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Algebra proof, abstract structure

  1. Oct 20, 2015 #1
    1. The problem statement, all variables and given/known data
    Let [itex](S,\cdot )[/itex] be an algebraic structure where the operation [itex]\cdot [/itex] is associative and commutative and also the following axiom is satisfied:
    [tex]
    \forall x,y\in S,\exists z\in S: zx = y\ \ (1)
    [/tex]
    Prove that if for every [itex]a,b,c\in S, ac = bc[/itex], then [itex]a=b[/itex].
    2. Relevant equations


    3. The attempt at a solution
    Assume by contradiction that [itex]a\neq b[/itex]. Then for every [itex]c\in S[/itex] either [itex]ac = bc[/itex] or [itex]ac\neq bc[/itex]:

    1. if [itex]ac\neq bc[/itex] then the proof is concluded.

    2. provided [itex]a\neq b[/itex] and for every [itex]c\in S, ac=bc[/itex]. By axiom [itex](1)[/itex] there exist [itex]x,y\in S[/itex] such that [itex]ax=c[/itex] and [itex]by=c[/itex]. It follows that [itex]a(ax)=b(by)[/itex], which is the same as [itex]a(by) = (ab)y = b(ax) = (ba)x = (ab)x[/itex]. We wind up with the equality [itex](ab)y = (ab)x\ \ (2)[/itex].

    Stuck, I feel like I have reached nowhere with the last statement, wandering in circles. If [itex]x(ab) = y(ab)[/itex], then [itex]x\neq y[/itex] as I have assumed at the beginning of 2. but this is a circle. The function [itex]\cdot : S\times S\to S[/itex] need not be injective, in other words if [itex]x\neq y[/itex] it doesn't necessarely mean that [itex]kx\neq ky[/itex] for some [itex]k\in S[/itex], in fact I have assumed if [itex]x\neq y[/itex], then [itex]kx=ky[/itex], somehow this has to yield a contradiction.

    Idea: Maybe I can show that the function [itex]\cdot [/itex] is, in fact, injective.
     
    Last edited: Oct 20, 2015
  2. jcsd
  3. Oct 21, 2015 #2

    RUber

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    I like the work that you have done so far. But you are right, you need to find something to say that a = b.

    You could prove the existence of an identity and inverse.

    Using (1), show that for all x there exists a y (denoted ##I##) such that ##x \cdot I = x##. (2)
    Using (1) and (2), show that for all x there exists a y (denoted ##x^{-1}##) such that ##x \cdot x^{-1} = I##. (3)

    From there, you can just apply the inverse then identity to conclude a = b.
     
  4. Oct 21, 2015 #3

    andrewkirk

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    I've been fiddling around with that approach on and off for the last day, because I felt there was an easy solution along that line and I might be able to give a hint. But I've got nowhere. Given an element ##x##, one can easily show the existence of element ##e_x## such that ##xe_x=x##, but I didn't manage to prove that ##\forall x,y:\ e_x=e_y##, which is what's needed for it to be an identity.

    I have a vague recollection of solving one like this recently. But I think it might have been for a finite structure, which allows the use of additional techniques.

    It's usually the case with algebra that when one finally finds the proof it's obvious. But right now I can't see it.
     
  5. Oct 21, 2015 #4
    I'm not a 100% sure but this is what I get:

    For any ##x\in S## there exist ## z_x \in S## such that ##x = x z_x ##.
    We have that ##z_{xy} = z_x z_y ## and then ##z_{x^2} = z_x z_x = z_x## as ##xz_xz_x = xz_x = x##

    If ##y = ax##, then ## ay = a y z_y = ayz_az_x = ay z_x##.
    So ##z_x = z_{ay} = z_a z_y = z_{a^2} z_x = z_a z_x ##. Multiplying by ##x##, you find that ##z_a = z_x##, and therefore ## z_a = z_x = z_y ##

    So ##ac = bc \Rightarrow z_{ac} = z_b = z_c \text{ and } z_{bc} = z_a = z_c \Rightarrow z_a = z_b = z_c ##
    Furthermore, there exists ##z\in S## such that ##a = (ac) z\Rightarrow cz = z_a = z_b = z_c ##.
    Therefore, ## a c z = a z_a = a##, ## bc z = bz_b = b##, and ##acz = bcz## since ##ac = bc##. Therefore ##a = b##
     
  6. Oct 22, 2015 #5
    What I wrote is just wrong because I'm assuming unicity of ##z_x##, it is circular reasoning o0). Furthermore it is badly writen.

    But what is sure is that there exist ##z,z'\in S## such that ##a = (ac)z ## and ## b = (bc)z'##.
    So ##cz## and ##cz'## are neutral for ##a## and ##b## respectively.

    But ##ac = bc \Rightarrow acz = bcz \text{ and } ac z' = bc z' \Rightarrow a = b(cz) \text{ and } b = a (cz') \Rightarrow a =(acz')(cz) = a ( cz') ##

    So ## cz'## is also neutral for ##a##, and since we had ## b = a(cz')##, then ##a = b##
     
  7. Oct 22, 2015 #6
    By given axiom:
    [itex]\forall x,y\in S\ \exists z\in S: zx=y[/itex]
    We can interprete this as E: no we can't
    [itex]\forall x\in S, \exists e\in S: xe = ex = x[/itex], which is the definition of a unit element. Analogously we can show every element has an inverse. So we have an abelian and the proof henceforth is trivial.
    My question is, is it safe to do so or we lose generality in some way?
    EDIT: nope, disregard this, if for some [itex]x,y\in S, xy = x[/itex], it doesn't necessarely mean that for some other [itex]z\in S, zy = z[/itex].
     
    Last edited: Oct 22, 2015
  8. Oct 22, 2015 #7

    PeroK

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    Try this:

    ##\exists z## s.t. ##za = b## and ##\exists w## s.t. ##wa^2 = a##

    ##a^2 = ba = ab = b^2## (using the given property of ##a, b##)

    ##a = wa^2 = wb^2 = b(bw) = a(bw) = (aw)b = (aw)za = zwa^2 = za = b##
     
    Last edited: Oct 22, 2015
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