The Undirected Power Graph of a Finite Group

Abstract

The power graph ${\fancyscript{P}}(G)$ of a group $G$ is the graph which has a vertex set of the group elements and two elements are adjacent if one is a power of the other. Chakrabarty, Ghosh, and Sen proved the main properties of the undirected power graph of a finite group. The aim of this paper is to generalize some results of their work and presenting some counterexamples for one of the problems raised by these authors. It is also proved that the power graph of a $p$-group is $2$-connected if and only if the group is a cyclic or generalized quaternion group and if $G$ is a nilpotent group which is not of prime power order then the power graph ${\fancyscript{P}}(G)$ is $2$-connected. We also prove that the number of edges of the power graph of the simple groups is less than or equal to the number of edges in the power graph of the cyclic group of the same order. This partially answers to a question in an earlier paper. Finally, we give a complete classification of groups in which the power graph is a union of complete graphs sharing a common vertex.

1 Introduction

There are number of constructing graphs by groups. The directed power graph of a group $G$ (and a semigroup $S$, as well) was first defined by Kelarev and Quinn as the digraph $\overrightarrow{\fancyscript{P}}(G)$, ($\overrightarrow{\fancyscript{P}}(S)$), with vertex set $G$, ($S$), in which there is an arc from $x$ to $y$ if and only if $x \ne y$ and $y = x^{m}$ for some positive integer $m$. Some interesting combinatorial properties of these graphs were investigated by the authors in [8–11]. Motivated by the work of Kelarev and Quinn, Chakrabarty et al. [6] introduced and studied some properties of undirected power graphs of semigroups and groups. The reader is encouraged to see Cameron’s work on the undirected power graphs of finite groups [3, 4]. Also in the recent works [12, 13], some properties of the undirected power graphs of finite groups are investigated. All groups and graphs in this paper are assumed to be finite. Also, we consider power graphs as simple graphs (with no loops or repeated edges). Throughout the paper, we follow the terminology and notation of [14] for groups and [15] for graphs.

Suppose $G$ is a finite group. The undirected power graph ${\fancyscript{P}}(G)$ is a graph in which $V({\fancyscript{P}}(G)) = G$ and two distinct elements $x$ and $y$ are adjacent if and only if one of them is a power of the other. If $G$ is a finite group then it is easy to see that the power graph ${\fancyscript{P}}(G)$ is a connected graph of diameter at most 2. In [6], it is proved that for a finite group $G$, ${\fancyscript{P}}(G)$ is complete if and only if $G$ is a cyclic group of order $1$ or $p^m$, for some prime number $p$ and positive integer $m$.

Suppose $G$ is a finite group and $x, y \in G$. The “cyclic subgroup of $G$ generated by $x$” is denoted by $\langle x \rangle$ and $\deg (x)$ denotes the degree of $x$ in ${\fancyscript{P}}(G)$. The distance between $x$ and $y$ in ${\fancyscript{P}}(G)$ is defined as the length of a minimal path connecting them. The girth of $\Gamma$, $g(\Gamma )$, is the length of a shortest cycle within the graph.

We need to know some graph theoretical properties of power graphs of finite groups to be able to handle them comprehensively. To fulfill this goal the results of the removal of identity element and the graph remaining connected have been studied. Also the classification of $p$-groups based on this assumption is presented in Sect. 2.2.

A cut vertex in a graph $\Gamma$ is any vertex whose removal increases the number of $\Gamma$-components. The graph $\Gamma$ is said to be $2$-connected if $\Gamma$ does not have a cut vertex. A trail in $\Gamma$ containing every edge exactly once is called an Eulerian trail. $\Gamma$ is called Eulerian, if it has an Eulerian cycle, or equivalently, every vertex of $\Gamma$ is of even degree. A Hamiltonian path in $\Gamma$ is a path that visits each vertex exactly once. A Hamiltonian cycle is a Hamiltonian path that is a cycle. A graph containing a Hamiltonian cycle is called a Hamiltonian graph.

Let $G$ be a finite group and $x \in G$. Since we assumed the power graph of a group to be simple, we have

$$\begin{aligned} \deg (x) = |\{g \in G \ | \ \langle x \rangle \le \langle g \rangle \ or \ \langle g \rangle \le \langle x \rangle \}| - 1. \end{aligned}$$

In the following result, a characterization of Eulerian power graphs is presented. Throughout $\phi (k)$ shall denote the Euler totient function of the natural number $k$.

Lemma 1.1

Let $G$ be a finite group. The power graph ${\fancyscript{P}}(G)$ is Eulerian if and only if $|G|$ is odd.

Proof

Choose $x \in G$. The number of elements $g$ such that $\langle g \rangle \le \langle x \rangle$ is $o(x)$ and the number of elements $g$ such that $\langle x \rangle < \langle g \rangle$ is $\sum _{\langle x \rangle < \langle g \rangle }\phi (o(g))$. From the fact that the graph is simple, we have

$$\begin{aligned} \deg (x) = o(x) + \sum _{\langle x \rangle < \langle g \rangle }\phi (o(g)) - 1. \end{aligned}$$

Since for each odd positive integer $n$, $\phi (n)$ is even, $o(x) - 1 + \sum _{\langle x \rangle < \langle g \rangle } \phi (o(g))$ is even. Therefore, the degree of each vertex is even and so ${\fancyscript{P}}(G)$ is Eulerian. Since $\deg (1) = |G| - 1$, the converse is obvious. $\square$

Lemma 1.2

The power graph of a finite group $G$ is a tree if and only if $G$ is an elementary abelian $2$-group.

Lemma 1.3

$g({\fancyscript{P}}(G)) = 3$ if and only if $G$ is not an elementary abelian $2$-group. Moreover, if ${\fancyscript{P}}(G)$ is 2-connected then $g({\fancyscript{P}}(G) - \{e\}) = 3$.

Proof

If $g({\fancyscript{P}}(G)) = 3$ then ${\fancyscript{P}}(G)$ has a cycle of length 3 and so it is not an elementary abelian $2$-group. Conversely, if $G$ is not an elementary abelian $2$-group then it has at least one element of order 4 or an odd prime $p$, as desired. To prove the second part, we notice that $G$ has an element of order $\ge 4$. $\square$

2 Main Results

In this section, we first present some counterexamples to a conjecture given in [6]. Then, we focus on the classification problem of 2-connected power graphs and the power graphs which is a union of complete graphs share the identity.

2.1 Counterexamples

Suppose $U_n$ denotes the unit group of the ring $\mathbb {Z}_n$. In [6], Chakrabarty, Ghosh, and Sen have asked about the values of $n$ for which ${\fancyscript{P}}(U_n)$ is Hamiltonian. They have written: in this context our conjecture is that ${\fancyscript{P}}(U_n)$ is Hamiltonian for all values of $n \ge 3$ except for $n = 2^mp_1p_2 \ldots p_k$, where $p_1$, $p_2$, $\ldots$, $p_k$ are distinct Fermat primes, $m$ and $k$ are non-negative integers, $m \ge 2$ for $k = 0, 1$ and $k \ge 2$ for $m = 0, 1$. We now present some sequences of counterexamples to prove that this conjecture is incorrect.

By [15, Proposition 7.2.3], if a graph $G$ has a vertex subset $S$ such that $V(G)- \ S$ has at least $|S| + 1$ components then $G$ is not Hamiltonian. We apply this result to provide a counterexample for the mentioned conjecture.

Counterexample 2.1

If $n = 2^v \times 3^2$, $v \ge 3$, then ${\fancyscript{P}}(U_n)$ does not have a Hamiltonian cycle.

Proof

We have $U_n \cong \mathbb {Z}_2 \times \mathbb {Z}_2 \times \mathbb {Z}_3 \times \mathbb {Z}_{2^{v-2}}$. Suppose $x, y, t$ and $z$ are elements of $U_n$ such that $o(x) = 2, o(y) = 2, o(t) = 3$, $x$ and $y$ are distinct, and do not lie in $\langle z \rangle \$ and $o(z) = 2^{v-2}$. If $S = \{ 1, t, t^2\}$ then $\{ x, xt, xt^2\}$, $\{ y, yt, yt^2\}$, $\{ xy, xyt, xyt^2\}$ and $\{ xyz^{2^{(v-3)}}$, $xyz^{2^{(v-3)}}t$, $xyz^{2^{(v-3)}}t^2\}$ are connected components of ${\fancyscript{P}}(U_n)-S$. $\square$

Counterexample 2.2

If $n = 2^t \times 7$, $t \ge 2$, then ${\fancyscript{P}}(U_n)$ does not have a Hamiltonian cycle.

Proof

The proof is similar to Counterexample 2.1. $\square$

Counterexample 2.3

If $n = 2^2 \times 3^2 \times p$, where $p$ is a Fermat prime, then ${\fancyscript{P}}(U_n)$ does not have a Hamiltonian cycle.

Proof

The proof follows from the same argument as Counterexample 2.1. $\square$

Other counterexamples exist as well, such as: $n=2^2 \times 3^2 \times 17$, $2^2 \times 3 \times 13$ and more counterexamples of this form can be constructed using the same method. Thus, the main question of Chakrabarty, Ghosh, and Sen [6] to determine those values of $n$ for which ${\fancyscript{P}}(U_n)$ is Hamiltonian, remains still open.

2.2 2-Connectivity of Power Graphs

Now, we discuss $2$-connectivity of power graphs. Suppose $G$ is a finite group such that ${\fancyscript{P}}(G) - \{ e \}$ is connected. Then one can easily see that ${\fancyscript{P}}(G)$ is $2$-connected. In what follows, we investigate the structure of $2$-connected power graphs. This is a generalized approach to what was proved by Chakrabarty et al. in [6].

Theorem 2.1

Suppose $G$ is a $p$-group. The power graph ${\fancyscript{P}}(G)$ is $2$-connected if and only if $G$ is a cyclic or generalized quaternion group.

Proof

Suppose $G$ is a $p$-group and ${\fancyscript{P}}(G)$ is $2$-connected. We first prove that for each pair of distinct elements $x, y \in G$ there exists a cycle of length 3 or 4 containing $x$ and $y$. Since $G$ is $2$-connected there exists a cycle containing $x$ and $y$. Suppose $C$ is a minimal cycle containing these elements. If $xy \in E({\fancyscript{P}}(G))$ then we have the cycle $e, x, y, e$, as desired. So, we can assume that $d(x,y) > 1$. Set

$$\begin{aligned} C: e, x, u_1, \ldots , u_k, y, e. \end{aligned}$$

By definition $\langle x \rangle \le \langle u_1 \rangle$ or $\langle u_1 \rangle \le \langle x \rangle$. Suppose $\langle x \rangle \le \langle u_1 \rangle$. If $\langle u_1 \rangle \le \langle u_2 \rangle$ then $x$ and $u_2$ are adjacent, contradicting the minimality of $C$. Thus, $\langle u_2 \rangle \le \langle u_1 \rangle$. Since $\langle u_1 \rangle$ is cyclic, $\langle x \rangle \le \langle u_2 \rangle$ or $\langle u_2 \rangle \le \langle x \rangle$, which is a contradiction. This shows that $g({\fancyscript{P}}(G)) = 3$. We now assume that $\langle u_1 \rangle \le \langle x \rangle$. Again from the minimality of $C$, we have: $\langle u_1 \rangle \le \langle u_2 \rangle$. If $k \ge 3$ then a similar argument as above leads to a contradiction. So, we have a cycle of length 4 containing $x$ and $y$. Hence, $G$ has a cycle of form $e, x, y, e$ or $e, x, u, y, e$.

Next, we prove that $G$ has a unique non-trivial subgroup contained in every subgroup of $G$. Assume $x, y, z$ are three non-identity elements of $G$. We show that $\langle x \rangle \cap \langle y \rangle \cap \langle z \rangle$ is non-trivial. We consider four different cases in our main proof, as follows:

Case 1

There is a cycle of length 3 containing $x, y$ and a cycle of length 3 containing $y, z$. Suppose $\langle z \rangle \le \langle y \rangle$. Since $\langle x \rangle \le \langle y \rangle$ or $\langle y \rangle \le \langle x \rangle$, we have: $\langle z \rangle$, $\langle x \rangle$ $\le \langle y \rangle$ or $\langle z \rangle \le \langle y \rangle \le \langle x \rangle$. In the fourth case, by cyclicity of $\langle y \rangle$, $\langle x \rangle \le \langle z \rangle \le \langle y \rangle$ or $\langle z \rangle \le \langle x \rangle \le \langle y \rangle$. The case of $\langle y \rangle \le \langle z \rangle$ is similar. This means that one of the elements of $\{ x, y, z\}$ is contained in cyclic subgroups generated by the other two elements.

Case 2

There is a cycle of length 3 containing $x, y$ and a cycle of length 4 containing $y, z$. Choose an element $t$ such that $\langle t \rangle \le \langle y \rangle$,$\langle z \rangle$. If $\langle x \rangle \le \langle y \rangle$ then $\langle t \rangle$, $\langle x \rangle \le \langle y \rangle$. Therefore, $\langle t \rangle \le \langle x \rangle$ or $\langle x \rangle \le \langle t \rangle$. Thus, one of the elements of $\{x, y, z, t\}$ is contained in cyclic subgroups generated by the other three elements.

Case 3

There is a cycle of length 4 containing $x, y$ and a cycle of length 3 containing $y, z$. This case is similar to the case 2.

Case 4

There is a cycle of length 4 containing $x, y$ and a cycle of length 4 containing $y, z$. Choose elements $u$ and $t$ such that $\langle u \rangle \le \langle x \rangle$, $\langle y \rangle$ and $\langle t \rangle \le \langle y \rangle$, $\langle z \rangle$. Then $\langle u \rangle$, $\langle t \rangle$, $\le \langle y \rangle$. Subsequently, $\langle u \rangle$ $\le \langle x \rangle$, $\langle y \rangle$, $\langle z \rangle$, $\langle t \rangle$ or $\langle t \rangle$ $\le \langle x \rangle$, $\langle y \rangle$, $\langle z \rangle$, $\langle u \rangle$.

Therefore, by an inductive argument, we can find a non-trivial subgroup of $G$ contained in each non-trivial subgroup of $G$. Now by [2, Theorems 8.5, 8.6], $G$ is cyclic or generalized quaternion. Conversely, from the fact that in the power graphs of cyclic groups all vertices are joined and in the generalized quaternion group $Q_{2^n}$ there exists a unique involution that is joined to all other vertices, the result follows. $\square$

As a consequence of what was proven above, we now know that:

Corollary 2.1

Suppose $G$ is a finite $p$-group. Then ${\fancyscript{P}}(G)$ has a Hamiltonian cycle if and only if $|G| \ne 2$ and $G$ is cyclic.

Suppose $G = Q_{2^n}$ and $t$ is the unique element of $G$ of order 2. Since the number of connected components of ${\fancyscript{P}}(G) - \{e, t\}$ is greater than 2, it is not Hamiltonian. If $p$ is odd, we have the following characterization of Hamiltonian power graphs of finite $p$-groups.

Corollary 2.2

If $p$ is a odd prime then the power graph of a $p$-group is $2$-connected if and only if it is Hamiltonian.

We are now ready to investigate the structure of $2$-connected power graphs in general. We start by nilpotent groups.

Theorem 2.2

Suppose $G$ is a nilpotent group. If $G$ is not a $p$-group then the power graph ${\fancyscript{P}}(G)$ is $2$-connected.

Proof

Suppose $G = P_1 \times P_2 \times \cdots \times P_k$, where $P_i$’s are $p_i$-group and $p_1$, $\ldots ,$ $p_k$, $k \ge 2$, are distinct primes. Choose two non-identity arbitrary elements $x, y \in G$. Obviously, $o(x) = p_1^{\alpha _1} \ldots p_k^{\alpha _k}$ and $o(y) = p_1^{\beta _1} \ldots p_k^{\beta _k}$, where $\alpha _i$’s and $\beta _i$’s are non-negative integers. We have two different cases as follows:

Case 1

There exists $i$, $1 \le i \le k$, such that $\alpha _i \ne 0$ and $\beta _i = 0$. Assume that $o(x) = p_i^{\alpha _i}t$ and $o(y) = r$. Then $o(x^t) = p_i^{\alpha _i}$ and $(r,p_i^{\alpha _i}) = 1$. Since $G$ is nilpotent, $[x^t,y] = 1$ and so $x^t = (x^ty)^{vr}$ and $y = (x^ty)^{up_i^{\alpha _i}}$, for some $u, v \in Z$. Hence $x, x^t, x^ty, y$ is a path connecting $x$ and $y$ in ${\fancyscript{P}}(G)$, as desired.

Case 2

$o(x)$ and $o(y)$ have the same prime divisors. We can assume that $o(x) = p_1^{\alpha _1} \ldots p_s^{\alpha _s}$ and $o(y) = p_1^{\beta _1} \ldots p_s^{\beta _s}$, where $\alpha _i$’s and $\beta _i$’s are positive integers. Then $o(x^{p_2^{\alpha _2} \ldots p_s^{\alpha _s}}) = p_1^{\alpha _1}$ and $o(y^{p_2^{\beta _2} \ldots p_s^{\beta _s}}) = p_1^{\beta _1}$. Choose $t \in G$ of order $p_2$ and define

$$\begin{aligned} u = x^{p_2^{\alpha _2} \ldots p_s^{\alpha _s}} ; \quad v = y^{p_2^{\beta _2} \ldots p_s^{\beta _s}}. \end{aligned}$$

Therefore, $(vt)^{o(v)} = t^{o(v)}$ and $(ut)^{o(u)} = t^{o(u)}$. So, we have the following path connecting $x$ and $y$ in ${\fancyscript{P}}(G)$:

$$\begin{aligned} x, x^{p_2^{\alpha _2} \ldots p_s^{\alpha _s}}, x^{p_2^{\alpha _2} \ldots p_s^{\alpha _s}}t, t, y^{p_2^{\beta _2} \ldots p_s^{\beta _s}}t, y^{p_2^{\beta _2} \ldots p_s^{\beta _s}}\!, y, \end{aligned}$$

proving the theorem. $\square$

The converse of Theorem 2.2, is not necessarily correct. To investigate this, it is enough to consider the group $G = \mathbb {Z}_5 \times \mathbb {S}_3$. Then $G$ is $2$-connected non-nilpotent group. Moreover, the dihedral group $D_{2p}$, $p$ is prime, is a solvable group that its power graph is not $2$-connected. On the other hand, we have seen that if $G$ is a nilpotent group with at least two prime factors in its order then ${\fancyscript{P}}(G)$ is $2$-connected, but nilpotency of groups does not yield that their power graphs are Hamiltonian. The group $G = \mathbb {Z}_3 \times D_8$ is an example of such groups.

Lemma 2.1

If $A$ and $B$ are groups of coprime orders such that $A$ is cyclic of prime order then ${\fancyscript{P}}(A \times B)$ is $2$-connected.

Proof

The proof is similar to that given in Theorem 2.2.

The previous lemma shows that $2$-connectivity does not imply solvability or nilpotency of groups.

2.3 Power Graphs of Simple Groups

Now the power graphs of simple groups are taken into account.

Lemma 2.2

Let $G$ be a finite group that is not of prime order. If ${{\mathrm{Max}}}\omega (G) = p$, where $p$ is prime and $\omega (G)$ is the set of all element orders of $G$, then ${\fancyscript{P}}(G)$ is not $2$-connected.

Proof

Suppose $G$ is $2$-connected. Then ${\fancyscript{P}}(G)-\{ e \}$ is connected. Choose an element $g$ of order $p$. Since $G$ is not of a prime order, there is an element of a coprime order to $p$ in $G$, say $t$. Now the connectivity of ${\fancyscript{P}}(G) - \{ e \}$ implies that $g$ and $t$ lie on a path. But, $\langle g \rangle$ does not have a non-trivial subgroup and since $g$ has the maximum order, $g$ is not contained in any cyclic subgroup of $G$. Therefore, there is no path from $g$ to $t$ in ${\fancyscript{P}}(G)-\{ e \}$, which is a contradiction. $\square$

Simple groups $\mathbb {A}_p$, $p = 5, 6, 7$, $L_2(p),$ for prime $p$, $p \ge 5$, and the sporadic groups, $M_{11}$, $M_{12}$, $M_{22}$, $M_{23}$, $M_{24}$, $J_1$, $O'N$, $Ly$, $J_3$, $Ru$, are examples of groups that satisfy the condition of Lemma 2.2 and so their power graphs are not $2$-connected.

Our calculations with finite groups of small orders suggest the following question:

Question 2.1

Determine all non-abelian simple groups with $2$-connected power graphs.

In the following theorem, we prove that the number of edges in the power graphs of a simple group of order n is at most the number of edges in the power graph of the cyclic group of order $n$. Recall that for a graph $\Gamma$, $\Delta (\Gamma )$ and $\delta (\Gamma )$ are the maximal and the minimal degree of $\Gamma$, respectively.

Theorem 2.3

If $G$ is a finite simple group of order $n$ then $|E({\fancyscript{P}}(G))| \le |E({\fancyscript{P}}(\mathbb {Z}_n))|$.

Proof

We first prove that if $x \in \mathbb {Z}_n$ then

$$ \begin{aligned} \deg (x) = o(x) + \displaystyle \sum _{ko(x) | n \ \& \ k \ne 1} \phi (ko(x)) - 1. \end{aligned}$$

To do this, it is enough to note that the number of subgroups of $\mathbb {Z}_n$ containing $\langle x \rangle$ as a proper subgroup is $\sum _{ko(x) | n \ \& \ k \ne 1} \phi (ko(x)).$ Next, we show that $\delta (\mathbb {Z}_n) \ge n/p$, where $p$ is the largest prime factor of $n$. For a proof it is enough to note that by [1, Lemma C] if $n$ is a positive integer and $p$ is the largest prime factor of $n$ then $\phi (n) \ge n/p$, and thus $\deg (x) \ge o(x) - 1 + n/p$ for each $x \in \mathbb {Z}_n$, as required. Define

$$\begin{aligned} s = {{\mathrm{Max}}}\{ \deg (g) \ | \ g \in {\fancyscript{P}}(G) - \{e\}\}. \end{aligned}$$

From the Euler’s theorem in elementary graph theory, if $\Delta ({\fancyscript{P}}(G)- \{e\}) \le \delta ({\fancyscript{P}}(\mathbb {Z}_n))$ then $|E({\fancyscript{P}}(G))| \le |E({\fancyscript{P}}(\mathbb {Z}_n))|$.

To complete our argument, we prove that if $G$ is a finite group and $x \in G$ such that $\deg _{{\fancyscript{P}}(G)- \{e\}}(x) > |G|/p$ then $G$ has a non-trivial normal subgroup $N$. Set

$$\begin{aligned} A = \{ g \in G - \{e\} \ | \ \langle x \rangle < \langle g \rangle \ \mathrm{or} \ \langle g \rangle < \langle x \rangle \}. \end{aligned}$$

Our assumption implies that $|A| > n/p$ and thus we have $r = |G : C_G(x)| < p$. If $r=1$ then $x \in Z(G)$, as desired. Otherwise $r \ne 1$. If $G$ is simple then $G$ can be embedded into $\mathbb {A}_r$. This implies that $p | r!$, which is impossible. This completes the proof. $\square$

2.4 A Classification Theorem

In this subsection, a complete classification of groups in which the power graph is a union of complete graphs that share the identity is presented. The following general result is needed for our main result of this section.

Lemma 2.3

Let $G$ be a finite group. If ${\fancyscript{P}}(G)$ is a union of complete graphs that share the identity then the power graph of each Sylow subgroup of $G$ has the same structure.

Proof

Suppose $Q$ is a Sylow $p$-subgroup of $G$ of order $p^s$. By [6, Proposition 5.4], ${\fancyscript{P}}(Q)$ is an induced subgraph of ${\fancyscript{P}}(G)$ and so ${\fancyscript{P}}(Q) \subseteq \cup X$, where each $X$ is a complete subgraph of ${\fancyscript{P}}(G)$. Suppose $\cup _{i=1}^rX_i$ is an irredundant cover of ${\fancyscript{P}}(Q)$, i.e., no proper subset is also a cover. Hence, each $X_i$ contains an element $y_i$ such that $p | o(y_i)$. We first show that $X_i$ is a subgroup of $G$. Set $X_i = \{ e = x_1, \cdots , x_n\}$. Choose $x_i, x_j \in X_i$. Since $x_i$ and $x_j$ are adjacent, $x_i = x_j^u$ or $x_j = x_i^v$. Therefore, $x_ix_j = x_j^{u+1}$ or $x_ix_j = x_i^{v+1}$ and so $x_ix_j \in X_i$, as desired. Since each $X_i$ has an element of $Q$, by [6, Theorem 2.12] that states that the power graph is complete if and only if the group is cyclic of prime power order, $X_i$ has prime power order.

Define $Y_i = \{ x \in X_i \ | \ x \in Q\}$, $1 \le i \le r$. Since $P(Q)$ is the induced subgraph of ${\fancyscript{P}}(G)$ and $X_i$ is a complete graph, then $Y_i = {\fancyscript{P}}(G) \cap X_i$ is always complete. $\square$

The following lemma is crucial in the last result of this section.

Lemma 2.4

[12, Theorem 12] ${\fancyscript{P}}(G)$ is a union of complete graphs which share the identity element of $G$ if and only if $G$ is an EPPO-group and for every maximal cyclic subgroup $A$ and $B$ with $A \ne B$, $A\cap B=e$.

In [4, Examples], P. J. Cameron proved that if $G$ is a finite group of order $n$ and exponent 3 then ${\fancyscript{P}}(G)$ consists of $(n-1)/2$ triangles sharing a common vertex. In the following theorem, we classify such $p$-groups.

Theorem 2.4

Let $G$ be a finite $p$-group. Then the power graph ${\fancyscript{P}}(G)$ is a union of complete subgraphs which share the identity element of $G$ if and only if $G$ is isomorphic to a cyclic group, $p$-group of exponent $p$ or a dihedral group.

Proof

Let $|G|=p^n$ with $\exp (G)\ne p$. Suppose that $P(G)\subseteq \cup _{i=1}^r \Gamma _i$ is an irredundant cover of complete subgraphs of ${\fancyscript{P}}(G)$. The vertex set $X_i=V(\Gamma _i)$ is a subgroup of $G$. Since $\exp (G)\ne p$, there exists a cyclic subgroup $T=X_j$ of order grater than $p$. Assume that $T= \langle t\rangle$ and $g\in G\backslash T$ such that $[t,g]=1$, then $o(tg)\ne p$. Apply Lemma 2.4 to prove that $\langle t\rangle \cap \langle tg\rangle =e$. If $g$ is a central element of order $p$ of $G$ then $\langle t\rangle \cap \langle tg\rangle \ne e$ and hence $g\in T$. Therefore, $\Omega _1(Z(G)) \le T$ and so $Z(G)$ is cyclic. Also, $T$ is a unique subgroup of order grater than $p$, which deduces that $T\lhd G$ and $Z(G)\le T$. Suppose $Z(G)=\langle z\rangle$. If $o(z)> p$ then for all $g\in G$, we have $o(zg)>p$, therefore $zg\in T$ and so $g\in G$. This implieas that $G=T$ is cyclic.

Let $|Z(G)|=p$ and $G$ is not cyclic. We also assume that $|T|=\exp (G)=p^{\ell }$ and $g\in G\backslash T$. Consider the subgroup $H = T\langle g\rangle$. Clearly, $H$ is non-abelian. Let $p$ is odd. Since $g$ is of order $p$ and ${{\mathrm{Aut}}}(T)$ has just one subgroup of order $p$, $t^g=t^s$ where $s={1+p^{\ell -1}}$. Set $s_i=1+ip^{\ell -1}$, then $t^{s^i}=t^{s_i}$ and hence

$$\begin{aligned} (tg)^p=\prod _{i=0}^{p-1}{t^{s^i}}=\prod _{i=0}^{p-1}{t^{s_i}}=t^{\sum _{i=0}^{p-1}s_i}=t^{p+p^{\ell -1}[p(p-1)/2]}=t^p, \end{aligned}$$

which is impossible. So $p=2$ and $T\langle g\rangle$ is isomorphic to a dihedral, semidihedral, or quasidihedral group. Since semidihedral and quasidihedral groups have more than one subgroup of order greater than $2$, $T\langle g\rangle \cong D_{2^{\ell +2}}$. Now, if $x\in G\backslash H$, then $T\langle x\rangle$ must be dihedral and so for both $x$ and $g$ we have $t^x=t^g=t^{-1}$. Therefore, $[t,xg]=1$ and $T\langle xg\rangle$ is abelian, which is impossible. We conclude that $G=T\langle g\rangle$ is of dihedral type. $\square$