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To clarify, I am not looking for a classification but rather for well-researched examples of families of (finitely generated) groups generated by a single conjugacy class.

A collection of examples, some from the answers and comments, is as follows.

  1. Let $W$ be a Coxeter group with presentation $$\langle s_1, \dots, s_n \;|\; (s_is_j)^{m_{ij}} \rangle,$$ with $m_{ii} = 1$ and $2 \leq m_{ij} = m_{ji} \leq \infty$ whenever $i \neq j$ (the relation $(s_is_j)^\infty$ stands for "no relation"). Suppose that the graph $G$ with vertices $s_1, \dots, s_n$ and edges between $s_i$ and $s_j$ whenever $m_{ij}$ is finite and odd, is connected. Then, all the $s_i$ are conjugate and thus, $W$ is generated by a single, somehow distinguished conjugacy class. Specific groups in this family include dihedral groups $D_m$ for odd $m$ (the groups obtained when $n=2$) and symmetry groups $S_n$ (by letting $m_{i(i+1)} = 3$ and all other $m_{ij} = 2$).
  2. Braid groups $B_n$ (or more generally Artin groups for which the same criterion on the off-diagonal weights as above holds) also satisfy that all standard generators are conjugate.
  3. Mapping class groups of surfaces are generated by finitely many conjugate Dehn twists around non-separating curves.
  4. Knot groups are generated by finitely many meridians.

Are there any other families that come to mind?

Levi
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  • Dihedral groups $D_n$ are not generated by a single conjugacy class if $n$ is even. I'm sure you know that, but it's not clear from the text of your question. – Matt Samuel Oct 27 '19 at 16:13
  • @MattSamuel It technically is because I require $n = m_{12}$ to be odd, but yes, this might be confusing. I will add this. – Levi Oct 27 '19 at 16:14
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    You might be interested in this recently asked question and answers. One example on their is knot groups. –  Oct 27 '19 at 16:44
  • @PaulPlummer You smelled well, knot groups are precisely the reason why I am asking this question :) – Levi Oct 27 '19 at 16:50
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    There's some objects called quandles which are algebraic structures that are sort of like conjugation in a group and there's a notion of a "connected quandle" which is like a conjugacy class. There's also some ways to turn quandles into groups that are "generated" by the quandle somehow. I don't know very much about the theory, but I'm pretty sure that there are at least some answers for questions like yours with quandles. (...but I don't think quandles are classified very well, especially noting that knots have "fundamental quandles" which are classifying invariants...) – Milo Brandt Oct 28 '19 at 03:21
  • Your description of Coxeter groups leaves out the relation $s_i^2$ for all $i$. Without that, as you have written it, what you get are Artin groups. – Lee Mosher Nov 04 '19 at 11:37
  • @LeeMosher I believed that this was implied by using the word Coxeter group. I also do not mention that the $m_{ij}$ are not allowed to be $1$ but they are allowed to be $\infty$. Should I define what Coxeter groups are when giving such an example? – Levi Nov 04 '19 at 11:40
  • @LeeMosher In any case, I have edited the question according to your suggestion - do you think it is more clear now? – Levi Nov 04 '19 at 11:44
  • Well, that's a fine convention in a context where only Coxeter groups were under discussion. All I'm saying is that in a context where both Coxeter and Artin groups are under discussion, and furthermore in 2 you are actually inviting your reader to examine your Coxeter group presentation while pondering the Artin group, that convention invites confusion. – Lee Mosher Nov 04 '19 at 11:46
  • @LeeMosher You are right. Thanks for the constructive and useful criticism! – Levi Nov 04 '19 at 11:48
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    No problem. This is a nice series of questions by the way. – Lee Mosher Nov 04 '19 at 11:48

3 Answers3

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The usual(?) Humphries generators for the mapping class group are Dehn twists about non-separating curves, and thus all conjugate.

I believe something similar ought to be true about Out$(F_n)$, the outer automorphism group of a free group of rank $n$. The Nielsen generators are not all conjugate (some have finite order while others do not), but I would be surprised if some smaller subset of the Nielsen generators didn’t suffice.

If $G$ is a torsion-free group, a result of Higman, Neumann and Neumann says that $G$ embeds into a larger group that has one conjugacy class of nontrivial elements.

Note that the statement that generators $g$ and $h$ are conjugate abelianizes to the statement that the image of $g$ is equal to the image of $h$. So a necessary condition for $G$ to be generated by a single conjugacy class is that $G$ has cyclic abelianization. (Thompson’s group $F$ is thus not an example.) I’d be curious to know if there are examples where $G$ has cyclic abelianization but $G$ is not generated by a single conjugacy class.

Robbie Lyman
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5

Given the mapping class group of a closed surface, there are many elements whose normal closure is the full group.

See this article by Lanier and Margalit for more details.

Santana Afton
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2

For every $n\geq2$, Denis Osin constructed an uncountable class of finitely generated groups each of which contains precisely $n$ conjugacy classes of elements. Therefore, the $n-1$ non-trivial conjugacy classes generate the group, as they are the entire group, minus the identity! This is Corollary 1.4 of the paper Osin, Denis. "Small cancellations over relatively hyperbolic groups and embedding theorems." Annals of Mathematics 172.1 (2010): 1-39.

Taking $n=2$ then gives a class of groups which are generated by a single conjugacy class. Here the groups are two-generated (in the usual sense; see his Corollary 1.3) and have precisely two conjugacy classes, so all the non-trivial elements are conjugate. Proving such finitely generated groups exist was an open problem at the time.

user1729
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