The Lagrange theorem states that divides for every subgroup of a finite group , with . The left cosets partition , and multiplication by a representative bijects with each coset, proving the formula.
The intersection contains the identity and is closed under , so it is a subgroup. Its order divides both and ; if these are coprime, .
The order is the least positive with . Division , , shows that exactly when , so exactly when .
If commute and have coprime orders , then . Conversely implies lies in , so and . Hence .
Cauchy theorem for groups says that every prime divisor of occurs as the order of an element of . For , the Sylow -subgroup is unique and normal, and Cauchy's theorem supplies a complement . Thus . The homomorphism is either trivial or has the unique image of order two, inversion. These give exactly and the dihedral group of order .
Differentiation of , gives
The two radial face areas are and . Each constant- face has area , and each constant- face has area
The volume is this last expression times .
Only the radial faces contribute to the flux of . Dividing their difference by the volume and taking gives
at .
A field is irrotational when . In define
where is any path from to . The difference between two choices is the circulation around a closed loop. By the Stokes theorem it is the flux of through a spanning surface and hence zero. Thus is well-defined, , and differentiation gives .
If , every solution is
for an arbitrary differentiable scalar , since the difference is curl-free. A necessary condition is , because the divergence of a curl vanishes.
Existence follows from the three-transitivity of Möbius transformations: one can send three distinct source points to and then send those to the three distinct target points.
For uniqueness, suppose and have the same values at three distinct points. Then fixes those points. If
the finite fixed points satisfy . A nonidentity Möbius map therefore has at most two fixed points on the Riemann sphere. Since has three, it is the identity, and .
Expanding with the Levi-Civita symbol and the product rule gives
The Stokes theorem states
where the boundary orientation follows the right-hand rule about .
Apply it to . On , and . Substitution of the vector identity and contraction with leaves
The boundary integrand satisfies , proving the formula. The vector is the outward co-normal in the tangent plane of .
For the hemisphere, and . The projected divergence is , so the left side is
On the positively oriented equator, , hence and the boundary integral is also .

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