If $A\subseteq C$ then $AB\cap C=A(B\cap C)$? $A,B,C$ can be anything (e.g. I proved it for Galois extensions of $\Bbb Q$).  Can someone find a proof or a counterexample (for any class of $A,B,C$, e.g. groups, rings) that if $A\subseteq C$ then $AB\cap C\subseteq A(B\cap C)$?  (since the $\supseteq$ is obvious).
My proof is really really long, by induction on the power of prime factors dividing $[B:B\cap C]$; however, I'm pretty sure I've seen an elementary proof back in high school (which obviously wasn't in the context of Galois extensions).
EDIT: AB is the compositum of A and B.
 A: We have $A(B\cap C) \subset AB\cap AC$, but the reverse containment will only hold if the lattice of subobjects is distributive.  For subfields of a Galois extension, this will happen only if the Galois group is locally cyclic.
You are asking about the case where $A\subset C$, equivalently $AC=C$.  This case is equivalent to the lattice of subobjects being modular, which is a more subtle condition.
But there are Galois groups with non-modular subgroup lattices, notably $D_8$.  So we have the following counterexample for finite extensions of fields, arising from the splitting field of $X^4-2$:
$A = \mathbb{Q}[\sqrt{2}], B=\mathbb{Q}[\sqrt[4]{-2}], C=\mathbb{Q}[\sqrt[4]{2}]$.  Then $AB\cap C = C$, but $A(B\cap C) = A$.
However, the case where all the extensions are Galois follows from the general fact that the lattice of normal subgroups of a group is modular, which is an elementary result.
A: You didn't specify but I shall assume that $A,B,C$ are substructures of some structure $M$ such that $\def\wi{\subseteq}$$A \wi C$, and that "$ST$" denotes the structure generated by (finite combinations from) $S,T$ for any $S,T \wi M$.
$\def\qq{\mathbb{Q}}$
$\def\gen#1{\langle#1\rangle}$
Then obviously $A(B \cap C) \wi AB \cap C$, because every element in $A(B \cap C)$ is in $AB$ and $AC = C$ by definition of generation.
The reverse containment does not hold, at least for groups. Let $A = \gen{(1,2,3,4,5)}$ (generated by this 5-cycle) and $B = \gen{(1,2)}$ (generated by this swap) and $C = A_5$ (alternating group on $\{1..5\}$). Then $A(B \cap C) = A\gen{} = A$ but $AB \cap C = S_5 \cap C = C$.
