[FBT]

> ... Does distributivity of inversion ~ over operation ⋆ follow from the other Abelian group axioms / properties? If so, how?

It does. For all x and y:

  (1) ~x ⋆ x = 0 (definition of the inverse)
  (2) ~y ⋆ y = 0 (definition of the inverse)
  (3) (~x ⋆ x) ⋆ (~y ⋆ y) = 0 ⋆ 0 = 0 (from (1) and (2))
  (4) (~x ⋆ ~y) ⋆ (x ⋆ y) = 0 (via associativity and commutativity)

In (4) we see that (~x ⋆ ~y) is the inverse of (x ⋆ y). That is to say, ~(x ⋆ y) = (~x ⋆ ~y). QED.

[stephencanon]

Right. Another way to see this is that for a general (possibly non-Abelian) group, the inverse of xy is y⁻¹x⁻¹ (because xyy⁻¹x⁻¹ = x1x⁻¹ = xx⁻¹ = 1 [using "1" for the identity here, as is typical for general groups], or more colloquially, "the inverse operation of putting on your socks and shoes is taking off your shoes and socks"). For an Abelian group, y⁻¹x⁻¹ = x⁻¹y⁻¹, and we're done.

[OjotCewIo]

Awesome, thanks! :)