The order of two subgroups product is not greater than the product of two subgroup orders $|AB|\le |A|\times|B|$ Let $G$ be a group, and let $A,B\le G$ be finite subgroups of $G$.
The product $AB$ is defined as following :
$AB =\{a\times b | a\in A, b\in B\} $
Show that $|AB|\le |A|\times |B|$.
My attempt so far:
Observe that $AB=\bigcup_{a\in A} aB$, and by Lagrange theorem we know that $|aB| = |B|$.
We also know that for any $x,y\in G$ either $xK\cap yK=\emptyset$ or $xK=yK$.
How can I continue from here?
 A: You can show that $|AB||A\cap B|=|A||B|$. There is a map from $A\times B$ to $AB$ that sends $(a,b)\to ab$. This is obviously surjective, which gives away your inequality. Now note that for any $c\in A\cap B$, $(ac,c^{-1}b)$ maps to $ab$. Now supose $(a',b'),(a,b)$ are such that $a'b'=ab$. Then note that $c=a^{-1}a'=bb'^{-1}\in A\cap B$, and $(a',b')=(ac,c^{-1}b)$. Thus any fibre has cardinality $|A\cap B|$.
Recall that if $f:X\to Y$ is any map, we can define an equivalence relation $\sim $ on $X$ by $x\sim y$ whenever $f(x)=f(y)$. The class of an element $x$ is $[x]=f^{-1}((f(x))$, and we can define $\tilde f: \widetilde X\to Y$ by $\tilde f[x]=f(x)$. Then $\tilde f$ is injective, and thus gives a bijection between $\widetilde X$ and $f(X)$. But $X$ has cardinality equal to $n_1+\cdots+n_k$ where $n_i=|[x_i]|$ for $1\leqslant i\leqslant k=|\widetilde X|$ and each $x_i$ is a representative of the classes of $\widetilde X$. If each class has equal cardinality $N$, then $|f(X)|= |\widetilde X|=|X|/N$. 
In the above, $N=|A\cap B|$, $|f(X)|=|Y|=|AB|$ and $|C|=|A\times B|=|A||B|$, whence the formula. 
