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Thicken the planar graph underlying a knot diagram. Its boundary supports a Heegaard diagram with face disks on one side and crossing-tunnel disks on the other. The face generators and crossing curves give a Dehn presentation of a knot group.

Use the region labels and curve orientations from part (a). A Dehn presentation of a knot group generator goes from the basepoint above the projection plane through its region and returns through the unbounded region. Equivalently it is dual to the correspondingly labeled $\alpha$ disk of the Heegaard diagram. A crossing relator records the signed intersections of its $\beta$ curve with these disks. Setting $o=1$ givesFor example, the overpassing strand at $A$ equates the meridian of a knot words $d$ and $a{e}^{-1}$; at $B$ it equates $a{d}^{-1}$ and $b{c}^{-1}$. The other two crossings equate $b{a}^{-1}$ with $c{e}^{-1}$ and $c{d}^{-1}$ with $e$.

Choose the region-generator direction so that $e$ is the shown meridian of a knot $\mu$. The Alexander numbering of the regions, starting from zero outside and increasing by one on crossing an oriented strand from right to left, isOne can also obtain this directly by abelianization of the relators: $a=d+e$, $c=d+e$, $b=a-d+c$, and $b-a+e-c=0$. The last two imply $d=e$, and the resulting first homology group is freely generated by $e$. Thus $H_1(X_8;\mathbb{Z})\cong \mathbb{Z}$, with no extra torsion.

Label the unbounded region $r_0$ and the three lobe regions $a,b,c$ in cyclic order, and label the central region $d$. The Dehn presentation of a knot group Heegaard diagram is obtained by thickening the projection graph: take one alpha compressing curve for each of $a,b,c,d$, one beta curve for each crossing, and place the basepoint in $r_0$. Orient and order the crossing curves so that their cyclic region words areThis intersection data, together with $r_0=1$, specifies the requested pointed Heegaard diagram up to the usual isotopies and handle slides.

Give each bounded region $R_i$ a generator $a_i$ and give the unbounded region $R_0$ the identity generator. At every crossing, read the four incident regions cyclically as $a,b,c,d$ and imposeUsing the opposite cyclic convention inverts all such relators and gives the same group. One crossing relation is redundant, leaving $n$ generators and $n-1$ relators; this is the Dehn presentation of a knot group.