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Math Help - question on order of element

  1. #1
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    question on order of element

    G be a group. [tex]a,b /in G[\math]
    Order of a = m
    Order of b = n

    Prove there exists an element in G with order = LCM (m,n). Let this LCM be l

    My approach:
    Consider ab
    Let order of ab be o
    Obviously o|l (as (ab)^l=1)
    Only two cases are possible
    Case 1: m|o AND n|o - we are done here
    Case 2: m doesn't divide o AND n doesn't divide o
    In this case we have a^x = b^y where 0<x<m and 0<y<n
    Stuck here for quite some time.

    Is my approach correct? If yes, hint on way fwd please?
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  2. #2
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    Quote Originally Posted by aman_cc View Post
    G be a group. [tex]a,b /in G[\math]
    Order of a = m
    Order of b = n

    Prove there exists an element in G with order = LCM (m,n). Let this LCM be l

    My approach:
    Consider ab
    Let order of ab be o
    Obviously o|l (as (ab)^l=1)
    Only two cases are possible
    Case 1: m|o AND n|o - we are done here
    Case 2: m doesn't divide o AND n doesn't divide o
    In this case we have a^x = b^y where 0<x<m and 0<y<n
    Stuck here for quite some time.

    Is my approach correct? If yes, hint on way fwd please?
    this is true in abelian groups only!
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  3. #3
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    Quote Originally Posted by NonCommAlg View Post
    this is true in abelian groups only!
    Apologies - G is abelian as well.
    Any hints? Am I on the right track?
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  4. #4
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    Quote Originally Posted by aman_cc View Post
    Apologies - G is abelian as well.
    Any hints? Am I on the right track?
    first prove it for the case \gcd(m,n)=1. (very easy!) can you do that?
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  5. #5
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    Quote Originally Posted by NonCommAlg View Post
    first prove it for the case \gcd(m,n)=1. (very easy!) can you do that?
    Yes I can do that. In that case required element is ab
    If order is o
    We have m|on and n|om
    Under the \gcd(m,n)=1 condition I get
    m|o and n|o and we are done (as per case 1 in my logic above). Correct?
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  6. #6
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    Quote Originally Posted by aman_cc View Post
    Yes I can do that. In that case required element is ab
    If order is o
    We have m|on and n|om
    Under the \gcd(m,n)=1 condition I get
    m|o and n|o and we are done (as per case 1 in my logic above). Correct?
    [so mn \mid o(ab) and we also have (ab)^{mn}=a^{mn}b^{mn}=1 and so o(ab) \mid mn. thus o(ab)=mn=\text{lcm}(a,b).] now we can prove the claim for the general case: clearly we may assume that m \nmid n.

    let \text{lcm}(m,n)=\prod_{j=1}^r p_j^{k_j} be the prime factorization of \text{lcm}(m,n). let A=\{j: \ p_j^{k_j} \mid m \} \neq \emptyset, because m \nmid n. let s=\prod_{j \in A} p_j^{k_j} and t=\frac{\text{lcm}(m,n)}{s}. obviously s \mid m, \ t \mid n, and \gcd(s,t)=1. therefore,

    since o(a^{m/s})=s and o(b^{n/t})=t, we have o(a^{m/s}b^{n/t})=st=\text{lcm}(m,n).
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  7. #7
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    Quote Originally Posted by NonCommAlg View Post
    [so mn \mid o(ab) and we also have (ab)^{mn}=a^{mn}b^{mn}=1 and so o(ab) \mid mn. thus o(ab)=mn=\text{lcm}(a,b).] now we can prove the claim for the general case: clearly we may assume that m \nmid n.

    let \text{lcm}(m,n)=\prod_{j=1}^r p_j^{k_j} be the prime factorization of \text{lcm}(m,n). let A=\{j: \ p_j^{k_j} \mid m \} \neq \emptyset, because m \nmid n. let s=\prod_{j \in A} p_j^{k_j} and t=\frac{\text{lcm}(m,n)}{s}. obviously s \mid m, \ t \mid n, and \gcd(s,t)=1. therefore,

    since o(a^{m/s})=s and o(b^{n/t})=t, we have o(a^{m/s}b^{n/t})=st=\text{lcm}(m,n).
    Thanks I get this. In the case where gcd(m,n) <> 1 we found a element with order equal to LCM. Can we comment on o(ab) as well? (In terms of prime factorization you did)


    All I could show was

    \frac{mn}{\gcd(m,n)^2}|o(ab)

    Is the above correct? Can we get more specific?

    Thanks
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