TITLE: Freiheitssatz implies a finitely generated one relator group embeds in a two-generator one relator group?
QUESTION [10 upvotes]: 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 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.

REPLY [12 votes]: 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, 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.