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16.1 (2006)

Partially Solved

Let $G$ be a finite non-abelian group, and $Z(G)$ its centre. One can associate a graph $\Gamma_G$ with $G$ as follows: take $G \setminus Z(G)$ as vertices of $\Gamma_G$ and join two vertices $x$ and $y$ if $xy \neq yx$. Let $H$ be a finite non-abelian group such that $\Gamma_G \cong \Gamma_H$.
$\qquad$ a) If $H$ is simple, is it true that $G \cong H$?
$\qquad$ b) If $H$ is nilpotent, is it true that $G$ is nilpotent?
$\qquad$ c) If $H$ is solvable, is it true that $G$ is solvable?

Contributor: A. Abdollahi, S. Akbari, H. R. Maimani

Progress

a) Comment of 2009: This is true for groups with disconnected prime graphs (L. Wang, W. Shi, Commun. Algebra, 36 (2008), 523–528).
a) Yes, it is mod CFSG (Ch. Khan', G. Ch`en', S. Go, Siberian Math. J., 49, no. 6 (2008), 1138–1146, for sporadic simple groups; A. Abdollahi, H. Shahverdi, J. Algebra, 357 (2012), 203–207, for alternating groups; R. M. Solomon, A. J. Woldar, J. Group Theory, 16, no. 6 (2013), 793–824, for simple groups of Lie type).

b) Comment of 2009: This is true if $|H| = |G|$ (A. Abdollahi, S. Akbari, H. R. Maimani, J. Algebra, 298 (2006), 468–492).
b) Comments of 2023: This is true if $|Z(H)| \geqslant |Z(G)|$ (H. Shahverdi, J. Algebra, 642 (2024), 60–64), using the earlier result under the additional assumption that the centralizer of every non-central element in $H$ is abelian (V. Grazian, C. Monetta, J. Algebra, 633 (2023), 389–402).

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