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A NEW MENON’S IDENTITY FROM GROUP ACTIONS
- Part of
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- Journal:
- Bulletin of the Australian Mathematical Society / Volume 99 / Issue 3 / June 2019
- Published online by Cambridge University Press:
- 28 November 2018, pp. 369-375
- Print publication:
- June 2019
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- Article
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- You have access
- Export citation
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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 where
$$\begin{eqnarray}\mathop{\sum }_{x\in H_{k}}(x-1,p^{\unicode[STIX]{x1D6FC}})=\frac{\unicode[STIX]{x1D711}(p^{\unicode[STIX]{x1D6FC}})}{k}\bigg(1+k(\unicode[STIX]{x1D6FC}-\unicode[STIX]{x1D6FD})+u\frac{p^{\unicode[STIX]{x1D6FD}}-1}{p-1}\bigg),\end{eqnarray}$$
$\unicode[STIX]{x1D711}(n)$ is the Euler totient function.
