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The classical Neumann lemma states that if a group is covered by finitely many cosets, then at least one of these cosets is the coset of a subgroup of finite index. (Actually, the lemma says more, namely that the group is covered by the cosets of subgroups of finite index.)

I wonder if there is an infinitary version of the lemma in the following sense: Suppose that $G$ is a group and $G=\bigcup_{i<\alpha}x_iH_i$, where the $H_i$ are subgroups and $\alpha$ is an ordinal less than some big cardinal $\kappa$. (Maybe even $\kappa$ is strongly inaccessible.) Is it necessarily the case that there is some $i$ for which the index of $H_i$ in $G$ is less than $\kappa$?

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The statement as you have written it is not true for any uncountable cardinal $\kappa$. To see this, let $G$ be any group of size $\aleph_{\beta+\omega}$, where $\kappa=\aleph_\beta$. Every such group is the union of a countable increasing family of subgroups $G=\bigcup_n G_n$, where $G_n$ has size $\aleph_{\beta+n}$. This is simply because every set of size $\aleph_{\beta+\omega}$ is the union of sets of size $\aleph_{\beta+n}$. But each $G_n$, being so small, must have index $\aleph_{\beta+\omega}$ in $G$, which is much larger than $\kappa$. Thus, $G$ is the union of countably many subgroups, each of index far larger than $\kappa$. And so it is a counterexample to the statement you make.

(But perhaps the statement could become true if you consider only groups $G$ of size $\kappa$, in the case when $\kappa$ is a large cardinal or at least regular, and perhaps this is what you meant.)

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  • $\begingroup$ I see. In the situation that I am interested in, I know in addition that each $H_i$ has the same cardinality as $G$. Now the above proof does not apply. Perhaps the result could be true in this situation? $\endgroup$
    – Isaac Goldbring
    Commented Oct 23, 2010 at 1:58
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    $\begingroup$ The doesn't appear to be enough, since we can just add a copy of some very large group as a direct summand to all the groups, which will make them all the same size while preserving the essence of the calculation above as a counterexample. $\endgroup$
    – Joel David Hamkins
    Commented Oct 23, 2010 at 2:09
  • $\begingroup$ O.K. However, going back to your earlier parenthetical remark, I am fine assuming that the group is of size $\kappa$ for a big $\kappa$, say $\kappa$ strongly inaccessible. In addition, let us also assume that the $H_i$'s are also of size $\kappa$. $\endgroup$
    – Isaac Goldbring
    Commented Oct 23, 2010 at 2:16
  • $\begingroup$ That is fine with me, if it makes the statement true, but it is a serious retreat, since the original theorem was not just about countable groups. Such a statement would not generalize the original theorem in the case $\kappa=\omega$. $\endgroup$
    – Joel David Hamkins
    Commented Oct 23, 2010 at 2:21
  • $\begingroup$ Let me add one result of similar kind where there is an infinite version: Podoski proved in 2001 that if $\kappa$ is an infinite cardinal and $G$ is covered by $\kappa$ Abelian group, then the index of $Z(G)$ is at most $2^\kappa$. $\endgroup$
    – Péter Komjáth
    Commented Oct 23, 2010 at 5:50
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So it appears that my question is false. Here is a counterexample communicated to me by Ehud Hrushovski, although he attributes the counterexample to Macpherson and Neumann: Fix an uncountable cardinal $\kappa$ and let $G$ be the group of permutations of $\kappa$ with finite support. Let $G_n$ be the subgroup of $G$ consisting of those permutations which fix $n\in \omega$. Then $G=\bigcup G_n$ and each $G_n$ has index $\kappa$ in $G$.

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  • $\begingroup$ Aha! Very good. So even the big retreat mentioned in the other comments is not enough. $\endgroup$
    – Joel David Hamkins
    Commented Oct 25, 2010 at 10:40
  • $\begingroup$ I initially understood there's a typo, but it's just that one thinks of $\omega$ as subset of $\kappa$. $\endgroup$
    – YCor
    Commented May 22, 2020 at 15:01
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Edit: I assume that the cardinality of the group is much bigger than the cardinality of the set of cosets.

Look at Tomkinson, M. J., Groups covered by finitely many cosets or subgroups. Comm. Algebra 15 (1987), no. 4, 845–859. It gives a quantitative version of Neumann's result. It's methods, I think, generalize to the infinite case (at least to "very large" cardinals).

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  • $\begingroup$ Regarding your edit, the issue isn't that the group should be large relative to the number of cosets, but rather whether it is large relative to the desired index bound $\kappa$. After all, in my counterexample, we have a very large group as the union of a small number of subgroups, each of very large index. $\endgroup$
    – Joel David Hamkins
    Commented Oct 23, 2010 at 3:00
  • $\begingroup$ @Joel: Isn't $k$ is supposed to be inaccessible? $\endgroup$
    – user6976
    Commented Oct 23, 2010 at 3:04
  • $\begingroup$ My counterexample works for any uncountable $\kappa$, including inaccessible $\kappa$. But perhaps you are assuming that $\kappa$ is also the size of $G$? This was not part of the OP's original statement. $\endgroup$
    – Joel David Hamkins
    Commented Oct 23, 2010 at 3:13
  • $\begingroup$ @Joel: I would assume (following Neumann) that the size of $G$ is "very large", and the number of cosets is "very small". If needed, yes, one should assume, say, that card$(G)$ is inaccessible. $\endgroup$
    – user6976
    Commented Oct 23, 2010 at 3:32
  • $\begingroup$ So let me clarify the question once again: Let us assume that $G$ is a group of cardinality $\kappa$, where $\kappa$ is inaccessible. Furthermore, let us assume that we have a collection $(H_i:i<\alpha)$ of subgroups of $G$, where $|\alpha|<\kappa$. I am also fine in assuming that each $|H_i|<\kappa$. Finally, assume that we have $(x_i:i<\alpha)$ so that $G=\bigcup_{i<\alpha}x_iH_i$. Then is it the case that there is $H_i$ with index $<\kappa$? $\endgroup$
    – Isaac Goldbring
    Commented Oct 23, 2010 at 14:23

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