 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 thatThus the failure of a word problem for a group to be decidable does not require torsion.