Free groups and automorphism groups of infinite structures (Q2879421)

It is known that every infinite group \(G\) is the automorphism group of a \(|G|\)-structure, where for an infinite cardinal \(\kappa\), a \(\kappa\)-structure is a structure such that the corresponding language and the domain of the structure are both of cardinality at most \(\kappa\). In contrast, for every infinite \(\lambda\), there is a group of cardinality \(\lambda^+\) that is not the automorphism group of a \(\lambda\)-structure.NEWLINENEWLINE Here, the authors deal with free groups and investigate the following question: Given an infinite cardinal \(\lambda\), is there a free group of rank greater that \(\lambda\) that is the automorphism group of a \(\lambda\)-structure?NEWLINENEWLINEIt was shown by \textit{S. Shelah} [Bull. Lond. Math. Soc. 35, No. 1, 1--7 (2003; Zbl 1033.20002)] that a free group of uncountable rank is not the automorphism group of an \(\aleph_0\)-structure.NEWLINENEWLINE The main result of the paper is:NEWLINENEWLINE Theorem 1.4. Let \(\lambda\) be a cardinal with \(\lambda = \lambda^{\aleph_0}\). Then, the free group of rank \(2^{\lambda}\) is the automorphism group of a \(\lambda\)-structure.NEWLINENEWLINE The authors show that the cardinal assumption \(\lambda = \lambda^{\aleph_0}\) is not necessary. They derive large cardinal strength from the nonexistence of certain automorphism groups. So, they show that if \(\lambda\) is singular with \(\mathrm{cf}(\lambda) > \aleph_0\), and there is no free group of rank greater than \(\lambda\) that is the automorphism group of a \(\lambda\)-structure, then there is an inner model with a Woodin cardinal.NEWLINENEWLINE The techniques developed for the proof of Theorem 1.4 can be applied in various varieties of groups. So, the free abelian group of rank \(2^{\lambda}\) is the automorphism group of a \(\lambda\)-structure whenever \(\lambda=\lambda^{\aleph_0}\).