If G is a group and A,B,C are subsets of G, prove that (AB)C=A(BC) $A,B$ are defined as follows:
$$AB = \big\{ ab \mid a\,\, \text{belongs to}\,\, A, \text{b belongs to B}\big\}\tag{1}$$
Where do I start?
 A: Let $x \in (AB)C$.  So $x = y c$, where $y \in (AB)$ and $c \in C$.  So $y = ab$, where $a \in A$ and $b \in B$.  So $x = yc = (ab)c = a(bc)$ by applying associativity of the group operation.  So $x \in A(BC)$.  This gives $(AB)C \subset A(BC)$.  Can you do the reverse inclusion?
A: 
(1) Let $G$ be a group, and let $A\subset G, B\subset G, C\subset G$.
  (2) Define the set $AB = \{ab \mid a \in A, b \in B\}.$

To show that $(AB)C = A(BC)$, we must show:
 $$(AB) C \subseteq A(BC)\tag{i}$$$$A(BC) \subseteq (AB)C\tag{ii}$$
We'll prove $\;\;$(i): $\;(AB)C\subset A(BC)$
$x\in (AB)C$ means $x = kc$ where $k\in AB $ and $c\in C.$
$k \in AB$ means there exist $a \in A, b\in B$ such that $k = ab$. So $x \in (AB)C$ means $x \in \{kc=(ab)c \mid k = ab \in AB; c \in C\}.\tag{iii}$
From premise $(1)$ we know $a, b, c \in G$. $G$ is a group, so associativity holds. Then $(ab)c = a(bc);$ with $\text{(iii)}$ this gives us $x \in \{a(bc) | a\in A, bc\in BC\}\implies x \in A(BC)\tag{iv}$
Thus we know that $x \in (AB)C \implies x\in A(BC)$. That is, $(AB)C \subseteq A(BC)$.

Set inclusion (ii) follows the same logic, for the same reasons.
A: Hint: The universal strategy. You prove that (i) if $x$ is an element of the left-hand side, then $x$ is an element of the right-hand side and (ii) the other way. 
If $x$ is an element of the left-hand side, there are objects $a\in A$, $b\in B$, $c\in C$ such that $x=(ab)c$. Continue. 
