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It is well-known that a finite group can be generated by arbitrary representatives of each conjugacy class, see e.g.: https://mathoverflow.net/questions/26979/generating-a-finite-group-from-elements-in-each-conjugacy-class

Now, for infinite groups this fails in general, for example the upper triangular matrices are a conjugate-dense subgroup of $GL_n(K)$ if $K$ is an algebraically closed field. So in this example there is some choice of representatives of conjugacy classes such that these elements do not generate the whole group. I am looking now for other examples:

  1. Is there an infinite group $G$ such that for any choice $S$ of representatives of conjugacy classes, the subgroup generated by $S$ is proper.
  2. As in 1. and additionally requiring that $G$ is finitely generated.
  3. As in 2. and additionally requiring that $G$ has finitely many conjugacy classes.
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Timm von Puttkamer
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1 Answers1

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One easy way to get examples of type (1) is to take an uncountable group with countably many conjugacy classes. For instance, let $S$ be an uncountable set and take the group $G$ of finite-support permutations of $S$. Then $G$ has only countably many conjugacy classes, since permutations with the same cycle structure are conjugate. So if you pick one element from each conjugacy class, they can only generate a countable subgroup of $G$.

There also exist infinite groups in which all non-identity elements are conjugate, so a subgroup generated by representatives of the conjugacy classes would be cyclic and not the whole group. Here's the idea: start with an infinite torsion-free group $G_0$, then for each pair of non-identity elements of $G_0$, adjoin a new generator to your group which conjugates one into the other. This gives a new group $G_1$, which can be shown to still be torsion free. Repeat this process infinitely many times and take the direct limit, and you get a group in which any two non-identity elements are conjugate. See Infinite group with only two conjugacy classes for a bit more detail and some references.

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Eric Wofsey
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