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Do there exist a group $G$ and its proper subgroup $H$, such that $G$ is covered by conjugates of $H$?

It is not hard to see, that if such $G$ and $H$ exist, then $|G:H| = \infty$.

Indeed, consider first the case, when $G$ is finite and $A \subset G$ is a left transversal of $H$. Then $|\bigcup_{g \in G} gHg^{-1}| = |\bigcup_{a \in A} aHa^{-1}| \leq \sum_{a \in A}|aHa^{-1}| - |\bigcap_{g \in G} aHa^{-1}| = |A||H| - |\bigcap_{g \in G} aHa^{-1}| = |G| - |\bigcap_{g \in G} aHa^{-1}| \leq |G|-1$.

Now, suppose $|G:H| < \infty$. Then $H$ contains a finite-index normal subgroup $N$ of $G$. Then if $G$ is covered by conjugates of $H$, then finite group $\frac{G}{N}$ is covered by conjugates of its proper subgroup $\frac{H}{N}$, which is proved above to be impossible.

However, I have no idea what to do about the case when $|G:H| = \infty$.

Chain Markov
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3 Answers3

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Let $G$ be the group of invertible $n \times n$ complex matrices and $H$ be the subgroup of upper triangular matrices. By the Jordan decomposition, conjugates of $H$ cover $G$.

Joshua P. Swanson
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Yes. In fact, Denis Osin proved that the "extreme" situation is possible: there exist (infinite) groups which are not $\mathbb{Z/2Z}$ and which have precisely two conjugacy classes of elements. Hence, the group is covered by conjugates of $\{1, g\}$ for any $g\in G$. Hence, it is also covered by conjugates of any non-trivial subgroup.

Proving this extreme case could occur was a well-known open problem at the time, and Osin's proof is (highly) non-trivial. The reference is: Osin, Denis. "Small cancellations over relatively hyperbolic groups and embedding theorems." Annals of mathematics (2010): 1-39. (doi, arXiv)

user1729
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Try this:

Take the group of permutations of the positive integers which fix all but finitely many points.

Take as your subgroup the elements which fix the integer $1$.

Mark Bennet
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