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Group definition for finite groups

  1. May 15, 2013 #1
    Was wondering if the only required definition for finite groups is closure (maybe associativity as well). It seems that is all that is necessary. The inverse and identity necessarily seem to follow based on the fact that if I multiply any element by itself enough times, I have to repeat back to the original element again (because of closure and finiteness). The fact that I have repeated means there is an inverse for each element and thus an identity.
     
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  3. May 15, 2013 #2

    Office_Shredder

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    Does it though?


    Imagine the following "group"

    Four elements a,b,c,d. Multiplication x*y is defined as x*y=a for any choice of x and y.

    It's associative - (x*y)*z = a*z = a. x*(y*z) = x*a = a.

    It's certainly closed under multiplication
     
  4. May 15, 2013 #3
    how does that mean there is no identity or inverse? it seems you have simply called the same element four different things and you are really just working with a group of one element, the identity element in your case seems to be a.

    for example, by your definition every element i multiply by any other element is a.

    however, the element needs to repeat back on itself after a certain number of multiplications. this is just what's called the "order" of the element.

    so if i multiplied b enough times, then that means i should reach b again. by your multiplication rule, i cannot do that unless i realize that b is actually a is actually c and d.
     
  5. May 15, 2013 #4

    Office_Shredder

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    No, I have four distinct guys, a b c and d. As another example maybe I have the 3x3 matrices a,b,c,d with a the zero matrix and b,c,d picked to be projections onto the x,y, and z axis. Then it's closed under multiplication - multiplying two projections will give me zero, unless I multiply a projection by itself in which case I get the projection back, and it's associative, since all matrix multiplication is.

    You only know this is true if what you have is a group with inverses and identities! Your proof that you get a group relies on knowing it's a group already
     
  6. May 16, 2013 #5
    For the record, for your first multiplication rule you didn't have four distinct guys, you had one element, how could you distinguish between the elements except by your arbitrary labeling?

    Your projection operators don't satisfy your initial multiplication rule.

    That said, the projection operators seem like a natural explanation of why closure alone does not satisfy the group definition as a whole, they almost by definition don't have inverses because inverses allow me to undo what was done and projection operators just give me a one way street. But, what about inverses? inverses + closure certainly guarantee that I have an identity. So why give the identity as a separate axiom?

    anyway, all this indicates that dresselhaus' proof of the theorem that says I can get an identity element seems to be lacking
     
    Last edited: May 16, 2013
  7. May 16, 2013 #6

    Office_Shredder

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    In the group S3, the transpositions (1 2) and (2 3) are indistinguishable (because there is a group automorphism taking one to the other) but that doesn't mean we call them the same element. Groups have elements that are indistinguishable algebraically all the time, but they have different labels so they are distinguished.


    I realize that, I was giving a completely different example of a closed set with associative multiplication. Notice that in this example also b,c and d are indistinguishable from one another, but they aren't the same element in any meaningful way.

    What is your definition of an inverse that does not involve an identity element existing already?

    I'm unfamiliar with what you're talking about here, if you post it we can work through what's going on in detail
     
  8. May 16, 2013 #7

    pwsnafu

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    No. What you describe is a semigroup. The simplest semigroup that isn't a group has two elements, hence finite.
     
  9. May 16, 2013 #8
    I believe Cauchy and Galois define groups in terms of closure only. But it seems they were restricting themselves to permutations so that is all you need to define a group for permutations.

    I see now, so you can arbitrarily label a number of elements and fill in their multiplication table however you wish. But it seems there should be some systematic way of knowing whether we are really talking about another element or just calling the same element multiple things. I mean for your projection operators, b, c, d I see they are physically different in the sense that they project in orthogonal directions. In short, how do we justify calling them something different if there is no physical or matrix representation of what's going on?I don't know, maybe I'll try to see how you could represent your initial multiplication rule, where every element gives only one of the elements.

    So if I multiply two elements AB, an inverse exists for A if I can multiply it by AB to give me B again. With closure, this necessarily means that an inverse implies there is an element such that I multiply it by any other element, I just get that other element, so we have an identity...voila.

    Well, In Dresselhaus' book the theorem is stated that...

    If in a finite group, an element X is multiplied by itself enough times (n), the identity XN = E is eventually recovered

    Proof: If the group is finite, and any arbitrary element is multiplied by itself repeatedly, the product will eventually give rise to a repetition. For example, for P(3) which has six elements, seven multiplications must give a repetition. Let Y represent such a repetition:

    Y = Xp=Xq, where p > q

    Then let p = q + n so that

    Xp = Xq+n=XqXn = Xq=XqE

    so it follows that Xn=E


    It seems I can do the same manipulations with your set of projections. What I would end up with is a bunch of identities that are orthogonal to each other and therefore not actually identities.So there must be a subtle point in the proof that it must have the effect of an identity for all elements in the group or something like that.
     
  10. May 16, 2013 #9

    Office_Shredder

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    I don't really understand this. It seems like you're saying that an inverse for A exists if A(AB) = B, or perhaps (AB)A = B. Is B an arbitrary element of the group, or a specific element?

    This is key the step. You have Xp = Xq Xn. You also have p=q. So let the element A = Xp = Xq. Then you have
    A = A Xn

    Multiply both sides on the left by A-1 to get E = Xn. So he is using the existence of inverses, or if you want to get really weak about what group properties are being used, the fact that if you have an element E, and some element A such that A = AE, then BE = B for all group elements B and also EB = B for all group elements B
     
  11. May 16, 2013 #10
    Why do I need to multiply anything by any inverse?

    If I can show that A = AXn and A = XnA... then Xn has the property of the identity element anyway. If I can show this for any element, then each element individually has it's own "identity"

    for example, something that acts like an identity for your a null matrix is a,b,c,d... something that acts like an identity for the b projection is b, something that acts like an identity for c is c... and so on.

    But they don't act like identities if I multiply them by other elements. for example CC = C but CB = A.

    I think there is an extra step necessary to show that Xn = Yn' where n is the order of element X and n' is the order of element Y for arbitrary element X and Y.
     
  12. May 16, 2013 #11

    Office_Shredder

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    Because that lets you finish solving the problem as opposed to spending an hour trying to figure out how to prove that Xn acts like the identity for every element of the group
     
  13. May 16, 2013 #12
    So by having the inverse explicitly you prove that Xn = Yn' for arbitrary elements x and y.

    Ok, fine.

    So now my only issue is with the identity AND the inverse. If I have an inverse for at least one element, then I have an identity element by closure. Why state an identity as separate axiom? Are you saying you need to define the inverse by the identity element?

    An inverse of A can be defined as the element A-1 that can undo what was done by A, such that A can also undo what was done by A-1. By closure, their product must be in the group, and their product already satisfies what we define as the identity element. So why state this axiom that there needs to be an identity element.

    I could understand if it was just necessary to define what is meant by identity element, but to require that it exists after already requiring that the inverse exists seems unnecessary.
     
  14. May 16, 2013 #13

    Office_Shredder

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    You could say for each A there is an A-1 such that AA-1B = A-1AB = BA-1A = BAA-1 = B.

    Now you also need to figure out if A A-1 = B B-1 for each A and B, or if that needs to be a separate axiom by itself.

    Even if you were OK without the second statement, this formulation is pretty unwieldy and it's easier to just say there's an identity element. It doesn't really matter if you have one long axiom or two short axioms anyway
     
  15. May 16, 2013 #14

    AlephZero

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    If you define what you mean by "can undo what was don't by A", that IS a definitionn of an identity element.

    Your definition willl be something similar to '##A^{-1}AX = X## for any (or every) element ##X##'.
     
  16. May 16, 2013 #15
    Oh yes I agree AlephZero, so back to my original question, it seems as though the inverse guarantees an identity element... so why the separate axiom?

    Office_Shredder, I think I see what you are saying. Stating there exists an identity element in the group allows us to DEFINE the combination for arbitrary A, AA-1 = A-1A = E (the defined identity). This axiom sets all of those combinations equal to each other, in the sense that AA-1 = BB-1 for arbitrary A and B in the group. So that all of these combinations act like the identity for all elements in the group.

    But I can't think of any case where BB-1 would not equal AA-1
     
    Last edited: May 16, 2013
  17. May 16, 2013 #16
    I think I get the role of both axioms. It's just a simple way to state that the inverse A-1 can undo the effect of A on any element in the group. Stating that A-1A = E, IS saying that the inverse can undo the effect of any for any element, because of how we have defined E for any element in the group. Okay, the essence of these axioms is getting a little more clear to me. Thanks for the help.
     
  18. Jun 12, 2013 #17

    lurflurf

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    You need all the axioms, it is a standard exercise to prove that they are independent. It is possible to combine all the axioms into one so that associativity, inverse, identity, and totality all follow from one law, but that law is different and I think more confusing.
     
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