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  1. #1
    Senior Member Sampras's Avatar
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    element

    Suppose  a,b \in G and  ab = ba . If  |a| and  |b| are relatively prime, prove that  ab has order  |a||b| .

    So suppose  a^m = e and  b^n = e . Then  \gcd(m,n) = 1 . Thus  1 = \frac{\text{lcm}(m,n)}{mn} or  mn = \text{lcm}(m,n) . So  (ab)^{mn} = (a^{m})^{n} \cdot (b^{n})^{m} = e^{n}e^{m} = e . Thus the order of  ab is  |a||b| .

    Is this correct?
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Sampras View Post
    Suppose  a,b \in G and  ab = ba . If  |a| and  |b| are relatively prime, prove that  ab has order  |a||b| .

    So suppose  a^m = e and  b^n = e . Then  \gcd(m,n) = 1 . Thus  1 = \frac{\text{lcm}(m,n)}{mn} or  mn = \text{lcm}(m,n) . So  (ab)^{mn} = (a^{m})^{n} \cdot (b^{n})^{m} = e^{n}e^{m} = e . Thus the order of  ab is  |a||b| .

    Is this correct?
    I suppose that is fine. Are you allowed to use that result? It makes the proof kind of trivial.
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  3. #3
    Senior Member Sampras's Avatar
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    Quote Originally Posted by Drexel28 View Post
    I suppose that is fine. Are you allowed to use that result? It makes the proof kind of trivial.
    What would you suggest?
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  4. #4
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    Quote Originally Posted by Sampras View Post
    Suppose  a,b \in G and  ab = ba . If  |a| and  |b| are relatively prime, prove that  ab has order  |a||b| .

    So suppose  a^m = e and  b^n = e . Then  \gcd(m,n) = 1 . Thus  1 = \frac{\text{lcm}(m,n)}{mn} or  mn = \text{lcm}(m,n) . So  (ab)^{mn} = (a^{m})^{n} \cdot (b^{n})^{m} = e^{n}e^{m} = e . Thus the order of  ab is  |a||b| .

    Is this correct?

    I'm afraid it's not: you only proved that (ab)^{mn}=e which is pretty trivial. The interesting part is in showing that there is not power of ab less than mn which equals the unit...

    Tonio
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  5. #5
    Senior Member Sampras's Avatar
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    Quote Originally Posted by tonio View Post
    I'm afraid it's not: you only proved that (ab)^{mn}=e which is pretty trivial. The interesting part is in showing that there is not power of ab less than mn which equals the unit...

    Tonio
    You have to use fact that ab=ba, because this isn't true if this condition is not met? So  (ab)^{mn} = (ba)^{mn} . Suppose  (ab)^{mn- \varepsilon} = e = (ba)^{mn- \varepsilon} . Then  \frac{(ab)^{mn}}{(ab)^{\varepsilon}} = e  = \frac{(ba)^{mn}}{(ba)^{\varepsilon}} for some  \varepsilon \in \mathbb{Z} .


    Or  \frac{a^{mn}b^{mn}}{a^{\varepsilon}b^{\varepsilon}  } = e . Or  \frac{e}{a^{\varepsilon} b^{\varepsilon}} = e so that  (ab)^{\varepsilon} = e . Contradiction?
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  6. #6
    MHF Contributor Drexel28's Avatar
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    NOTE: I am actually not sure if this is correct. Let another member validate this. I feel as though I may have made a stupid incorrect assumption.

    Problem: Let (m,n)=1. Prove that if |a|=m,|b|=n that |ab|=mn.

    Proof: Clearly \left(ab\right)^{mn}=e. Now suppose that |ab|=\omega<mn then (ab)^{\omega}=e\implies a^{\omega}=b^{-\omega} and b^{\omega}=a^{-\omega}. Thus \left(b^{-\omega}\right)^m=\left(a^{\omega}\right)^m=e\impli  es a^{\omega\cdot m-\omega\cdot n}=e. Similarly b^{\omega\cdot n-\omega\cdot m}=e. And since |a|=m we see that m|\omega\cdot m-\omega\cdot n\implies m|\omega n and since (m,n)=1 this is only true if m|\omega. Similarly n|\omega\cdot n-\omega\cdot m and since |b|=n this means n|\omega\cdot n-\omega\cdot m\implies n|\omega\cdot m\ and by previous reasoning n|\omega. Therefore m,n|\omega which means that \text{lcm}(m,n)|\omega but since (m,n)=1 we have that \text{lcm}(m,n)=mn and mn|\omega which is a contradiction since 0<\omega<mn.
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  7. #7
    Senior Member Sampras's Avatar
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    Quote Originally Posted by Drexel28 View Post
    NOTE: I am actually not sure if this is correct. Let another member validate this. I feel as though I may have made a stupid incorrect assumption.

    Problem: Let (m,n)=1. Prove that if |a|=m,|b|=n that |ab|=mn.

    Proof: Clearly \left(ab\right)^{mn}=e. Now suppose that |ab|=\omega<mn then (ab)^{\omega}=e\implies a^{\omega}=b^{-\omega} and b^{\omega}=a^{-\omega}. Thus \left(b^{-\omega}\right)^m=\left(a^{\omega}\right)^m=e\impli  es a^{\omega\cdot m-\omega\cdot n}=e. Similarly b^{\omega\cdot n-\omega\cdot m}=e. And since |a|=m we see that m|\omega\cdot m-\omega\cdot n\implies m|\omega n and since (m,n)=1 this is only true if m|\omega. Similarly n|\omega\cdot n-\omega\cdot m and since |b|=n this means n|\omega\cdot n-\omega\cdot m\implies n|\omega\cdot m\ and by previous reasoning n|\omega. Therefore m,n|\omega which means that \text{lcm}(m,n)|\omega but since (m,n)=1 we have that \text{lcm}(m,n)=mn and mn|\omega which is a contradiction since 0<\omega<mn.
    Suppose  a^{r}b^{r} = e for some  r \in \mathbb{Z}^{+} . Then  b^{nr-r}b^{r} = b^{nr} = e . Let  t = nr-r . Then  a^{r} = b^t . So  (a^r)^n = (b^t)^n so that  a^{rn} = e . So  m divides  rn but  (m,n) = 1 . Thus  m|r so that  m divides  |a|b| . Similarly,  n divides  |a|b| .
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  8. #8
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Sampras View Post
    Suppose  a^{r}b^{r} = e for some  r \in \mathbb{Z}^{+} . Then  b^{nr-r}b^{r} = b^{nr} = e . Let  t = nr-r . Then  a^{r} = b^t . So  (a^r)^n = (b^t)^n so that  a^{rn} = e . So  m divides  rn but  (m,n) = 1 . Thus  m|r so that  m|ab . Similarly  n|ab .
    So wait haha. Are you agreeing or disagreeing with me?
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  9. #9
    Senior Member Sampras's Avatar
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    Quote Originally Posted by Drexel28 View Post
    So wait haha. Are you agreeing or disagreeing with me?
    It's essentially the same proof. Except just choosing an arbitrary r to begin with.
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  10. #10
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Sampras View Post
    It's essentially the same proof. Except just choosing an arbitrary r to begin with.
    Ok. Good. Hope that helped.
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