The isomorphism problem for abelian projective planes
Let \(G\) be a cyclic group of order \(v\), and let \(\Pi\) and \(\Pi'\) be two cyclic incidence structures with point set \(V=G\), i.e., \(G\) acts by right translation as an automorphism group of both \(\Pi\) and \(\Pi'\). Then \(\Pi\) and \(\Pi'\) are called \textit{multiplier equivalent} if some automorphism \(\alpha\) of \(G\) induces an isomorphism between \(\Pi\) and \(\Pi'\). Then \(\alpha\) has the form \(\alpha:x\mapsto tx\), where \(t\) is a positive integer with \(\gcd(v,t)=1\). Moreover, if \(G\) is an arbitrary finite group, then \(\Pi\) and \(\Pi'\) are called \textit{equivalent with respect to \(G\)} if some group isomorphism \(\alpha\) of \(G\) induces an isomorphism between \(\Pi\) and \(\Pi'\). This article proves that if \(G\) is an abelian group of order \(n^2+n+1\), then two \(G\)-invariant projective planes of order \(n\) are isomorphic if and only if they are equivalent with respect to \(G\). As a corollary, this result shows that two cyclic projective planes of order \(n\) are isomorphic if and only if they are multiplier equivalent, that is, if and only if the associated difference sets in \(Z_{n^2+n+1}\) are equivalent. The motivation of the author is the Bays-Lambossy theorem. The Bays-Lambossy theorem states that if \(p\) is a prime, and \(\Pi\) and \(\Pi'\) are two cyclic incidence structures on \(p\) points, where the point sets of both structures are the cyclic group \(Z_p\), then if \(\Pi\) and \(\Pi'\) are isomorphic, then they are multiplier equivalent. In this way, the author obtains an equivalent to the Bays-Lambossy theorem for finite projective planes.
scientific article; zbMATH DE number 5314403