How to show that a set operation is associative? I have a problem to solve, that goes like that:
(roughly translated from German)
$G$ is a set with associative composition. $e$ is the neutral element, so that for each $g\in G$, $e\cdot g=g\cdot e=g$. From the power set $\mathcal P(G)$ of $G$, we define the composition:
$$\mathcal P(G)\times\mathcal P(G)\to\mathcal P(G)$$
$$(A,B)\mapsto A\cdot B :=\{g\in G|\exists a\in A,b\in B : g = a\cdot b\}$$
Show that
a) this operation is associative
b) there is a neutral element in that operation. Which one?
I've already tried several approaches to that problem, but I struggle to understand what is the "composition" and what then should be associative? Wikipedia says "function composition is the pointwise application of one function to the result of another to produce a third function", which I imagine refers to $f(g(x))$ for example. But how does that apply here?
Thank you very much for your hints.
J.

The term composition is slightly misleading here (which might be a translation issue).
What you are dealing with is not function composition (which is the "default" thing we think about when we say composition. It is the composition of elements in a set, i.e. you are given a mapping that, for every pair of elements $x,y\in G$, gives you the element $x\cdot y$. This is more commonly refered to as "operation".
Given that operation, you then define a new operation on the set $\mathcal P(G)$.
What you must do now is prove that, for any three subsets of $G$ (i.e., elements of $\mathcal P(G)$, the set
$$(A\cdot B)\cdot C$$
is equal to the set $$A\cdot (B\cdot C)$$
