(Dualising a plane cone)
In the plane the dual of \(\sigma = \Cone(v_1, v_2)\) is found by rotating each generator a quarter turn and orienting the result inwards. Rotation by \(\pi/2\) is \[ e_1 \mapsto e_2, \quad e_2 \mapsto -e_1, \qquad R = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix} . \] For \(v_1 = (2,1)\) and \(v_2 = (1,4)\) this gives \(R v_1 = (-1,2)\) and \(R v_2 = (-4,1)\). The second normal points out of the cone, so flip it to \((4,-1)\), and \[ \dualof{\sigma} = \Cone\big( (4,-1),\ (-1,2) \big) . \] Check the two pairings that must vanish: \(\inp{(4,-1)}{(1,4)} = 4 - 4 = 0\) and \(\inp{(-1,2)}{(2,1)} = -2 + 2 = 0\), so each of these is the inward normal to one wall.
(The semigroup and its equations)
The Hilbert basis of \(S_\sigma = \dualof{\sigma} \intersect M\) is \[ (4,-1), \quad (1,0), \quad (0,1), \quad (-1,2) , \] found by walking from one extremal ray to the other and keeping the lattice points that are not sums of earlier ones. Write \(u = x^4 \inverseof{y}\), \(v = x\), \(w = y\), \(t = \inverseof{x} y^2\) for the corresponding characters.
Now read the additive relations straight off the exponent vectors: \[ (4,-1) + (0,1) = (4,0) , \qquad (1,0) + (-1,2) = (0,2) , \qquad (4,-1) + (-1,2) = (3,1) , \] that is \(uw = v^4\), \(vt = w^2\), and \(ut = v^3 w\). Hence \[ X_\sigma = \Spec k[x^4 \inverseof{y}, x, y, \inverseof{x} y^2] = \Spec \frac{k[u,v,w,t]}{(uw - v^4,\ vt - w^2,\ ut - v^3 w)} , \] a surface in \(\AA^4\). The three equations are the \(2 \times 2\) minors of \[ \begin{bmatrix} v & w & u \\ w & t & v^3 \end{bmatrix} , \] so the singularity is determinantal, as every two-dimensional toric singularity is.
The index \(\abs{\det \begin{bmatrix} 2 & 1 \\ 1 & 4 \end{bmatrix}} = 7\) is the whole story: \(v_1, v_2\) generate a sublattice of index \(7\) in \(N\), so \(X_\sigma\) is the cyclic quotient singularity \(\AA^2 / \mu_7\) and \(\Cl(X_\sigma) = \ZZ/7\). The computation has two steps: rotate to obtain \(\dualof{\sigma}\), then list its lattice points and read the additive relations among them.