$U\neq H< G$ infinite groups, $p$ prime, $|G:U|=p=|G:H|$; is $|G:H\cap U|= p^2$? I know that if we got two finite index subgroups $U,H$ of a group $G$. Then the index of the intersection $H\cap U$ of both is smaller than the product of the indices of $H,U$. 
We proof this with an injective map from the cosets of $H\cap U$ in H to the cosets of $U$ in $G$. But is there a chance to show that this map is surjective in some cases? Especially in the case:
$U\neq H< G$ infinite groups, $p$ prime number. $|G:U|=p=|G:H|$. Then $|G:H\cap U|= p^2$?
Or when the indices are coprime?
Or can we show that the index of the intersection $H\cap U$ in $G$ divides the product of the indices of $H$ and $U$ in $G$.
If we can't do this for infinite groups. Do anyone have an example where $|G:H\cap U|$  does not divide $|G:U|\cdot|G:H| $?
Thanks for help.
 A: $S_4$ has two (in fact, three) subgroups of index 3, but their intersection is not of index 9. Yes, I know, $S_4$ is a finite group, but just do $S_4\times G$ for some infinite group, $G$. 
A: Finiteness or infiniteness is irrelevant; a finite example yields an infinite one by taking a direct product with an infinite group. And every infinite example can be turned into a finite example by moding out by a normal subgroup of $G$ of finite index contained in $H\cap U$. 
In general, if $G$ is a group, $H$ and $K$ are subgroups, then
$$[H\colon H\cap K] \leq [G:K]$$
in the sense of cardinalities; but you may or may not get equality. In particular, if $[G:H]=[G:K]=p$, then $[G:H\cap K]\leq p^2$, and is a multiple of $p$, but you may or may not have equality to $p^2$.
Proposition. Let $G$ be a group, $H$ and $K$ subgroups. Then
$$[H:H\cap K] \leq [G:K]$$
in the sense of cardinalities, with equality if and only if $HK=G$.
Proof. We define a map from the left cosets of $H\cap K$ in $H$ to the left cosets of $K$ in $G$ by mapping $h(H\cap K)$ to $hK$. I claim the map is well-defined and one-to-one:
$$\begin{align*}
h(H\cap K) = h'(H\cap K) &\iff h'^{-1}h\in H\cap K\\
&\iff h'^{-1}h\in K\quad\text{(it is always in }H\text{)}\\
&\iff hK = h'K.
\end{align*}$$
The map is onto if and only if for every $g\in G$ there exists $h\in H$ such that $gK=hK$, if and only if for every $g\in G$ there exists $h\in H$, $k\in K$ such that $g=hk$, if and only if $G\subseteq HK$, if and only if $G=HK$. $\Box$
In particular, for you situation, if $HU$ is a subgroup of $G$ (it need not be!) then $HU=G$, so you will have $[G:H\cap U] = [G:H][H:H\cap U] = [G:H][G:U] = p^2$. But if $HU$ is not a subgroup, then the best you have is that $p$ divides $[G:H\cap U]$, and $p\lt [G:H\cap U] \leq p^2$; so it could be any of $2p$, $3p,\ldots,(p-1)p,p^2$.
In the example Gerry Myerson gives, for instance, the intersection has index $6=2p$ ($p=3$).
