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There is a question with same title, but the problem, I will post here, is about the solution by Arturo Magidin.

A weaker version of HNN's theorem would be, given a group $H$ and $x,y\in H$ of same order, there exist a group $G$ such that $G$ contains $H$ and $x,y$ are conjugate in $G$.

As A. Magidin proceeds: consider $H=\mathbb{Z}$. Then $1,2\in \mathbb{Z}$ have same order, there is a group $G_1$ such that $\mathbb{Z}\subseteq G_1$ and $1,2$ are conjugate in $G_1$. In the group $G_1$ constructed, we have that the conjugacy class of $1$ contains $1,2,4,8,\cdots$, but not $3$, so we should construct a new containing $G_1$ in which $1$ and $3$ will be conjugate.

To say in short, the above procedure should be repeated for countably many times to make all torsion-free elements of $\mathbb{Z}$ conjugate in a bigger group. That is to say, in the solution by A. Magidin, there would be countably infinite groups between $\mathbb{Z}$ and $\mathbb{G_1}$.

Then, again, to make all torsion free elements outside $\mathbb{Z}$ conjugate in a bigger group, we proceed in above manner, infinitely many times.

How could finally we achieve the group with only two conjugacy classes?

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So $G_1$ is formed by making all nontrivial elements of $\Bbb Z$ conjugate. And $G_2$ is formed by making all of the nontrivial elements of $G_1$ conjugate. And $G_3$ and $G_4$ and so on are all formed in a similar manner. This creates an infinite linear direct system $G_1\hookrightarrow G_2\hookrightarrow G_3\hookrightarrow\cdots$ of embeddings.

Take the direct limit $G=\varinjlim G_n$. Informally, this is just the union $G=\bigcup G_n$ of all of them.

Then if $x,y\in G\setminus0$, we have $x\in G_i$ and $y\in G_j$ for some $i,j$, so $x,y\in G_k$ with $k\ge i,j$. Thus $x,y$ are conjugate in $G_{k+1}$ which is a subgroup of $G$, so $x,y$ are conjugate in $G$.

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