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Thread: Group Theory - order of element

  1. #1
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    Group Theory - order of element

    Determine the order of $\displaystyle a^m$ where $\displaystyle a$ is an element of order $\displaystyle n$ in a group G

    I don't really know what they book is looking for here. It seems to me this will break down into a bunch of cases depending on what $\displaystyle m$ and $\displaystyle n$ are, but maybe I'm wrong.

    $\displaystyle a^n=1$ (the identity)

    and yeah that's pretty much all I got so far
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by artvandalay11 View Post
    Determine the order of $\displaystyle a^m$ where $\displaystyle a$ is an element of order $\displaystyle n$ in a group G

    I don't really know what they book is looking for here. It seems to me this will break down into a bunch of cases depending on what $\displaystyle m$ and $\displaystyle n$ are, but maybe I'm wrong.

    $\displaystyle a^n=1$ (the identity)

    and yeah that's pretty much all I got so far
    We kind of need this lemma

    Lemma: Let $\displaystyle g\in G$ with $\displaystyle |g|=n$. If $\displaystyle g^{m}=e$ then $\displaystyle n|m$.

    Proof: By the division algorithim $\displaystyle m=qn+r\quad q\in\mathbb{Z},0\le r<n$. So then $\displaystyle g^m=g^{qn+r}=\left(g^n\right)^qg^r=g^r=e$ and since $\displaystyle 0\le r<n$ and $\displaystyle n$ is the least positive integer such that $\displaystyle g^k=e$ we must have that $\displaystyle r=0$. Therefore $\displaystyle m=qn\quad q\in\mathbb{Z}$ and the conclusion follows. $\displaystyle \blacksquare$

    So basically we need the least number $\displaystyle k$ such that $\displaystyle n|mk$, i.e. we need $\displaystyle \left|a^m\right|=\text{lcm}(m,n)$
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    Quote Originally Posted by Drexel28 View Post
    We kind of need this lemma

    Lemma: Let $\displaystyle g\in G$ with $\displaystyle |g|=n$. If $\displaystyle g^{m}=e$ then $\displaystyle n|m$.

    Proof: By the division algorithim $\displaystyle m=qn+r\quad q\in\mathbb{Z},0\le r<n$. So then $\displaystyle g^m=g^{qn+r}=\left(g^n\right)^qg^r=g^r=e$ and since $\displaystyle 0\le r<n$ and $\displaystyle n$ is the least positive integer such that $\displaystyle g^k=e$ we must have that $\displaystyle r=0$. Therefore $\displaystyle m=qn\quad q\in\mathbb{Z}$ and the conclusion follows. $\displaystyle \blacksquare$

    So basically we need the least number $\displaystyle k$ such that $\displaystyle n|mk$, i.e. we need $\displaystyle \left|a^m\right|=\text{lcm}(m,n)$
    Hi - Is your final result correct?
    $\displaystyle \left|a^m\right|=\text{lcm}(m,n)$
    I trust $\displaystyle \left|a^m\right|=n/(m,n)$ where (m,n) is gcd(m,n)
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  4. #4
    Newbie lepton's Avatar
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    $\displaystyle \text{lcm}(m,n)=(mn)/\gcd(m,n)$
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    Quote Originally Posted by lepton View Post
    $\displaystyle \text{lcm}(m,n)=(mn)/\gcd(m,n)$
    Right. But what I'm saying is that order is $\displaystyle n/\gcd(m,n)$
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  6. #6
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by aman_cc View Post
    Right. But what I'm saying is that order is $\displaystyle n/\gcd(m,n)$
    Thanks, typo. It was supposed to be $\displaystyle \frac{\text{lcm}(m,n)}{m}=\frac{n}{(m,n)}$
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