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Free groups.

  1. Dec 23, 2007 #1
    Hello all.

    I am reading again about free vector spaces over a set. In the Theory of Groups by Kurosh part of the construction of a free group is to construct a set of elements inverse to the those of the original set which can effectively "cancel" each other out if juxtaposed in a word made from elements of these two sets. I understand the mechanics of the constuction and the reduction of words but as a first question i would like to ask what is the inverse of an element of a set, if such a thing exists, apart from having this defined "cancellation" property

    Matheinste.
     
  2. jcsd
  3. Dec 23, 2007 #2

    mathwonk

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    what is the inverse of 1? we call it -1.
     
  4. Dec 24, 2007 #3
    Thanks for your reply Mathwonk.

    I see what you are saying but as the original set as far as i can see can be anything, and as being a set it has no structure defined upon it, how do we manage to always define the (unique?) inverse of any element. I am aware that in this case closure is not a requirement because in this case the inverses are to form a seperate set.

    If two of the elements of the original set were a square and a circle how would we define their inverses. I suppose there are ways of doing this, but what if one of the members of the original set was say, the letter A itself, not standing for anything else, merely a symbol, what then.

    Do we, for this construction of free groups, define the inverse of a symbol or letter A as that which when put next to a symbol or letter A in a "word" formed by members of the original set and its inverses, produces, by definition, the identity element.

    In other references to free groups i have seen the members of the inverse set merely refered to symbolically as primed elements of the original set with no further explanation of what they are apart from the property of "cancelling out" the relevant elements of the original set. Is this therefore THE defining property.

    The concept of free groups as with any mathematical structure is rigorously defined and the lack of understanding is mine. This is just an attempt on my part to correct my lack of knowledge.

    I hope, but doubt, that i have explained my problem with regards to free groups adequately.

    Thanks to anybody for any help. Matheinste.
     
  5. Dec 24, 2007 #4

    HallsofIvy

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    I'm not sure I understand your question. If we have a set with elements {a, b, c}, then we define "inverse" elements {-a, -b, -c} (or, with "multiplicative notation", {a-1,b-1,c-1}) as well as an identity element. you can't say that they "are" something in particular since they don't exist before you define them.
     
  6. Dec 24, 2007 #5
    Thanks HallsofIvy.

    At the abstract level it is not a problem. Perhaps i must stop thinking of concrete examples. I suppose the abstractness is the essence.

    To recap, so far in making a free group we have made a structure from our original and inverse sets. This structure has an identity but we do not have a binary operation defined. I believe this may be what the free group structure is all about, making the group in some way independent of any particular structure, although presumably it must have an assumed group axiom structure even if the group operation is not specifically defined. Am i going in the right direction.

    Thanks Matheinste.
     
  7. Dec 24, 2007 #6

    mathwonk

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    let S be any set, and define V to be the set of all functions from S to the reals R. then an element x of S corresponds to the function whose value is 1 at x and 0 elsewhere.

    the inverse of a function f, is the function -f, so the inverse of the function corresponding to the element x, is the function whose value is -1 at x and 0 elsewhere.
     
  8. Dec 24, 2007 #7

    mathwonk

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    to make a non commutative, free, group, start from a set S, and consider the product set Sx{1,-1}. the pairs of form {x,1} correspond to the elements x of S, and the pairs {x,-1} correspond to their inverses. introduce a new element called e which will be the identity.

    then consider all functions from the positive integers into the set Sx{1,-1} union {e}, whose value is eventually just e, above some point.

    think of these objects as words spelled with the letters x,..., -x,....,e.

    add them by juxtaposing them and cancelling adjacent inverse elements into an e, an canceling e with any other adjacent element, until no e's are left which are followed by other elements.

    gacckkkk!!! i have no patience for this.
     
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