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This question is related to

Are contranormally closed subgroups normal subgroups?

where one can also find the definition of a contranormally closed subgroup.

Given a group $G$, a contranormally closed subgroup $H$ of $G$ and a contranormally closed subgroup $K$ of $H$, is $K$ a contranormally closed subgroup of $G$? I conjecture that this is in general not true (similar as for normal subgroups).

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I think this property is transitive. Let $L$ be the contranormal closure of $K$ in $G$. Then the normal closure of $H$ in $\langle H,L \rangle$ contains $H$ and $L$, and so is equal to $\langle H,L \rangle$.

But then since $H$ is contranormally closed in $G$, we get $\langle H,L \rangle=H$, so $L \le H$, and hence, since $K$ is contranormally closed in $H$, $L=K$.

Incidentally, for finite groups I think that a subgroupis contranormally closed if and only if it is subnormal.

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  • $\begingroup$ You are right, thank you! $\endgroup$ – Paul Frost Jan 30 '18 at 23:27
  • $\begingroup$ In that case you should accept (and probably upvote) the answer. $\endgroup$ – Derek Holt Feb 1 '18 at 8:29
  • $\begingroup$ You are right again, sorry that I forgot to do that! $\endgroup$ – Paul Frost Feb 2 '18 at 9:43

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