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Math Help - HNN extension

  1. #1
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    HNN extension

    I have a question.

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

    Is it true that \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 G=\langle b,t | t^{-1}b^{\beta}tb^{\beta} \rangle be a HNN extension of base group \langle b \rangle with stable letter t.

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

    I think is it true.

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

    As t is a "foreign" letter to the group <b> then clearly <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 G=\langle b,t | t^{-1}b^{\beta}tb^{\beta} \rangle be a HNN extension of base group \langle b \rangle with stable letter t.

    Is it true that \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 G^{\ast} = <G, t: A^t = A\phi> then G = <G> and clearly 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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