Four Modern Algebra ProblemsDate: 08/14/97 at 09:48:19 From: Cat Fyle Subject: Modern Algebra problems I have my final next week in Modern Algebra. I have been lost in that class since day one but have managed to maintain a B with your help. Could you please get me started on the following homework problems: 1) a) Show that (1-i)^2i = (2^i)(e^1.570796) b) For all points z = x+iy in the right half-plane, (i.e.,x>0); show that the principal value of ln(z) equals: ln(z) = 1/2ln(x^2 + y^2) + itan^-1(y/x). 2) Prove the following statement: If the integers a and b are relatively prime, then there exist integers m and n such that 1 = ma+nb. 3) Let H be a subgroup of a finite group G. Show that the order of H is a divisor of the order of G. 4) Let G be a set of four elements that is closed under an associative binary operator which satisfies the conditions: a) There exist and element e in G such that a*e = a for all a in G. b) Given that a is in G, there exists a mapping f(a) = b in G such that f(a)*a = b*a = a. Is this set a group? Thank you for any help you can give me, CAT. Date: 08/15/97 at 13:47:40 From: Doctor Rob Subject: Re: Modern Algebra >1) a) Show that (1-i)^2i = (2^i)(e^1.570796) Write 1-i in the form r*e^(i*t) for some r and t. Do the same for 2. Now substitute the first one in the left side and the second one in the right side of the above equation. > b) For all points z=x+iy in the right half-plane, (i.e.,x>0); >show that the principal value of ln(z) equals: >ln(z) = 1/2ln(x^2 + y^2) + itan^-1(y/x). Write z = x + i*y in the form r*[cos(t) + i*sin(t)] = r*e^(i*t) for some r and t. Now take ln of the resulting expression, and back- substitute to get rid of r and t, replacing them with expressions in x and y. >2) Prove the following statement: If the integers a and b are >relatively prime, then there exist integers m and n such that >1 = ma+nb. Let d be the smallest positive integer of the form m*a + n*b (this exists by the Well-Ordering Principle: every nonempty set of positive integers has a least element; you have to show that the set of positive integers of this form is nonempty). Divide a = q*d + r, 0 <= r < d. Then r = a - q*d = a - q*(m*a + n*b) = (a - q*m)*a + (-q*n)*b is of the same form. Since 0 <= r < d, and d was the smallest positive one, we must have r = 0, so a = q*d. Similarly b = s*d. Then d is a common divisor of a and b. Since a and b are relatively prime, d = 1. >3) Let H be a subgroup of a finite group G. Show that the order of >H is a divisor of the order of G. Consider the left cosets of H: 1H, g[2]H, ..., g[k]H. Show that they are all the same size, since they can be put into one-to-one correspondence with H. (You have to say what the correspondence is, and prove that it is one-to-one and onto.) They are disjoint (prove this), and together their union is all of G. There are k of them, all of size #H, so k*(#H) = #G, and #H | #G. >4) Let G be a set of four elements that is closed under an >associative binary operator which satisfies the conditions: >a) There exist and element e in G such that a*e=a for all a in G. >b) Given that a is in G, there exists a mapping f(a)=b in G such that > f(a)*a=b*a=a. Is this set a group? Are you sure that the last equation shouldn't say "f(a)*a=b*a=e"? I will call this (b'). You need to show that the group axioms are satisfied. You already have closure and associativity. You need identity and inverses. The natural candidate for the identity is e and for the inverse of a would be f(a). You are given by (a) above that e is a right identity. You need to show that e is also a left identity. You are given by (b') above that f(a) is a left inverse of a. You need to show that f(a) is also a right inverse of a. Alternatively, you could construct an operation table for * which demonstrates that G doesn't have to be a group. -Doctor Rob, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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