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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 meridian of a knot $\mu$ and Seifert longitude $\lambda$ of the unknot on its boundary torus. The unknot's knot exterior is a solid torus whose disk-bounding curve is $\lambda$. A genus-one Heegaard diagram is thereforeHere the horizontal coordinate is $\mu$ and the vertical coordinate is $\lambda$. The drawing uses a small translate of $\alpha$ to avoid intersections on the edge of the square; opposite edges are identified. The algebraic intersection number of curves on an oriented surface has absolute value five.

Starting with $\Sigma \times [0,1]$, attach a three-dimensional two-handle along $\alpha \times \{0\}$, using its surface framing, and cap the resulting sphere with a three-handle. This gives the solid torus on that side. Attach a two-handle along $\beta \times \{1\}$, again using the surface framing, and cap its sphere with a three-handle. The second solid torus has its disk-bounding curve identified with $5\mu +3\lambda$, so the resulting oriented lens space is precisely the specified $5/3$ Dehn filling.

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.