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Math Help - Free Group

  1. #1
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    Free Group

    Hi,
    I've following task:

    Prove that every free group is torsion-free and is non-abelian if its rank is \geq 2.

    The second part is easy.
    What about the first part? I think that we should suppose that this free group F[X] is torsion (but I'm not sure).
    Thanks for any help.
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  2. #2
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    Quote Originally Posted by Arczi1984 View Post
    Hi,
    I've following task:

    Prove that every free group is torsion-free and is non-abelian if its rank is \geq 2.

    The second part is easy.
    What about the first part? I think that we should suppose that this free group F[X] is torsion (but I'm not sure).
    Thanks for any help.
    This is an easy task if you know already about the normal form of an element in a free group...besides this, we

    can work with F(2) = the free group of rank two, which contains an isomorphic copy of the free

    group of any rank between 2 and countably infinite.

    If you don't know yet about the normal form, then you can do as follows: suppose w:=w(x,y) is a

    non-trivial word on x,y in the free group F(x,y), and ord(w) =k<\infty . We can

    carry on now by induction on the length of this word: if l(w)=1\Longrightarrow w(x,y) =x,y,x^{-1},\,\,or\,\,y^{-1} , suppose

    w=x\Longrightarrow defining the function f:\{x,y\}\rightarrow C_{k+1}=\langle c\rangle the cyclic group of

    order k+1, by f(x):=c\,,\,f(y)=1 , we know (the universal property of free groups) that there

    exists a unique group homomorphism \phi(F(x,y))\rightarrow C_{k+1} extending

    the function f. But this is impossible since then it must be \phi(x)=c\Longrightarrow 1\neq c^k=\phi(x)^k=\phi(x^k)=\phi(1)=1 ,

    thus getting a straighfroward contradiction. Try now to extend this idea when l(w)>1

    Tonio

    Ps. Or read about the normal reduced form for an element in a free group. The proof then is

    almost trivial.
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  3. #3
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    Thanks for help. I'll try to do the rest

    Could You show how can You do this task using normal reduced form? It will be nice to see the second solution.
    BTW. Where I can find something about this definition?
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  4. #4
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    Quote Originally Posted by Arczi1984 View Post
    Thanks for help. I'll try to do the rest

    Could You show how can You do this task using normal reduced form? It will be nice to see the second solution.
    BTW. Where I can find something about this definition?

    If w(x,y) is in reduced normal form, then we can assume that its first letter is no the inverse of its

    last one (why?), so we get at once that l(w)=m\Longrightarrow l(w^r)=rm , so

    if the length of the word is possitive also its multiples also have positive length and are thus

    not the unit element.

    Tonio
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  5. #5
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    We can assume this because this is cyclically reduced word (am I right?).

    I'm still do not see how this proof the fact that every free group is torsion-free .

    Could You show this with details , of course if You have time. (I'm thinking with difficulty so this is unclear for me.) I'll be glad for this.
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  6. #6
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    Quote Originally Posted by Arczi1984 View Post
    We can assume this because this is cyclically reduced word (am I right?).

    I'm still do not see how this proof the fact that every free group is torsion-free .

    Could You show this with details , of course if You have time. (I'm thinking with difficulty so this is unclear for me.) I'll be glad for this.

    Well, you know that ONLY the empty word (of length 0 = zero) in free generators represents the unit element, so if ANY non trivial element's

    word increases its length as we multiply this element by itself then its length won't ever be zero and this

    means the element raised to no (non-zero) power will be the unit element <==> the element has infinite order.

    Tonio
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  7. #7
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    Ok, I'll look at this tomorrow. Now it's a little late
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