Why $U\le A \times B$ does not imply $U=\left(A\cap U\right) \times \left( B \cap U \right)$? Let $A,B$ be groups.Can you explain why $U\le A \times B$ does not imply $U=\left(A\cap U\right) \times \left( B \cap U \right)$ 
this is an exercise in the book of the theory of finite groups an introuduction written by H.Kurzweil. the meaning of each symbol may as follows.
$A\times B$ is the direct product of $A$,$B$. $U$ is the subgroup of $A\times B$ , thus $U= \lbrace\left(a,b\right)|a \in A,b\in B \rbrace$, $A \cap U=\lbrace a_1|\left(a_1,b_1\right) \in U,a_1\in A  \rbrace$,in the same way we could know $\left( B \cap U \right)$
How about a simple example such as $A=B\ne1$ and $U=\{(x,x)\mid x\in A\}$?
Think about it intuitively. 

Take $A,B = \mathbb{R}$. The LHS $U$ is "some set of points in the plane" whereas the RHS is "all possible x-coordinates from points in $U$ with all possible y-coordinates from points in $U$".

Clearly $U\subseteq$ RHS but the RHS can be bigger.

For a specific example let $U = \{(0,1),(1,0)\}$. Then the RHS is $\{(0,0),(0,1),(1,0),(1,1)\}$.

The point is that the object on the RHS is combining things componentwise that might not have existed when we chose "special" points to be in $U$.