= Solution
The factors $\mathbb Z^2$ and $\mathbb Z$ are <torsion-free groups>, and so is their <free product> $K$. Indeed every element of a <free product> is conjugate either into one factor or to an alternating word whose first and last syllables lie in different factors. A word of the latter kind has nonempty reduced powers of every positive length, so has infinite order. A nonidentity element of either factor also has infinite order.
By the finite-order result in part (b), every finite-order element of the multiple <HNN extension> $K_{\mathcal M}$ is conjugate into the torsion-free base $K$, and hence is the identity. Apply the same result once more to $G_{\mathcal M}$, whose base is $K_{\mathcal M}$. It follows that
$$
\boxed{G_{\mathcal M}\text{ is torsion-free}.}
$$
Thus the failure of a <word problem for a group> to be decidable does not require torsion.
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