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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 $g_1'=g_1g_n, g_2'=g_2,\ldotsh_1=g_1g_n^{-1},h_2=g_2,\ldots, h_n=g_n$. The group with this set of generators has presentation $\langle h_1,\ldots, g_n'=g_n$h_n | (h_1h_n)^k=1\rangle$. One may apply the previous construction to this presentation. 1 Yes, this follows from the Freiheitssatz. Assume that the 1-relator group is defined by$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$, then do a Nielsen transformation$g_1'=g_1g_n, g_2'=g_2,\ldots, g_n'=g_n\$.