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.
 A: The composition is the operation which takes in a pair of subsets $A$ and $B$ of $G$ and spits out the set $$A*B=\{g\in G: \exists a\in A, b\in B\mbox{ such that } g=a*b\}$$ (using "$*$" for the binary operations). The problem is asking you to show that this operation is associative - that is, $(A*B)*C=A*(B*C)$ for every $A, B, C\subseteq G$ (note: "$\subseteq G$" means the same thing as "$\in\mathcal{P}(G)$," and might be easier to think about) - and has a neutral element - that is, there is some $E\subseteq G$ such that $A*E=E*A=A$ for all $A\subseteq G$.
Note that there's a bit of abuse of notation going around: we're using the same symbol, "$*$," to refer to the binary operation on $G$ and the induced "composition" operation on $\mathcal{P}(G)$. Technically we should really use a different symbol, but meh.

Before tackling this problem, if you're still confused it might help to play with some more concrete problems. For instance:


*

*Suppose $G$ is the set of all integers, and $*$ (on $G$) is the operation of addition. What is $$\mbox{$\{$evens$\}*\{$odds$\}$?}$$

*Let $A\subseteq G$. Knowing nothing about $A$ or $G$, what can we say about $\emptyset*A$?

*Suppose additionally $G$ has inverses - for every $a\in G$ there is a $b\in G$ such that $a*b=b*a=e$. What can you say about $G*A$ for $A\subseteq G$ nonempty?
A: 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 $$A\cdot (B\cdot C)$$
