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The Generalized Power Graph of a Finite Group

Graphs associated with groups and other algebraic structures have been actively investigated, since they have valuable applications in data mining. For a finite group G, let ΓG be the graph with the non-identity elements of G as the vertex set, and two vertices are adjacent if they respectively lie in two conjugate proper subgroups of G. ΓG is called the generalized power graph with respect to G. This paper explores how the graph theoretical properties of ΓG can affect on the group theoretical properties of G.


Introduction
Graphs associated with groups and other algebraic structures have been actively investigated, since they have valuable applications (cf. [1]) and are related to automata theory (cf. [2,3]), such as, Cayley graphs as classifiers for data mining (cf. [4]). Also, study of algebraic structures by graphs associated with them gives rise to many recent and interesting results in the literature. This field is relatively new, and over the years different types of graphs derived from a group have been defined, such as the prime graph [5], the non-commuting graph [6], the power graph, and of course Cayley graphs, which have a long history.
The non-commuting graph was first considered by Neumann [6] and this graph are studied by many authors. Recently, Erfanian et al. [7] introduced the relative n-th non-commuting graph of a finite group, which is a generalized non-commuting graph. The directed power graph of a semigroup was defined by Kelarev and Quinn [8]. Chakrabarty et al. [9] introduced the (undirected) power graph of a finite group.
Let G be a finite group with identity element e. The power graph P G of G is a simple graph whose vertex set is G and two vertices x and y in G are adjacent if and only if x=y m or y=x m for some positive integer m. Moghaddamfar et al. [10] discussed the subgraph of P G induced by G\{e}, where e is the identity element of G. Now consider the graph Γ G as follows: Take G\{e} as the vertex set and join two vertices if they respectively lie in two conjugate proper subgroups of G. For example, Fig. 1 is the generalized power graph of the quaternion group Q 8 with 8 elements and Fig. 2 is the generalized power graph of the symmetric group S 3 on 3 letters. By the definition, if G is non-cyclic then the subgraph of P G induced by G \{e} is a subgraph of Γ G . If G is a cyclic group with order n, then Γ G has at least Φ(n) isolate vertices which are all generators of G (In fact, if n=2p for some prime p, then the number of the isolate vertices of Γ G is Φ(n)+1; otherwise it is Φ(n)), and denoted by the set S, here Φ is the Euler totient function. In the case, the subgraph induced by G \{e}\S is also a subgraph of Γ G . So call Γ G the generalized power graph of G. Let x, y  G \{e}. If <x, y> ≠ G, then x is adjacent to y in Γ G . In general, x ~ y does not imply that <x, y> is a proper subgroup of G (an example is showed in Fig. 2, (1,2) ~ (1,3) but <(1,2), (1,3)> = S 3 ). However, if G is restricted on Dedekind group (G is called a Dedekind group if its every subgroup is normal, for example, every abelian group is Dedekind), then x and y are joined by an edge of Γ G if and only if there exists a proper subgroup H of G such that x, yH.
This paper explores how the graph theoretical properties of Γ G can affect on the group theoretical properties of G. Section 2 discusses the connectivity of generalized power graphs. In Section 3, all groups whose generalized power graphs are empty, bipartite, planar and complete are obtained.

The connectivity of Γ G
The exponent of group G is defined as the least common multiple of the orders of all elements of G, is denoted by exp(G). Denote by π(n) the set of the prime divisors of a positive integer n. For convenience, write π(G) and π(x) instead of π(|G|) and π(|x|), respectively. In this section, for the groups G with |π(G)|=1 or 2, some sufficient and necessary conditions for which Γ G is connected are given. It also shows that if |π(G)|=3 then Γ G is connected.

Theorem 2.1 Let G be a p-group. Then Γ G is connected if and only if G is non-cyclic and G is not isomorphic to Z p × Z p .
Proof. First let Γ G be connected. It is clear that G is not cyclic. Assume, to the contrary, that G is isomorphic to Z p × Z p . Then exp(G)=p. Thus, for some non-identity element g of G, <g> is a maximal subgroup of G and hence Γ G is constructed by some complete subgraphs isomorphic to K p-1 . Now let a, bG, and let Γ <a> and Γ <b> be two distinct complete subgraphs of Γ G induced by the subgroups <a> and <b>, respectively. If there is an edge a i ~ b j between Γ <a> and Γ <b> , where 1≤i, j ≤ p-1, then <a i , b j > is a proper subgroup of G, which is impossible since <a i > = <a> and <b i > = <b>. It follows that Γ G =Γ Zp × Zp is isomorphic to (p+1)K p-1 . Certainly, Γ G is not connected, a contradiction.
Conversely, suppose that G is non-cyclic and is not isomorphic to Z p × Z p . Since H is self-conjugate for every proper subgroup H of G, the subgraph of Γ G induced by H\{e} is complete. If the Frattini subgroup of G, is non-trivial, then Γ G is connected; if not, G is an elementary abelian p-group. Therefore G is isomorphic to (Z p ) n , where n ≥ 3. In the case, it is easy to see that < a, b> is a proper subgroup of G, for all a,bG. It means that Γ G is complete.

Theorem 2.2 Let p and q be two distinct prime integers, G a group with π(G)={p, q}. Then Γ G is connected if and only if G is non-cyclic and has a proper subgroup S such that π(S)=π(G).
Proof. Suppose that Γ G is connected. It is straightforward that G is non-cyclic. Since π(G)={p, q} there exist two elements x and y of G\{e} such that |x|=p and |y|=q. In view of the connectivity of Γ G , there is a path: x ~ v 1 ~ v 2 ~ … ~ v n ~ y. Since x is adjacent to v 1 , there exist two conjugate proper subgroups S and K in G such that xS and yK. If π(S)={p, q}, then this completes the proof. Thus, now assume that S is a p-group and next, consider the edge v 1 ~ v 2 . Similarly, there is a proper subgroup S i of G, such that v i  S i and π(S i )={p, q}, where 1 ≤ i ≤ n. This procedure shall prove the result. Now suppose that G is non-cyclic and has a proper subgroup S such that π(S)=π(G). Assume that x and y are two distinct vertices of Γ G . Since π(G)={p, q}, π(x) has three probabilities: {p, q}, {p} and {q}, and so is π(y). Without loss of generality, consider four cases. Case 1. π(x)={p} and π(y)={p}. Clearly, G has two Sylow p-subgroups P 1 and P 2 (here P 1 and P 2 could be equal) such that xP 1 and yP 2 ; particularly, P 1 and P 2 are conjugate in G. Thus x and y are joined by an edge of Γ G . Case 2. π(x)={p,q} and π(y)={p}. Choose a p-subgroup P 0 of <x> and let v be a non-identity element of P 0 . Case 1 shows that y ~ v. Note that <x> is a proper subgroup of G, x ~ v. So x ~ v ~ y, as desired.
Case 3. π(x)={p} and π(y)={q}. Assume that x 1 and y 1 are two p-element and q-element of S respectively. Therefor x 1 ~ y 1 . Case 1 means that x 1 ~ x and y 1 ~ y. Thus there exists a path x ~ x 1 ~ y 1 ~ y, as required. Case 4. π(x)={p, q} and π(y)={p, q}. Clearly, there are p-element x 1 and q-element y 1 , such that x 1~ x and y 1~ y. Now Case 3 ends the proof.
Let P i be the Sylow p i -subgroup of G, where p i is prime for every 1≤ i ≤ s. {P 1 , P 2 ,...,P s } is called a Sylow basis of G provided that P i P j =P j P i for all 1≤ i, j≤ s. A well-known theorem of P. Hall shows that G is solvable if and only if G possesses a Sylow basis. Theorem 2.3 Let G be a non-cyclic group such that |π(G)|= 3. Then Γ G is connected.
Proof. First assume that G is solvable. Let π(G)={p 1 , p 2 , p 3 } and let {P 1 , P 2 , P 3 } be a Sylow basis of G. Now choose x in G\{e}, if π(x)=π(G), then an argument similar to the one used in Theorem 2.2 ends the proof. Therefore, now assume that π(x) is a proper subset of π(G), and let y be any vertex of Γ G . If π(x)  π(y) ≠  , then there exists at least a prime number p in π(G), such that P x and P y are two psubgroups of <x> and <y> respectively. Let u be a non-identity element belonging to P x and let v be a non-identity element belonging to P y . Case 1 of Theorem 2.2 indicates that u and v is adjacent. So x ~ u ~ v ~ y is a path in Γ G , as required. Now suppose that π(x)  π(y)=  . Then pick u,v in G\{e} such that u in P x ≤ P u =P i g and v in P y ≤ Pv=P j h , where P x is a p i -subgroup of <x>, P y is a p j -subgroup of <y>, P u and P v are Sylow p i -subgroup and Sylow p j -subgroup of G respectively, g, h  G and 1≤i, j≤ 3. Set now w P i \{e} and z  P j \{e}. Thus, as assuming that {P 1 , P 2 , P 3 } is a Sylow basis, w ~ z. It means that x ~ u ~ w ~ z ~ v ~ y is a path. It also follows that Γ G is connected. Now suppose that G is non-solvable. Corollary 4 of [11] tells us that, if G is a non-solvable group with |π(G)|= 3, then one of the following groups is involved in G: L 2 (4), L 2 (7), L 2 (8), L 2 (17), L 3 (3). By GAP, it follows that the generalized power graphs of these groups above are connected. The proof is now complete.

Groups characterized by their generalized power graphs
In the section, all groups whose generalized power graphs are empty, bipartite, planar and complete are obtained.

Theorem 3.1 Let G be a group. Then Γ G is empty if and only if G is isomorphic to one of the groups: Z 2 × Z 2 , Z 4 , Z p for some prime p.
Proof. It is easy to check that each of the generalized power graphs of Z 2 × Z 2 , Z 4 and Z p is empty, where p is a prime. Now suppose that Γ G is empty. Let x be an isolate vertex of Γ G . If <x>=G then G is cyclic. Thus, suppose now that <x>≠G. Since both x and x -1 belong to <x>, |x|=2. If there exists an element g in G such that <x> ≠ <x> g , then <x> g =<x g >. Therefore, x is adjacent to x g , a contradiction. It means that <x> is normal in G. On the other hand, clearly, <x> is a maximal subgroup of G. Thus, assume now that |G|=2p n , for some odd prime p and some integer n. If G is a 2-group then |G|=4. Otherwise, |G|=2p and so G is isomorphic to Z 2 × Z p , which is a cyclic group. It means that G is isomorphic to Z 2 × Z 2 or a cyclic group. If G is a cyclic group with order composite number, then n=4. Now it follows that G is one of the groups: Z 2 × Z 2 , Z 4 , Z p, where p is a prime.

Theorem 3.2 Let G be a group. Then Γ G is bipartite if and only if G is isomorphic to one of the groups: Z 2 × Z 2 , Z 4 , Z 6 , Z 3 × Z 3 , Z 9 , Z p for some prime p.
Proof. The sufficiency is obvious. Now assume that Γ G is a bipartite graph. It is well known that a graph is bipartite if and only if the graph contains no odd cycles. Thus, G has no proper subgroups of order greater than or equal 4. If |π(G)|=1 then |G|=2 2 , 3 2 or Z p , where p is prime. It is easy to verify that G is one group of the groups: Z 2 × Z 2 , Z 4, Z 3 × Z 3 , Z 9 , Z p .
If |π(G)|=2 then the order of G equal to 6. Since Γ S3 contains a triangle induced by {(1,2), (1,3), (2,3)}, which is non-bipartite. So G is isomorphic to Z 6 , as desired. Proof. First assume that Γ G is planar. If there exists a proper subgroup H of G of order greater than or equal 6, then the subgraph of Γ G induced by H\{e} is isomorphic to the complete graph K |H\{e}| , which is a contradiction since K 5 is non-planar. It follows that the order of any proper subgroup of G must be less than or equal to 5. It follows that π(G) is a subset of {2, 3, 5} by Sylow theorems. Now consider three cases. In this case, it is easy to see that |G|=2×3×5 or 2 2 ×3×5. If |G|=2×3×5, then G has at least a normal Sylow subgroup P. By Burnside's p a q b -theorem, it follows that G/P is solvable and so is G. Since solvable group possesses Hall π-subgroups for all sets π of primes of π(G), G has at least a subgroup of order 15, a contradiction. Now suppose that |G|=2 2 ×3×5. If G is simple, then G is isomorphic to A 5, the alternating group on 5 letters. It is well known that < (3,4,5), (1,2)(4,5)> is a subgroup of A 5 of order 6. Hence, It follows that Γ G is non-planar. Thus now assume that G is not simple. Let H and P be a non-trivial normal subgroup and a Sylow 5-subgroup of G, respectively. Considering the subgroup HP, it is clearl that |HP|=5|H| if H  P =1. If HP is a proper subgroup of G, then |HP|>5, a contradiction; otherwise G=HP and so |H|= 12, a contradiction again. It means that H  P=P. If P is a proper subgroup of H, then, a contradiction as |P|=5. This implies H=P, that is, P is normal in G. It follows that G is solvable. Similarly as above, also a contradiction.

Theorem 3.3 Let G be a group. Then Γ G is planar if and only if G is isomorphic to one of the groups
Case 2. |π(G)|=2. Obviously, |G|=2×3, 3×5, 2×5, 2 2 ×3 or 2 2 ×5. It is easy to verify that Γ Z6 , Γ Z10 , Γ S3 and Γ Z15 are planar. Also, it is easy to see that Γ D10 is non-planar since the subgraph induced by the set of all elements of order 2 is K 5 .
Let |G|= 2 2 ×5. Then, in view of Sylow theorem, the Sylow 5-subgroup P is normal in G. Let H be a subgroup of G of order 2. One can see from (|H|, |P|)=1 that |HP|=10, a contradiction.
Suppose that |G|=2 2 ×3. It is clear that G is isomorphic to one of the groups: Z 12 , Z 2 ×Z 6 , D 12 , Q 12 , A 4 . If G is not isomorphic to A 4 , then there exists at least a subgroup of G of order 6, which is impossible. Considering A 4 , a fact is that A 4 has eight Sylow 3-subgroups, and they are conjugate each other in G. It is easy to see that K 8 is a subgraph of Γ A4 , a contradiction. Case 3. π(G)|=1. It is straightforward that in this case, G is isomorphic to one of Z 4 , Z 2 × Z 2 ,, Z 9 , Z 3 × Z 3 , Z 25 , Z 5 × Z 5 , Z_8, D 8 , Q 8 , Z 2 × Z 2 × Z 2 , Z 2 × Z 4 , Z p, where p is a prime. By checking, it is easy to see that the generalized power graph of Z 2 × Z 2 × Z 2 is isomorphic to K 8, which is non-planar, others are planar.
The proof of the converse is clear. The proof is now complete. Denote by d(G) the minimum possible size of all generating sets for G. Clearly, if d(G) ≥ 3 for any group G, then Γ G is complete.

Corollary 3.4 Let G be a Dedekind group. Then Γ G is complete if and only if d(G) ≥ 3.
By the fundamental theorem for finitely generated abelian groups, if G is isomorphic to <r 1 > × <r 2 > × … × <r s >, where |r i+1 | is a divisor of |r i | for all i=1, 2,..., s-1, then r s is uniquely determined by G, and d(G) ≥ r s . Now, in view of Corollary 3.4, the following result is obtained. Clearly, if a group G is isomorphic to a group H then Γ G is isomorphic to Γ H . However, the converse is not true in general. For example, it is easy to see that Q 8 and D 8 are not isomorphic, but their generalized power graphs are isomorphic. Moreover, Theorem 3.1 also presents an example, the generalized power graph of Z 4 is isomorphic to the generalized power graph of Z 2 × Z 2, however, Z 4 and Z 2 × Z 2 are clearly no-isomorphic. Now, using generalized power graphs, characterize all cyclic groups except for the group with order 4.

Theorem 3.6 Let G be a group and n ≠ 4. Then Γ G is isomorphic to the generalized power graph of Z n if and only if G is isomorphic to Z n .
Proof. The sufficiency follows trivially. Now suppose that Γ G is isomorphic to the generalized power graph of Z n . Then there exists at least a vertex x of Γ G such that deg(x)=0. It is clear that |G|=n. By the proof of Theorem 3.1, it follows that G is isomorphic to Z 2 × Z 2 or a cyclic group. Since n ≠ 4, G is cyclic, as desired.
Finally, an applications of Theorem 3.6 is given. Proof. The proof of the sufficiency is straightforward. In order to prove the necessity, assuming that x is an end-vertex of Γ G and y is the one vertex of Γ G such that x ~ y belongs to E(Γ G ). So there are two conjugate subgroups H and K such that xH and yK.
It is clear that <x> is a normal subgroup of G. Now it follows that |G|=9 or 3q, where q is a prime number such that q ≠ 3. If |G|=9, then G = Z 9 or Z 3 × Z 3 . Now let |G|=3q. If q=2 then G = S 3 or Z 6 . Otherwise, it follows from Sylow theorems that G is isomorphic to Z 3 × Z q . It follows that G is one of the groups: S 3 , Z 3 × Z 3 , Z 3p for some prime number p.
If G = Z 3p , then Theorem 3.6 completes the proof. If G = S 3 , then R is Z 6 or S 3 . Since the generalized power graph of S 3 has no isolate vertices, R = S 3 , as desired. Similarly, if G is isomorphic to Z 3 × Z 3, then R is also isomorphic to Z 3 × Z 3, as desired.

Conclusion
In this paper, the generalized power graph of a finite group is introduced. Based on the results and discussions presented above, the conclusions are obtained as below: (1) Let G be a p-group. Then, Γ G is connected if and only if G is non-cyclic and G is not isomorphic to Z p × Z p .
(2) Let p and q be two distinct prime integers, and let G be a group with π(G)={p, q}. Then, Γ G is connected if and only if G is non-cyclic and has a proper subgroup S such that π(S)=π(G).
(4) All finite groups whose generalized power graphs are empty, bipartite, planar and complete were classified.