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). 



Magnus hierarchy of 1-related groups 
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](http://en.wikipedia.org/wiki/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
$$\langle a\_0, a\_1, a\_2, t | a\_1^{a\_2a\_0}= a\_1^2, a\_0^t = a\_1, a\_1^t=a\_2\rangle=\langle a, t | (a^t)^{a^{t^2}a} = (a^t)^2\rangle.$$


