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Prove that a set with following properties is a group

  1. Jul 22, 2011 #1
    1. The problem statement, all variables and given/known data
    I'm self-studying Abstract Algebra from baby Herstein. This is an exercise in its problem sets:

    Suppose that G is closed under an associative operation such that
    1. given a,y in G, there is an x in G such that ax = y, and
    2. given a,w in G, there is a u in G such that ua = w.
    show that G is a group.

    3. The attempt at a solution

    well, I first tried to show that the identity element exists in the set. using the 1st property, we can let y=a where a can be any arbitrary element (I'm fixing one of the two). then the first one tells us that an element x exists s.t. ax = a for any a in G. doing the same thing, we can show that there is an element u such that ua = a using the second one. therefore we've proved that there are right and left identity elements contained in the set. Now I must show that the right and left identity elements are equal.

    Now, first, before I go ahead, I need to prove that eR (the right identity element) = eL (the left identity element) which means that the right and left identity elements are in fact the same element in G.

    eR = eL.eR (because eL is a left identity element it doesn't change anything if we left multiply eR by it) = eL (because eR is a right identity element).

    therefore if e is a right identity element it is also a left identity element. Now It's easy to prove that e must be unique. (Herstein itself has proved it in the very beginning of the chapter).

    well, now let e be the identity element. first I'll show that every element in G has at least a right and a left inverse element.
    well, this time take y to be the identity element in G. in both cases the first postulate assumes that there exists an x in G that satisfies ax=e for any a in G. therefore for any a in G there is at least one right inverse for it. the same reasoning can show that for any a in G there is at least one left inverse for it using the second property.

    Now that we've proved that the identity element is in the set and is uniquely determined we can prove that for every element in G there is a unique inverse. Now I wanna switch to additive notion for convenience. (It's easier to type + and - than powers of a number by a standard keyboard, I hope it doesn't bother you). what we need to show is that for every element in G there is a unique inverse for it, namely -a that when a is in G we have: a+(-a)=e. (I don't use 0 here just to reduce the need of switching to different notions a lot).
    well, we denote the right inverse of a by -aR and the left inverse of a by -aL.
    we have: (-aL) + a + (-aR) = e + (-aR) & (-aL) + a + (-aR) = (-aL) + e.
    e is commutative (we already proved that in previous step). therefore from e + (-aR) = (-aL) + e it yields -aR = -aL. that proves that the left and right inverses of a are the same. Now It's easy to prove that the inverse of an element of G is unique (Again Herstein itself has proved it).

    well, the closure property is assumed although it can be proved without the need of assuming it I guess. since for any a in G there is a unique -a in G s.t. a + (-a) =e we can conclude that for any z,y in G there exists an x in G such that z + y = x.
    well, let first take a=(-z): (-z) + x = y (the first property) => x = z + y. the other case that for any y,z in G y + z equals some x' in G can be handled the same way using the second property.

    Associativity is assumed as well, therefore the set G satisfies all fundamental properties of a group and is a group.

    well, Have I proved it correctly or there are flaws in my argument?
     
    Last edited: Jul 22, 2011
  2. jcsd
  3. Jul 22, 2011 #2

    micromass

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    Hi AdrianZ! :smile:

    Your proof looks ok, there's just one gap:

    This is not true. You only have that, for a, there exists an x such that ax=a. But this x can depend on a!!
    That is, if b is another element, then it is not necessarily the case that bx=b. All you know is that there is an element y such that by=b. But nothing tells us that x=y...
     
  4. Jul 23, 2011 #3
    Yea, I was also thinking about that part first but later forgot to come back to it. I mean what you're saying is that there might be millions of left or right identity elements depending on what a I choose; How did I know that there is only one right and one left identity element. right?
    well, I asked myself the same question and then I thought because a can be any element in G then x should work for any a that I plug in. so what I claimed was that there exists an x in G that for any a we have: ax=a. but what you're saying is that for any a there exists an x in G that we have: ax=a and this x is not necessarily independent of the choice of a.
    well, my proof is incomplete if I can't prove that this x is not independent of a. What should I do?
     
  5. Jul 23, 2011 #4

    micromass

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    Well, for a certain a, there exists an x such that ax=a. Now, prove that this x is good for all elements in our "group". For example, can you show that this x is idempotent?
     
  6. Jul 23, 2011 #5
    How can I prove that this x is good for all elements in our group?
    And what should I do after if I succeeded to show that x is idempotent?
     
  7. Jul 23, 2011 #6

    micromass

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    You must use (2) in such a way to show that x is good for all elements in the "group".
     
  8. Jul 23, 2011 #7
    I have no idea how to do that. Would you give me a clear strategy of what I should follow that I could show that the x we're talking is good for all elements in the set G? It's not easy for me to see that I guess.
     
  9. Jul 23, 2011 #8

    micromass

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    It's very easy once you know it, so I'm careful not to give away to much information.
    Let's say you have ax=a. Is there a way to use (2) to change the a into b?
     
  10. Jul 23, 2011 #9
    well, I can use (2) and say there is a u in G such that ua=b for all a,b in G. then I can write down u(ax) = u(a). therefore (ua)x=(ua) and that yields bx=b. does that prove it or I'm lost? and If yes, is my proof complete now?
     
  11. Jul 23, 2011 #10

    micromass

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    That's good! :smile:
     
  12. Jul 23, 2011 #11
    so Now I'm done and can write down the proof completely without flaws?

    If I ever take a real abstract algebra course, will the proof be so long as this one? I'm asking because if yes then it'll be so boring and tiring to write down such a long proof in your exam paper!
     
  13. Jul 23, 2011 #12

    micromass

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    No, I don't think that this exercises is a great reflection for what exam questions will be. This is indeed too long, and perhaps a bit too technical.
    However, you will meet quite a few boring proofs in abstract algebra, but you'll need to get through those to reach the fun results :smile:
     
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