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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.

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up vote 1 down vote accepted

$N=\cap H^g$ 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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Do you not need finitely generated for $\cap H^g$ to be of finite index? – user1729 May 22 '12 at 9:13
Nice. Thank you. – Peter May 22 '12 at 10:12
@user1729: You just need $H$ to be of finite index. When $H$ has index $n$, there is a homomorphism from $G$ to $S_n$ (coset action) where the kernel is $\cap H^g$ (the normal core). By the first isomorphism theorem $G/\cap H^g$ is finite since $S_n$ is. – Mikko Korhonen May 22 '12 at 14:24
@m.k.: Ah yes, I was thinking you needed to intersect all the subgroups of index $n$ (which will give you characteristic). – user1729 May 22 '12 at 14:47

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! – Peter May 22 '12 at 10:01

If $U$ is an infinite subgroup of $G$ with and $H$ is a finite-index subgroup of $G$ then $U\cap H\neq 1$.


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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It is not assumed that $U$ has infinite index in $G$. – Brian M. Scott May 22 '12 at 8:55
@BrianM.Scott seems I can't read. Ah well, I don't use that fact anywhere anyway! – user1729 May 22 '12 at 8:58
Why does $uv^{-1}\in H$ implies $u,v\in H$? – Peter May 22 '12 at 10:17
@Peter: No, that's a typo. I meant $u, v\in U$. This means that $uv^{-1}\in U$, so $uv^{-1}\in U\cap H$. – user1729 May 22 '12 at 10:20
Ah. Okay. It seems, that I can't read, too. ;-) – Peter May 22 '12 at 10:30

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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Clear exhibition of the use of cosets, but $H$ is not supposed to be normal, is it? If it is not normal, how can one define the map $f$? So in any case, it appears similar to the above answer. And why is the intersection of $U$ with a point $y$ equal to a coset of $U$ intersection $H$? – awllower May 24 '12 at 5:51
@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. – Marc van Leeuwen May 24 '12 at 8:02

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