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20.40 (2022)

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Let $G$ be a $\kappa$-existentially closed group of cardinality $\lambda > \kappa$, where $\kappa$ is a regular cardinal. Is it true that $|\text{Aut}(G)| = 2^\lambda$?

Contributor: B. Kaya, M. Kuzucuoğlu

Progress

The answer is known to be affirmative if $\lambda = \kappa$ (B. Kaya, M. Kuzucuoğlu, P. Longobardi, M. Maj, J. Algebra, 666 (2025), 840–849).

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