LAUSR

Let $n$ be a positive integer. We obtain new Menon’s identities by using the actions of some subgroups of $(\mathbb{Z}/n\mathbb{Z})^{\times }$ on the set $\mathbb{Z}/n\mathbb{Z}$ . In particular, let $p$ be an odd prime and let $\unicode[STIX]{x1D6FC}$ be a positive integer. If $H_{k}$ is a subgroup of $(\mathbb{Z}/p^{\unicode[STIX]{x1D6FC}}\mathbb{Z})^{\times }$ with index $k=p^{\unicode[STIX]{x1D6FD}}u$ such that $0\leqslant \unicode[STIX]{x1D6FD}<\unicode[STIX]{x1D6FC}$ and $u\mid p-1$ , then $$\begin{eqnarray}\mathop{\sum }_{x\in H_{k}}(x-1,p^{\unicode[STIX]{x1D6FC}})=\frac{\unicode[STIX]{x1D711}(p^{\unicode[STIX]{x1D6FC}})}{k}\bigg(1+k(\unicode[STIX]{x1D6FC}-\unicode[STIX]{x1D6FD})+u\frac{p^{\unicode[STIX]{x1D6FD}}-1}{p-1}\bigg),\end{eqnarray}$$ where $\unicode[STIX]{x1D711}(n)$ is the Euler totient function.