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Let $G$ be a group and let $(K_n : n \in \mathbb N)$ where $K_n \ne \{e\}$ for all $n$ be a descending central filtration of $G$ with the trivial intersection. Can anyone please give me examples of more or less general conditions on $G$ and $(K_n)$ (or examples in some 'not-too-small' classes of groups) under which any common supplement to all terms of the filtration, that is, $H \le G$ with

$$H K_n = G \qquad (n \in \mathbb N),$$

must be $G$ itself?

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I think this needs a more descriptive title. –  Andy Putman Jul 28 '10 at 16:29
If you don't get a good answer here, you might try asking this question on the Group Pub Forum: people.bath.ac.uk/masgcs/gpf.html –  Dan Ramras Jul 28 '10 at 19:16
Thanks, I will. –  VladAr Jul 28 '10 at 19:33
It is probably a good idea to mention in the question that the condition "G is nilpotent" works. See mathoverflow.net/questions/16777 for a related question. –  Jack Schmidt Jul 29 '10 at 20:08
You are right, for I certainly meant the situation where all terms of the filtration are nontrivial. Otherwise, only G is evidently the only common complement for all $K_n.$ –  VladAr Jul 29 '10 at 21:00

1 Answer 1

What is wanted, I suppose, are examples where each of the subgroups $K_i$ has a proper supplement but where no common proper supplement exists. Such examples occur in the infinite cyclic group $C$. First, note that each nonidentity subgroup K of C has a proper supplement. To see thus, observe that the index $|C:K|$ is finite. If we choose a prime $p$ not dividing $|C:K|$ then the (unique) subgroup $H$ having index $p$ in $C$ is a proper supplement for $K$.

Now for the example. Enumerate the prime numbers $p_1,p_2,p_3,\ldots$ in arbitrary order, and let $K_n$ be the unique subgroup of $C$ having index $p_1p_2\cdots p_n$. Then $K_n \supseteq K_{n+1}$, and we argue that $D = \bigcap K_n$ is trivial. Otherwise, $|C:D|$ is some integer $r$, and yet $|C:K_n|$ exceeds $r$ for sufficiently large $n$.

Finally to show that the only common supplement is the whole group $C$, suppose that $H$ is a proper common supplement, and note that $H > 1$, so $H$ has finite index. Let $q$ be a prime divisor of $|C:H|$, and let $n$ be such that $p_n = q$. Then $q$ divides the index of both $H$ and $K_n$, and thus each of these subgroups is contained in the subgroup of $C$ of index $q$. This contradicts $HK_n = C$.

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@Marty Isaacs: I appreciate your input. Indeed, the situation with the additive group Z of integers is quite clear: if K_n=m_n Z where m_n is in Z then all terms K_n can be simultaneously supplemented if and only if there is integer k which is coprime to all m_n. F.e. 2Z+3^nZ = Z for all natural numbers n > 1. Information on other, less well-behaving, classes of groups might be more intriguing, though. –  VladAr Jul 13 '12 at 14:40

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