# 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 $\left( B \cap U \right)$

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Is this really what you mean? What do $A\cap U$ and $B\cap U$ mean in this context? E.g., if $A=B=\mathbb Z$, and $U=A\times B=\mathbb Z\times\mathbb Z$, then $A\cap U=\mathbb Z\cap (\mathbb Z\times \mathbb Z)=$?? –  Jonas Meyer Dec 27 '12 at 9:28
@JonasMeyer: Other than naming projection maps, is there a standard say of denoting $\{ a \in A\, :\, (a,b) \in U\ \text{for some}\ b \in B\}$? I presume this is what $A \cap U$ denotes. –  Clive Newstead Dec 27 '12 at 9:44
What does $A\cap U$ means ? $U$ is a set of ordered pairs, $A$ is not, so the intersection is always an empty set –  Belgi Dec 27 '12 at 9:49
It is standard in a direct product $A \times B$ to identify $A$ and $B$ with the subgroups $\{(a,1) \mid a \in A\}$ and $\{(1,b) \mid b \in B \}$ of $A \times B$. –  Derek Holt Dec 27 '12 at 9:55
@Clive: That is one possible interpretation. Notice that Derek has another interpretation, which would have been my guess if I had to guess. But if user53587 won't tell us what it means, I have no interest in guessing. –  Jonas Meyer Dec 27 '12 at 17:30

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 $U$.

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thanks for your interpretation –  Song Dec 28 '12 at 9:42
If you like it then choose it as the best answer by clicking on the tick at the side. –  fretty Dec 28 '12 at 11:27
But $U$ is not a subgroup of $A \times B$. –  Derek Holt Dec 28 '12 at 16:52
I didn't know we were looking for groups with this property. To be fair the question had been edited. Anyway, it is easy to tweak this example to provide subgroups which dont work...it is really the fact that it doesn't work as sets that makes it not work as groups. –  fretty Dec 29 '12 at 9:37
@fretty: There is a convention widely used in the group theory community that if $G$ is a group then $H \le G$ means that $H$ is a subgroup of $G$, whereas $H \subseteq G$ means that it is a subset and not necessarily a subgroup. But of course not everyone might be aware of that! –  Derek Holt Dec 29 '12 at 9:56

How about a simple example such as $A=B\ne1$ and $U=\{(x,x)\mid x\in A\}$?

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Is this really an example of when they aren't equal? Here $(A \cap U) = A$ and $(B\cap U) = B = A$ so the RHS is $A\times A$ which is exactly $U$. –  fretty Dec 28 '12 at 11:36
@fretty: That is incorrect. $U\neq A\times A$. For example, if $x\neq 1$ and $1$ is the identity in $A$, then $(x,1)\in (A\times A)\setminus U$. –  Jonas Meyer Dec 28 '12 at 16:55
Stupid me, I think I might have had an off day with this question. –  fretty Dec 29 '12 at 9:35