Kourovka Notebook Problem 10.49

Definitions and mathematical statement

Let $G$ be a group with identity $1$. A group is called simple if it is nontrivial and has no normal subgroups other than $\{1\}$ and itself. For $x\in G$, its conjugacy class is $x^G=\{gxg^{-1}:g\in G\}$. For a subset $X\subseteq G$ and a positive integer $r$, define \[ X^r=\{x_1\cdots x_r:x_1,\ldots,x_r\in X\}. \] Thus this notation means a product of exactly $r$ factors, with repetitions allowed. An abelian subgroup $H\leq G$ is maximal abelian if no strictly larger abelian subgroup of $G$ contains it. Write \[ N_G(H)=\{g\in G:gHg^{-1}=H\},\qquad C_G(h)=\{g\in G:gh=hg\}. \] The source asks whether there exists a group $G$ satisfying the following conditions, with integers $n,m$ interpreted as positive finite product lengths:

1. $G$ is simple, and there exists $n\geq 1$ such that \[ G=(x^G)^n\qquad\text{for every }x\in G. \] The integer $n$ is uniform over all conjugacy classes.
2. For any maximal abelian subgroups $H,K\leq G$, there exists $g\in G$ such that $K=gHg^{-1}$.
3. For every maximal abelian subgroup $H\leq G$, \[ N_G(H)=H,\qquad C_G(h)=H\quad\text{for every }h\in H\setminus\{1\}. \]
4. There exists $m\geq 1$ such that, for every maximal abelian subgroup $H\leq G$ and every $a\in G\setminus H$, \[ G=(aH)^m, \qquad aH=\{ah:h\in H\}. \] Equivalently, for every such $H,a$ and every $g\in G$, there exist $h_1,\ldots,h_m\in H$ with $g=(ah_1)\cdots(ah_m)$. The integer $m$ is uniform over $H$, $a$, and $g$.

No finiteness or infinitude assumption on $G$ is supplied. There is an unresolved qualification in condition (1): the source says "any conjugacy class," without excluding the identity class $\{1\}$. The displayed formulation retains that literal quantifier. Its compatibility with the usual nontrivial meaning of simplicity requires clarification; an intended restriction to nontrivial conjugacy classes has not been silently inserted. The source also does not explicitly specify the signs of the integers $n,m$; positive lengths are the standard interpretation of the stated products.

Short English statement

B. Poizat asks for a simple group with a uniform bound on products from each conjugacy class, conjugate maximal abelian subgroups that are self-normalizing and equal the centralizers of their nonidentity elements, and a uniform exact product length for every coset $aH$ with $a\notin H$. The source leaves the identity conjugacy class unexcluded, so the intended first condition needs clarification. A supplied 2005 comment reports a partial solution.

Sources

• Local UnsolvedMath registry, supplied complete source record: permanent ID $700664$, exact source problem code catalogue entry, title Kourovka Notebook Problem 10.49. The supplied registry status is open; this is a source label, not an independently verified mathematical or literature assessment.
• The Kourovka Notebook (Unsolved Problems in Group Theory), original problem 10.49, 10th issue (1986); proposed by B. Poizat. The supplied background cites the 21st issue edition (2026), arXiv:1401.0300v48. The supplied collection description names E. I. Khukhro and V. D. Mazurov as editors.
• Supplied Comment of 2005: a partial solution is reported in E. Jaligot and A. Ould Houcine, J. Algebra 280 (2004), 772--796. The supplied record gives neither an article title nor the scope of the partial solution. This historical report has not been independently verified here.