Analysis QuestionsDate: 12/07/96 at 23:57:34 From: Yue Chen Subject: Abstract Algebra Hi, I'm just trying to get ready for finals and I have been going over some of my professor's old exams and I can't seem to figure out some of the questions. I was wondering if you could verify some of my answers and help with a few questions. I hope you can help. The questions that I am having difficulty with go like this: (1) Let G= <a> be a cyclic group of order 15: (a) Show that G has subgroup H of order 5 (b) List all the cosets of H in G for the subgroup H that you found in part (a) My answer for part (a) was H = <a^3 >= {a^3, a^6, a^9, a^12, a^15 = e} For part (b), I have no idea. (2) Describe two subgroups H and K of (Z, +) such that HUK is not a subgroup of (Z, +). I think it has something to do with -1 and 1. That's about as much as know. (3) If G is a subgroup of order 15 and H is a non-trivial subgroup of G, what are all the possible orders for H? My answer was 1 and 15. (4) Let A be a finite set with at least two elements. Is it possible to define a mapping from A -> AxA that is (i) one-to-one? (ii) onto? The problem that I have with this question is, I can't see how this can be a mapping. Is this similary to a binary operation where AxA -> A? How can a single element map to 2 elements? (5) If x, y, z, w are integers with x = yx + w, prove that (x, z) = (z, w). Should I use relations to prove this? Date: 12/11/96 at 14:18:01 From: Doctor Ceeks Subject: Re: Abstract Algebra Hi, (1) When you have a subgroup H of a group G, the group is naturally partitioned into cosets of H, which are subsets of G of the form xH (for left cosets...for right cosets, you use Hx, but in an abelian group left and right cosets are the same). So to answer (b), you have to look at your group G and pick an element g and look at the set gH. Next, find an element of G not in gH, say x, and look at the set xH. Next, find an element of G not in gH or xH...say y, and look at the set yH. Keep going until you every element of the group is accounted for. In this case, the answer is H, aH, and a^2 H. Note: gH means the set of elements of the form gh where h is in H. (2) This doesn't have to do with -1 and 1. Instead, the point is that a union of subgroups is generally not closed under the group law. Consider H = 2Z and K = 3Z. Note that 2 and 3 are in H U K, but 2+3 = 5 is not. Therefore H U K cannot be a subgroup of Z. (3) The order of a subgroup always divides the order of the group. Without any additional information, it is possible for a group to have subgroups of every order which divides G (for example, if G is cyclic, this will be true). In this case, 15 = 3*5, and by Sylow's theorem, there exist subgroups of order 3 and 5. So in this case, there are subgroups of orders 1,3,5, and 15. (4) AxA is the set of ordered pairs (x,y) where x and y are in A. These ordered pairs are the elements of A x A by definition, so a map which maps into AxA has values elements of AxA, or, in other words, values which are ordered pairs (x,y) with x and y in A. Example: A = {a,b}. Then AxA = { (a,a), (a,b), (b,a), (b,b)} A map from A -> AxA is the same thing as assigning to a an element in AxA and assigning to b an element in AxA. For example: f(a)=(a,b), f(b)=(b,b). Since A and AxA are finite, the questions you are being asked are really questions about the relative sizes of the domain and range. Since the order of AxA is the order of A squared, and since it is assumed that the order of A excedes 1, we know that the order of AxA is greater than the order of A. So it is possible to have one to one maps from A to AxA, but it is impossible to have an onto map from A to AxA. (This conclusion changes in the case where the order of A is 1, or if A has infinite order.) (5) I assume that (a,b) denotes the greatest common divisor of a and b. If (x,z)=(z,w), it means that every d which divides both x and z must also divide w, and conversely, every d which divides both w and z must also divide x. Let d divide both x and z. Since w = x-yx = x(1-y), we see d divides w. Let d divide both w and z. From x(1-y) = w, it cannot be concluded that d divides x, because some factors of d may divide 1-y only. This suggests a way of arriving at a counterexample: let x = 3, y = 3, w = -6, z = 2. -Doctor Ceeks, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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