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Thread: Factor group order

  1. #1
    Senior Member sfspitfire23's Avatar
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    Factor group order

    If $\displaystyle K\lhd G$ and $\displaystyle |g|=n$, $\displaystyle g\in G$, show that the order of $\displaystyle Kg$ in $\displaystyle G/K$ divides $\displaystyle n$.


    So I know that $\displaystyle G=\{1,g,\cdots g^n\}$ and will $\displaystyle K=\{1,g, \cdots g^{n-1}\}$ ? From Lagrange's theorem K|G?



    Thanks guys
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  2. #2
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    Quote Originally Posted by sfspitfire23 View Post
    If $\displaystyle K\lhd G$ and $\displaystyle |g|=n$, $\displaystyle g\in G$, show that the order of $\displaystyle Kg$ in $\displaystyle G/K$ divides $\displaystyle n$.


    So I know that $\displaystyle G=\{1,g,\cdots g^n\}$ and will $\displaystyle K=\{1,g, \cdots g^{n-1}\}$ ? From Lagrange's theorem K|G?



    Thanks guys
    Two easy lemmas for you to remember/prove:

    1) In a finite group the order of any element divides the order of the group;

    2) If $\displaystyle f: G\rightarrow H$ is a group homomorphism, then for any $\displaystyle g\in G\,,\,\,ord(f(g))\mid ord(g)$ , and if $\displaystyle ord(g)=\infty \,\,then\,\, ord(f(g))=\infty \,\,or\,\,else\,\,f(g)=1$

    Tonio
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  3. #3
    Member alunw's Avatar
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    I'm confused about your claim in the second part of your second lemma.
    Let G be the group of integers under addition and H the integers modulo n under addition. That is surely a homomorphism that is sending torsion free elements to elements with finite order that are not the identity. Am I misunderstanding your notation?
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    Quote Originally Posted by alunw View Post
    I'm confused about your claim in the second part of your second lemma.
    Let G be the group of integers under addition and H the integers modulo n under addition. That is surely a homomorphism that is sending torsion free elements to elements with finite order that are not the identity. Am I misunderstanding your notation?

    Nop, you're right and I got confused. It should be that if $\displaystyle ord(g)=\infty$ then all the options are open.
    Anyway, and for your problem, only the finite case matters.

    Tonio
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  5. #5
    Senior Member sfspitfire23's Avatar
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    What I have-

    For any element g in G $\displaystyle Kg=\{kg|k\in K\}$ defines the right coset of K. Each k in K will have a different product when multiplied by any g. Thus, each element of K will create a corresponding unique element of Kg. So, Kg will have the same number of elements as K (same order as K). Now, the order of K will be $\displaystyle k^m=1, |k|=m$ and the order of G is $\displaystyle g^m=1$. They each produce the identity element and thus divide each other.....?
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