Symmetric harmoniousness of odd-order groups
Mohammad Javaheri
TL;DR
The paper proves that a finite group $G$ is symmetric harmonious if and only if $|G|$ is odd. A central tool is a lifting lemma for normal extensions: if $H$ is a normal subgroup of $G$ with $|H|$ and $|G/H|$ odd and both $H$ and $G/H$ are symmetric harmonious, then $G$ is symmetric harmonious. The authors construct a symmetric harmonious sequence in $G$ by combining a symmetric harmonious sequence in $H$ with one in $G/H$, taking care to align coset representatives to preserve the permutation property. Using Feit–Thompson, they complete the converse and derive consequences for $R^*$-sequenceable groups, linking the main result to the Hall–Paige condition and broadening the catalog of symmetric harmonious and R*-sequenceable groups.
Abstract
We prove that every odd-order group is symmetric harmonious: there exists a permutation $g_0,g_1,\ldots, g_{\ell-1}$ of elements of $G$ such that the consecutive products $g_0g_1,g_1g_2,\ldots, g_{\ell-1}g_0$ also form a permutation of elements of $G$ and $g_{\ell-i}=g_i^{-1}$ for all $1\leq i \leq \ell-1$. We apply this result to obtain new examples of R*-sequenceable groups.