The Stable Subgroup Graph

In this paper we introduce stable subgroup graph associated to the group G. It is a graph with vertex set all subgroups of G and two distinct subgroups H1 and H2 are adjacent if StG(H1)∩H2 6= 1 or StG(H2)∩H1 6= 1, where StG(Hi) = {g ∈ G : H i = Hi}, i = 1, 2. The planarity of the stable subgroup graph of solvable groups has been discussed. Finally, the induced subgraph of stable subgroup graph with vertex set whole non-normal subgroups is considered and its planarity is verified for some certain groups.


Introduction
Algebraic graph theory is part of algebraic methods in applied problems about graphs.One of the main branches of algebraic graph theory, involving the use of group theory, and the study of graph invariants.A graph represents the information about the relations among nodes which is a very efficient way of describing a structure.
Recently, mathematicians constructed very interesting graphs which are assigned to an algebraic structure by different methods.In this paper we consider a graph with vertex set of all subgroups of a group so let us name some graphs which has a connection with the graph that is defined here, for instance subgroup graphs, subgroup lattice and intersection graphs.The subgroup graph of a group is the graph whose vertices are the subgroups of the group and two vertices, H 1 and H 2 , are connected by an edge if and only if H 1 ≤ H 2 and there is no subgroup K such that H 1 < K < H 2 (see [1]).Starr and Turner [11] were the first to study groups G with planar subgroup graph and classified all planar abelian groups.Also, Schmidt [9,10], Bohanon and Reid [1] simultaneously classified all finite planar groups.
The intersection graph of a group G is an undirected graph without loops and multiple edges defined as follows: the vertex set is the set of all proper non-trivial subgroups of G, and there is an edge between two distinct vertices H and K if and only if H ∩ K = {1} where 1 denotes the trivial element of the group G. Already, planarity of subgroup graphs, subgroup lattice, intersection graphs are studied [1,6,9,10,11].
In this paper, we introduced stable subgroup graph Γ G with vertex set all subgroups of the finite group G and two vertices H 1 and H 2 join by an edge if H x 2 = H 2 or H y 1 = H 1 , for some non-identity elements x ∈ H 1 , y ∈ H 2 .This graph has a narrow connection with subgroup graphs and intersection graphs.For instance subgroup graph and intersection graph are induced subgraph of the stable subgroup graph.The stable subgroup graph is Γ G is connected graph.If the vertex H intersects Z(G) non-trivially, then has a complete degree in the graph.We observe that subgroups in the same conjugacy classes have similar properties in the stable subgroup graph for instance their degrees are equal.It is not hard to deduce that stable subgroup graph is regular if and only it is a complete graph.Moreover, the stable subgroup graph of Dedekind groups is a complete graph.We verify the planarity of stable subgroup graph for abelian groups, p-groups, nilpotent groups and supersoluble groups directly with out using the planarity of subgroup graphs or intersection graphs.Moreover, we clarify all soluble groups whose stable subgroup graphs are planar by the fact that subgroup graph is its induced subgraph.It is clear that normal subgroups have complete degree by definition of the stable subgroup graph so we consider the induced subgraph of stable subgroup graph by omitting its normal subgroups.Let us denote it by Γ G\N s .It is a connected graph whenever Z(G) = 1.If G is a p-group of order p 3 , then Γ G\N s is planar.The stable non-normal subgroup graph of dihedral groups D 8 , D 12 , D 16 , D 18 and D 2p are planar, where p is an odd prime number.
Throughout the paper, graphs are simple and all the notations and terminologies about the graphs are found in [2,4].

Stable subgroup graph
Let us start with the definition of stable subgroup graph associated to the group G.
Definition 2.1.The stable subgroup graph Γ G is a graph with vertex set all the subgroups of the group G and two subgroups H 1 and H 2 are adjacent whenever The adjacency condition of H 1 and H 2 can be presented by the following equivalent statement.One can use each of them depends on the situation in the argument.
The following three items are equivalent and (i) is deduced from them but not the converse.
(iv) There exists at least a non-identity element in one of the subgroups such that it commutes with all the elements of the other one.If the intersection of two subgroups is non-trivial, then these two subgroups are adjacent.Thus intersection graph is induced subgraph of stable subgroup graph.
Let H, K be two subgroups of the G such that they are adjacent in subgroup graph of G. Then H and K are adjacent in stable subgroup graph of G.This fact implies that the subgroup graph is an induced subgraph of Γ G .
We denote the conjugacy class subgroup of H in G by cl G (H) which is the set of all subgroups of G conjugate to H.In the following lemma we observe that subgroups in the same conjugacy class share some similar properties.
Proof: (i) Assume x ∈ H t ∩ Z(G).As H t ∈ cl G (H) we have x ∈ H g and x = h g , for some g ∈ G.By hypothesis we conclude that Then by adjacency condition we have H x = H or K y = K, for some x ∈ K, y ∈ H.If the first case happened, then x g ∈ K g exists such that H x g t = H t .Suppose K y = K, for some y ∈ H. Therefore, y g ∈ H t and (K g ) y g = K g .These argument imply that K g and H t are adjacent.Hence we can define a bijection between the set of the neighbors of H and H t .✷

B. Tolue
Lemma 2.2.If A and B are two adjacent subgroups which belong to two different conjugacy classes, then for every If H and K are two adjacent vertices, then K join to all conjugates of H by K.The converse holds if K join to at least a conjugate of H by K.
Proof: Suppose H and K are two adjacent vertices, so by definition of adjacency in the graph we have Theorem 2.1.If G is abelian or the direct product of a quaternion group of order 8, an elementary abelian 2-group and an abelian group with all its elements of odd order, then Γ G is a complete graph.
Proof: Suppose G is a group which is mentioned in the theorem, thus by Theorem 5.3.7 in [8] it is a Dedekind group which means all its subgroups are normal.✷ If G is a group satisfies Theorem 2.1, then ω(Γ G ) and χ(Γ G ) are the number of subgroups of G.
An element a ∈ G is said to be a persistence element if a = 1 and a is contained in all nontrivial subgroups of G.A group G is persistent if it has a persistence element.
If G is a finite persistent group, then it is either cyclic or isomorphic to a generalized quaternion group.An example of infinite persistent group is Z p ∞ .
Let us denote the number of conjugacy classes of non-normal subgroups of G by ν(G).Clearly, ν(G) = 0 if and only if G a is Dedekind group.R. Brandl in [3] and H. Mousavi in [7] classified finite groups which have respectively just one or exactly two conjugacy classes of non-normal subgroups.By their main theorems groups whose associated graphs have at most three kind of degrees are specified.
The group with exactly one and two non-trivial subgroup are Z p 2 and Z pq , Z p 3 respectively.Therefore one could deduce the following result.Proposition 2.5.Let G be a cyclic group.Then or Z pq , where p, q are prime numbers.
Proof: If we consider other abelian groups, then they have more than three nontrivial subgroups so the graph associated to them is not planar.✷ Example 2.1.In this example we verify the planarity of some non-abelian groups.
(ii) Γ Sn and Γ An are not planar graphs, where n ≥ 4.
) and (iv) The graph associated to dicyclic group Q 4n = a, x : a 2n = 1, a n = x 2 , a x = a −1 is not planar, where n ≥ 2. By the subgroups a , ax , a 2 x , the center of the group a n , trivial subgroup and the group itself we can form K 3,3 .
Theorem 2.2.Let G be a p-group, for a prime number p.
Proof: Suppose G is a p-group of order p n , and S the set of all subgroups of order p m .By Sylow theorem, we see Card(S) congruent to 1 modulo p.There is at least one normal subgroup for every power of p up to the order of the group.Therefore a lower bound for the total number of normal subgroups of G is n + 1.
Since normal subgroups join to all other subgroups in this graph, all these normal subgroups form a clique.Assume Γ G is planar.Thus n ≤ 3 and the planarity of the groups {1}, , where p, q are prime numbers.
therefore G is a 2-group and Γ G is not planar by Theorem 2.2.(ii) G ′ ∼ = Z p , then G is a group of order 4p.If p = 2, then Γ G is not planar by Theorem 2.2.Consider p is a prime number greater than 2. If we denote the number of its Sylow 2-subgroups by N 2 , then N 2 = 1 or p.For N 2 = 1, as intersection of distinct Sylow p-subgroups is trivial, we have N p = 4, where N p is the number of Sylow p-subgroups.Since the Sylow 2-subgroup is normal it is adjacent to 3 Sylow p-subgroups and we have K 3,3 by the vertices G and {1}.Moreover, if N 2 = p, then in the worst case 4p = |G| = p(4 − 1) + N p p which implies that N p = 1.By normality of the Sylow p-subgroup, Γ G contains K 3,3 .Moreover, if Sylow 2subgroups have non-trivial intersection, then we can form K 5 by use of intersection of two Sylow 2-subgroups which has the maximal order.(iii) Analogously, for G ′ ∼ = Z p 2 , Z p 3 we obtain a non-planar Γ G .(iv) G ′ ∼ = Z pq , then |G| = 4pq.If p or q are equal to 2, then Γ G is not planar.Suppose p, q are odd prime numbers.If G is a non-simple group, then it has at least one non-trivial normal subgroup N .Furthermore, there are x 1 , x 2 , x 3 ∈ G of order 2, p and q, respectively.If the cyclic group generated by them is not equal to N , then K 3,3 can be form by the vertices {{1}, G, N, x 1 , x 2 , x 3 }.Otherwise by the Sylow t-subgroup, Γ G contains K 3,3 , where t = 2, p or q depends on the situation.If G is a simple group of order 4pq, then |G| = 60 and every simple group of order 60 is isomorphic to A 5 .By Example 2.1, Γ A5 is not planar.Now, if G/G ′ ∼ = Z p , Z p 2 , Z p 3 , Z pq , then planarity of Γ G can be discussed similarly.One of the cases which its argument is to some extent different from the previous cases is when G/G ′ ∼ = Z p and G ′ ∼ = Z q , where p and q are different prime numbers.If G is extension of a cyclic group of order p by a cyclic group of order q, then G = a, b|a p = 1, b q = a t , bab −1 = a r , where b i ∈ a 0 < i < q, r q ≡ 1 and rt ≡ t (mod p), such a group exists for every choice of integers r, t with these property (see [5,Theorem 12.9]).By argument about the number of Sylow p-subgroups and Sylow q-subgroups we deduce that p or q is less or equal than 2. Thus G is a group of order 2p, which is cyclic or dihedral group of order 2p.Hence, the only non-abelian group with this structure which its associated graph is planar is S 3 .Hence the result is clear.✷ Recall that a group is planar if its subgroup graph is planar [1].
Proposition 2.8.[1] There are no soluble planar groups whose orders have more than three distinct prime factors.
Proposition 2.9.Let G be a finite soluble group.Then Γ G is planar if and only if , where p, q are prime numbers.
Proof: Since the subgroup graph is induced subgraph of Γ G , Proposition 2.8 implies that the order of G is divisible by at most three distinct prime numbers.If G is a nilpotent group, then we discussed about the planarity of Γ G in Proposition 2.7.Therefore, suppose G is non-nilpotent of order p α q β r γ , where p, q and r are prime numbers and α, β, γ ∈ N ∪ {0}.Since the product of Sylow subgroups is a Hall subgroup of G, we can find K 5 by vertices {{1}, G, S p , S q , S p S q } as induced subgraph of Γ G , where S i is Sylow i-subgroup of G, i = p, q, r.Thus |G| = p α q β .As Γ Si is induced subgraph of Γ G , we conclude that 0 ≤ α, β ≤ 3 by Theorem 2.2.
In the case α or β are greater than 2, Γ G is not planar.Since K 5 is the induced subgraph of Γ G by vertices {{1}, G, S p S q , S i , H}, where S i is Sylow i-subgroup of G of order i γ and H is the cyclic subgroup of G generated by x ∈ S i , |x| = i, γ ≥ 2 and i = p or q.Hence α, β = 1 or 0 and the assertion follows.✷ Since every supersoluble group is soluble group, Theorem 2.3 can be deduced from the Proposition 2.9.

Stable non-normal subgroup graph
In general let us denote the induced subgraph of stable subgroup graph of the group G which is obtained by omitting the set X from the the set of vertices by Γ V (G)\X .Let us denote the subgraph of stable subgroup subgraph with vertex set whole non-normal subgroups of G, by Γ G\N s .In this section, we focus on some properties of Γ G\N s .
Proof: Suppose H 1 , H 2 are two non-normal subgroups of G which are non-adjacent in Γ G\N s .We claim that H 1 Z(G) is a proper subgroup of G, since otherwise St G (H 1 ) = G and the adjacency of H 1 and H 2 are deduced, which is a contradiction.Therefore H 1 Z(G) is a vertex which is adjacent to both H 1 and H 2 .

B. Tolue
Thus diam(Γ G\N s ) = 2. Now assume H 1 and H 2 are two adjacent vertices.If H i Z(G) = G, then both H 1 and H 2 are adjacent to it, for i = 1 or 2. Let H i Z(G) = G, for i = 1 and 2. Thus H ∩ St G (H i ) = H for all H < G, i = 1, 2. This implies that H 1 , H 2 join H and the assertion is clear.✷ The direct result of the above proposition is that Example 3.1.In this example we observe that if Z(G) = 1, then Γ G\N s may be non-connected.
(i) Γ S3\N s is an empty graph with 3 vertices.
(ii) Γ S4\N s is a connected graph which is not planar.The vertices This shows every non-normal subgroup is adjacent to its stabilizer.This fact implies that there is no enough distinct vertices to form K 3,3 or K 5 .✷ In the following results the structure of stable subgroup graph of dihedral group of order 2n is verified.If D 2p is a dihedral group of order 2p, then Γ D2p\N s is a graph with p isolated vertices.The proof of the following proposition is very similar to the previous one so we omit it.By the above two proposition we deduce that the stable non-normal subgroup graph of D 8 , D 12 , D 16 , D 18 and D 2p are planar, where p is an odd prime number.As we mentioned Γ A5\N s is a connected graph which is not planar.Thus Γ An\N s is not planar for n ≥ 5.

133Proof:
The result follows by the lattice of subgroups of the group G. ✷ Proposition 2.6.Let G be an abelian group.Then Γ G is planar if and only if 2)(3 4) of A 4 together with A 4 and trivial subgroup we have K 3,3 .In Example 3.1 we will observe that after omitting normal subgroups from the vertices of stable subgroup graph of A 4 the new graph is planar.Moreover, the induced subgraph of stable graph of the simple group A 5 , which is obtained after omitting A 5 and identity subgroup from the vertices of Γ A5 is still non-planar.(iii)Let D 2n = a, b : a n = b 2 = 1, a b = a −1be the dihedral group of size 2n, n ≥ 4. Γ D2n is not a planar graph.If n is an even number, then the subgroups a , a n/2 , a n/2 , b and a n/2 , ab together with D 2n and trivial subgroup form K 3,3 .By similar argument we can find K 3,3 in Γ D2n , whenever n is an odd number.

Proposition 3 . 3 .
Let D 2n be dihedral group of order 2n, n ≥ 4 an even number and all the notions are the same as the Proposition 3.2.Then (i) |V (Γ D2n\N s )| = σ(n) − 3. (ii) The subgroups a i b and a j b are adjacent whenever j = (n/2) + i, 0 ≤ i, j ≤ n − 1 and i = j.Moreover, a d , a i b join to a i b and a d+i b , where d is a divisor of n. (iii) The subgroups a d , a i b join to a d , a j b , where 0 ≤ i, j, ≤ d− 1.Furthermore, a d , a i b join to a d ′ , a i b but a d , a i b does not join a d ′ , a j b , where d, d ′ are distinct divisor of n. (iv) By third part a d , a r b form a component which may have a connection with the component a d ′ , a r ′ b by the possible vertices such that r = r ′ , 0 ≤ r, r ′ ≤ d − 1. (v) Γ D2n\N s is planar if and only if n = 2 2 , 6 or 2 3 .
It is clear that G and its trivial subgroup join to all other subgroups.Therefore, if two non-adjacent subgroups exist, then they join via G or trivial subgroup.Thus diam(Γ G ) ≤ 2 and girth(Γ G ) = 3.Moreover, G and its trivial subgroup have the maximum degrees which implies that Γ G is a regular graph if and only if Γ G is a complete graph.Proposition 2.1.If H is a subgroup of G such that intersects the center of G non-trivially, then H is adjacent to all other subgroups.Consider the subgroup H of G such that H ≤ Z(G) or 1 = Z(G) ≤ H, then H joins to all other subgroups.Assume G is an abelian group.Then Γ G is a complete graph by Proposition 2.1.
which implies K and H k ′ are adjacent.By similar computation the second assertion follows.
✷Proposition 2.3.All normal subgroups of G are of complete degree.
D 8 and Q 8 should be checked.By Proposition 2.6 and Example 2.1, it is enough to verify the planarity of Γ G , where and Z pq .Proof: By the above argument and the fact that every finite nilpotent group is the direct product of p-groups, the assertion is clear by Proposition 2.6 and Theorem 2.2.Theorem 2.3.Let G be a finite supersoluble group.Then Γ G is planar if and only ✷For finite nilpotent group G, ω(Γ G ) ≥ n + 2, where n is the number of its Sylow p-subgroups of G.
Theorem 3.1.If G is a p-group of order p 3 , then Γ G\N s is planar.By considering Theorems 2.2 and 2.1, Γ G\N s is planar for Dedekind groups G ∼ = D 8 , Q 8 and abelian groups of order p 3 .It is enough to discuss about the planarity of Γ G\N s , where G ∼ = (Z p × Z p ) ⋊ Z p , Z p 2 ⋊ Z p which are extra special pgroup.Every subgroup of index p is normal so non-normal subgroups of these two groups are of order p.The intersection of distinct non-normal subgroups are trivial, also intersection of them with the center of the group is trivial too.Thus two nonnormal subgroups H i , H j are adjacent if St G