www_math_umn_edu__stanton_5245_oldprob5245_html

# www_math_umn_edu__stanton_5245_oldprob5245_html - Homework...

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pdfcrowd.com open in browser PRO version Are you a developer? Try out the HTML to PDF API Homework #2 Solutions Here are solutions to some selected problems. p. 20: 43,44,45: For a relation which is reflexive, symmetric, but not transitive, take the set {a,b,c}, and put aRa, bRb, cRc, aRb, bRc, bRa, cRb. It does not include aRc. For a relation which is reflexive, transitive, but not symmetric, take the set {a,b}, and put aRa, bRb, aRb. For a relation which is symmetric, transitive, but not reflexive, take the set {a,b}, and put aRa. p. 49 18: To prove G is Abelian we must show that ab=ba for all a and b in G. Suppose, by contradiction, there is a pair a,b of elements of G with aba^(-1) not equal to b. Then aba^(-1)=c for some c not equal to b. Multiplying this equation on the right by a gives ab=ca. By hypothesis, we have c=b, which is a contradiction. p. 50 24: Several people were confused on this problem. The book asks for the inverse of a product of n elements, a1*a2*. ..*a_n, which is the product of the inverses in the opposite order, a_n^(-1)*. ..*a2^(-1)*a1^(-1). p. 50 25: If you have a group containing 5 and 15 mod 56 under multiplication, you must have all powers and products of these numbers too. This is how the book gets the answer. p. 50 30: By hypothesis, abab=(ab)^2=ab, so multiplying on the left by a^(-1) and the right by b^(-1) gives ba=e, so ab=e=ba also. p. 51 38: Let a and b be in G, we must show that ab=ba. By hypothesis, (ab)^2=abab=e, and we know that a^2=e=b^2. So multiplying abab=e on the left by a and the right by b gives ba=ab. Homework #3 Solutions p. 63 4: Suppose that g has order n, so that g^n=e. Taking inverses, we see that g^(-n)=e also, so the order of g^(-1),

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pdfcrowd.com open in browser PRO version Are you a developer? Try out the HTML to PDF API call it m, must be <=n. If g^(-m)=e, taking inverses we have g^(m)=e, so that m>=n. So m=n. If the order of g in infinite, then again taking inverse says the order of g^(-1) must not be finite. 14. Let H and K be subgroups of G. Clearly e is in the intersection. Suppose that a,b are in H intersect K. Since a,b are in H, a subgroup, a*b is in H. Since a,b are in K, a subgroup, a*b is in K. Thus a*b is in H intersect K. The same argument shows a^(-1) must be in both H and K. 20. Again we show that a,b in C(H) implies a*b is in C(H). Let h be in H. (a*b)*h=a*(b*h)=a*(h*b)=(a*h)*b=(h*a)*b=h*(a*b), so a*b is in C(H). Since H is closed under inverses, a*h^(-1)=h^(- 1)*a, for any h. Taking inverses we have h*a^(-1)=a^(-1)*h, which shows a^(-1) is in C(H), so C(H) is closed under inverses. 42. We need 3 elements of U(40) not equal to 1 whose squares are all 1, and which are closed under multiplication. Try -1=39, 19, 21, so the subgroup is {1,19,21,39}.
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## This note was uploaded on 01/18/2012 for the course INFORMATIK 2011 taught by Professor Phanthuongcang during the Winter '11 term at Cornell University (Engineering School).

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www_math_umn_edu__stanton_5245_oldprob5245_html - Homework...

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