Free groups as HNN-extensions

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Let $G$ be the free group with two generators, say $a$ and $b$.
Then the Cayley of $G$ with respect with the generating set is 4 regular tree.
If we contract every edge with the label $b$, then we end up with a tree $T$ such that $G$ acts on the edges of $T$ transitively.



Applying Bass-Serre theory implies that $G$ is the free product $langle arangle $ and $langle brangle $.



My question is the following:



How can I modify the Cayley graph of $G$ in order to get an HNN-exentension?






geometric-group-theory







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    up vote
    2
    down vote

    favorite












    Let $G$ be the free group with two generators, say $a$ and $b$.
    Then the Cayley of $G$ with respect with the generating set is 4 regular tree.
    If we contract every edge with the label $b$, then we end up with a tree $T$ such that $G$ acts on the edges of $T$ transitively.



    Applying Bass-Serre theory implies that $G$ is the free product $langle arangle $ and $langle brangle $.



    My question is the following:



    How can I modify the Cayley graph of $G$ in order to get an HNN-exentension?






    geometric-group-theory







    share|cite|improve this question

























      up vote
      2
      down vote

      favorite









      up vote
      2
      down vote

      favorite











      Let $G$ be the free group with two generators, say $a$ and $b$.
      Then the Cayley of $G$ with respect with the generating set is 4 regular tree.
      If we contract every edge with the label $b$, then we end up with a tree $T$ such that $G$ acts on the edges of $T$ transitively.



      Applying Bass-Serre theory implies that $G$ is the free product $langle arangle $ and $langle brangle $.



      My question is the following:



      How can I modify the Cayley graph of $G$ in order to get an HNN-exentension?






      geometric-group-theory







      share|cite|improve this question















      Let $G$ be the free group with two generators, say $a$ and $b$.
      Then the Cayley of $G$ with respect with the generating set is 4 regular tree.
      If we contract every edge with the label $b$, then we end up with a tree $T$ such that $G$ acts on the edges of $T$ transitively.



      Applying Bass-Serre theory implies that $G$ is the free product $langle arangle $ and $langle brangle $.



      My question is the following:



      How can I modify the Cayley graph of $G$ in order to get an HNN-exentension?






      geometric-group-theory




      geometric-group-theory






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      share|cite|improve this question













      share|cite|improve this question




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      edited Sep 4 at 14:53

























      asked Sep 4 at 12:04









      Jivid

      1817




      1817




















          1 Answer
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          Actually, the tree that you have described is already an HNN extension tree, because its quotient graph of groups is a graph with one vertex and one edge. The vertex is labelled with the cyclic group $langle b rangle$, and the edge is labelled with the trivial group $E$.



          The group element $a$ is the stable letter of that quotient edge, leading to the HNN presentation
          $$F_2 = langle a,b mid aea^-1=e rangle
          $$
          where $e$ is the identity element of $langle b rangle$.



          Although there are other seemingly less trivial HNN extensions of $F_2$ that can be described, and although this may seem like a somewhat trivial example of an HNN extension, it is nonetheless an example.






          share|cite|improve this answer




















          • Thanks, another question can be raised here: Can every infinite group split over $mathbb Z$? My observation: Every manifolds contains a closed curve.
            – Jivid
            Sep 6 at 13:31











          • That's a nice question. However, instead of burying that question in a comment of an answer of your last question ---- where it will be essentially invisible to the community --- you should instead ask it in a new question where the whole community will see it.
            – Lee Mosher
            Sep 7 at 16:14










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          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes








          up vote
          2
          down vote



          accepted










          Actually, the tree that you have described is already an HNN extension tree, because its quotient graph of groups is a graph with one vertex and one edge. The vertex is labelled with the cyclic group $langle b rangle$, and the edge is labelled with the trivial group $E$.



          The group element $a$ is the stable letter of that quotient edge, leading to the HNN presentation
          $$F_2 = langle a,b mid aea^-1=e rangle
          $$
          where $e$ is the identity element of $langle b rangle$.



          Although there are other seemingly less trivial HNN extensions of $F_2$ that can be described, and although this may seem like a somewhat trivial example of an HNN extension, it is nonetheless an example.






          share|cite|improve this answer




















          • Thanks, another question can be raised here: Can every infinite group split over $mathbb Z$? My observation: Every manifolds contains a closed curve.
            – Jivid
            Sep 6 at 13:31











          • That's a nice question. However, instead of burying that question in a comment of an answer of your last question ---- where it will be essentially invisible to the community --- you should instead ask it in a new question where the whole community will see it.
            – Lee Mosher
            Sep 7 at 16:14














          up vote
          2
          down vote



          accepted










          Actually, the tree that you have described is already an HNN extension tree, because its quotient graph of groups is a graph with one vertex and one edge. The vertex is labelled with the cyclic group $langle b rangle$, and the edge is labelled with the trivial group $E$.



          The group element $a$ is the stable letter of that quotient edge, leading to the HNN presentation
          $$F_2 = langle a,b mid aea^-1=e rangle
          $$
          where $e$ is the identity element of $langle b rangle$.



          Although there are other seemingly less trivial HNN extensions of $F_2$ that can be described, and although this may seem like a somewhat trivial example of an HNN extension, it is nonetheless an example.






          share|cite|improve this answer




















          • Thanks, another question can be raised here: Can every infinite group split over $mathbb Z$? My observation: Every manifolds contains a closed curve.
            – Jivid
            Sep 6 at 13:31











          • That's a nice question. However, instead of burying that question in a comment of an answer of your last question ---- where it will be essentially invisible to the community --- you should instead ask it in a new question where the whole community will see it.
            – Lee Mosher
            Sep 7 at 16:14












          up vote
          2
          down vote



          accepted







          up vote
          2
          down vote



          accepted






          Actually, the tree that you have described is already an HNN extension tree, because its quotient graph of groups is a graph with one vertex and one edge. The vertex is labelled with the cyclic group $langle b rangle$, and the edge is labelled with the trivial group $E$.



          The group element $a$ is the stable letter of that quotient edge, leading to the HNN presentation
          $$F_2 = langle a,b mid aea^-1=e rangle
          $$
          where $e$ is the identity element of $langle b rangle$.



          Although there are other seemingly less trivial HNN extensions of $F_2$ that can be described, and although this may seem like a somewhat trivial example of an HNN extension, it is nonetheless an example.






          share|cite|improve this answer












          Actually, the tree that you have described is already an HNN extension tree, because its quotient graph of groups is a graph with one vertex and one edge. The vertex is labelled with the cyclic group $langle b rangle$, and the edge is labelled with the trivial group $E$.



          The group element $a$ is the stable letter of that quotient edge, leading to the HNN presentation
          $$F_2 = langle a,b mid aea^-1=e rangle
          $$
          where $e$ is the identity element of $langle b rangle$.



          Although there are other seemingly less trivial HNN extensions of $F_2$ that can be described, and although this may seem like a somewhat trivial example of an HNN extension, it is nonetheless an example.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Sep 5 at 12:58









          Lee Mosher

          46.1k33579




          46.1k33579











          • Thanks, another question can be raised here: Can every infinite group split over $mathbb Z$? My observation: Every manifolds contains a closed curve.
            – Jivid
            Sep 6 at 13:31











          • That's a nice question. However, instead of burying that question in a comment of an answer of your last question ---- where it will be essentially invisible to the community --- you should instead ask it in a new question where the whole community will see it.
            – Lee Mosher
            Sep 7 at 16:14
















          • Thanks, another question can be raised here: Can every infinite group split over $mathbb Z$? My observation: Every manifolds contains a closed curve.
            – Jivid
            Sep 6 at 13:31











          • That's a nice question. However, instead of burying that question in a comment of an answer of your last question ---- where it will be essentially invisible to the community --- you should instead ask it in a new question where the whole community will see it.
            – Lee Mosher
            Sep 7 at 16:14















          Thanks, another question can be raised here: Can every infinite group split over $mathbb Z$? My observation: Every manifolds contains a closed curve.
          – Jivid
          Sep 6 at 13:31





          Thanks, another question can be raised here: Can every infinite group split over $mathbb Z$? My observation: Every manifolds contains a closed curve.
          – Jivid
          Sep 6 at 13:31













          That's a nice question. However, instead of burying that question in a comment of an answer of your last question ---- where it will be essentially invisible to the community --- you should instead ask it in a new question where the whole community will see it.
          – Lee Mosher
          Sep 7 at 16:14




          That's a nice question. However, instead of burying that question in a comment of an answer of your last question ---- where it will be essentially invisible to the community --- you should instead ask it in a new question where the whole community will see it.
          – Lee Mosher
          Sep 7 at 16:14

















           

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