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ALGEBRA PROBLEMS - (abstract)

  1. Apr 9, 2006 #1
    I decided to group my threads into one, as I'm finding more and more problems that I need help on. I hope I haven't hurt anyone by doing so.

    PROBLEMS I NEED HELP WITH:


    1) find an integral domain where NOT every element (not a unit) is expressible as a finite product of irreducibles.

    2) ZXZ = <(a,b), (c,d)> iff ad-bc = +/- 1 , Z= the integers

    3) find a finite group with elements a,b, such that (a^2)(b^2)= (b^2)(a^2), and
    (a^3)(b^3)= (b^3)(a^3), but ab doesn't equal ba [EDIT: I originally posted the question improperly. Thanks AKG for the correction. This is how it is suppose to be]

    4) Prove that for any infinite set A, there is always a 1-1 and onto function from A to AxA.

    5) Show that (RXR, +) ~ (R, +)
     
    Last edited: Apr 9, 2006
  2. jcsd
  3. Apr 9, 2006 #2

    AKG

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    For number 2, it's just basic linear algebra, plus the fact that you're dealing with integers. The second question basically says that for any pair of integers (u,v), there exists integers x, y such that x(a,b) + y(c,d) = (u,v), that is:

    u = ax + cy
    v = bx + dy

    Do you see why this is? Now just apply linear algebra. Doing so will make it clear why if ad-bc = +/- 1, then (a,b) and (c,d) generate ZxZ. The reason why, if ad-bc is not +/- 1, then (a,b) and (c,d) do not generate ZxZ is because, although as long as ad-bd is non-zero we are guaranteed a solution for x and y, if ad-bc is not +/- 1, then the solution we get may not be an integer solution.

    For questions 4 and 5, see my post in your previous thread with those questions. It wasn't all that helpful, but it did have one or two ideas that may end up working.

    For question 3, it makes a difference if the questions asks "find a finite group with (ab)² = (ba)²...." versus one that asks "find a finite group with a²b² = b²a²...." In fact, if (ab)² = (ba)² and (ab)³ = (ba)³, then ab necessarily equals ba. Rewrite the question exactly as given, and show your work.
     
  4. Apr 9, 2006 #3
    I don't have much on any of the questions, which is why I'll need some help..any sort of nudge will do. I know many mentors frown upon this, but I will do my part in answering the questions.

    1) I have no clue where to start

    2) < x, y> is the set generated by x and y...and here as well, I have no clue where to start.

    3) Here, I was thinking about looking at a set of 2X2 matrices, and seeing what turns out. In my next post, i'll tell about where this goes. Hopefully it works out

    4) First off, I understand that this would have to be the case (that A is infinite), as A would have n elements and AxA would have n*n elements in the finite case. Is there a theorem which states that |AxA| = |A| when A is infinite? I think there is, but I also doubt it is needed, as it wasn't discussed on in class.

    5) Well, the profs hint for this question was "use the idea of basis for vector spaces", but I'm not too sure how to (not yet, anyway...perhaps in a day or so). If you looked in my other thread, you saw that I reckoned to use the Pi function defined by PI(a,b)= a , for (a,b) in RXR. Does this work?


    I will post as much progress as I can ("as I can" because I'm also studying for finals..so please bear with me), to show that I am actually working on it. Thanks
     
  5. Apr 9, 2006 #4

    AKG

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    If, for number 4, there were a theorem that |AxA| = |A| when A is infinite, then the entire problem would be trivial. What I mean is that there is a theorem, but if you were allowed to use it then the entire problem would be trivial. You are essentially being asked to prove this theorem.
     
  6. Apr 9, 2006 #5
    5) according to a certain topology Q+A board... considering R and RXR as Vector spaces over Q, both have dimension |R|. So there exists bases of equal cardinality in both spaces. "This defines a Q-linear, and so
    additive, isomorphism between them." - I suppose I'll just fiddle with this a bit.
     
  7. Apr 9, 2006 #6
    Thanks AKG...I understand the question regarding det[abcd] now.

    Also, you're absolutely right. I'm not sure what I was thinking saying that PI was injective. =\

    As for the question regarding the bijection from R to RXR, you answered that we could use the decimal expansion of numbers. It does make sense. In that thread, I had mistyped and posted that I needed a bijection from RXR to R. So I assume that your expansion involved (A,B) in RXR, such that A= a1 a2 a3 ...
    B= b1 b2 b3 ...

    to "GO TO" C= a1 b1 a2 b2 a3 b3 ...

    Seeing how I meant for the function to go from R to RXR, I suppose I can take any decimal expansion C= c1 c2 c3..., and define A= c2 c4 c(2n), and B= c1 c3 c(2n+1) : A reversed construction works, right?
     
    Last edited: Apr 9, 2006
  8. Apr 9, 2006 #7

    AKG

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    For number 5), there's nothing to fiddle with. What you've said in post #5 is a sufficient proof.

    Bijections are two-ways, so if you have a bijection from R to RxR, then you have one from RxR to R (it's just the inverse of the one from R to RxR). The problem comes when R is any arbitrary infinite set, and not just R in particular.

    If you really need to, then yes, the reverse construction works. You have to be careful with cases where a number has distinct decimal expansions, e.g. 0.5 = 0.49999... But the question you posted is for arbitrary infinite sets A, not [itex]\mathbb{R}[/itex] in particular. So this is irrelevant.
     
  9. Apr 9, 2006 #8
    If it works for R, then can we utilize some sort of "similarity" for infinite sets to close the deal?
     
  10. Apr 10, 2006 #9
    re: question 3

    I tried the set of 2X2 matrices. It didn't work- Something I feel I should have noticed from the start. I suppose I only thought of it, because I couldn't think of many groups where ab does not equal ba.

    I also tried certain permutation groups S_1 to S_4, and their subgroups. I found nothing to work there as well. S_1 was obviously wrong, S_2, which consists of {(1), (12)} always has ab=ba, since (12) is it's own inverse. In S_3, (13)(12) doesn't equal (12)(13) and there squares produce (1)(1)=(1)(1) [a^2 * b^2=b^2 * a^2]. However, there cubes bring them back to ab not equalling ba. The same types of scenarios occur for subgroups of these groups.

    O well. I'll keep trying...
     
  11. Apr 10, 2006 #10

    matt grime

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    then you need to rethink what a group is; almost no groups have this property (abelianicity) so you should try thinking of more groups.

    oh, and if you considered S_3, as you claim, and didn't find an answer then you evidently haven't understood the question correctly (or written it correctly).
     
  12. Apr 10, 2006 #11
    For some reason, when I chose a,b, I chose both as 2-cycles when considering the squared values, and both as 3-cycles when looking at the cubes, which scared me off. I did understand the question earlier. I suppose I just didn't "consider" S_3, truly. It is, of course, obvious now, since cubed 3-cycles lead to the identity and squared 2-cycles lead to the identity, that I should have chose a and b such that one was 3-cycled and the other 2-cycled. My apologies...
     
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