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By Magnus-Moldavansky theorem (see Wilhelm Magnus, Abraham Karrass, Donald Solitar, Combinatorial group theory. Presentations of groups in terms of generators and relations, Reprint of the 1976 second edition, Dover Publications, Inc., Mineola, NY, 2004 and http://arxiv.org/PS_cache/math/pdf/0608/0608635v3.pdf), every 1-related group is an HNN extensions of a "smaller" 1-related group. Say, if one letter occurs in the relator with total exponent 0, one can take this letter as the free letter in the HNN extension.

Question. Is it true that in this HNN extension one of the associated subgroups is always undistorted? Is it possible that one can always represent the group as an HNN extension satisfying that property (one of the associated subgroups is undistorted)?

For example, in some sense the "worst" 1-related group is the Baumslag group $$\langle a,t\mid a^{a^t}=a^2\rangle=\langle a,b,t\mid a^b=a^2, a^t=b\rangle.$$ In that case $t$ is a free letter, the base group is $\langle a,b \mid a^b=a^2\rangle$, the associated subgroups are $\langle a\rangle$ (exponentially distorted in the base group) and $\langle b \rangle$ (undistorted in the base group).

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Can you define what you mean by "associated subgroup"? –  Ian Agol Sep 6 '11 at 0:11
    
@Ian: If $G$ is an HNN extension of $H$ and $t$ is the free letter conjugating subgroup $A<H$ to subgroup $B<H$ (so that $G=\langle H, t\mid A^t=B\rangle$, then $A,B$ are called the associated subgroups of the HNN extension. –  Mark Sapir Sep 6 '11 at 1:21
    
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1 Answer

up vote 6 down vote accepted

You can reverse engineer distortion. Consider the group $\langle a_0, a_1, a_2 |a_1^{a_2a_0}= a_1^2\rangle$. By the Freiheitssatz, the subgroups generated by $\langle a_0, a_1\rangle$ and $\langle a_1, a_2\rangle$ are free. But both are distorted, since $a_1$ is conjugate to $a_1^2$ (if I understand the notion of distortion correctly).

Now, take the HNN extension of these two free subgroups to get a 1-relator group $$\langle a_0, a_1, a_2, t | a_1^{a_2a_0}= a_1^2, a_0^t = a_1, a_1^t=a_2\rangle=\langle a, t | (a^t)^{a^{t^2}a} = (a^t)^2\rangle.$$

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@Ian: Thank you! –  Mark Sapir Sep 6 '11 at 2:18
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