Questions in Modern Algebra
Date: 2/3/96 at 17:47:39 From: Anonymous Subject: Abstract Algebra I have three questions. I hope you are able to help. I am studying Modern Algebra (grad level course) and would appreciate your help in answering the questions below: 1. I need to think of two examples of an abstraction or generalization. I have come up with one already and need help in thinking of another. The one I have come up with deals with integers. That is, 312 (in base 10) can be written 3x10^2 + 1x10^1 + 2x10^0. Extending this to a parallel structure, 101 (in base 2) can be written as 1x2^2 +0x2^1 +1x2^0. I believe this is an example of an abstraction. I cannot think of another example. Can you help? 2. I am to construct a Cayley table for the 8th roots of unity. This I have done. Also I am to determine the order of each element. This too I have done. Then I am to find all the subgroups and determine which are isomorphic. Here is where I am having trouble. I know that a subgroup is a subset of a group which is itself a group (has an inverse, identity, is associative and closed). I am confused as to a method as to how to find the subgroups. Can you help? 3. I am to solve x^6 = 1 and find ALL roots, real and complex. I think I need to use the unit circle and De Moivres theorem (?) taking the cos and sin of (2xpi)/6. But the professor goes so fast that this is all I know. How does this information help me find the six roots? Can you help please. In closing, I am a mathematics teacher, 7-12, currently looking for work. I finished my bachelors in Industrial Engineering in 1987 and so have not been at this level of academia for a while. Any help you could provide will be much appreciated.
Date: 3/21/96 at 22:53:29 From: Doctor Steven Subject: Re: Abstract Algebra Well, let's try: #1. Give an abstraction. Angel Flights - Think of angels. Angels fly. Confine their flights to straight lines. Now an angel's flight has direction, sense, and length. If a flight is parallel to a given straight line then it has that line's direction. If it goes from point A to point B then it has that sense - from point B to A it has the opposite sense. A flight from point A to B and from point C to D are the same if they agree in direction, sense, and length. Consider the flight a from A to B and flight b from B to C. Then d would be the flight from A to C, and d = a()b, where () is the binary operation of angel flight addition. I got this example of an abstraction of the idea of vectors from an essay in _The World of Mathematics_ by Cassius J Keyser. The book was edited by James R. Newman. #2. The subgroups of the group formed by the eighth roots of unity, and the isomorphism between them. Usually the nth roots of unity are denoted with a Greek letter, but since that isn't possible here I'll use z_n to denote an nth root unity. The group formed by the eighth roots of unity should be: z_8, (z_8)^2, (z_8)^3, (z_8)^4, (z_8)^5, (z_8)^6, (z_8)^7, (z_8)^8 = 1. To find the subgroups of this, look at the group generated by z_8 (i.e. the powers of z_8). It's the whole group. Now look at the group generated by (z_8)^2. Its (z_8)^2, (z_8)^4, (z_8)^6, (z_8)^8 = 1. Now look at the group generated by (z_8)^3. It's the whole group again, so the group generated by z_8, and the group generated by (z_8)^3 are isomorphic. Continue this process until you've looked at the groups generated by all the elements, and when you find two groups that have the same elements in them, they're isomorphic. #3. Find all the roots to x^6 = 1. You could do this with De Moivres' theorem, but cyclotomic polys would be easier I think. Basically you'd use this theorem: x^n - 1 = product for d divides n of ( f_d(x) ), where f_d(x) is the dth cyclotomic polynomial. For your case x^6 - 1 = 0 we get: x^6 - 1 = f_1(x)f_2(x)f_3(x)f_6(x). Now all we have to do is find the cyclotomic polys and their roots. So using the theorem above we get: x^1 - 1 = f_1(x), so f_1(x) = x - 1. x^2 - 1 = f_1(x)f_2(x), so f_2(x) = x + 1. x^3 - 1 = f_1(x)f_3(x), so f_3(x) = x^2 + x + 1. So x^6 - 1 = (x-1)(x+1)(x^2+x+1)(x^2-x+1). Now use the quadratic equation to find the roots and you've got the sixth roots of unity. Hope this helps. -Doctor Steven, The Math Forum
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