I've read that every finitely generated one relator group embeds in a two generator one relator group, and that this fact follows from the Freiheitssatz. 


Unfortunately, the only proof I can find of this fact applies [B.H. Neumann's proof](http://www.jstor.org/stable/91573) for denumerable n-relator groups, and it doesn't seem to use the Freiheitssatz. Further, I haven't found any mention of this in Lyndon and Schupp, but it's possible I overlooked a more general theorem from which this follows.


My question is: does this fact truly follow from the Freiheitssatz? Is the proof trivial? I apologize if it is; unfortunately I am new to one relator groups.



Freiheitssatz implies a finitely generated one relator group embeds in a two-generator one relator group?
Yes, this follows from the Freiheitssatz. Assume that the 1-relator group is defined by $G=\langle g\_1,\ldots, g\_n | R(g\_1,\ldots,g\_n)\rangle$, such that the relator $R(g\_1,\ldots, g\_n)$ is cyclically reduced, and involves the generators $g\_1$ and $g\_n$ non-trivially. By the [Freiheitssatz](http://en.wikipedia.org/wiki/Freiheitssatz), the subgroups $\langle g\_1,\ldots, g\_{n-1}\rangle$ and $\langle g\_2,\ldots g\_n\rangle$ are free groups of rank $n-1$ freely generated by these elements. Then embed $G$ in an HNN extension $\langle g\_1,\ldots, g\_n, t | R(g\_1,\ldots,g\_n), tg\_it^{-1} = g\_{i+1}, i=1,\ldots ,n-1 \rangle = \langle g\_1,t | R(g\_1, tg\_1t^{-1},\ldots, t^{n-1}g\_1t^{1-n}\rangle$ by eliminating generators and relators. 


By permuting the labels, one may guarantee that the relator involves $g\_1, g\_n$ unless the relator involves only one generator. In that case, if the relator is of the form $g\_1^k$, then do a Nielsen transformation $h\_1=g\_1g\_n^{-1},h\_2=g\_2,\ldots, h\_n=g\_n$. The group with this set of generators has presentation $\langle h\_1,\ldots, h\_n | (h\_1h\_n)^k=1\rangle$. One may apply the previous construction to this presentation. 


