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Thread: HNN extension

  1. #1
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    HNN extension

    I have a question.

    Let $\displaystyle G=\langle b,t | t^{-1}b^{\beta}tb^{\beta} \rangle$ be a HNN extension of base group $\displaystyle \langle b \rangle$ with stable letter $\displaystyle t$.

    Is it true that $\displaystyle \langle b \rangle \cap \langle t \rangle=1$?

    I think is it true.
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  2. #2
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    Quote Originally Posted by deniselim17 View Post
    I have a question.

    Let $\displaystyle G=\langle b,t | t^{-1}b^{\beta}tb^{\beta} \rangle$ be a HNN extension of base group $\displaystyle \langle b \rangle$ with stable letter $\displaystyle t$.

    Is it true that $\displaystyle \langle b \rangle \cap \langle t \rangle=1$?

    I think is it true.

    As an HNN extension , we have that in fact $\displaystyle G=\left<<b>,t\;\slash\;t^{-1}b^kt=\phi(b^k)\right>$ , for some isomorphism $\displaystyle \phi:<b>\rightarrow <b>$ .
    Since you give $\displaystyle G=\langle b,t | t^{-1}b^{\beta}tb^{\beta} \rangle$ , this means you chose $\displaystyle \phi(b)=b^{-1}$ = the inverse involution automorphism of $\displaystyle <b>$ , so your group is the amalgamated product (as any other HNN extension) of $\displaystyle <b>$ with itself via the above involution.

    As t is a "foreign" letter to the group $\displaystyle <b>$ then clearly $\displaystyle <b>\cap <t>=1$

    Tonio
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  3. #3
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by deniselim17 View Post
    I have a question.

    Let $\displaystyle G=\langle b,t | t^{-1}b^{\beta}tb^{\beta} \rangle$ be a HNN extension of base group $\displaystyle \langle b \rangle$ with stable letter $\displaystyle t$.

    Is it true that $\displaystyle \langle b \rangle \cap \langle t \rangle=1$?

    I think is it true.
    HNN-extensions were constructed to show that we can embed a group in another group in a special way (if we are given two isomorphic subgroups of a group then the group can be embedded in a bigger group such that these groups are conjugate. I can't remember who asked this question - anyone know? It's been bugging me since I started typing this!) So, without proving anything you know that if $\displaystyle G^{\ast} = <G, t: A^t = A\phi>$ then $\displaystyle G = <G>$ and clearly $\displaystyle t \notin G$.

    My point is that the result is clear and is the whole point of HNN-extensions! Although it does still require proof...
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