$U\leq G$ infinite subgroup, $H\leq G$ subgroup of finite index in $G$ $\Rightarrow$ $U\cap H\neq 1$ I have the following question:
If I have an infinite subgroup $U$ of a group $G$ and a subgroup $H$ of finite index in $G$ then how can I show that the intersection of $U\cap H$ is non-trivial???
Thanks for help.
 A: HINT: Let $K=U\cap H$. Suppose that $u\in U$. Clearly $uK\subseteq U\cap uH$. Suppose that $h\in H$ and $uh\in U\cap uH$; then $h=u^{-1}(uh)\in U\cap H=K$, so $uh\in uK$. Thus, $uK=U\cap uH$. This is a problem if $K$ is trivial; why?
A: If $U$ is an infinite subgroup of $G$ with and $H$ is a finite-index subgroup of $G$ then $U\cap H\neq 1$.

Proof:
There exists a finite-index subgroup of $G$, $H^{\prime}$ say, such that $H^{\prime}\lhd G$ and $H^{\prime}\leq H$. Clearly, if $H^{\prime}\cap U\neq 1$ then $H\cap U\neq 1$. This means that we can assume $H\lhd G$.
Now, look at $G/H$. As $U$ is infinite, we must have that there exists some $u, v\in U$ such that $uH=vH\Rightarrow uv^{-1}\in H$. As $u, v\in U$ we have $uv^{-1}\in U\cap H$ and we're done.

I believe this actually means that if $U\cap H=1$ $U$ then $G=K\rtimes V$ where $V$ is finite (so $K$ has finite index in $G$) and $U\leq V$, $H\leq K$. This is because you need to find a copy of $U$ in $G/H$, so it must somehow split...but the problem is the "somehow". Certainly, $H\rtimes U$ must have finite index in $G$, where $H$ is the finite-index normal subgroup we got above (and so $U$ must be finite).
A: $N=\cap H^g$ is normal and finite index in $G$, and by the second isomorphism theorem $U/U\cap N \cong UN/N\le G/N$, which is a finite group.
A: Restrict the map $G\to G/H$ to $U$ giving a map $f:U\to G/H$. For every left coset $y=uH\in f(U)\subseteq G/H$ with $u\in U$ one has $f^{-1}(y)=y\cap U=u(U\cap H)$ (the last equality is obtained by checking both inclusions). Now since $G/H$ is finite, $U=\bigcup_{y\in f(U)}f^{-1}(y)$ would be a finite union of finite sets if $U\cap H$ were finite, which cannot be true since $U$ is infinite. So $U\cap H$ is not finite, and even less trivial.
