Let $G$ be a finite group and let $C_1,...,C_k$ be conjugacy classes of $G$. We define the following set:

$\Sigma$ = $\{(g_1,...,g_k) \in \prod_{i=1}^{k}C_i \hspace{1mm} | \hspace{1mm} \prod_{i=1}^{k}g_i = 1 \text{ and } \langle g_1,...,g_k\rangle = G\}$,

where $1$ denotes the unit element of $G$ and $\langle g_1,...,g_k\rangle$ stands for the group generated by $g_1,...,g_{k-1}$ and $g_k$. The group $G$ acts on $\Sigma$ by conjugation.

Assume now that $G$ has a trivial center. Then the action of $G$ on $\Sigma$ is $\textit{free}$. For a proof of this fact, see paragraph 7.3 on page 70 of these notes of a course "Topics in Galois Theory" given by J-P. Serre at Harvard University in the Fall of 1988.

Assume $k > 1$. We call a $k$-tuple of conjugacy classes $(C_1,...,C_k)$ of $G$ $\textit{rigid}$ if $\Sigma \neq \emptyset$ and if $G$ acts transitively on $\Sigma$.

Question: does, for every finite group $G$ with trivial center, there exist a $k > 1$ (depending on the group) such that $G$ has a rigid $k$-tuple of conjugacy classes?

Or, put differently:

Question: Does there exist a finite centerless group $G$ that does not have a rigid $k$-tuple of conjugacy classes for all $k > 1$?


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