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Thread: Abstract Algebra: Orders of Groups

  1. #1
    is up to his old tricks again! Jhevon's Avatar
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    Abstract Algebra: Orders of Groups

    Hello All,

    I'm stuck on part (b)

    Problem:

    (a) Prove that if $\displaystyle a$ and $\displaystyle b$ are elements of an Abelian group $\displaystyle G$, with $\displaystyle \circ (a) = m$ and $\displaystyle \circ (b) = n$, then $\displaystyle (ab)^{mn} = e$ ($\displaystyle e$ denotes the identity element in my text). Indicate where you use the condition that $\displaystyle G$ is Abelian.

    (b) With $\displaystyle G,~a$ and $\displaystyle b$ as in part (a), prove that $\displaystyle \circ (ab)$ divides $\displaystyle \circ (a) \circ (b)$.

    (c) Give an example of an Abelian group $\displaystyle G$ and elements $\displaystyle a$ and $\displaystyle b$ in $\displaystyle G$ such that $\displaystyle \circ (ab) \ne \circ (a) \circ (b)$. Compare part (b)



    Things that might come in Handy:

    Let $\displaystyle G$ be a group and $\displaystyle a \in G$.

    We define the order of $\displaystyle a$, denoted $\displaystyle \circ (a)$, to be the smallest, positive integer $\displaystyle n$ such that $\displaystyle a^n = e$. If no such integer exists, we say $\displaystyle a$ has infinite order.

    $\displaystyle a^n$ of course means $\displaystyle a$ operating on itself $\displaystyle n$ times. that is, $\displaystyle a^n = \underbrace{aaaa \cdots a}_{n \mbox{ times}} = a*a*a*a* \cdots *a$

    Tell me if you guys need anymore information.




    What I have tried:

    (a) Proof: Let $\displaystyle a,~b,~G$ be as in part (a). Since $\displaystyle G$ is Abelian, $\displaystyle (ab)^{mn} = a^{mn}b^{mn}$

    Now, $\displaystyle a^{mn}b^{mn} = a^{mn}b^{nm} = (a^m)^n(b^n)^m = e^ne^n = ee = e$

    QED


    (b) This is where I'm stuck. Here's what I did so far.

    Let $\displaystyle a,~b,~G$ be as in part (a). And let $\displaystyle \circ (ab) = p$.

    Then $\displaystyle m,~n, \mbox{ and }p$ are all the smallest, positive integers such that: $\displaystyle a^m = e$, $\displaystyle b^n = e$ and $\displaystyle (ab)^p = a^pb^p = e$. We want to show that $\displaystyle p|mn$, that is, $\displaystyle mn = kp$ for some $\displaystyle k \in \mathbb{Z}$.

    Now, $\displaystyle a^mb^n = ee = e = a^pb^p$

    ...

    that's it, I'm not sure where to go from there. I've tried several things, but they all end up doing nothing.


    (c) My example was $\displaystyle (\mathbb{Z}_3, \oplus)$ with $\displaystyle a = [1]$ and $\displaystyle b = [2]$


    Thanks guys and gals
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  2. #2
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    Quote Originally Posted by Jhevon View Post
    (b) With $\displaystyle G,~a$ and $\displaystyle b$ as in part (a), prove that $\displaystyle \circ (ab)$ divides $\displaystyle \circ (a) \circ (b)$.
    Just use a theorem. Suppose $\displaystyle x$ is a group (any group not just abelian) $\displaystyle G$ and $\displaystyle x$ has finite order $\displaystyle k$. If $\displaystyle n$ is any positive integer so that $\displaystyle x^n = e$ then $\displaystyle k|n$. Now apply this theorem here.

    (c) Give an example of an Abelian group $\displaystyle G$ and elements $\displaystyle a$ and $\displaystyle b$ in $\displaystyle G$ such that $\displaystyle \circ (ab) \ne \circ (a) \circ (b)$. Compare part (b)
    Consider $\displaystyle G=\mathbb{Z}_5$ and $\displaystyle a=[2],b=[3]$.
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  3. #3
    Senior Member JaneBennet's Avatar
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    (b)
    If $\displaystyle \mathrm{o}(x)=p$ and $\displaystyle x^k=e$, write $\displaystyle k=qp+r$ where $\displaystyle 0\leq r<p$. Hence $\displaystyle e=x^k=(x^p)^qx^r=x^r$ and it follows that $\displaystyle r=0$ since p is the smallest positive integer such that x to that power is the identity.

    (c)
    Take a = b = any element of order 2. Then $\displaystyle \mathrm{o}(a\circ b)=1$ but $\displaystyle \mathrm{o}(a)\cdot\mathrm{o}(b)=4$.
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