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Math Help - A question about Group

  1. #1
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    A question about Group

    Hi I am new to Group Theory and I am confused by a problem :

    Let  G be any group of order n and  k be an integer prime to  n .

    Is it true that  x^k = y^k ~ \implies x = y ~~~ x,y \in G ?

    I know there exists  A,B ~\in \mathbb{N} such that  Ak - Bn = 1

    Then  (x^k)^A = (y^k)^A

     x^{Ak} = y^{Ak}

     x^{1 + Bn} = y^{1 + Bn}

     x (x^n)^B = y (y^n)^B

    I think if  a^n = e , then i can prove it but is it really true ? Or my proof has some mistakes ? Actually , it is what i guess and i don't know if it is correct .

    Thanks !
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  2. #2
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by simplependulum View Post
    Hi I am new to Group Theory and I am confused by a problem :

    Let  G be any group of order n and  k be an integer prime to  n .

    Is it true that  x^k = y^k ~ \implies x = y ~~~ x,y \in G ?

    I know there exists  A,B ~\in \mathbb{N} such that  Ak - Bn = 1

    Then  (x^k)^A = (y^k)^A

     x^{Ak} = y^{Ak}

     x^{1 + Bn} = y^{1 + Bn}

     x (x^n)^B = y (y^n)^B

    I think if  a^n = e , then i can prove it but is it really true ? Or my proof has some mistakes ? Actually , it is what i guess and i don't know if it is correct .

    Thanks !
    Looks fine to me!
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  3. #3
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    Quote Originally Posted by chiph588@ View Post
    Looks fine to me!

    But how to prove it true ?  a^n =e

     e is the neutral element of  G
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  4. #4
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by simplependulum View Post
    But how to prove it true ?  a^n =e

     e is the neutral element of  G
    See here.
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  5. #5
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    Quote Originally Posted by chiph588@ View Post
    See here.

    Oh i see . If the group is finite , then we would find that

     \{\ e,a,a^2,... \}\ contains some repeated elements and write  a^s = a^t ~ \implies a^{s-t} = e ~~ s>t

     \{\ e , a , a^2 ,..., a^{s-t-1} \}\ is a subgroup of  G then use Lagrange's Theorem to show that  a^n = e .

    Thank you very much
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  6. #6
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by simplependulum View Post
    Hi I am new to Group Theory and I am confused by a problem :

    Let  G be any group of order n and  k be an integer prime to  n .

    Is it true that  x^k = y^k ~ \implies x = y ~~~ x,y \in G ?

    I know there exists  A,B ~\in \mathbb{N} such that  Ak - Bn = 1

    Then  (x^k)^A = (y^k)^A

     x^{Ak} = y^{Ak}

     x^{1 + Bn} = y^{1 + Bn}

     x (x^n)^B = y (y^n)^B

    I think if  a^n = e , then i can prove it but is it really true ? Or my proof has some mistakes ? Actually , it is what i guess and i don't know if it is correct .

    Thanks !
    The property 1 \neq x^n = y^n \Rightarrow x=y is called the `unique roots property', and it does not always hold. It does hold in, for example, free groups (infinite groups that can be viewed as the sort of top-down building blocks of group theory).

    Such groups have the property that g^mh^n = h^ng^m \Rightarrow gh=hg.

    However, this property does not always hold. A finite example would be the elements \alpha = (123)(45) and \beta = (123). Then \alpha^2 = \beta^2 but \alpha \neq \beta.

    What I believe you proved in your post is that if |G| = n and gcd(n, k) = 1 then if a^k=b^k we have that a=b.

    (An infinite example of a group without the unique roots property is the group given by the presentation (rules) \langle a, b; a^2=b^2\rangle. This is non-trivial and infinite as it has a homomorphism onto the infinite dihedral group, C_2 \ast C_2)
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