Let $H$ be an infinite subgroup of a discrete group $G$. $H$ is called nearly normal if it is commensurable with a normal subgroup $K$ of $G$, that is $H\cap K$ is a finite index subgroup of both $H$ and $K$. If $H$ contains a finite index subgroup which is also a normal subgroup of $G$, then $H$ is obviously nearly normal. Is the converse also true? Or does there exist an example of a nearly normal subgroup $H$ which does not contain any finite index subgroup which is normal in $G$? If there is such a counterexample, can we at least show that every nearly normal subgroup contains a non-trivial normal subgroup of $G$? Or perhaps there is a counterexample for this question too?
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2$\begingroup$ When $H$ is finitely generated, the answer is 'yes'. Let $L$ be the intersection of all subgroups of $K$ of index equal to $|K:H\cap K|$. Then $L$ is clearly contained in $H$, and is characteristic in $K$, hence normal in $G$. $\endgroup$– HJRWFeb 3, 2014 at 11:06
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1$\begingroup$ Also, you ask 'If there is such a counterexample, can we at least show that every nearly normal subgroup contains a non-trivial normal subgroup of G?'. This can't hold. Suppose $L$ is the maximal such non-trivial subgroup. Then $G/L$ is a counterexample to the weaker question. $\endgroup$– HJRWFeb 3, 2014 at 11:10
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5$\begingroup$ And if $H$ is not f.g., then the answer is `no'. For example, one can let $G$ be the lamplighter group $\mathbb{Z}_2 wr \mathbb{Z}$, let $K\cong \mathbb{Z}_2^\infty$ be the kernel of the homomorphism from $G$ to $\mathbb{Z}$, and let $H$ be the index $2$ subgroup of $K$, which is the kernel of the projection from $K$ to one of the coordinates. Then the intersection of all conjugates of $H$ in $G$ is trivial. $\endgroup$– Ashot MinasyanFeb 3, 2014 at 11:12
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$\begingroup$ @HJRW: Thanks for your comment. But I am not a group theorist and I still don't understand: Where did you use the assumption that $H$ is finitely generated? And, is $L$ finite index in $H$? And, why is $L$ non-trivial? $\endgroup$– user23860Feb 3, 2014 at 11:14
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1$\begingroup$ @VahidShirbisheh - the fact that $H$ (and hence $K$) is finitely generated implies that there are only finitely many subgroups of $K$ of fixed index. (The number is bounded above by the number of such subgroups of the free group of appropriate rank.) Therefore, $L$ is an intersection of finitely many subgroups of finite index in $K$, and hence also of finite index in $K$ (and hence $H$). $\endgroup$– HJRWFeb 3, 2014 at 11:17
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Let $K$ be an infinite product of cyclic groups of order 2, and $H$ an index two subgroup of $K$. Let $A$ be the full automorphism group of $K$, and $G=K\rtimes A$.
Then since $G$ acts transitively on the non-trivial elements of $K$, $H$ has no non-trivial subgroup normal in $G$.
answered Feb 3, 2014 at 11:11
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5$\begingroup$ Of course, this is essentially the same as the example almost simultaneously posted in comments by Ashot Minasyan, but he wins 'cos his $G$ is finitely generated. $\endgroup$ Feb 3, 2014 at 11:23
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