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

TL;DR

This paper investigates the minimum degree of the power graph of finite cyclic groups, providing new results for cases with three or more prime divisors, including explicit vertex identification under certain conditions.

Contribution

It extends the understanding of the minimum degree of power graphs for cyclic groups with multiple prime divisors, identifying specific vertices related to this degree.

Findings

Minimum degree known for groups with one or two prime divisors.

Identification of vertices for minimum degree when r ≥ 3 under certain conditions.

Explicit identification of two vertices for r=3 or when n is a product of distinct primes.

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 $n$.

Equations

$$\mathrm{deg}(a)=\frac{n}{b}+\sum _{d|b,d\ne b}\varphi \left(\frac{n}{d}\right)-1=\frac{n}{b}+\sum _{d|b}\varphi \left(\frac{n}{d}\right)-\varphi \left(\frac{n}{b}\right)-1,$$

$$\delta (\mathcal{P}(C_n))=\min \{\mathrm{deg}(p_{r-1}{p}_r),\mathrm{deg}(p_r)\}.$$

$$\delta (\mathcal{P}(C_n))=\min \{\mathrm{deg}\left(p_s^{{\alpha }_s}\right):t\le s\le r\}.$$

$$\delta (\mathcal{P}(C_n))=\min \{\mathrm{deg}\left(p_2^{{\alpha }_2}\right),\mathrm{deg}\left(p_3^{{\alpha }_3}\right)\}.$$

$$\sum _{d|m}\varphi (d)=m$$

$$\mathrm{deg}\left(p_i^{{\alpha }_i}\right)-\mathrm{deg}\left(p_r^{{\alpha }_r}\right)>p_i^{{\alpha }_i-1}\left[(p_r-1)\varphi \left(\frac{n}{p_i^{{\alpha }_i}{p}_r^{{\alpha }_r}}\right)-\frac{n}{p_i^{{\alpha }_i-1}{p}_r^{{\alpha }_r}}\right].$$

$$\sum _{d|a,d<a}\varphi \left(\frac{n}{d}\right)-\sum _{d|b,d<b}\varphi \left(\frac{n}{d}\right)$$

$$=\sum _{u|\frac{a}{p_k},u<\frac{a}{p_k}}\varphi \left(\frac{n}{u{p}_k}\right)-\sum _{v|\frac{b}{p_l},v<\frac{b}{p_l}}\varphi \left(\frac{n}{v{p}_l}\right)$$

$$=\sum _{u|\frac{a}{p_k},u<\frac{a}{p_k}}\left[\varphi \left(\frac{n}{u{p}_k}\right)-\varphi \left(\frac{n}{u{p}_l}\right)\right]$$

$$=\left[\varphi (p_l)-\varphi (p_k)\right]\left[\sum _{u|\frac{a}{p_k},u<\frac{a}{p_k}}\varphi \left(\frac{n}{u{p}_k{p}_l}\right)\right].$$

$$\mathrm{deg}(a)-\mathrm{deg}(b)$$

$$>\sum _{d|a,d<a}\varphi \left(\frac{n}{d}\right)-\sum _{d|b,d<b}\varphi \left(\frac{n}{d}\right)$$

$$=\left[\varphi (p_l)-\varphi (p_k)\right]\left[\sum _{u|\frac{a}{p_k},u<\frac{a}{p_k}}\varphi \left(\frac{n}{u{p}_k{p}_l}\right)\right].$$

$$\sum _{d|y}\varphi \left(\frac{x}{d}\right)=\left(\prod _{\begin{array}{c}1\le i\le r\\ {\alpha }_i={\beta }_i\end{array}}{p}_i^{{\alpha }_i}\right)\left(\prod _{\begin{array}{c}1\le i\le r\\ {\alpha }_i>{\beta }_i\end{array}}\left(p_i^{{\alpha }_i}-p_i^{{\alpha }_i-{\beta }_i-1}\right)\right).$$

$$\sum _{0\le {\gamma }_j\le {\beta }_j}\varphi \left(p_j^{{\alpha }_j-{\gamma }_j}\right)=p_j^{{\alpha }_j}\text{mbox}or{p}_j^{{\alpha }_j}-p_j^{{\alpha }_j-{\beta }_j-1}$$

$$\sum _{d|y}\varphi \left(\frac{x}{d}\right)=\sum _{\begin{array}{c}1\le j\le r\\ 0\le {\gamma }_j\le {\beta }_j\end{array}}\varphi \left(p_1^{{\alpha }_1-{\gamma }_1}{p}_2^{{\alpha }_2-{\gamma }_2}\cdots p_r^{{\alpha }_r-{\gamma }_r}\right)=\prod _{1\le j\le r}\left(\sum _{0\le {\gamma }_j\le {\beta }_j}\varphi \left(p_j^{{\alpha }_j-{\gamma }_j}\right)\right),$$

$$\mathrm{deg}(p_{k_1}\cdots p_{k_{t-1}}{p}_{k_t})\ge \mathrm{deg}(p_{k_1}\cdots p_{k_{t-1}}{p}_r)\ge \cdots \ge \mathrm{deg}(p_{r-t+1}\cdots p_{r-1}{p}_r).$$

$$p_s{p}_{s+1}\cdots p_r\ge \left(\frac{p_1{p}_2\cdots p_{s-2}}{\varphi (p_1{p}_2\cdots p_{s-2})}-1\right)\varphi (p_{s-1})+1.$$

$$\sum _{d|(p_{s-1}{p}_s\cdots p_r)}\varphi \left(\frac{n}{d}\right)-\sum _{d|(p_s{p}_{s+1}\cdots p_r)}\varphi \left(\frac{n}{d}\right)=\varphi (p_1{p}_2\cdots p_{s-2})p_s{p}_{s+1}\cdots p_r.$$

$$\varphi \left(\frac{n}{p_s{p}_{s+1}\cdots p_r}\right)-\varphi \left(\frac{n}{p_{s-1}{p}_s\cdots p_r}\right)=\varphi (p_1{p}_2\cdots p_{s-2})(\varphi (p_{s-1})-1).$$

$$\mathrm{deg}(p_{s-1}{p}_s\cdots p_r)-\mathrm{deg}(p_s\cdots p_r)$$

$$=-p_1{p}_2\cdots p_{s-2}\varphi (p_{s-1})+\varphi (p_1{p}_2\cdots p_{s-2})p_s{p}_{s+1}\cdots p_r+\varphi (p_1{p}_2\cdots p_{s-2})(\varphi (p_{s-1})-1)$$

$$=\varphi (p_1{p}_2\cdots p_{s-2})p_s{p}_{s+1}\cdots p_r-\left[p_1{p}_2\cdots p_{s-2}-\varphi (p_1{p}_2\cdots p_{s-2})\right]\varphi (p_{s-1})-\varphi (p_1{p}_2\cdots p_{s-2}).$$

$$\mathrm{deg}(p_{s-2}{p}_{s-1}\cdots p_r)>\mathrm{deg}(p_{s-1}{p}_s\cdots p_r).$$

$$\varphi (p_1{p}_2\cdots p_{s-2})(p_s{p}_{s+1}\cdots p_r-1)\ge \varphi (p_{s-1})\left[p_1{p}_2\cdots p_{s-2}-\varphi (p_1{p}_2\cdots p_{s-2})\right].$$

$$\mathrm{deg}(p_{s-2}{p}_{s-1}\cdots p_r)>\mathrm{deg}(p_{s-1}{p}_s\cdots p_r).$$

$$\varphi (p_1{p}_2\cdots p_{s-3})(p_{s-1}{p}_s\cdots p_r-1)>\varphi (p_{s-2})\left[p_1{p}_2\cdots p_{s-3}-\varphi (p_1{p}_2\cdots p_{s-3})\right].$$

$$\varphi (p_1{p}_2\cdots p_{s-2})(p_s{p}_{s+1}\cdots p_r-1)$$

$$<\varphi (p_1{p}_2\cdots p_{s-3})(p_{s-1}{p}_s\cdots p_r-1).$$

$$\varphi (p_{s-1})\left[p_1{p}_2\cdots p_{s-2}-\varphi (p_1{p}_2\cdots p_{s-2})\right]$$

$$=p_1{p}_2\cdots p_{s-3}\left[\varphi (p_{s-2})+1\right]-\varphi (p_1{p}_2\cdots p_{s-2})$$

$$>\varphi (p_{s-2})\left[p_1{p}_2\cdots p_{s-3}-\varphi (p_1{p}_2\cdots p_{s-3})\right].$$

$$r-3\ge \frac{p_1{p}_2\cdots p_{r-3}}{\varphi (p_1{p}_2\cdots p_{r-3})}-1.$$

$$p_{r-1}{p}_r>(r-2)\varphi (p_{r-2})\ge (r-3)\varphi (p_{r-2})+1\ge \left(\frac{p_1{p}_2\cdots p_{r-3}}{\varphi (p_1{p}_2\cdots p_{r-3})}-1\right)\varphi (p_{r-2})+1.$$

$$\mathrm{deg}(p_2{p}_3\cdots p_r)>\mathrm{deg}(p_3{p}_4\cdots p_r)>\cdots >\mathrm{deg}(p_{r-2}{p}_{r-1}{p}_r)>\mathrm{deg}(p_{r-1}{p}_r).$$

Full text

Key words. Power graph, Cyclic group, Minimum degree, Edge connectivity, Euler’s totient function.

AMS subject classification. 05C25, 05C07, 05C40

1 Introduction

Let $\Gamma$ be a simple graph with at least two vertices. The edge connectivity $\kappa^{\prime}(\Gamma)$ of $\Gamma$ is the minimum number of edges whose deletion from $\Gamma$ gives a disconnected subgraph of $\Gamma$. The vertex connectivity $\kappa(\Gamma)$ of $\Gamma$ is the minimum number of vertices which need to be removed from $\Gamma$ so that the induced subgraph of $\Gamma$ on the remaining vertices is disconnected or has only one vertex. The latter case arises only when $\Gamma$ is a complete graph. The minimum degree of $\Gamma$, denoted by $\delta(\Gamma)$, is the minimum of the degrees of vertices of $\Gamma$. The study of vertex/edge connectivity is an interesting problem in graph theory. It is known that $\kappa(\Gamma)\leq\kappa^{\prime}(\Gamma)\leq\delta(\Gamma)$, and $\kappa^{\prime}(\Gamma)=\delta(\Gamma)$ if the diameter of $\Gamma$ is at most $2$, see Theorem 4.1.9 and Exercise 4.1.25 in [19].

1.1 Power graph

The notion of directed power graph of a group was introduced in [11], which was further extended to semigroups in [13, 12]. Then the undirected power graph of a semigroup, in particular, of a group was defined in [2]. Many researchers have investigated both directed and undirected power graphs of groups from different view points. More on these graphs can be found in the survey paper [1] and the references therein.

Let $G$ be a finite group. The power graph of $G$, denoted by $\mathcal{P}(G)$, is the simple undirected graph with vertex set $G$, in which two distinct vertices are adjacent if one of them can be written as an integral power of the other. Since $G$ is finite, the identity element of $G$ is adjacent to all other vertices. So $\mathcal{P}(G)$ is a connected graph and its diameter is at most 2.

By [2, Theorem 2.12], $\mathcal{P}(G)$ is a complete graph if and only if $G$ is a cyclic group of prime power order. It is proved in [7, Theorem 1.3] and [8, Corollary 3.4] respectively that, among all finite groups of a given order, the cyclic group of that order has the maximum number of edges and has the largest clique in its power graph. By [9, Theorem 5] and [14, Corollary 2.5], the power graph of a finite group is perfect, in particular, the clique number and the chromatic number coincide. Explicit formula for the clique number of the power graph of a finite cyclic group is given in [15, Theorem 2] and [9, Theorem 7]. The full automorphism group of the power graph of a finite group is described in [10, Theorem 2.2].

For a positive integer $n$, let $C_{n}$ denote the finite cyclic group of order $n$. The vertex connectivity of $\mathcal{P}(C_{n})$ is studied in [3, 4, 5, 17] and the exact value of $\kappa(\mathcal{P}(C_{n}))$ is obtained in the following cases: (i) $n$ is a product of distinct primes, (ii) $n$ is divisible by at most three distinct primes, (iii) $n$ is divisible by the square of its largest prime factor, and (iv) the smallest prime divisor of $n$ is greater than or equal to the number of distinct prime divisors of $n$. The above articles also provide some sharp upper bounds for $\kappa(\mathcal{P}(C_{n}))$. It is proved in [18, Theorem 6.7] that the vertex connectivity and the minimum degree of $\mathcal{P}(C_{n})$ coincide if and only if either $n$ is a prime power or $n$ is twice of an odd prime power. For these values of $n$, the relation $\kappa(\mathcal{P}(C_{n}))\leq\kappa^{\prime}(\mathcal{P}(C_{n}))\leq\delta(\mathcal{P}(C_{n}))$ implies that the vertex connectivity and the edge connectivity of $\mathcal{P}(C_{n})$ are equal. Since the diameter of $\mathcal{P}(C_{n})$ is at most $2$, the edge connectivity and the minimum degree of $\mathcal{P}(C_{n})$ coincide for every $n$. Thus, in order to determine the edge connectivity of $\mathcal{P}(C_{n})$, it is enough to find the minimum degree of $\mathcal{P}(C_{n})$.

1.2 Minimum degree of $\mathcal{P}(C_{n})$

Throughout the paper, we shall identify $C_{n}$ with $\mathbb{Z}_{n}=\{0,1,2,\ldots,n-1\}$, the group of integers modulo $n$. The degree of a vertex $a\in C_{n}$ is denoted by $\deg(a)$. By [16, Lemma 3.4] (also see [7, Lemma 2.7]), we have the following formula for $\deg(a)$:

[TABLE]

where $\phi$ is the Euler’s totient function and $b$ is the greatest common divisor of $a$ and $n$. If $a=0$ or $a$ is a generator of $C_{n}$, then $\deg(a)=n-1$.

To determine $\delta(\mathcal{P}(C_{n}))$, our objective will be to identify a vertex of $\mathcal{P}(C_{n})$ having minimum degree and then the degree of that vertex can be calculated using (1). The formula (1) implies that $\deg(a)=\deg(b)$. Thus the degree of a given non-zero vertex of $\mathcal{P}(C_{n})$ is equal to the degree of some element of $C_{n}$ which is a divisor of $n$. Therefore, in order to identify a vertex of $\mathcal{P}(C_{n})$ of minimum degree, we need to compare the degrees of all possible vertices which are divisors of $n$.

If $n$ is a prime power, then $\mathcal{P}(C_{n})$ is a complete graph and so $\delta(\mathcal{P}(C_{n}))=n-1=\deg(a)$ for every vertex $a\in C_{n}$. If $n>1$ is not a prime power, then $\mathcal{P}(C_{n})$ is not a complete graph and so $\delta(\mathcal{P}(C_{n}))<n-1$. Hence the minimum degree of $\mathcal{P}(C_{n})$ will be equal to the degree of a vertex which is a proper111A positive integer $a$ is called a proper divisor of $n$ if $a$ divides $n$ and $a\notin\{1,n\}$. divisor of $n$. For certain values of $n$, a vertex of $\mathcal{P}(C_{n})$ of minimum degree was obtained in [18, Theorem 4.6] which we mention below.

Proposition 1.1**.**

[18]** Let $p_{1},p_{2},p_{3},p_{4}$ be prime numbers with $p_{1}<p_{2}<p_{3}<p_{4}$. Then the following hold:

1. (i)

If $n=p_{1}^{\alpha_{1}}p_{2}^{\alpha_{2}}$ for some positive integers $\alpha_{1},\alpha_{2}$, then $\delta(\mathcal{P}(C_{n}))=\deg\left({p_{2}^{\alpha_{2}}}\right)$. 2. (ii)

If $n=p_{1}p_{2}p_{3}$, then $\delta(\mathcal{P}(C_{n}))=\deg({p_{3}})$. 3. (iii)

Let $n=p_{1}p_{2}p_{3}p_{4}$. If $n$ is odd or $p_{4}\geq p_{3}+\displaystyle\frac{2(p_{3}-1)}{p_{2}-1}$, then $\delta(\mathcal{P}(C_{n}))=\deg({p_{4}})$, otherwise, $\delta(\mathcal{P}(C_{n}))=\deg({p_{3}p_{4}})$.

In this paper, we generalize the results stated in Proposition 1.1 to several other values of $n$. In view of Proposition 1.1(i), if necessary, we may assume that $n$ is divisible by at least three distinct prime numbers.

The following theorem is proved in Section 3 for the minimum degree of $\mathcal{P}(C_{n})$ when $n$ is a product of distinct prime numbers.

Theorem 1.2**.**

Let $n=p_{1}p_{2}\cdots p_{r}$, where $r\geq 3$ and $p_{1},p_{2},\ldots,p_{r}$ are prime numbers with $p_{1}<p_{2}<\cdots<p_{r}$. Then

[TABLE]

Further, $\delta(\mathcal{P}(C_{n}))=\deg(p_{r})$ if and only if $\phi(p_{r})\geq\left(\dfrac{p_{1}p_{2}\cdots p_{r-2}}{\phi(p_{1}p_{2}\cdots p_{r-2})}-1\right)\phi(p_{r-1})$. In particular, if $\phi(p_{r})\geq(r-2)\phi(p_{r-1})$, then $\delta(\mathcal{P}(C_{n}))=\deg({p_{r}})$.

For general $n$, under certain conditions involving its prime divisors, the following theorem is proved in Section 4 on the minimum degree of $\mathcal{P}(C_{n})$.

Theorem 1.3**.**

Let $n=p_{1}^{\alpha_{1}}p_{2}^{\alpha_{2}}\cdots p_{r}^{\alpha_{r}}$, where $r\geq 2$, $\alpha_{1},\alpha_{2},\ldots,\alpha_{r}$ are positive integers and $p_{1},p_{2},\ldots,p_{r}$ are prime numbers with $p_{1}<p_{2}<\cdots<p_{r}$. Suppose that any of the following two conditions holds:

1. (i)

$2\phi(p_{1}p_{2}\cdots p_{r})\geq p_{1}p_{2}\cdots p_{r}$, 2. (ii)

$\phi(p_{i+1})\geq r\phi(p_{i})$* for each $i\in\{1,2,\ldots,r-1\}$.*

If $t\in\{2,3,\ldots,r\}$ is the largest integer such that $\alpha_{t}\geq\alpha_{j}$ for $2\leq j\leq r$, then

[TABLE]

As an application of Theorem 1.3, we prove the following corollary in Section 4 which can be used to determine $\delta(\mathcal{P}(C_{n}))$ for many values of $n$.

Corollary 1.4**.**

Let $n=p_{1}^{\alpha_{1}}p_{2}^{\alpha_{2}}\cdots p_{r}^{\alpha_{r}}$, where $r\geq 2$, $\alpha_{1},\alpha_{2},\ldots,\alpha_{r}$ are positive integers and $p_{1}<p_{2}<\cdots<p_{r}$ are prime numbers. Suppose that any of the following two conditions holds:

1. (i)

$p_{1}\geq r+1$* and $p_{r}>rp_{r-1}$,* 2. (ii)

$p_{i+1}>rp_{i}$* for each $i\in\{1,2,\ldots,r-1\}$.*

Then $\delta(\mathcal{P}(C_{n}))=\deg(p_{r}^{\alpha_{r}})$.

For $r=3$, the following theorem is proved in Section 5 which shows that the conclusion of Theorem 1.3 holds good without any condition involving the prime divisors of $n$.

Theorem 1.5**.**

Let $n=p_{1}^{\alpha_{1}}p_{2}^{\alpha_{2}}p_{3}^{\alpha_{3}}$, where $\alpha_{1},\alpha_{2},\alpha_{3}$ are positive integers and $p_{1},p_{2},p_{3}$ are prime numbers with $p_{1}<p_{2}<p_{3}$. Then

[TABLE]

1.3 Remark

We remark that Proposition 1.1 can be obtained from Theorems 1.2 and 1.3.

1. $\bullet$

If $r=3$ and $n=p_{1}p_{2}p_{3}$, then $\phi(p_{r})=\phi(p_{3})>\phi(p_{2})=(r-2)\phi(p_{r-1})$ and so Proposition 1.1(ii) follows from the last part of Theorem 1.2. 2. $\bullet$

Suppose that $r=4$ and $n=p_{1}p_{2}p_{3}p_{4}$. If $n$ is odd, then $p_{1}\geq 3$ and so $2\phi(p_{1}p_{2})>p_{1}p_{2}$ by Lemma 2.1(ii). Then $\phi(p_{4})>\phi(p_{3})>\left(\dfrac{p_{1}p_{2}}{\phi(p_{1}p_{2})}-1\right)\phi(p_{3})$. If $n$ is even, then $p_{1}=2$ and so $1+\left(\dfrac{p_{1}p_{2}}{\phi(p_{1}p_{2})}-1\right)\phi(p_{3})=p_{3}+\displaystyle\frac{2(p_{3}-1)}{p_{2}-1}$. In this case, $p_{4}\geq 1+\left(\dfrac{p_{1}p_{2}}{\phi(p_{1}p_{2})}-1\right)\phi(p_{3})$ if and only if $p_{4}\geq p_{3}+\displaystyle\frac{2(p_{3}-1)}{p_{2}-1}$. Then it follows that Proposition 1.1(iii) can be obtained from Theorem 1.2. 3. $\bullet$

Finally, suppose that $n=p_{1}^{\alpha_{1}}p_{2}^{\alpha_{2}}$. If $p_{1}\geq 3$, then $2\phi(p_{1}p_{2})>p_{1}p_{2}$ by Lemma 2.1(ii). If $p_{1}=2$, then $\phi(p_{2})\geq 2=2\phi(p_{1})$. Thus condition (i) or (ii) of Theorem 1.3 is satisfied and hence Proposition 1.1(i) follows from Theorem 1.3.

2 Preliminaries

Recall that $\phi$ is a multiplicative function, that is, $\phi(ab)=\phi(a)\phi(b)$ for any two positive integers $a,b$ which are relatively primes. We have $\phi(p^{k})=p^{k-1}\phi(p)$ for any prime number $p$ and positive integer $k$. Also,

[TABLE]

for every positive integer $m$. We need the following two inequalities: the first one can be found in [6, Lemma 3.1] and the second one was proved in [4] while proving Corollary 1.4.

Lemma 2.1**.**

[4, 6]** Let $p_{1}<p_{2}<\cdots<p_{t}$ be prime numbers. Then the following hold:

1. (i)

$(t+1)\phi(p_{1}p_{2}\cdots p_{t})\geq p_{1}p_{2}\cdots p_{t}$, with equality if and only if $(t,p_{1})=(1,2)$ or $(t,p_{1},p_{2})=(2,2,3)$. 2. (ii)

If $p_{1}\geq t+1$, then $2\phi(p_{1}p_{2}\cdots p_{t})\geq p_{1}p_{2}\cdots p_{t}$, with equality if and only if $t=1$ and $p_{1}=2$.

Certain inequalities involving degree of vertices of $\mathcal{P}(C_{n})$ were proved in [18, Proposition 4.5]. From the proof of these inequalities, it can be seen that those inequalities are in fact strict and we have stated them accordingly in the following proposition.

Proposition 2.2**.**

[18]** Let $n=p_{1}^{\alpha_{1}}p_{2}^{\alpha_{2}}\cdots p_{r}^{\alpha_{r}}$, where $r\geq 2$, $\alpha_{1},\alpha_{2},\ldots,\alpha_{r}$ are positive integers and $p_{1},p_{2},\ldots,p_{r}$ are prime numbers with $p_{1}<p_{2}<\cdots<p_{r}$. Then the following strict inequalities hold in $\mathcal{P}(C_{n}):$

1. (i)

$\deg\left({p_{1}^{\alpha_{1}}}\right)>\deg\left({p_{r}^{\alpha_{r}}}\right)$. 2. (ii)

$\deg\left({p_{i}^{\gamma}}\right)>\deg\left({p_{i}^{\beta}}\right)$* for $1\leq i\leq r$ and $1\leq\gamma<\beta\leq\alpha_{i}$.* 3. (iii)

$\deg\left({p_{i}^{\beta}}\right)>\deg\left({p_{j}^{\beta}}\right)$* for $1\leq i<j\leq r$ and $1\leq\beta\leq\min\{\alpha_{i},\alpha_{j}\}$.* 4. (iv)

$\deg\left({p_{1}^{\beta_{1}}p_{2}^{\beta_{2}}\cdots p_{r}^{\beta_{r}}}\right)>\deg\left({p_{2}^{\beta_{2}}\cdots p_{r}^{\beta_{r}}}\right)$, where $1\leq\beta_{i}\leq\alpha_{i}$ for each $i\in\{1,2,\ldots,r\}$.

We need the strict inequality (3) stated in the following lemma while proving Corollaries 1.4 and 5.3.

Lemma 2.3**.**

Let $n=p_{1}^{\alpha_{1}}p_{2}^{\alpha_{2}}\cdots p_{r}^{\alpha_{r}}$, where $r\geq 2$, $\alpha_{1},\alpha_{2},\ldots,\alpha_{r}$ are positive integers and $p_{1},p_{2},\ldots,p_{r}$ are prime numbers with $p_{1}<p_{2}<\cdots<p_{r}$. For $i\in\{1,2,\ldots,r-1\}$, the following strict inequality holds in $\mathcal{P}(C_{n}):$

[TABLE]

Proof.

This follows from the proof of [18, Proposition 4.5(i)], in which we replace the subscript $1$ by $i$ and take $m=\dfrac{n}{p_{i}^{\alpha_{i}}p_{r}^{\alpha_{r}}}$. We note that the first inequality in the proof of [18, Proposition 4.5(i)] is strict. ∎

Lemma 2.4**.**

Let $n=p_{1}p_{2}\cdots p_{r}$, where $p_{1},p_{2},\ldots,p_{r}$ are prime distinct numbers and let $a,b$ be two distinct proper divisors of $n$ such that $\dfrac{a}{b}=\dfrac{p_{k}}{p_{l}}$ for some $k,l\in\{1,2,\ldots,r\}$. If $a<b$, then $\deg(a)>\deg(b)$.

Proof.

Note that both $a$ and $b$ have the same number of prime divisors, say $s$. Since $a<b$, we have $p_{k}<p_{l}$ and $1\leq s\leq r-1$. The lemma follows from Proposition 2.2(iii) if $s=1$. Assume that $s\geq 2$. We have

[TABLE]

The second last equality in the above holds as $\dfrac{a}{p_{k}}=\dfrac{b}{p_{l}}$. Using the formula (1), we then get

[TABLE]

Since $p_{l}>p_{k}$, we have $\phi(p_{l})-\phi(p_{k})>0$ and it follows from the above that $\deg(a)>\deg(b)$. ∎

Lemma 2.5**.**

Let $x=p_{1}^{\alpha_{1}}p_{2}^{\alpha_{2}}\cdots p_{r}^{\alpha_{r}}$, where $\alpha_{1},\alpha_{2},\ldots,\alpha_{r}$ are positive integers and $p_{1},p_{2},\ldots,p_{r}$ are prime numbers. If $y=p_{1}^{\beta_{1}}p_{2}^{\beta_{2}}\cdots p_{r}^{\beta_{r}}$, where $0\leq\beta_{i}\leq\alpha_{i}$ for $1\leq i\leq r$, then

[TABLE]

Proof.

If $\alpha_{i}=\beta_{i}$ for all $i\in\{1,2,\ldots,r\}$, then $y=x$ and the result follows from the fact that $\displaystyle\sum\limits_{d|x}\phi\left(\frac{x}{d}\right)=\displaystyle\sum\limits_{d|x}\phi\left(d\right)=x$. So assume that $\beta_{i}<\alpha_{i}$ for at least one $i\in\{1,2,\ldots,r\}$. For each $j\in\{1,2,\ldots,r\}$, observe that

[TABLE]

according as $\alpha_{j}=\beta_{j}$ or $\alpha_{j}>\beta_{j}$. We have

[TABLE]

and consequently the lemma follows using (4). ∎

3 Proof of Theorem 1.2

In this section, we take $n=p_{1}p_{2}\cdots p_{r}$, where $r\geq 3$ and $p_{1},p_{2},\ldots,p_{r}$ are prime numbers with $p_{1}<p_{2}<\cdots<p_{r}$.

Lemma 3.1**.**

$\delta(\mathcal{P}(C_{n}))=\min\{\deg(p_{s}p_{s+1}\cdots p_{r}):2\leq s\leq r\}$.

Proof.

Let $\{k_{1},k_{2},\ldots,k_{t}\}$ be a proper subset of $\{1,2,\cdots,r\}$ with $k_{1}<k_{2}<\cdots<k_{t}$. Applying Lemma 2.4 repeatedly, we get

[TABLE]

Since $\delta\left(\mathcal{P}(C_{n})\right)$ is equal to the degree of a vertex which is a proper divisor of $n$, it follows from the above that $\delta(\mathcal{P}(C_{n}))=\min\{\deg(p_{s}p_{s+1}\cdots p_{r}):2\leq s\leq r\}$. ∎

Lemma 3.2**.**

Let $3\leq s\leq r$. Then $\deg({p_{s-1}p_{s}\cdots p_{r}})\geq\deg({p_{s}p_{s+1}\cdots p_{r}})$ if and only if

[TABLE]

Further, $\deg({p_{s-1}p_{s}\cdots p_{r}})=\deg({p_{s}p_{s+1}\cdots p_{r}})$ if and only if equality holds in (5).

Proof.

We have $\dfrac{n}{p_{s-1}p_{s}\cdots p_{r}}-\dfrac{n}{p_{s}p_{s+1}\cdots p_{r}}=-p_{1}p_{2}\cdots p_{s-2}\phi(p_{s-1})$. Using (2), an easy calculation gives that

[TABLE]

We also have

[TABLE]

Then, using the degree formula (1), it follows that

[TABLE]

Now it can be seen that $\deg({p_{s-1}p_{s}\cdots p_{r}})\geq\deg({p_{s}p_{s+1}\cdots p_{r}})$ if and only if inequality (5) holds. Also, $\deg({p_{s-1}p_{s}\cdots p_{r}})=\deg({p_{s}p_{s+1}\cdots p_{r}})$ if and only if equality holds in (5). ∎

Lemma 3.3**.**

Let $3\leq s\leq r$. If $\deg(p_{s-1}p_{s}\cdots p_{r})\geq\deg(p_{s}p_{s+1}\cdots p_{r})$, then

[TABLE]

Proof.

If $s=3$, then $\deg(p_{1}p_{2}\cdots p_{r})=\deg(n)=\deg(0)=n-1>\dfrac{n\phi(p_{1})}{p_{1}}=\deg(p_{2}p_{3}\cdots p_{r})$ and the lemma follows. Assume that $4\leq s\leq r$. Observe that the inequality (5) in the statement of Lemma 3.2 is equivalent to the following inequality:

[TABLE]

Since $\deg(p_{s-1}p_{s}\cdots p_{r})\geq\deg(p_{s}p_{s+1}\cdots p_{r})$ by the given hypothesis, Lemma 3.2 then implies that the inequality (6) holds. We need to show that

[TABLE]

By Lemma 3.2 again, it is enough to show that

[TABLE]

Since $\phi(p_{s-2})<p_{s-1}$, we have

[TABLE]

Moreover,

[TABLE]

Now (7) follows from the inequalities (6), (3) and (3). ∎

Proof of Theorem 1.2.

If $r=3$, then $\delta(\mathcal{P}(C_{n}))=\min\{\deg(p_{2}p_{3}),\deg({p_{3}})\}$ by Lemma 3.1. Assume that $r\geq 4$. By Lemma 2.1(i), we have $(r-2)\phi(p_{1}p_{2}\cdots p_{r-3})\geq p_{1}p_{2}\cdots p_{r-3}$, and this gives

[TABLE]

Then

[TABLE]

So inequality (5) is satisfied with $s=r-1$ and hence $\deg({p_{r-2}p_{r-1}p_{r}})>\deg({p_{r-1}p_{r}})$ by Lemma 3.2. Then, using Lemma 3.3 repeatedly, we have

[TABLE]

Therefore, by Lemma 3.1, we get

[TABLE]

By Lemma 3.2, $\delta(\mathcal{P}(C_{n}))=\deg(p_{r})$ if and only if $\phi(p_{r})\geq\left(\dfrac{p_{1}p_{2}\cdots p_{r-2}}{\phi(p_{1}p_{2}\cdots p_{r-2})}-1\right)\phi(p_{r-1})$.

Now suppose that $\phi(p_{r})\geq(r-2)\phi(p_{r-1})$. Since $(r-1)\phi(p_{1}p_{2}\cdots p_{r-2})\geq p_{1}p_{2}\cdots p_{r-2}$ by Lemma 2.1(i), we have

[TABLE]

Therefore,

[TABLE]

Then $\deg(p_{r-1}p_{r})\geq\deg(p_{r})$ by Lemma 3.2 and hence $\delta(\mathcal{P}(C_{n}))=\deg(p_{r})$. This completes the proof of Theorem 1.2. ∎

Example 3.4**.**

The following examples shows that all possibilities can occur in Theorem 1.2 for the minimum degree of $\mathcal{P}(C_{n})$.

1. (i)

If $n=2\cdot 13\cdot 17\cdot 19$, then $\delta(\mathcal{P}(C_{n}))=\deg(17\cdot 19)<\deg(19)$. 2. (ii)

If $n=2\cdot 13\cdot 17\cdot 23$, then $\delta(\mathcal{P}(C_{n}))=\deg(23)<\deg(17\cdot 23)$. 3. (iii)

If $n=2\cdot 5\cdot 13\cdot 19$, then $\delta(\mathcal{P}(C_{n}))=\deg(13\cdot 19)=\deg(19)$.

Note that if $n=2\cdot 3\cdot p_{3}\cdot p_{4}$ with $p_{4}=2p_{3}-1$, then $\deg(p_{3}p_{4})=\deg(p_{4})$.

4 Proof of Thereom 1.3 and Corolary 1.4

In this section, we take $n=p_{1}^{\alpha_{1}}p_{2}^{\alpha_{2}}\cdots p_{r}^{\alpha_{r}}$, where $r\geq 2$, $\alpha_{1},\alpha_{2},\ldots,\alpha_{r}$ are positive integers and $p_{1},p_{2},\ldots,p_{r}$ are prime numbers with $p_{1}<p_{2}<\cdots<p_{r}$.

Proposition 4.1**.**

Let $m=p_{k_{1}}^{\beta_{k_{1}}}p_{k_{2}}^{\beta_{k_{2}}}\cdots p_{k_{s}}^{\beta_{k_{s}}}$, where $2\leq s\leq r$, $k_{1}<k_{2}<\cdots<k_{s}$ and $1\leq\beta_{k_{i}}\leq\alpha_{k_{i}}$ for $1\leq i\leq s$. Suppose that one of the following two conditions holds:

1. (i)

$2\phi(p_{1}p_{2}\cdots p_{r})\geq p_{1}p_{2}\cdots p_{r}$, 2. (ii)

$\phi(p_{j+1})\geq r\phi(p_{j})$* for each $j\in\{1,2,\ldots,r-1\}$.*

Then $\deg(m)>\deg\left(\dfrac{m}{p_{k_{i}}}\right)$ for $i\in\{1,2,\ldots,s-1\}$ in $\mathcal{P}(C_{n})$.

Proof.

Fix $i\in\{1,2,\ldots,s-1\}$. Using the degree formula (1), we have

[TABLE]

where

[TABLE]

First calculate \sum_{\begin{subarray}{c}d|m\end{subarray}}\phi\left(\frac{n}{d}\right)-\sum_{\begin{subarray}{c}d\big{|}\frac{m}{p_{k_{i}}}\end{subarray}}\phi\left(\frac{n}{d}\right). Define $n^{\prime}:=\dfrac{n}{p_{k_{1}}^{\alpha_{k_{1}}}p_{k_{2}}^{\alpha_{k_{2}}}\cdots p_{k_{s}}^{\alpha_{k_{s}}}}\times{p_{k_{i}}^{\alpha_{k_{i}}-\beta_{k_{i}}}}$. Then

[TABLE]

Taking $x=\dfrac{p_{k_{1}}^{\alpha_{k_{1}}}p_{k_{2}}^{\alpha_{k_{2}}}\cdots p_{k_{s}}^{\alpha_{k_{s}}}}{p_{k_{i}}^{\alpha_{k_{i}}}}$ and $y=\dfrac{p_{k_{1}}^{\beta_{k_{1}}}p_{k_{2}}^{\beta_{k_{2}}}\cdots p_{k_{s}}^{\beta_{k_{s}}}}{p_{k_{i}}^{\beta_{k_{i}}}}=\dfrac{m}{p_{k_{i}}^{\beta_{k_{i}}}}$ in Lemma 2.5, we get

[TABLE]

Next calculate $\phi\left(\dfrac{np_{k_{i}}}{m}\right)-\phi\left(\dfrac{n}{m}\right)$. We have

[TABLE]

where equality in the above holds if and only if $\alpha_{k_{i}}=\beta_{k_{i}}$. Using (4) and (4), we get

[TABLE]

Note that equality holds in (12) if and only if $\alpha_{k_{i}}=\beta_{k_{i}}$, which follows from (4). It can be seen that equality holds in (13) if and only if $\alpha_{k_{j}}>\beta_{k_{j}}$ for all $j\in\{1,2,\ldots,s\}$. Combining these two facts, we thus have

[TABLE]

Finally, we get

[TABLE]

Since $k_{s}>k_{i}$, we have $p_{k_{s}}>p_{k_{i}}$. So $\phi\left(p_{k_{s}}\right)>\phi\left(p_{k_{i}}\right)$ and hence $\phi\left(p_{k_{s}}\right)+\phi\left(p_{k_{i}}\right)>2\phi\left(p_{k_{i}}\right)$. If $2\phi(p_{1}p_{2}\cdots p_{r})\geq p_{1}p_{2}\cdots p_{r}$, then it follows from (4) that $\deg(m)-\deg\left(\frac{m}{p_{k_{i}}}\right)>0$. If $\phi(p_{j+1})\geq r\phi(p_{j})$ for each $j\in\{1,2,\ldots,r-1\}$, then $\phi(p_{k_{s}})\geq r\phi(p_{k_{i}})$ and so $\phi\left(p_{k_{s}}\right)+\phi\left(p_{k_{i}}\right)\geq(r+1)\phi\left(p_{k_{i}}\right)$. Using Lemma 2.1(i), it again follows from (4) that $\deg(m)-\deg\left(\frac{m}{p_{k_{i}}}\right)>0$. This completes the proof. ∎

Proof of Theorem 1.3.

Let $m$ be a proper divisor of $n$. We can write $m=p_{k_{1}}^{\beta_{k_{1}}}p_{k_{2}}^{\beta_{k_{2}}}\cdots p_{k_{s}}^{\beta_{k_{s}}}$ for some $s\in\{1,2,\ldots,r\}$, where $k_{1}<k_{2}<\cdots<k_{s}$ and $1\leq\beta_{k_{i}}\leq\alpha_{k_{i}}$ for $1\leq i\leq s$. If $s=1$, then $\deg(m)=\deg\left(p_{k_{1}}^{\beta_{k_{1}}}\right)\geq\deg\left(p_{k_{1}}^{\alpha_{k_{1}}}\right)$ by Proposition 2.2(ii). So assume that $s\geq 2$. Then applying Proposition 4.1 repeatedly, we find that $\deg(m)>\deg\left(p_{k_{s}}^{\beta_{k_{s}}}\right)\geq\deg\left(p_{k_{s}}^{\alpha_{k_{s}}}\right)$. Here the last inequality holds again by Proposition 2.2(ii). Thus

[TABLE]

By Proposition 2.2(i), we have $\deg\left(p_{1}^{\alpha_{1}}\right)>\deg\left(p_{r}^{\alpha_{r}}\right)$. Let $t\in\{2,3,\ldots,r\}$ be the largest integer such that $\alpha_{t}\geq\alpha_{j}$ for $2\leq j\leq r$. Then, by Proposition 2.2(ii) and (iii), we have $\deg\left(p_{j}^{\alpha_{j}}\right)>\deg\left(p_{t}^{\alpha_{t}}\right)$ for $2\leq j\leq t-1$ (if $t\geq 3$). It now follows that

[TABLE]

This completes the proof. ∎

Example 4.2**.**

Let $n=2\cdot 3\cdot 5\cdot 11$. Then the minimum degree of $\mathcal{P}(C_{n})$ is equal to $\deg(11)$ by Theorem 1.2, but none of the two conditions mentioned in Theorem 1.3 is satisfied. Thus each of the two conditions stipulated in Theorem 1.3 is sufficient but not necessary.

Example 4.3**.**

Let $n=2^{2}\cdot 7\cdot 11\cdot 13$. By Proposition 2.2(iii), we have $\deg(13)<\deg(11)<\deg(7)$. Using the degree formula (1), it can be seen that $\deg(11\cdot 13)<\deg(13)$. This shows that if none of the two conditions stated in Theorem 1.3 is satisfied, then the minimum of degree of $\mathcal{P}(C_{n})$ may not be equal to the degree of $p_{i}^{\alpha_{i}}$ for any $i\in\{2,3,\ldots,r\}$.

Proof of Corollary 1.4.

If $p_{1}\geq r+1$, then $2\phi(p_{1}p_{2}\cdots p_{r})\geq p_{1}p_{2}\cdots p_{r}$ by Lemma 2.1(ii). If $p_{i+1}>rp_{i}$ for each $i\in\{1,2,\ldots,r-1\}$, then $\phi(p_{i+1})=p_{i+1}-1>rp_{i}-1>rp_{i}-r=r\phi(p_{i})$ for $1\leq i\leq r-1$. So, by Theorem 1.3, the minimum degree of $\mathcal{P}(C_{n})$ is equal to $\deg\left({p_{i}^{\alpha_{i}}}\right)$ for some $i\in\{2,3,\ldots,r\}$ and the result follows for $r=2$. Assume that $r\geq 3$. For $2\leq i<r$, let $m=\dfrac{n}{p_{i}^{\alpha_{i}}p_{r}^{\alpha_{r}}}$. By Lemma 2.1(ii), $\phi\left(\dfrac{p_{1}p_{2}\cdots p_{r}}{p_{i}}\right)\geq\dfrac{p_{1}p_{2}\cdots p_{r}}{rp_{i}}$ and so

[TABLE]

From the given conditions in both cases, we have $p_{r}>rp_{r-1}\geq rp_{i}$ and then the result follows from (3) and (4). ∎

5 Proof of Theorem 1.5

In this section, take $n=p_{1}^{\alpha_{1}}p_{2}^{\alpha_{2}}p_{3}^{\alpha_{3}}$, where $\alpha_{1},\alpha_{2},\alpha_{3}$ are positive integers and $p_{1},p_{2},p_{3}$ are prime numbers with $p_{1}<p_{2}<p_{3}$.

Lemma 5.1**.**

Let $i,j\in\{1,2,3\}$ with $i<j$. If $1\leq\beta_{i}\leq\alpha_{i}$, then $\deg\left(p_{i}^{\beta_{i}}p_{j}^{\beta_{j}}\right)>\deg\left(p_{i}^{\beta_{i}-1}p_{j}^{\beta_{j}}\right)$ for $\beta_{j}\geq 2$.

Proof.

Using the degree formula (1), we have

[TABLE]

Let $\{k\}=\{1,2,3\}\setminus\{i,j\}$. Then

[TABLE]

and

[TABLE]

where equality holds if and only if $\alpha_{j}>\beta_{j}$. We also have

[TABLE]

where equality holds if and only if $\alpha_{i}=\beta_{i}$. From (16), (17) and (18), we get

[TABLE]

Since $\beta_{j}\geq 2$ and $j>i$ by the given hypotheses, we have $p_{j}^{\beta_{j}}+\cdots+p_{j}+\phi(p_{i})>3p_{i}$. So $\phi\left(p_{j}p_{k}\right)\left(p_{j}^{\beta_{j}}+\ldots+p_{j}+\phi(p_{i})\right)-p_{i}p_{j}p_{k}>3p_{i}\phi\left(p_{j}p_{k}\right)-p_{i}p_{j}p_{k}\geq 0,$ where the last inequality follows using Lemma 2.1(i). Hence the lemma follows. ∎

Lemma 5.2**.**

Let $\{i,j,k\}=\{1,2,3\}$ with $i<j$. If $(p_{i}+p_{j})\phi\left(p_{j}p_{k}\right)-p_{i}p_{j}p_{k}>0$, then $\deg\left(p_{i}^{\beta_{i}}p_{j}\right)>\deg\left(p_{j}\right)$ for $1\leq\beta_{i}\leq\alpha_{i}$.

Proof.

Using the degree formula (1), we get

[TABLE]

We have

[TABLE]

and

[TABLE]

where equality holds if and only if $\alpha_{j}>1$. We also have

[TABLE]

From (19), (5) and (21), we get

[TABLE]

Since $(p_{i}+p_{j})\phi\left(p_{j}p_{k}\right)-p_{i}p_{j}p_{k}>0$, it follows from the above that $\deg\left(p_{i}^{\beta_{i}}p_{j}\right)>\deg\left(p_{j}\right)$. ∎

Proof of Theorem 1.5.

If $p_{1}\geq 4$, then $2\phi(p_{1}p_{2}p_{3})\geq p_{1}p_{2}p_{3}$ by Lemma 2.1(ii). So $\delta(\mathcal{P}(C_{n}))=\min\{\deg\left({p_{2}^{\alpha_{2}}}\right),\deg\left({p_{3}^{\alpha_{3}}}\right)\}$ by Theorem 1.3. Now assume that $p_{1}=2$ or $3$. In view of Proposition 2.2(i), (ii) and (iv), the minimum degree of $\mathcal{P}(C_{n})$ can be attained at the vertex $p_{2}^{\alpha_{2}}$ or $p_{3}^{\alpha_{3}}$, or at a vertex of the form $p_{i}^{\beta_{i}}p_{j}^{\beta_{j}}$ for some $i,j\in\{1,2,3\}$ with $i<j$, where $1\leq\beta_{i}\leq\alpha_{i}$ and $1\leq\beta_{j}\leq\alpha_{j}$.

Consider the vertices of the form $p_{i}^{\beta_{i}}p_{j}^{\beta_{j}}$ with $i<j$. We show that $\deg\left(p_{i}^{\beta_{i}}p_{j}^{\beta_{j}}\right)>\deg\left(p_{j}^{\beta_{j}}\right)$. Then Proposition 2.2(ii) implies that $\deg\left(p_{j}^{\beta_{j}}\right)\geq\deg\left(p_{j}^{\alpha_{j}}\right)$ and this would complete the proof.

If $\beta_{j}\geq 2$, then applying Lemma 5.1 repeatedly we find that $\deg\left(p_{i}^{\beta_{i}}p_{j}^{\beta_{j}}\right)>\deg\left(p_{j}^{\beta_{j}}\right)$. Suppose that $\beta_{j}=1$. Let $\{k\}=\{1,2,3\}\setminus\{i,j\}$. We show that

[TABLE]

Then Lemma 5.2 implies that $\deg\left(p_{i}^{\beta_{i}}p_{j}\right)>\deg\left(p_{j}\right)$. Clearly, (22) holds using Lemma 2.1(i) if $p_{j}>2p_{i}$. Since $p_{j}\neq 2p_{i}$, assume that $p_{j}<2p_{i}$. We have the following two cases.

1. $\bullet$

$i=1$: Since $p_{1}\in\{2,3\}$ and $p_{j}<2p_{1}$, we have $j=2$ and $(p_{1},p_{2})=(2,3)$ or $(3,5)$. If $(p_{1},p_{2})=(2,3)$, then $\left(p_{1}+p_{2}\right)\phi\left(p_{2}p_{3}\right)-p_{1}p_{2}p_{3}=10\phi\left(p_{3}\right)-6p_{3}=4p_{3}-10>0$ as $p_{3}\geq 5$. If $(p_{1},p_{2})=(3,5)$, then $p_{3}\geq 7$ and $\left(p_{1}+p_{2}\right)\phi\left(p_{2}p_{3}\right)-p_{1}p_{2}p_{3}=32\phi\left(p_{3}\right)-15p_{3}=17p_{3}-32>0$. 2. $\bullet$

$(i,j)=(2,3)$: Here $k=1$ and $p_{3}\geq p_{2}+2$. If $p_{k}=p_{1}=2$, then $\left(p_{2}+p_{3}\right)\phi\left(p_{1}p_{3}\right)-p_{1}p_{2}p_{3}=p_{3}^{2}-\left(p_{2}+p_{3}\right)-p_{2}p_{3}=p_{3}(p_{3}-p_{2}-1)-p_{2}>0$. If $p_{k}=p_{1}=3$, then $\left(p_{2}+p_{3}\right)\phi\left(p_{1}p_{3}\right)-p_{1}p_{2}p_{3}=2p_{3}^{2}-2\left(p_{2}+p_{3}\right)-p_{2}p_{3}=p_{3}(2p_{3}-p_{2}-2)-2p_{2}>0.$

This completes the proof. ∎

Corollary 5.3**.**

If $p_{3}\geq 2p_{2}+1$, then $\delta(\mathcal{P}(C_{n}))=\deg\left(p_{3}^{\alpha_{3}}\right)$.

Proof.

By (3), we have $\deg\left({p_{2}^{\alpha_{2}}}\right)-\deg\left({p_{3}^{\alpha_{3}}}\right)>p_{2}^{\alpha_{2}-1}\left[(p_{3}-1)\phi\left(p_{1}^{\alpha_{1}}\right)-p_{2}p_{1}^{\alpha_{1}}\right]$. Since $p_{3}\geq 2p_{2}+1$, it follows that $\deg\left({p_{2}^{\alpha_{2}}}\right)-\deg\left({p_{3}^{\alpha_{3}}}\right)>p_{1}^{\alpha_{1}-1}p_{2}^{\alpha_{2}}\left(2\phi(p_{1})-p_{1}\right)\geq 0$ and so $\delta(\mathcal{P}(C_{n}))=\deg\left(p_{3}^{\alpha_{3}}\right)$ by Theorem 1.5. ∎

Example 5.4**.**

Take $n=2\cdot 3^{3}\cdot 5$. Then $\delta(\mathcal{P}(C_{n}))=\deg(3^{3})=113<125=\deg(5)$. It follows that if $p_{3}<2p_{2}+1$, then $\delta(\mathcal{P}(C_{n}))=\deg\left(p_{2}^{\alpha_{2}}\right)<\deg\left(p_{3}^{\alpha_{3}}\right)$ may occur.