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Thread: isomorphism problem

  1. #1
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    isomorphism problem

    Show that <a, b|a^5=b^2=e, ba=(a^2)b> is isomorphic to Z2.
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  2. #2
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    Quote Originally Posted by aabsdr View Post
    Show that <a, b|a^5=b^2=e, ba=(a^2)b> is isomorphic to Z2.
    Let the game begin... :

    $\displaystyle bab^{-1}=a^2\Longrightarrow ba^2b^{-1}=\left(bab^{-1}\right)^2=a^4$ , but also $\displaystyle ba^2b^{-1}=b\left(bab^{-1}\right)b^{-1}=b^2ab^{-2}=a$ , so we get $\displaystyle a=a^4\Longrightarrow a^3=1$ , which together with the given data $\displaystyle a^5=1$

    means that $\displaystyle a=1$ ,and thus in fact $\displaystyle \left<a,b\;/\;a^5=b^2=1\,,\,ba=a^2b\right>=\left<b\;/\;b^2=1\right>\cong\mathbb{Z}_2$

    Tonio
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  3. #3
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    b(a^2)b^-1= (bab^-1)^2 due to that bab^-1 is a conjugate, and this is the same for ba^2b^-1= b(bab^-1)b^-1 is because of the definition of conjugate as well, right?
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  4. #4
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by aabsdr View Post
    b(a^2)b^-1= (bab^-1)^2 due to that bab^-1 is a conjugate, and this is the same for ba^2b^-1= b(bab^-1)b^-1 is because of the definition of conjugate as well, right?
    The first part of your statement/question is right, but the second part is not true `because of the definition of conjugation'. Tonio has substituted in $\displaystyle a^2 = bab^{-1}$ which can be done as $\displaystyle ba=a^2b \Rightarrow bab^{-1} = a^2$.

    How about...$\displaystyle baba=a^2a=a^3$ (as $\displaystyle bab = a^2$ by $\displaystyle ba=a^2b$). This gives us that $\displaystyle bab=1$ by cancelling the $\displaystyle a$s and so we have that $\displaystyle <a, b; a^5 = b^2=1, ba=a^2b> = <b; b^2>$ as required.
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