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Let $G$ be an infinite locally finite (non-solvable) group, let $\{H_i\}_{i\in\mathbb{N}}$ be a strictly totally ordered family of subgroups of $G$ and let $H$ be the intersection of that family. If $H$ is normal in $G$, it is clear that $G/H$ is an infinite locally finite group and hence there is no bound on the cardinality of the finite subgroups $F$ such that $\langle H,F\rangle$ is a proper subgroup of $G$. The same holds if $H$ is finite. Is this true for a not necessarily normal (or finite) subgroup $H$, intersection of such a family of proper subgroups of $G$? Under which conditions could this hold?

(Clearly, the only case to consider is that in which the family is ordered as the negative integers with $<$)

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I think the following example works. Let $X$ be any finite group and let $H = \times_{k \in{\mathbb N}} X_i$ be the (restricted) direct product of countably many copes of $X$. Let $G = H \wr C_2$.

So $G$ is an extension of the base group $H_1 \times H_2$ with $H_i \cong H$ by a group $F = \langle t \rangle$ of order $2$, where $H_1^t=H_2$.

Then $G = \langle H_1,F \rangle$, and it is straightforward to write $H_1$ as the intersection of an infinite strictly descending chain of subgroups.

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  • $\begingroup$ I'm now thinking that the title may be misleading. The question, posed in terms of chains, should be the following: "Is $H$, intersection of an infinite strictly descending chain of subgroups of a locally finite group $G$, contained in an infinite strictly ascending chain of subgroups of $G$ (just as in the cases in which $H$ is normal or finite)? Are there additional conditions for this to hold?" $\endgroup$ – Alex Doe Aug 17 '18 at 8:17
  • $\begingroup$ Which is clearly the case of your example, since $H_1$ is contained in the base group. But is it true in general? $\endgroup$ – Alex Doe Aug 17 '18 at 8:24
  • $\begingroup$ The example shows that the answer to the question in the title is yes. But I will think about your more detailed question. $\endgroup$ – Derek Holt Aug 17 '18 at 9:18
  • $\begingroup$ I know, but, trying to simplify my question for the title, I overdid it getting to a misleading title. The true question is in the text, as it should be. Sorry for the bungle! Maybe I should change the title. $\endgroup$ – Alex Doe Aug 17 '18 at 9:47
  • $\begingroup$ I've just changed the title, which now is general and hopefully not misleading! $\endgroup$ – Alex Doe Aug 20 '18 at 8:09

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