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Thread: Abelian Group Question

  1. #1
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    Abelian Group Question

    G is an abelian group.

    a,b are in G
    order of a = m
    order of b = n

    what is this order of ab ?

    Struggling with this. I suspect it is lcm(m,n) but unable to prove it.

    Any help/pointers please?
    Last edited by aman_cc; Oct 17th 2009 at 05:28 AM.
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    Quote Originally Posted by aman_cc View Post
    G is an abelian group.

    a,b are in G
    order of a = m
    order of b = n

    what is this order of ab ?

    Struggling with this. I suspect it is lcm(m,n) but unable to prove it.

    Any help/pointers please?
    it's not necessarily lcm(m,n). the only thing we can say is that $\displaystyle o(ab)$ divides lcm(m,n). the point is that $\displaystyle o(ab)$ doesn't only depend on $\displaystyle o(a)$ and $\displaystyle o(b)$. it also depends on the relationship between $\displaystyle a$

    and $\displaystyle b$. for example if $\displaystyle b=a^{-1},$ then $\displaystyle o(a)=o(b)=m$ but $\displaystyle o(ab)=1.$
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    Quote Originally Posted by NonCommAlg View Post
    it's not necessarily lcm(m,n). the only thing we can say is that $\displaystyle o(ab)$ divides lcm(m,n). the point is that $\displaystyle o(ab)$ doesn't only depend on $\displaystyle o(a)$ and $\displaystyle o(b)$. it also depends on the relationship between $\displaystyle a$

    and $\displaystyle b$. for example if $\displaystyle b=a^{-1},$ then $\displaystyle o(a)=o(b)=m$ but $\displaystyle o(ab)=1.$
    Thanks! I get it.

    I had worked out the following some time back
    $\displaystyle
    \frac{mn}{\gcd(m,n)^2}|o(ab)$

    Does it make sense?
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    Quote Originally Posted by aman_cc View Post
    Thanks! I get it.

    I had worked out the following some time back
    $\displaystyle
    \frac{mn}{\gcd(m,n)^2}|o(ab)$

    Does it make sense?
    that's correct. so we have $\displaystyle \frac{\text{lcm}(m,n)}{\gcd(m,n)} \mid o(ab) \mid \text{lcm}(m,n).$ this is not bad but to find $\displaystyle o(ab),$ as i mentioned, we need to know how $\displaystyle a,b$ are related.
    Last edited by NonCommAlg; Oct 17th 2009 at 08:52 PM.
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