On the symmetric difference of two sets in a group
In the proof of certain combinatorial theorems on sums of sets in a general group the results on symmetric differences of two sets play an important role. The first main result of the paper is as follows: Suppose X and Y are finite subsets of a group G, H and K are subgroups, \(X+H=X\), \(Y+K=Y\), \(X+K\neq X\), and \(Y+H\neq Y\). Then \(| X\setminus Y| +| Y\setminus X| \geq | H| +| K| -2| H\cap K|\). This removes the hypothesis that H, K are normal subgroups in the known results. This removal leads, however, to two possible modifications of the hypothesis. The second main result of the paper contains the alternative in which the subgroups H, K fix the sets X, Y on the opposite sides, i.e. \(H+X=X\) and \(Y+K=Y\). Further, in an erlier paper [J. Number Theory 18, 110-120 (1984; Zbl 0524.10043)] the author proved that if \(C=A+B\), where A, B are finite subsets in a group G, then there is a subset S of C such that \(| S| \geq | A| +| B| - | H|\). In the present paper he proves that among the sets S satisfying this conclusion there is a unique maximal one that include all the others. In the last part of the paper a conjecture of \textit{G. T. Diderrich} [Proc. Am. Math. Soc. 38, 443-451 (1973; Zbl 0266.20041)] is disproved by constructing a pair of sets A, B in a group G of order 144 satisfying \(| A+B| <| A| +| B| -1\), but no one of the relations \(g+A+B=A+B\), \(A+g+B\subseteq A+B\), \(A+B+g=A+B\) has a solution \(g\neq 0\) in G.
scientific article; zbMATH DE number 3961624
scientific article