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Thread: Is parity preserved when multiplying two group elements

  1. #1
    Newbie diehardwalnut's Avatar
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    Is parity preserved when multiplying two group elements

    I want to show $H$ is a subgroup of $G$, an Abelian group, where $H=\{x \in G | |x| \ is \ odd \}$. If I assume $a,b\in H$ can I be certain $ab^{-1} \in H$, that is to say does multiplying two elements with odd order in an Abelian group produce an element with odd order? I've tried it out with a few specific groups and it seems to be the case but I haven't been able to come up with a proof.
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  2. #2
    MHF Contributor
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    Re: Is parity preserved when multiplying two group elements

    Hey diehardwalnut.

    Can you use Lagrange's theorem in this case?
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  3. #3
    Newbie diehardwalnut's Avatar
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    Re: Is parity preserved when multiplying two group elements

    I don't think so, Lagrange's equation doesn't occur in the book until chapter 5 and this question occurs in chapter 3.
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  4. #4
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    Re: Is parity preserved when multiplying two group elements

    Here are the facts you need:

    1. In any group $G$ and $x\in G$, if $x^n=e,\text{ the identity of }G$, the order of $x$ divides $n$. If the order of $x$ is $m$ and $k$ is any integer, $x^{mk}=e$.

    2. In an abelian group $G$ with $a,\,b\in G$ and $n$ any integer, $(ab)^n=a^nb^n$.

    For your problem, let $a,\,b\in H\text{ with }|a|=m\text{ and }|b|=n$. Show $(ab^{-1})^{mn}=e$. Conclude that the order of $ab^{-1}$ is odd.
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