Let $G$ be a second countable profinite group, $g\in G$ and $g^G:=\{hgh^{-1}~|~h\in G\}$ the conjugacy class of $g$ in $G$. Theorem 3.2 in Wesolek's [Conjugacy class conditions in locally compact second countable groups](https://www.ams.org/journals/proc/2016-144-01/S0002-9939-2015-12645-7/) gives a characterization of a profinite group having an open conjugacy class. For example, $g^G$ is open if and only if the Haar measure on $g^G$ is greater than $0$. In particular, if $g^G$ is open, then $g$ must has finite order in $G$, see Lemma 3.1 of the paper. I'm interested in the following case:


**Question:** Let $G$ be a second countable infinite profinite group, $g\in G$ of finite order such that the conjugacy class $g^G$ of $g$ generates $G$ as an abstract group. Is $g^G$ open in $G$?


Since infinite profinite group is uncountable, the set $g^G$ must be infinite. Moreover, $g^G$ and $G$ have the same cardinality by the question [Does a countable set generate a countable group?](https://math.stackexchange.com/questions/346244/does-a-countable-set-generate-a-countable-group?stw=2). In this sense, one can expect that the Haar measure on $g^G$ should be greater than $0$, and hence it should be true.



Open conjugacy classes in a second countable profinite group
Yes, there exist profinite groups $G$ with a conjugacy class of empty interior and consisting of elements of finite order, generating $G$ as an abstract group.


Let $H$ be a nonabelian finite simple group, and $S$ a nontrivial conjugacy class in $H$. Consider $G=H^\mathbf{N}$ and $T=S^\mathbf{N}$. Then $T$ is a single conjugacy class in $G$ and has empty interior. To be done, it is enough to check that $T$ generates $G$.


This is equivalent to asking that $S^n=H$ for some $n$ (where $S^n$ is the set of products of *exactly* $n$ elements of $S$). I don't know if it's always true. However it is enough to show that it's true in a single case.


Indeed, using that for a prime $p$ in $4\mathbf{N}+1$, every nonzero element is a sum of two nonzero squares, we see that every upper or lower unipotent matrix in $\mathrm{SL}\_2(\mathbf{F}\_p)$ is product of two conjugates of the matrix $u=\begin{pmatrix}1 & 1 \\ 0 & 1\end{pmatrix}$, and hence that every element of $\mathrm{SL}\_2(\mathbf{F}\_p)$ is product of exactly 6 $\mathrm{SL}\_2(\mathbf{F}\_p)$-conjugates of $u$. (For other odd primes it still holds that every element is sum of three nonzero squares and hence every element of $\mathrm{SL}\_2(\mathbf{F}\_p)$ is product of exactly 9 $\mathrm{SL}\_2(\mathbf{F}\_p)$-conjugates of $u$.) Hence $H=\mathrm{PSL}\_2(\mathbf{F}\_p)$ works.


[NB The proof also works, in the same way, for the product $\prod\_n\mathrm{PSL}\_2(\mathbf{F}\_{p^n})$, which has the additional requirement of being topologically finitely generated. Note that in this case the element $u$ still has order $p$.]




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Edit (given follow-up question in comments): for an odd prime $p$, $\mathrm{SL}\_3(\mathbf{Z}\_p)$ has finite Prüfer rank and is generated by the conjugacy class of the diagonal matrix $d$ given by the diagonal $(1,-1,-1)$.


Indeed, first of all, $\mathrm{SL}\_n$ over a Euclidean domain is generated by elementary matrices. So it is enough to show that the matrix $\begin{pmatrix}1 & x & 0\\ 0 & 1 & 0 \\ 0 & 0 & 1\end{pmatrix}$ is a product of conjugates of $d$. It is indeed the product $dd'$ where $d'=\begin{pmatrix}1 & x & 0\\ 0 & -1 & 0 \\ 0 & 0 & -1\end{pmatrix}$. Now $d'$ is conjugate to $d$, namely $d'=udu^{-1}$ with $u=\begin{pmatrix}1 & -x/2 & 0\\ 0 & 1 & 0 \\ 0 & 0 & 1\end{pmatrix}$.


