I am not a mathematics researcher but I am concerned that this question, posed with slightly different wording on math.stackexchange, may be too esoteric for that forum since it concerns the details of a proof that I don't think is standard in the graduate level curriculum. At least, nobody has answered it so far.

Let $G$ be a group with presentation $<x_1,x_2...,x_m|R>$ and let $G'=G \ast<y_0>$.

Now define $y_i=y_{0}x_i$. Notice that $G'=<y_0,y_1,...y_m|R'>$ for appropriately defined $R'$ and that $<y_i>$ is free on $y_i$ for all $i$.

Consider the group

$$H = <y_0,y_1,...y_m,t_1,t_2,...t_m,|R',t_iy_it^{-1}_i=y_{i}^2>$$

Either take my word for it that $G'$ embeds (isomorphically) into $H$, or notice that, since $<y_i>$ is free on $y_i$ for all $i$, $H$ is an iterated HNN extension on $G'$ and therefore embeds $G'$ by $m$ applications of Britton's Lemma.

Now let $w$ be a nonidentity element in G. Define $H^{'}$ as $H$ subject to the relation $(w^{-1}y_{0}^{-1}wy_0)t_i(w^{-1}y_{0}^{-1}wy_0)^{-1} = t_{i}^2$ for each $t_i$.

I can show that $w^{-1}y_{0}^{-1}wy_0$ is infinite order since it is cyclically reduced word not containing the identity and length $\geq 2$. $t_i$ is also infinite order in $H$ since it is the pivot letter in an HNN extension. So it would seem that I'm not introducing any torsion for the elements $y_0,t_i$.

I also don't think I'm introducing any torsion for elements of $G$ since it does not reference them at all.

In particular it seems like $G'$ embeds into $H'$. I have no idea how to prove this, but I have done pages of calculations in this group to try and disprove this.

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Is this true? Here's why I think no, in spite of my best efforts to (dis)prove it:

In the proof of the Adian-Rabin theorem, $G$ is the free product of some group with unsolvable word problem $U$ with a group $G_{-}$, which has the property that any group it embeds into does not satisfy some property $M$. (A $G_{-}$ with such a property is given by hypothesis).

The proof constructs $H$, and then uses $H$ in the construction of another group, call it $K$. $K$ has the property that it embeds $G_{-}$ iff $w\neq1$, and is trivial otherwise. Proving that $K$ depends on $w$ in this way is the crux of the theorem. However, I think that my group $H'$ also has the desired property w.r.t. $w$, which would mean that the proof could be simplified somewhat. (The move by which I construct $H'$ from $H$ is similar to the final step of the construction of $K$, and so the steps by which one shows that $w=1$ implies $H'$ is trivial are similar.)

All proofs of Adian-Rabin that I have seen go as far as constructing $K$, they don't stop at $H$ or construct $H'$, so there must be something about my construction that makes it not necessarily embed $G_{-}$. I cannot figure out what that could be, for reasons outlined above.

I am not a mathematics researcher but I am concerned that this question, posed with slightly different wording on math.stackexchange, may be too esoteric for that forum since it concerns the details of a proof that I don't think is standard in the graduate level curriculum. At least, nobody has answered it so far.

Let $G$ be a group with presentation $<x_1,x_2...,x_m|R>$ and let $G'=G \ast<y_0>$.

Now define $y_i=y_{0}x_i$. Notice that $G'=<y_0,y_1,...y_m|R'>$ for appropriately defined $R'$ and that $<y_i>$ is free on $y_i$ for all $i$.

Consider the group

$$H = <y_0,y_1,...y_m,t_1,t_2,...t_m,|R',t_iy_it^{-1}_i=y_{i}^2>$$

Either take my word for it that $G'$ embeds (isomorphically) into $H$, or notice that, since $<y_i>$ is free on $y_i$ for all $i$, $H$ is an iterated HNN extension on $G'$ and therefore embeds $G'$ by $m$ applications of Britton's Lemma.

Now let $w$ be a nonidentity element in G. Define $H^{'}$ as $H$ subject to the relation $(w^{-1}y_{0}^{-1}wy_0)t_i(w^{-1}y_{0}^{-1}wy_0)^{-1} = t_{i}^2$ for each $t_i$.

I can show that $w^{-1}y_{0}^{-1}wy_0$ is infinite order since it is cyclically reduced word not containing the identity and length $\geq 2$. $t_i$ is also infinite order in $H$ since it is the pivot letter in an HNN extension. So it would seem that I'm not introducing any torsion for the elements $y_0,t_i$.

I also don't think I'm introducing any torsion for elements of $G$ since it does not reference them at all.

In particular it seems like $G'$ embeds into $H'$. I have no idea how to prove this, but I have done pages of calculations in this group to try and disprove this.

+++++++++++++++++

Is this true? Here's why I think no, in spite of my best efforts to (dis)prove it:

In the proof of the Adian-Rabin theorem, $G$ is the free product of some group with unsolvable word problem $U$ with a group $G_{-}$, which has the property that any group it embeds into does not satisfy some property $M$. (A $G_{-}$ with such a property is given by hypothesis).

The proof constructs $H$, and then uses $H$ in the construction of another group, call it $K$. $K$ has the property that it embeds $G_{-}$ iff $w\neq1$, and is trivial otherwise. Proving that $K$ depends on $w$ in this way is the crux of the theorem. However, I think that my group $H'$ also has the desired property w.r.t. $w$, which would mean that the proof could be simplified somewhat.*

All proofs of Adian-Rabin that I have seen go as far as constructing $K$, they don't stop at $H$ or construct $H'$, so there must be something about my construction that makes it not necessarily embed $G_{-}$. I cannot figure out what that could be, for reasons outlined above.

*NB: In $H'$, $w=1$ implies $w^{-1}y_{0}^{-1}wy_0=1$ implies the $t_i$'s and thus the $y_i$'s all equal 1. The proof that $K$ is trivial goes exactly like this too, except it involves the extra relations used to construct $K$.