider U(200) = U(2^3)+U(5^2) = Z4+Z20 and U(4)+U(50) = U(4)+U(5^2)+U(2) = Z2+Z20+Z0 = Z2+Z20 2. Thus, |U(200)|=|Z4+Z20|=4x20=80 and |U(4)+U(50)|=|Z2+Z20|=2x20=40. 3. Isomorphisms preserve order but 80 does not equal 40. Thus, this is not an isomorphism.

Let G be a finite group and let H be a normal subgroup of G. Prove that the order of the element gH in G/H must divide the order of g in G.

1. Consider g in G and gH in G/H. Say that |g| = m and |gH| = n. 2. Consider (gH)^n = (g^n)H = eH = H, which is the identity element of G/H. 3. By Cor. 2 to Thm 4.1, since |gH| = m and (gH)^n = H, m must divide n.

Let G = {ax^2 + bx + c | a,b,c E Z3}. Add elements of G as you would polynomials with integer coefficients, except use modulo 3 addition. Prove that G is isomorphic to Z3+Z3+Z3. Generalize.

1. Consider phi: G --> Z3+Z3+Z3, defined by phi(ax^2 + bx + c) = (a,b,c). Note that phi is clearly a bijection. 2. Consider g1=ax^2+bx+c and g2=a'x^2+b'x+c'. Then phi(g1+g2) = (a+a', b+b', c+c') = (a,b,c)+(a',b',c') = phi(g1) + phi(g2) and it follows that phi is operation-preserving. 3. Thus, phi is an isomorphism and G is isomorphic to Z3+Z3+Z3. 4. In general, any G = {a_{n-1} x^{n-1} + .... + a0 | a_{n-1},...,a0 in Z3} under addition modulo m is isomorphic to the external direct product of n Zm's.

Determine Aut(Z2+Z2).

1. Consider the group Z2+Z2 = {(0,0), (1,0), (0,1), (1,1)}. For any automorphism phi, phi(0,0) = (0,0). By permutating the remaining elements, |Aut(Z2+Z2)| = 6. 2. Consider phi1, phi2: Z2+Z2 --> Z2+Z2. Define phi1 = {(0,0)=(0,0), (0,1)=(0,1), (1,1)=(1,0), and (1,0)=(1,1)}. Define phi2 = {(0,0)=(00), (1,1)=(0,1), (1,0)=(1,0), and (0,1)=(1,1)}. 3. Consider phi1(phi2(1,0)) = (1,1), which does not equal phi2(phi1(1,0)) = (0,1). 4. Thus, Aut(Z2+Z2) is not Abelian. 5. Consider S3, which has order 6 and is non-Abelian. 6. Aut(Z2+Z2) is isomorphic to S3.

If a group has exactly 24 elements of order 6, how many cyclic subgroups of order 6 does it have?

1. Each cyclic group of order 6 has two elements of order 6. So, the 24 elements of order 6 yield 12 cyclic subgroups of order 6. 2. In general, if a group has 2n elements of order 6, it has n cyclic subgroups of order 6. (Recall from Cor. to Thm 4.4 that if a group has a finite number of elements of order 6 the number is divisible by phi(6)=2.)

Why does the fact that A4 has no subgroup of order 6 imply that |Z(A4)| = 1.

1. If |Z(A4)|>1, then A4 would have an element of order 2 or 3 that commutes with every element. But any subgroup generated by an element of order 2 and an element of order 3 that commute has order 6. 2. This contradicts the fact that A4 has no subgroup of order 6.

Determine the order of (Z+Z)/<(2,2)>. Is the group cyclic?

1. Infinity. 2. The group is not cyclic. (1,1)+<(2,2)> has order 2 whereas in an infinite cyclic group, every non-identity element has infinite order.

Suppose that H and K are subgroups of a group with |H| = 24, |K| = 20. Prove that HnK is Abelian.

1. It is clear that HnK is a subgroup of both H and K. Thus, by Lagrange's Thm, |HnK| divides |H| and |K|. 2. Common divisors of |H| and |K| are 1,2, and 4. 3. Case 1. If |HnK| = 1, then HnK=<e>, which is obviously Abelian. 4. Case 2. If |HnK| = 2, then HnK is a group of prime order which is cyclic and thus Abelian by Cor. 3 to Lagrange's Thm. 5. Case 3. If |HnK| = 4 = 2^2, then HnK is a group of order p^2, where p is a prime. Thus, HnK is Abelian by Cor. to Thm 9.7.

How many elements of order 9 does Z3+Z9 have?

1. Let (a,b) be in Z3+Z9, where a is in Z3 and b is in Z9. 2. By Lagrange's Thm, possible orders of a are 1,3 and possible orders of b are 1,3,9. 2. By Thm 8.1, 9 = |(a,b)| = lcm(|a|,|b|). For this to be true, |a|=1,3 and |b|=9. 3. Consider |a|=1,|b|=9. The number of elements (a,b) that satisfy is phi(1)phi(9) = 1x6 = 6. Consider |a|=3,|b|=9. The number of elements (a,b) that satisfy is phi(3)phi(9) = 2x6 = 12. 4. In total, there are 6+12=18 elements (a,b) in Z3+Z9 with order 9.

Give an example of a group G and subgroups H and K such that HK = {h E H, k E K} is not a subgroup of G.

1. Let G = D3 = {R0, R120, R240, F1, F2, F3}, let H = {R0, F1}, and let K = {R0, F2}. 2. Thus, HK = {R0R0, R0F2, F1R0, F1F2} = {R0, F2, F1, R240}. 3. Consider R240^2 = R120, which is not in HK. Thus, HK is not closed under the binary operation. 4. Hence, HK is not a subgroup of G.

Let G be a group with |G| = pq, where p and q are prime. Prove that every proper subgroup of G is cyclic.

1. Let H a proper subgroup of G. 2. By Lagrange's Thm, |H| divides |G|, but H is proper in G so |H| = {1,p,q}. 3. Case 1: Suppose |H| = 1. Then H = {e} and the case is trivial. 4. Case 2: Suppose |H| = p or q. By Cor. 3 to Lagrange's Thm, any group of prime order is cyclic. 5. Hence, every proper subgroup of G is cyclic.

Prove that every subgroup of Dn of odd order is cyclic.

1. Let H be a subgroup of Dn with odd order. By Cor. 2 of Lagrange's Thm, the order of each element of H must divide the order of H. 2. Let F (which is any reflection) be in H. Reflections have order 2. However, 2 cannot divide the odd order of H. It follows that H consists entirely of rotations. 3. Consider the subgroup of all rotations of Dn, call it R, which is cyclic and generated by <R_2pi/n>. Since H consists entirely of rotations, it is a subgroup of R and is thus cyclic.

Let H be a subgroup of a group G with the property that for all a and b in G, aHbH = abH. Prove that H is a normal subgroup of G.

1. Let a be in G (and not H) and b be in H. 2. Consider (aba^-1)H = (ab)H(a^-1)H = a(bH)a^-1(H) = a(H)a^-1(H) = (aa^-1)H = eH = H. 3. Thus, aba^-1 is in H and by the Normal Subgroup Test, H is normal in G.

Suppose that a finite Abelian group G has at least three elements of order 3. Prove that 9 divides |G|.

1. Let g1, g2, g3 be in G, where each g has order 3. 2. Consider <g1>, which is a proper subgroup in G and has (at most) 2 elements of order 3, say g1 and g2. 3. WLOG: g3 is not in <g1>. Thus, <g1>n<g3> = {e} and |<g1>n<g3>|=1. 4. By Thm 7.2, |<g1><g3>| = |<g1>||<g3>|/|<g1>n<g3>| = 9. 5. By FSGT, <g1><g3> is in G. 6. Thus, by Lagrange's Thm, 9 divides |G|.

Show that G+H is Abelian if and only if G and H are Abelian. State the general case.

1. Let g1,g2 be in G, h1,h2 be in H, and (g1,h1),(g2,h2) be in G+H. 2. (-->) Suppose G+H is Abelian. Then (g1,h1)(g2,h2) = (g2,h2)(g1,h1). However, (g1,h1)(g2,h2) = (g1g2,h1h2) and (g2,h2)(g1,h1) = (g2g1,h2h1). Thus, (g1g2,h1h2) = (g2g1,h2h1) which implies g1g2=g2g1 and h1h2=h2h1. Hence, G and H are Abelian. 3. (<--) Suppose G and H are Abelian. Then g1g2=g2g1 and h1h2=h2h1. It follows that in G+H, we have (g1g2,h1h2) = (g2g1,h2h1). However, note that (g1g2,h1h2) = (g1,h1)(g2,h2) and (g2g1,h2h1) = (g2,h2)(g1,h1). Thus, (g1,h1)(g2,h2) = (g2,h2)(g1,h1) and it follows that G+H is Abelian. 4. In general, G1+G2+...+Gn is Abelian if and only if Gi is Abelian for every i=1,2,...,n.

Prove that if H has index 2 in G, then H is normal in G.

1. Let x be in G. 2. Suppose x is in H. Then xH=H and Hx=H so xH=Hx. 3. Suppose x is not in H. Then xH is the set of elements in G but not H. However, since H has index 2 and H is one coset, then xH=Hx. 4. Therefore, for all x in G, xH=Hx and thus H is normal in G.

Let |G| = 15. If G has only one subgroup of order 3 and only one of order 5, prove that G is cyclic. Generalize to |G| = pq, where p and q are prime.

1. Let |H|=3, |K|=5. Thus, H has 2 elements of order 3 and K has 4 elements of order 5. Hence, HuK contains 1+2+4=7 elements. G has 15 elements, so there exist elements in G that are not in HuK. 2. Let x be in G and not in HuK. By Lagrange's Thm, |x|=1,3,5, or 15. We cannot have |x|=1,3, or 5, because then x would be in HuK. Thus, |x|=15 and it follows that G=<x>. 3. In general, if |H|=p,|K|=q, the HuK has 1+(p-1)+(q-1)<pq elements. Thus there exists an x in G that is not in HuK. By Lagrange's Thm, |x|=1,p,q, or pq. It follows that |x| must be pq and thus G=<x>.

Let G be a group of order pqr, where p,q, and r are distinct primes. If H and K are subgroups of G with |H| = pq and |K| = qr, prove that |HnK| = q.

1. Note that HnK is a subgroup of H and a subgroup of K. Thus, by Lagrange's Thm, |HnK| divides |H| and |K|. 2. Note that gcd(|H|,|K|) = q, so |HnK| divides q. Thus, |HnK| = {1,q}. 3. Suppose |HnK| = 1. Then, by Thm 7.2, |HK| = |H||K|/|HnK| = pq^2r/1 > pqr = |G|, which implies that |HK| > |G|. 4. However, by closure, HK is a subset of G and thus by Lagrange's Thm, |G| is greater than or equal to |HK|. (--><--). 5. Hence, |HnK| = q.

What is the largest order of any element in U(900)?

1. Note that U(900) = U(2^2)+U(3^2)+U(5^2) = Z2+Z6+Z20. 2. Since isomorphisms preserve order, the largest order of any element in U(900) is equal to the largest order of any element in Z2+Z6+Z20. 3. Consider (a,b,c) in Z2+Z6+Z20. By Lagrange's Thm, the largest order of any a is 2, the largest order of any b is 6 and the largest order of any c is 20. 4. Thus, for the maximum order of (a,b,c), we have lcm(2,6,20)=60.

Let a and b be elements of a group G and H and K be subgroups of G. If aH = bK, prove that H=K.

1. Note that aH=bK implies H = a^-1bK. 2. By Ch. 7 Lemma, we know a is in aH. Thus, by aH=bK, there is some k in K such that a=bk. Hence, a=bk --> ak^-1=b --> k^-1=a^-1b. 3. Thus, H = a^-1bK = k^-1K, and since k^-1 is in K since K is a subgroup, we have k^-1K = K. 4. It follows that H=K.

Let H = {0, +/-3, +/-6, +/-9, ...}. Find all the left cosets of H in Z.

1. Note that every integer n of Z can be written as n=3q+r, for 0≤r<3. 2. Thus, n+H = (3q+r)+H = r+H (because 3q is in H), where r=0,1,2. 3. Hence, all left cosets of H in Z are H, 1+H, and 2+H.

If G is a group and |G:Z(G)| = 4, prove that G/Z(G) is isomorphic to Z2+Z2.

1. Note that |G:Z(G)| = |G|/|Z(G)| = |G/Z(G)| = 4, so the factor group G/Z(G) has order 4. A group of order 4 is isomorphic to either Z4 or Z2+Z2. 2. ABWOC that G/Z(G) is isomorphic to Z4. Z4 is cyclic so then G/Z(G) is cyclic. By Thm 9.3, it follows that G is Abelian. 3. However, if G is Abelian, then |G:Z(G)| = 1 and we have reached a contradiction. 4. Hence, G/Z(G) is isomorphic to Z2+Z2.

Prove that a factor group of a cyclic group is cyclic.

1. Suppose G = <a>. Then any g = a^i. 2. Any element of G/H is of the form gH for some g in G. 3. Thus, gH = (a^i)H. 4. By the factor group operation, (a^i)H = (aH)^i. 5. Thus, G/H = <aH>.

Prove that a factor group of an Abelian group is Abelian.

1. Suppose G is Abelian and that G/H is any factor group of G. 2. Let a,b be in G and aH,bH be in G/H. 3. By definition of factor groups, (aH)(bH) = (ab)H. Since G is Abelian (ab)H = (ba)H. Again by definition, (bH)(aH) = (ba)H. 4. Hence (aH)(bH) = (bH)(aH) and G/H is Abelian.

Let |G| = 33. What are the possible orders for elements of G? Show that G must have an element of order 3.

1. The possible orders are 1, 3, 11, and 33. 2. If |x|=33, then |x^11| = 3. 3. By Cor. to Thm 4.4, the number of elements of order 11 is a multiple of 10 so they account for 0,10,20, or 30 elements of the group. The identity accounts for one or more. So, at most, we have accounted for 31 elemetns. By Cor. 2 to Lagrange's Thm, the elements unaccounted for have order 3.

Determine the number of elements of order 15 and the number of cyclic subgroups of order 15 in Z30+Z20.

1. Using Lagrange's Theorem, determine possible orders a and b of (a,b) in Z30+Z20. 2. Using Thm 8.1, solve |(a,b)| = lcm(|a|,|b|) = 15. Note that solutions include: (15,1), (15,5), and (3,5). 3. Using the Euler Totient Function, calculate the number of elements of order 15, which equals 48. 4. Each cyclic group <(a,b)> of order 15 contains 8 distinct elements of order 15. If subgroups H and K are distinct, H intersect K = {e}. 5. Thus, since there are 48 distinct elements of order 15, there are 48/8 = 6 distinct cyclic subgroups of order 15.

Find an integer n such that U(n) is isomorphic to Z2+Z4+Z9.

1. Z2+Z4+Z9 is isomorphic to Z2+Z18. 2. Z2+Z18 is isomorphic to U(5)+U(27). 3. U(5)+U(27) is isomorphic to U(135).

Let G = {(1), (12)(23), (1234)(56), (13)(24), (1432)(56), (56)(13), (14)(23), (24)(56)}. Find the stabilizers and orbits of 1, 3, and 5.

1. stab(1) = {(1), (24)(56)}. orb(1) = {1,2,3,4} 2. stab(3) = {(1), (24)(56)}. orb(3) = {3,4,1,2} 3. stab(5) = {(1), (12)(34), (13)(24), (14)(23)}. orb(5) = {5,6}

Express Aut(U(25)) in the form Zm+Zn.

Aut(U(25)) = Aut(Z20) = U(20) = U(4)+U(5) = Z2+Z4.

Let G be a group of rotations of a plane about a point P in the plane. Thinking of G as a group of permutations of the plane, describe the orbit of a point Q in the plane.

It is the set of permutations that carry face 2 to face 1.

Let H = { [a b, 0 d] | a,b,d E R, ad does not equal 0}. Is H a normal subgroup of GL(2,R)?

No. Let A = [1 0, 0 -1] and B = [1 0, 1 1]. Then A is in H and B is in GL(2,R) but BAB^-1 is not in H.

Express U(165) as an external direct product of U-groups in four different ways.

U(165) is isomorphic to U(15)+U(11), U(3)+U(55), U(5)+U(33), and U(3)+U(5)+U(11).

Let H be a normal subgroup of a finite group G and let a be an element of G. Complete the following statement: the order of the element aH in the factor group G/H is the smallest positive integer n such that a^n is ___.

in H.