On the minimum degree of the power graph of a finite cyclic group

TLDR: In this article, it was shown that for any integer n ≥ 3, under certain conditions involving the prime divisors of the cyclic group of order n, one can identify at most n-1 vertices such that the degree of at least one of these vertices is equal to that of the prime degree of the other vertices.

Abstract: The power graph $\mathcal{P}(G)$ of a finite group $G$ is the simple undirected graph whose vertex set is $G$, in which two distinct vertices are adjacent if one of them is an integral power of the other. For an integer $n\geq 2$, let $C_n$ denote the cyclic group of order $n$ and let $r$ be the number of distinct prime divisors of $n$. The minimum degree $\delta(\mathcal{P}(C_n))$ of $\mathcal{P}(C_n)$ is known for $r\in\{1,2\}$, see [18]. For $r\geq 3$, under certain conditions involving the prime divisors of $n$, we identify at most $r-1$ vertices such that $\delta(\mathcal{P}(C_n))$ is equal to the degree of at least one of these vertices. If $r=3$ or if $n$ is a product of distinct primes, we are able to identify two such vertices without any condition on the prime divisors of $n$.