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This question is in some sense equivalent to my question here. A proof would answer that question in the case when the base field is perfect.

Let $G$ be a profinite group of cardinality $\kappa$, where $\kappa$ is an infinite cardinal. Is it the case that for every infinite cardinal $\mu \leq \kappa$ there's a subgroup $H \subset G$ of cardinality $\mu$? As a follow-up does there need to be a such a subgroup that's normal (I'm just curious about this)?

I don't have a good intuition for infinite group theory and I have an even worse understanding of profinite groups, so I have made little progress on this question.

Edit: It seems to me that for a group $G$, a subgroup $H \subset G$ and $g \in G \setminus H$ we have that $|\langle g,H\rangle|\leq \max\{\aleph_0,|H|\}.$ This would seem to imply that a transfinite induction argument would answer the question yes in the general case.

Community
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JSchlather
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1 Answers1

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Let $G$ be an infinite group of cardinality $\kappa$. Then, over ZFC, I can show:

  1. For each infinite cardinal $\lambda < \kappa$ there is a proper subgroup of cardinality $\lambda$
  2. In general there is no proper subgroup of cardinality $\kappa$
  3. If $G$ is profinite, there are proper subgroups of all cardinalities $\le \kappa$

Proof: i) Choose a subset $X\subseteq G$ of cardinality $\lambda$. Then $H := \langle X \rangle$ is a subgroup of cardinality $\lambda$; in particular, it's a proper subgroup.

ii) $\mathbb{Z}_{p^\infty}:= \{z \in \mathbb{C}^\times\mid \exists n>0: z^{p^n}=1\}$ ($p$ a prime) is countably infinite, but all proper subgroups are finite (Rotman, Introduction to the Theory of Groups, Theorem 10.13).

iii) In the profinite case, $G=\varprojlim_N G/N$ where $G/N$ is finite. Hence the cardinality of $N$ is $\kappa$.

Ralph
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  • In 1. you need to assume that $\lambda$ is infinite, of course. Otherwise this is spot on. – Pete L. Clark Mar 19 '13 at 20:42
  • In discussing 2., one might also want to mention Tarski monster groups: http://en.wikipedia.org/wiki/Tarski_monster_group. – Pete L. Clark Mar 19 '13 at 20:45
  • Yes, of course. Thanks for the hint, Pete. – Ralph Mar 19 '13 at 20:57
  • @Ralph I am not yet good with infinite cardinals: expanding on (1), can we expect that $\langle g^{−1}Xg:g∈G\rangle$ will also be a subgroup of cardinality λ? (and thus a normal subgroup?) – Alexander Gruber Mar 20 '13 at 05:49
  • No, this can't be expected. As an example let $X \subset Y$ be infinite sets. Then, for a fix $x \in X$, ${yxy^{-1}\mid y \in Y}$ is a subset of of cardinality $|Y|$ of $G := F(Y)$. Hence the supset $\langle gXg^{-1}\mid g \in G\rangle$ has cardinality $|Y|$ as well. – Ralph Mar 20 '13 at 07:19
  • Also note, that there are simple groups of all cardinalities. Thus, proving the existence of normal subgroups, is at best possible for profinite groups (if it's true at all). – Ralph Mar 20 '13 at 07:44