If $X$ and $Y$ are subgroups of a group $G$ and if $|X|, |Y| < \infty$, show that $|XY| = \frac{|X||Y|}{|X \cap Y|}.$ I have a question;
$X$ and $Y$ are subgroups of a group $G$. If $|X|, |Y| < \infty$, show that
$$|XY| = \frac{|X||Y|}{|X \cap Y|}.$$
but I can't really picture what it is talking about to even get started.
Is it to do with looking for different ways so combine the elements {x, y, xy} ?
Thanks
 A: The point is that if $z\in X\cap Y$ then it is counted twice when you form $XY$, so when you consider the cardinality of $|XY|$ you must remove one of these occurrences. You do the numbers, and it turns out that when you remove them you end up with the formula you have given. To prove the formula, consider the natural map $f:X\times Y\rightarrow XY$ and work out its "kernel".
Addendum: As requested, I'll included some more detail. This detail will basically be the description of some technology and how to apply it. The proof is neater with the technology.
The technology: Define an equivalence relation on the set $X\times Y$ as follows: Set $(x_0, y_0)\sim (x_1, y_1)$ if and only if $x_0y_0=x_1y_1$. Note that $(x_0, y_0)\sim (x_1, y_1)$ is equivalent to $x_1^{-1}x_0=y_1y_0^{-1}$. It is therefore equivalent to the existence of an element $z\in X\cap Y$ such that $(x_0z, z^{-1}y_0)=(x_1, y_1)$. Denote the equivalence class containing $(x, y)$ by $\widehat{(x, y)}$.
Step 1: Prove that each equivalence class $\widehat{(x, y)}$ has cardinality $|X\cap Y|$.
Step 2: Prove that the map $\frac{X\times Y}{\sim}\rightarrow XY$, $\widehat{(x, y)}\mapsto xy$ is well-defined and a bijection (note that ${X\times Y}/{\sim}$ denotes the set of equivalence classes of $X\times Y$ under $\sim$).
Step 3: Conclude.
A: Can I summarize some comments? 
Cosets $xY$ (for $x \in X$) are either disjoint, or equal. Suppose $xY=x′Y$. It happens if and only if $x^{-1}x' \in Y$ or $x^{-1}x' \in X \cap Y$ (since $x,x' \in X$). So, $xY$ and $x′Y$ are disjoint iff $x$ and $x'$ are representatives of distinct cosets of $X$ by $X \cap Y$. And there are $|X|/|X \cap Y|$ such cosets. For every coset there are |Y| elements in $XY$, and the formula follows.
A: From what I know, in this theorem we are trying to remove the repetitions while defining $XY$.
For simplicity, let's assume that $X\cap Y = \phi$. In this case, the total number of combinations you will get for elements in $XY$ is, $|X|\times |Y|$ (For every element in $Y$, there are $|X|$ elements in $X$. 
Now if, there are $k$ elements in $X\cap Y$, then $$\forall x \in X-(X \cap Y),  \forall y \in Y-(X \cap Y), $$ Every element $xy \in XY$ can be written as $ xh_{1}h_{1}^{-1}y, xh_{2}h_{2}^{-1}y, xh_{3}h_{3}^{-1}y..., xh_{k}h_{k}^{-1}y$, where $h_{1}, h_{2}... h_{k} \in X \cap Y$. So to account for these repititions, we divide it it by $|X \cap Y|$.
