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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.

4 Answers 4

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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.

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    @awllower: The map $p:G\to G/H$ is defined for every subgroup $H$ of $G$, not necessarily normal. It is the canonical projection to the set $G/H$ of left cosets (oops, I got left and right wrong in my answer; will correct it), which sends every $g\in G$ to its left coset $gH$. As a consequence if $y\in G/H$ is a left coset, then $p^{-1}(y)=\{ g\in G\mid p(g)=y\}$ is _equal_ to $y$ (viewed now as a subset of $G$). In the restriction $f$ of $p$ to $U$ we must intersect these fibers with $U$ to get a subset of the domain of $f$. For $y\cap U=u(U\cap H)$, check $\subseteq$ and $\supseteq$ as said.2012-05-24
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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?

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    If K is trivial we get infinitely many left cosets and so the index of H would be infinte. Thanks!2012-05-22
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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).

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    @Peter: No, $y$ou read correctly. I edited my post a$f$ter your comment.2012-05-22
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$N=\cap H^g$ 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.

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    @m.k.: Ah yes, I was thinking you needed to intersect all the subgroups of index $n$ (which will give you characteristic).2012-05-22