# HNN extension

• February 23rd 2010, 03:16 AM
deniselim17
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.
• February 23rd 2010, 03:57 AM
tonio
Quote:

Originally Posted by deniselim17
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 $G=\left<,t\;\slash\;t^{-1}b^kt=\phi(b^k)\right>$ , for some isomorphism $\phi:\rightarrow $ .
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 $$ , so your group is the amalgamated product (as any other HNN extension) of $$ with itself via the above involution.

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

Tonio
• February 23rd 2010, 06:46 AM
Swlabr
Quote:

Originally Posted by deniselim17
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} = $ then $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...