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Difficulty checking group axioms

  1. Jul 20, 2013 #1
    Let ##G## be a set equipped with a binary associative operation ##\cdot##.

    In both of the following situations, we have a group:

    1) ##G## is not empty, and for all ##a,b\in G##, there exists an ##x,y\in G## such that ##bx=a## and ##yb=a##.

    2) There exists a special element ##e\in G## such that ##xe = x## for all ##x##. And (if we fix this ##e##), then for each ##x\in G##, there exists some ##x^\prime \in G## such that ##xx^\prime = e##.

    For reference, ##G## equipped with a binary associative operation is a group if there exists an element ##e\in G## such that ##xe=ex=x##. And for any ##x\in G##, there is an ##x^\prime\in G## such that ##xx^\prime = x^\prime x = e##.

    I've been thinking for hours and I really can't seem to figure out why either (1) or (2) forms a group.

    It is easy to show that if we have a group, then ##(1)## and ##(2)## hold.

    Furthermore, I know that if ##(1)## holds, then ##(2)## holds as well. Indeed, take ##a## a special element, then we can find an ##e## such that ##ae=a## (for this special ##a##). Now, take ##b## arbitrary, then there is an ##x## such that ##b = xa = xae = be##. So we have found the right element ##e##. The existence of ##x^\prime## for every ##x## is now obvious.

    So it suffices to show that ##(2)## implies that we have a group. But I really can't figure it out.
     
  2. jcsd
  3. Jul 20, 2013 #2

    HallsofIvy

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    Certainly, it is true that if either (1) or (2) holds, we have an identity and each element has an inverse. So the only thing left is the "associative law" for any a, b, and c in the set, (ab)c= a(bc). I don't see how that follows from (2).
     
  4. Jul 20, 2013 #3
    Associative is given, so there is no need to check that.

    I don't see why we have an identity, if ##(2)## holds. Sure, we have that ##xe= x## and ##xx^\prime = e##. But why do we also have ##ex= x## and ##x^\prime x = e##?
     
  5. Jul 20, 2013 #4
    Just for clarity - you want to show that given a right inverse, a right identity element, and associativity (conditions 1 and 2) that each member of a group i) commutes with its own right inverse and that ii) commutes with the right identity element?
     
  6. Jul 20, 2013 #5

    hilbert2

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    The facts that an element of a group commutes with both its inverse and the identity element are proved here: http://www.efgh.com/math/algebra/groups.htm (see theorems 1.1. and 1.2.)
     
  7. Jul 20, 2013 #6
    Awesome! Thanks a lot. That was a pretty annoying problem, haha.
     
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