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Thread: Finding the order of a group element

  1. #1
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    Finding the order of a group element

    I'm stuck on the following problem: if a and b are in a group G, and a^5=e and b^2=a*b*a^-1, what is the order of b (assuming b is not e). Any ideas where to start? I've tried a bunch of manipulations of the two relationships listed above, but I can't seem to get anywhere.
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  2. #2
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by Jessemath View Post
    I'm stuck on the following problem: if a and b are in a group G, and a^5=e and b^2=a*b*a^-1, what is the order of b (assuming b is not e). Any ideas where to start? I've tried a bunch of manipulations of the two relationships listed above, but I can't seem to get anywhere.
    Assume $\displaystyle b$ has even order, then find a contradiction (this is relatively easy, just re-arrange the formula you are given).

    Next, assume $\displaystyle b$ has odd order, $\displaystyle b^{2i+1}=1$, $\displaystyle b^j \neq 1 \text{ } \forall 1 \leq j < 2i+1$. Then, note that $\displaystyle 1=ab^{2i+1}a^{-1}=ab^{2i}a^{-1}aba^{-1}$ and see what pops out.

    You should get that the order of $\displaystyle b$ can be one of two things.

    I should point out that I didn't use the fact that $\displaystyle a^5=1$, but I can't see a problem in my proof...
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  3. #3
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    Quote Originally Posted by Swlabr View Post
    Assume $\displaystyle b$ has even order, then find a contradiction (this is relatively easy, just re-arrange the formula you are given).

    Next, assume $\displaystyle b$ has odd order, $\displaystyle b^{2i+1}=1$, $\displaystyle b^j \neq 1 \text{ } \forall 1 \leq j < 2i+1$. Then, note that $\displaystyle 1=ab^{2i+1}a^{-1}=ab^{2i}a^{-1}aba^{-1}$ and see what pops out.

    You should get that the order of $\displaystyle b$ can be one of two things.

    I should point out that I didn't use the fact that $\displaystyle a^5=1$, but I can't see a problem in my proof...
    @Swlabr - Hi. I follow your first part that we can't have an even order. But have been struggling with following -


    Next, assume $\displaystyle b$ has odd order, $\displaystyle b^{2i+1}=1$, $\displaystyle b^j \neq 1 \text{ } \forall 1 \leq j < 2i+1$. Then, note that $\displaystyle 1=ab^{2i+1}a^{-1}=ab^{2i}a^{-1}aba^{-1}$ and see what pops out.


    Any further help plz?
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  4. #4
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    A relatively simple to prove (I'm certain you could prove it on your own!) theorem states that:

    If G is a group, $\displaystyle a,b \in G : \ \exists i \in \mathbb{N} \ s.t. \ b^i = aba^{-1}$, then $\displaystyle b^{(i^n)} = a^nba^{-n}$

    Using this fact makes this question relatively simple.

    If you want to see it in a more "algebraic" way:

    $\displaystyle aba^{-1} = b^2$

    $\displaystyle a^2ba^{-2} = a(aba^{-1})a^{-1} = ab^2a^{-1} = (b^2)^2 = b^4$

    $\displaystyle a^3ba^{-3} = a(a^2ba^{-2})a^{-1} = ab^4a^{-1} = (b^2)^4 = b^8$

    $\displaystyle a^4ba^{-4} = a(a^3ba^{-3})a^{-1} = ab^8a^{-1} = (b^2)^8 = b^{16}$

    $\displaystyle a^5ba^{-5} = a(a^4ba^{-4})a^{-1} = ab^{16}a^{-1} = (b^2)^{16} = b^{32}$

    But $\displaystyle a^5=e \Rightarrow b = b^{32} \Rightarrow b^{31} = e$ and since 31 is prime, $\displaystyle o(b) = 31$
    Last edited by Defunkt; Nov 13th 2009 at 07:50 AM. Reason: fixed typo
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  5. #5
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by aman_cc View Post
    @Swlabr - Hi. I follow your first part that we can't have an even order. But have been struggling with following -


    Next, assume $\displaystyle b$ has odd order, $\displaystyle b^{2i+1}=1$, $\displaystyle b^j \neq 1 \text{ } \forall 1 \leq j < 2i+1$. Then, note that $\displaystyle 1=ab^{2i+1}a^{-1}=ab^{2i}a^{-1}aba^{-1}$ and see what pops out.


    Any further help plz?
    Well, my thought was that $\displaystyle 1=ab^{2i+1}a^{-1}=ab^{2i}a^{-1}aba^{-1} \Rightarrow ab^{2i}a^{-1}b^2 \Rightarrow b^{-2} = ab^{2i}a^{-1}$ and that this gave us a contradiction. However, it doesn't (I think I thought the $\displaystyle b^{-2}$ was actually a $\displaystyle b$). oops.
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Defunkt View Post
    A relatively simple to prove (I'm certain you could prove it on your own!) theorem states that:

    If G is a group, $\displaystyle a,b \in G : \ \exists i \in \mathbb{N} \ s.t. \ b^i = a^{-1}ba$, then $\displaystyle b^{(i^n)} = a^nba^{-n}$

    Using this fact makes this question relatively simple.

    If you want to see it in a more "algebraic" way:

    $\displaystyle aba^{-1} = b^2$

    $\displaystyle a^2ba^{-2} = a(aba^{-1})a^{-1} = ab^2a^{-1} = (b^2)^2 = b^4$

    $\displaystyle a^3ba^{-3} = a(a^2ba^{-2})a^{-1} = ab^4a^{-1} = (b^2)^4 = b^8$

    $\displaystyle a^4ba^{-4} = a(a^3ba^{-3})a^{-1} = ab^8a^{-1} = (b^2)^8 = b^{16}$

    $\displaystyle a^5ba^{-5} = a(a^4ba^{-4})a^{-1} = ab^{16}a^{-1} = (b^2)^{16} = b^{32}$

    But $\displaystyle a^5=e \Rightarrow b = b^{32} \Rightarrow b^{31} = e$ and since 31 is prime, $\displaystyle o(b) = 31$
    Nice. .

    @ OP-The only thing that may nee to be said is that the reasong why the text in red above is important is because if $\displaystyle b^n=e$ then $\displaystyle |b|\text{ }|\text{ }n$, but since $\displaystyle 31$ is prime the only positive divisors are $\displaystyle 1,31$ and since $\displaystyle b\ne e$ the only choice is that $\displaystyle |b|=31$
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  7. #7
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    Quote Originally Posted by Defunkt View Post
    A relatively simple to prove (I'm certain you could prove it on your own!) theorem states that:

    If G is a group, $\displaystyle a,b \in G : \ \exists i \in \mathbb{N} \ s.t. \ b^i = a^{-1}ba$, then $\displaystyle b^{(i^n)} = a^nba^{-n}$

    Using this fact makes this question relatively simple.

    If you want to see it in a more "algebraic" way:

    $\displaystyle aba^{-1} = b^2$

    $\displaystyle a^2ba^{-2} = a(aba^{-1})a^{-1} = ab^2a^{-1} = (b^2)^2 = b^4$

    $\displaystyle a^3ba^{-3} = a(a^2ba^{-2})a^{-1} = ab^4a^{-1} = (b^2)^4 = b^8$

    $\displaystyle a^4ba^{-4} = a(a^3ba^{-3})a^{-1} = ab^8a^{-1} = (b^2)^8 = b^{16}$

    $\displaystyle a^5ba^{-5} = a(a^4ba^{-4})a^{-1} = ab^{16}a^{-1} = (b^2)^{16} = b^{32}$

    But $\displaystyle a^5=e \Rightarrow b = b^{32} \Rightarrow b^{31} = e$ and since 31 is prime, $\displaystyle o(b) = 31$

    Thanks Defunkt. Guess there was a small typo error in what you wrote. Following should be the correct statement

    If G is a group, $\displaystyle a,b \in G : \ \exists i \in \mathbb{N} \ s.t. \ b^i = aba^{-1}$, then $\displaystyle b^{(i^n)} = a^nba^{-n}$

    I follow it now. Nice question and an equally nice solution. Thanks again
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  8. #8
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    Quote Originally Posted by aman_cc View Post
    Thanks Defunkt. Guess there was a small typo error in what you wrote. Following should be the correct statement

    If G is a group, $\displaystyle a,b \in G : \ \exists i \in \mathbb{N} \ s.t. \ b^i = aba^{-1}$, then $\displaystyle b^{(i^n)} = a^nba^{-n}$

    I follow it now. Nice question and an equally nice solution. Thanks again
    Thanks, and yes you're right of course, that was a typo :P I'll fix the original post now.
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