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Let $G$ be a group, and let $\mathcal C$ be the collection of conjugacy classes of $G$. Let $S$ be a traversal of $\mathcal C$ (that is $S$ contains exactly one element from each set in $\cal C$). Is it necessary that $G=\langle S\rangle$?

This is affirmative if $G$ is finite as answered here. The proof utilizes the fact that a finite group $G$ is not a union of the conjugate subgroups $gHg^{-1}$ for any proper subgroup $H\leq G$. This lemma of the proof can be generalized to an infinite group $G$ and a subgroup $H$ of finite index as exemplified here.

These generalizations give us pretty strict requirements on a counter-example. A counter-example to my question would have to be non-abelian (a traversal of the conjugacy classes is the whole group) and have a proper subgroup generated by a traversal without finite index. Overmore, suppose there does exists such a group $G$ and a proper subgroup $H$ such that $H$ traverses the conjugacy classes. Then $H$ cannot contain a subgroup of finite index in $G$ since the subgroups of finite index form a filter in the subgroup lattice.

Any help in coming up with a counter-example or proving the general claim is appreciated.

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2 Answers 2

up vote 6 down vote accepted

A simpler example is provided by ${\rm GL}(n,\mathbb{C})$ for any $n >1.$ We may choose a transversal to the conjugacy classes which consists completely of upper triangular matrices ( as every matrix in ${\rm GL}(n,\mathbb{C})$ is conjugate to an upper triangular matrix). This transversal generates a proper subgroup of $G,$ as the subgroup it generates consists exclusively of upper triangular matrices.

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thank you. that's much simpler. –  Bryan Jul 22 at 6:18

No. For example, there exist infinite groups with exactly two conjugacy classes. There might be easier counterexamples though.

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whoa ok. Yeah, I hope there is an easier counter-example. –  Bryan Jul 22 at 5:34

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