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Thread: 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 $\displaystyle G $ be any group of order $\displaystyle n $ and $\displaystyle k $ be an integer prime to $\displaystyle n $ .

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

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

    Then $\displaystyle (x^k)^A = (y^k)^A $

    $\displaystyle x^{Ak} = y^{Ak} $

    $\displaystyle x^{1 + Bn} = y^{1 + Bn} $

    $\displaystyle x (x^n)^B = y (y^n)^B $

    I think if $\displaystyle 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 $\displaystyle G $ be any group of order $\displaystyle n $ and $\displaystyle k $ be an integer prime to $\displaystyle n $ .

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

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

    Then $\displaystyle (x^k)^A = (y^k)^A $

    $\displaystyle x^{Ak} = y^{Ak} $

    $\displaystyle x^{1 + Bn} = y^{1 + Bn} $

    $\displaystyle x (x^n)^B = y (y^n)^B $

    I think if $\displaystyle 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 ? $\displaystyle a^n =e $

    $\displaystyle e $ is the neutral element of $\displaystyle 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 ? $\displaystyle a^n =e $

    $\displaystyle e $ is the neutral element of $\displaystyle 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

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

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

    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 $\displaystyle G $ be any group of order $\displaystyle n $ and $\displaystyle k $ be an integer prime to $\displaystyle n $ .

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

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

    Then $\displaystyle (x^k)^A = (y^k)^A $

    $\displaystyle x^{Ak} = y^{Ak} $

    $\displaystyle x^{1 + Bn} = y^{1 + Bn} $

    $\displaystyle x (x^n)^B = y (y^n)^B $

    I think if $\displaystyle 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 $\displaystyle 1 \neq x^n = y^n \Rightarrow x=y$ 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 $\displaystyle g^mh^n = h^ng^m \Rightarrow gh=hg$.

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

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

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