Using your Head is Permitted

UPDATE (3 May): You may find the following basic theorem in group theory, known as Cauchy's theorem, to be useful: if G is a finite group over a set of size n, and p is a prime divisor of n, then G has an element, w, not equal to the identity, such that w^p is the identity. (The theorem is usually worded somewhat differently, but was simplified here.)

This month's riddle comes from Serge Gautier. (Thanks, Serge!) It extends one of the very earliest Using your Head is Permitted riddles (December 2007), which is a personal favourite of mine, and one that has already been extended once (in June 2010).

A group is a set, S, and a binary operation, "*" over it. (Here, "binary" just means it acts on two elements, as in x*y.) The result of x*y for any x, y in S must also be in S. The operation must also satisfy the following conditions:

• Associativity: For every x, y, z in S, x*(y*z)=(x*y)*z.
• Identity: There exists an element, i in S, such that for every x in S, x*i=i*x=x. This is known as the group's identity element. (It is necessarily unique.)
• Inverse element: For every x in S there exists a y in S such that x*y=y*x=i, the identity element.

A group is called abelian if it furthermore satisfies commutativity: for every x, y in S, x*y=y*x.

Traditionally, the group operation of abelian groups is denoted "+".

We now repeat the December 2007 riddle:

Let n>1 be an integer, let S be the set of numbers {0, 1, ... , n-1} and let A, B and C be one-to-one functions to and from S.

Part 1:

Part 2:

The only difference from the original riddle is that now we are asking this question not only over standard multiplication and standard addition, but over any group operation "*" in Part 2 and any abelian group operation "+" in Part 1.

In other words, you are to answer whether for each possible choice of n one can find not only A, B and C, but also a corresponding group, to satisfy the equalities. The only constraint imposed on the group is that its underlying set should be S. (Note that i may be any element in S.)

As always, prove your answer.

A separate list of solvers will be kept for each of the two parts. Answer either or both.

List of solvers:

Part 1:

Part 2: