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Show G is a Group.

  1. Jul 25, 2008 #1
    Other than problem 2 being over a finite set, what is the fundamental difference in these questions (i.e. I see problem 2 has left / right cancellation properties but not sure how it changes the answer for both these questions)

    Problem 1:
    If G is a set closed under an associative operation such that:
    Given a, y [tex]\in[/tex] G, there is an x [tex]\in[/tex] G such that ax = y, and
    Given a, w [tex]\in[/tex] G, there is a u [tex]\in[/tex] G such that ua = w. prove that G is a group.

    Problem 2:
    If G is a finite set closed under an associative operation such that ax=ay forces x=y and ua=wa forces u=w, for every a,x,y,u,w [tex]\in[/tex] G, prove that G is a group.
  2. jcsd
  3. Jul 25, 2008 #2
    Remove "finite" from problem two and figure out a set and operation that satisfies the property given, but which is not a group. (hint:
    positive integers under multiplication
    ). Why does the condition given in the first problem eliminate the problem that arises? This should help you understand the reason why the conditions are nonequivalent.
  4. Jul 27, 2008 #3
    I'm sorry, but the more I study this the more confused I am getting. I posted the problem #2 ax=ay, ua=wa (reference link:http://www.artofproblemsolving.com/Forum/viewtopic.php?t=215804 From this I understand that there are semi-groups which may have the property of ac=bc (or ax=ay) but are not a group because there is no inverse or identity (i.e. for the set {N,+} the identity is zero and not included in N set; inverse is "-a" and is not included in N set since its not positive). I am still struggling with the finite vs not finite difference, and also the injective/surjective = bijective part. I also worked through the example in the Math Link site (ref:http://www.libraryofmath.com/mappings.html

    In particular, the Proposition (Mappings on Finite and Infinite Sets). I think I understand the alpha/beta example, and tried to apply this to the (N,x) example using the alpha function of a(g)=2g, and N = {1,2,3,4,5,6}. It appears to me that the set is injective but not surjective. thus not bijective (and it also appears in particular the inverse is not 1-1 since 10,5 map back to 5).

    In any case, I feel like I am still missing some key points, and in particular have a difficult time looking at these problems and understanding if there is a fundamental difference between what they are asking for. I have spent a lot of time already on this, and am quickly getting very frustrated. Any help you could offer would be greatly appreciated. There are 4 Herstein problems asking to prove that there is a group (given associative and closure).
    1) given ex=x, yx=e
    2) ab=ac, ba=ca therefore b=c
    3) ax=y, ua=w
    4) finite set--- ua=wa, ax=ay.
    I am taking this class on line (with no help from the instructor). I'm looking for someone who can just simply lay out what I am missing with understanding whats behind the 4 questions.
  5. Jul 27, 2008 #4

    Just play around until it becomes obvious. But surely it is the definition of bijective that it is something that is both injective and surjective.

    This makes no sense: what is N? Is it supposed to be Z/6Z with multiplication as the binary operation? I guess it is meant to be a group, so that ought to make it the non-zero elements of Z/7Z under multiplication. Note a 'set' isn't injective, a function is. What two residues do you think are sent to the same thing on multiplication by 2? Here's what multiplication by 2 does:


    That looks like an injection and a surjection to me.

    The fundamental difference between finite and infinite sets is that if S is finite and


    is an injection, then S is a bijection.

    This is not true for infinite sets.
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