On some graphs associated with the finite alternating groups

arXiv:1412.7324v1 [math.GR] 23 Dec 2014

2-CONNECTIVITY OF THE POWER GRAPH OF FINITE ALTERNATING GROUPS D. BUBBOLONI, MOHAMMAD A. IRANMANESH AND S. M. SHAKER

Abstract. For a finite group G, let P0 (G), O0 (G) and P0 (T (G)) be the proper power graph, the proper order graph and the proper power type graph of G respectively. In this paper we find the number of connected components of those graphs when G = An . As a consequence we deduce that the power graph P (An ) is 2-connected if and only if n = 3 or n, n−1, n−2, n2 and n−1 are not primes. This 2 gives an infinite family of counterexamples to a conjecture posed by Pourgholi et al. in [3].

1. Introduction Let G be a finite group. The power graph P (G) is the graph with vertex set G and edge set E, where {x, y} ∈ E for two distinct vertices x, y ∈ G if one of them is a power of the other ([1]). In [2] we focused on the 2-connectivity of P (G). After having observed that P (G) is 2-connected if and only if P0 (G), the 1G -cut subgraph of P (G), is connected, we exposed a method to find the number c0 (G) of connected components of P0 (G) for a wide class of groups. We exhibited, in particular, a concrete procedure to deal with the fusion controlled permutation groups (see Definition 1.1 in [2]) and applied that to the symmetric group Sn , for n ≥ 2. We computed c0 (Sn ) and showed that P (Sn ) is 2-connected if and only if n = 2 or both n, n − 1 are not prime (Theorem A and Corollary C in [2]). Here, we look now for similar results about An . Since P0 (A2 ) is the empty graph, we assume throughout the paper, that n ≥ 3. The motivation for this research relies on a conjecture by MSC(2010): Primary: 05C25; Secondary: 20B30. Keywords: Power graph, alternating group. 1

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Bubboloni, Iranmanesh and Shaker

Pourgholi, Yousefi-Azari and Ashrafi (Conjecture 13 in [3]), stating that there is no non-abelian simple group with a 2-connected power graph. Let P be the set of prime numbers. For b, c ∈ N, we set bP + c = {x ∈ N : x = bp + c, for some p ∈ P } and define (1.1)

A = P ∪ (P + 1) ∪ (P + 2) ∪ (2P ) ∪ (2P + 1).

Our main result gives an infinite family of counterexamples to that conjecture. Theorem A. P (An ) is 2-connected if and only if n = 3 or n ∈ / A. In particular, there exist infinite n ∈ N such that P (An ) is 2-connected. Note that n = 16 is the minimum n for which P (An ) is 2-connected and An is non-abelian simple. Theorem A is a rephrase of Corollaries 5.10 and 5.11. Its proof is carried on similarly to the symmetric case in [2] but it is more deep and requires some more technicalities. In [2], approaching the problem of 2-connectivity for P (G), we were naturally driven to consider some further graphs. First of all we defined the quotient power graph Pe(G) as the quotient graph of P (G) with respect to the equivalence relation identifying x, y ∈ G if hxi = hyi. The vertex set of Pe(G) is [G] and is formed by the corresponding equivalence classes [x] ∈ [G], for x ∈ G. The edge set of Pe(G) is denoted by [E]. The corresponding proper quotient power graph Pe0 (G) is defined as the [1G ]-cut subgraph of Pe(G). Its vertex set is [G]0 = [G] \ {[1]} and its vertex set is denoted by [E]0 . By [2, Lemma 4.1], the number e c0 (G) of the connected components of Pe0 (G) is equal to c0 (G). We also considered the order graph O(G), that is, the graph O(G) with vertex set O(G) = {a ∈ N : ∃g ∈ G with o(g) = a} and edge set EO(G) , where for a, b ∈ O(G), {a, b} ∈ EO(G) if a 6= b and a divides b or b divides a. The proper order graph O0 (G) was defined as the 1cut subgraph of O(G). We proved that O0 (G) is a quotient of Pe0 (G) ( [2, Corollary 4.4]) which, usually, simplify in a too drastic way the complexity of Pe0 (G). Finally, for permutation groups, we defined the power type graph P (T (G)) and the corresponding proper graph P0 (T (G)) (see Section 5 in [2] and Section 2). It turns out that P0 (T (G)) is a quotient of

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Pe0 (G) ([2, Corollary 5.5]) and, in the fusion controlled case, the reduction of complexity given is very appropriate to well treat the number of connected components of Pe0 (G) ([2, Proposition 5.12]). In general, all the above mentioned graphs are strictly related and any information reached about one of them reflects onto the others. So it is not surprising that, in one go, we compute c0 (An ) as well as the number c0 (O(An )) of connected components of O0 (An ) and the number c0 (T (An )) of connected components of P0 (T (An )) (Theorems 3.2, 4.4, 5.2, 5.7). We heavily base our arguments on the general results obtained in [2] for quotient graphs. For short, we usually do not recall them explicitly but give at every use the exact reference. 2. Preliminaries In this section, we briefly recall the main notation and the definitions given in [2], strictly necessary for the purpose of the paper. For more details and properties the reader may refer to Sections 2.1, 4, 5.4 in [2]. Let n ∈ N. We denote by T (n) the set of partitions of n. Let T = [x1 , . . . , xr ] ∈ T (n) for some r ≤ n. For each m ∈ N with m ≤ n, let t(m) ∈ N ∪ {0} be the multiplicity of m in T , that is, the number of times in which m appears in the unordered list x1 , . . . , xr of terms of T . Then we write T = [1t(1) , 2t(2) , . . . , nt(n) ] and say that this is the normal form of T. We usually omit to write down those m such that t(m) = 0 and write T = [mt11 , ..., mtkk ], with distinct mj ∈ N and tj ≥ 1 for j ∈ {1, . . . , k}, where k ∈ N counts the number of distinct terms in the partition T . We usually also omit to write down the multiplicities equal to 1. The partition [1n ] is called the trivial partition. We define the order of T by o(T ) = lcm{mi }ki=1 . Given T = [mt11 , ..., mtkk ] ∈ T (n), the power of T of exponent a ∈ N, is defined as the partition of n " t1 gcd(a,m1 ) tk gcd(a,mk ) # m m 1 k , ..., . (2.1) T a = gcd(a, m1 ) gcd(a, mk ) We say that T a is a proper power of T if [1n ] 6= T a 6= T. We consider the symmetric group Sn and the alternating group An as naturally acting on the set N = {1, . . . , n}. Let ψ ∈ Sn . Then ψ is a vertex of P (Sn ) and [ψ] is a vertex of Pe(Sn ) which is a vertex of Pe0 (Sn ) if and only if ψ 6= id. We define the order of [ψ] as the order of the permutation ψ. If ψ decomposes into the product of r ∈ N pairwise

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disjoint cycles of lengths x1 , . . . , xr , then [x1 , . . . , xr ] ∈ T (n) is called the type of ψ (or of [ψ]) and denoted by Tψ . Note that o([ψ]) = o(ψ) = o(Tψ ). We also set Mψ = {i ∈ N : ψ(i) 6= i}. For X ⊆ Sn (or [X] ⊆ [Sn ]), we call T (X) = {Tψ : ψ ∈ X} the set of types admissible for X. We set also T0 (X) = T (X) \ {[1n ]}. Note that T (Sn ) = T (n) ⊃ T (An ). The number of permutations in An of type T = [mt11 , ..., mtkk ] ∈ T (An ) is the same of those in Sn and given by (2.2)

µT (An ) =

mt11

n! · · · mtkk t1 ! · · · tk !

while the number of vertices in [An ] of type T is given by (2.3)

µT [An ] =

µT (An ) , φ(o(T ))

where φ denotes the Euler totient function ([2, Lemma 5.10]). If G ≤ Sn is a permutation group, the power type graph of G, is the graph P (T (G)) with vertex set T (G) and edge set ET (G) , where {T, T ′ } ∈ ET (G) if T, T ′ ∈ T (G) are, in a suitable order, one the power of the other and T 6= T ′ . We denote by P0 (T (G)), the [1n ]-cut subgraph of P (T (G)), called the proper power type graph of G. The vertex set of P0 (T (G)) is T0 (G) and, given T, T ′ ∈ T0 (G) there is an edge between them if and only if they are, in a suitable order, one the proper power of the other. We denote the connected component of P0 (T (G)) having T ∈ T0 (G) as a vertex with C(T ). Recall that O0 (G) may be seen as a quotient of P0 (T (G)) through the complete homomorphism oe : Pe0 (G) → O0 (G) defined by oe([ψ]) = o(ψ) ([2, Lemma 4.3 and Proposition 5.7]). We deal with the set of graphs G = {Pe (An ), P (T (An )), O(An )} and their corresponding proper graphs G0 = {Pe0 (An ), P0 (T (An )), O0 (An )}.

If Γ ∈ G, then we denote the corresponding proper graph in G0 with Γ0 . Note that if Γ ∈ G, then Γ is connected because contains a vertex adjacent to any other and is 2-connected if and only if Γ0 is connected. We denote by Ce0 (An ), C0 (T (An )), C0 (O(An )) the sets of connected components of Pe0 (An ), P0 (T (An )), O0 (An ) respectively, and put c0 (An ) = |Ce0 (An )|, c0 (T (An )) = |C0 (T (An ))|, c0 (O(An )) = |C0 (O(An ))|. e

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By [2, Proposition 3.2], we have the following. Remark 1. Let n ≥ 3. Then c0 (O(An )) ≤ c0 (T (An )) ≤ e c0 (An ).

If C is a connected component of some graph in G0 , we denote its vertex set by VC . Note that a connected component is always an induced graph and thus it can be identified with its vertices. In particular, if x ∈ VC we sometimes briefly say that x is contained in C. Let C ∈ Ce0 (An ). We define T (C) by T (VC ). We also set oe(C) = oe(VC ), so that oe(C) is the set of orders of vertices in C. The set of connected components in Ce0 (An ) containing a permutation of type T ∈ T0 (An ) is denoted by Ce0 (An )T and its size by e c0 (An )T . Lemma 2.1. ([2, Remark 9]) Let T ∈ T0 (An ) and C ∈ Ce0 (An )T . Then: i) T (C) = VC(T ) . In particular the set of types admissible for a connected component of Pe0 (An ) is always the set of vertices of a connected component of P0 (T (An )). ii) The set oe(C) ⊆ O0 (An ) is connected and included in the connected component of O0 (An ) which contains o(T ).

If C ∈ Ce0 (An )T , the symbol kC (T ) denotes the multiplicity of T in C, that is, the number of vertices [ψ] ∈ VC such that Tψ = T (see [2, Definition 3.12]). From [2, Lemma 5.11], we know that (2.4)

c0 (An )T = e

µT [An ] . kC (T )

We will use several times the following results about isolated vertices, which follows immediately from [2, Corollary 5.5 iii) and Remark 8] . Lemma 2.2. i) The type T ∈ T0 (An ) is isolated in P0 (T (An )) if and only if each [ψ] ∈ [An ]0 of type T is isolated in Pe0 (An ). ii) If m ∈ O0 (An ) is isolated in O0 (An ), then each vertex of order m is isolated in Pe0 (An ). If, for some ψ ∈ An , [ψ] is isolated in Pe0 (An ), then o(ψ) is prime. By [2, Proposition 5.12] applied to the alternating group, we have the following reediting of the Procedure [2, Section 3.4.1] for counting the number of connected components of Pe0 (An ).

2.3. Procedure to compute e c0 (An )

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I) Selection of types Ti and connected components Ci Start : Pick T1 ∈ T0 (An ) and choose C1 ∈ Ce0 (An )T1 . Basic step : if T1 , . . . , Ti ∈ T0 (An ) are chosen together with C1 , . . . , Ci ∈ Ce0 (An ) such that Cj ∈ Ce0 (An )Tj , for j ∈ {1, . . . , i}, then choose S Ti+1 ∈ T0 (An ) \ ij=1 T (Cj ) and Ci+1 ∈ Ce0 (An )Ti+1 . Stop : the procedure stops in c0 (T (An )) steps. II) The value of e c0 (An ) Compute the integers (2.5)

c0 (An )Tj = e

µTj [An ] , kCj (Tj )

for j ∈ {1, . . . , c0 (T (An ))}, and sum them up to get e c0 (An ).

We emphasize that the complete freedom in the choice of the Cj ∈ Ce0 (An )Tj in the above procedure allows to compute each e c0 (An )Tj = µTj [An ] kCj (Tj ) ,

selecting Cj as the connected component containing [ψ], for [ψ]

chosen as preferred among those ψ ∈ An with Tψ = Tj . In the following, we will apply this fact with no further reference. Note also that the number c0 (T (An )) counts the steps in the procedure. 3. Some small degrees In this section, we point out some general properties and use them to determine e c0 (An ) = c0 (An ), c0 (T (An )) and c0 (O(An )) for 3 ≤ n ≤ 7. Since [A3 ]0 = {[(1 2 3)]}, we have e c0 (A3 ) = c0 (T (A3 )) = c0 (O(An )) = 1.

Lemma 3.1. Let p be a prime number and n ∈ {p, p + 1, p + 2}, with n ≥ 4. Then: i) p ∈ O0 (An ) and is isolated in O0 (An ). If [ψ] ∈ [An ] has order p, then [ψ] is isolated in Pe0 (An ) and Tψ is isolated in P0 (T (An )); ii) if n = p, then the number of connected components of Pe0 (Ap ) containing the elements of order p is (p − 2)!; iii) if n = p+1, then the number of connected components of Pe0 (Ap+1 ) containing the elements of order p is (p + 1)(p − 2)!; iv) if n = p+2, then the number of connected components of Pe0 (Ap+2 ) if p is odd containing the elements of order p is (p+2)(p+1)(p−2)! 2 and 3 if p = 2.

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Proof. i) Let n ∈ {p, p + 1, p + 2}, with n ≥ 4. Then p | n! 2 so that An admits elements of order p, but there exists no element with order kp for some k ≥ 2. This says that p is isolated in O0 (An ). By Lemma 2.2, it follows also that [ψ] is isolated in Pe0 (An ) and Tψ is isolated in P0 (T (An )), for all [ψ] ∈ [An ] with o(ψ) = p. ii)-iv) We consider the elements of order p in An . If n = p, they are those of type [p]; if n = p + 1, those of type [1, p]; if n = p + 2, those of type [12 , p] if p is odd and those of type [22 ] if p = 2. So the counting follows from (2.4) and (2.3). Theorem 3.2. The values of c0 (An ), c0 (T (An )) and c0 (O(An )) for 3 ≤ n ≤ 7, are given by the following table. Table 1. c0 (An ) and c0 (T (An )), for 3 ≤ n ≤ 7. n c0 (An ) c0 (T (An )) c0 (O(An ))

3 1 1 1

4 5 6 7 7 31 121 421 2 3 4 4 2 3 3 3

Proof. Let G = An , for 3 ≤ n ≤ 7, acting on N = {1, . . . , n}. We compute e c0 (An ) separately for each degree. Since Pe0 (A4 ) has only three vertices of type [22 ] and four of type [1, 3] and all of them are isolated, we have e c0 (A4 ) = 7, c0 (T (A4 )) = c0 (O(A4 )) = 2. In Pe0 (A5 ), by Lemma 3.1, all the vertices of type [12 , 3] and [5] are isolated. It is also immediately checked that also the vertices of type [1, 22 ] are isolated. In other words the graph Pe0 (A5 ) is totally disconnected. By (2.4) and (2.2), we then get c0 (A5 ) = µ[1,22 ] [A5 ] + µ[12 ,3] [A5 ] + µ[5] [A5 ] = 31, e

while c0 (T (A5 )) = 3. Since O0 (A5 ) = {2, 3, 5} we also have c0 (O(A5 )) = 3. In Pe0 (A6 ), by Lemma 3.1, all the vertices of type T1 = [1, 5] are contained in e c0 (A6 )T1 = 36 components which are isolated vertices. Also, having in mind [2, Corollary 5.6], it is clear that all the vertices of types T2 = [13 , 3] and T3 = [32 ] are isolated. Thus by (2.4) and (2.2), we get e c0 (A6 )T2 = e c0 (A6 )T3 = 20. Let T4 = [12 , 22 ] and consider the connected component C4 containing ψ = (1 2)(3 4). Since T4 admits

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no proper power, the edges incident to [ψ] are given only by {[ψ], [ϕ]} for those ϕ ∈ A6 such that ϕ2 = ψ. In particular ϕ = (a b)(c d e f ), where {a, b, c, d, e, f } = N and (c e)(d f ) = (1 2)(3 4). It follows that (a b) = (5 6), while either (c d e f ) = (1 3 2 4) or (c d e f ) = (3 1 4 2). On the other hand, by [2, Lemma 5.2], the type Tϕ = [2, 4] is not a proper power, because its terms are distinct. Thus C4 is the path: [(1 3 2 4)(5 6)], [(1 2)(3 4)], [(3 1 4 2)(5 6)] c0 (A6 )T4 = µ[12 ,22 ] [A6 ] = 45. Since all the and kC4 (T4 ) = 1. So, by (2.4), e types admissible for A6 have appeared, the application of the procedure 2.3 gives e c0 (A6 ) = 121, while c0 (T (A6 )) = 4. Moreover c0 (O(A6 )) = 3 with the connected components of O0 (A6 ) having as vertex sets {3}, {5} and {2, 4}. In Pe0 (A7 ), by Lemma 3.1, all the vertices of type T1 = [12 , 5] and T2 = [7] are isolated and e c0 (A7 )T1 = 126, e c0 (A7 )T2 = 120. Also, having in mind [2, Corollary 5.6], it is clear that all the vertices of types T3 = [1, 32 ] are isolated and, by (2.4) and (2.2), we get e c0 (A7 )T3 = 140. Let Ci for i ∈ {1, 2, 3} be a connected component admissible for Ti . Consider now the type T4 = [14 , 3] and the connected component C4 containing ψ = (1 2 3). Since T4 admits no proper power, the edges incident to [ψ] are given only by {[ψ], [ϕ]} for those ϕ ∈ A7 such that Tϕ = [22 , 3] and ϕ2 = ψ. Then ϕ = (a b)(c d)(e f g), where {a, b, c, d, e, f, g} = N and (e g f ) = (1 2 3), {a, b, c, d} = {4, 5, 6, 7}. It follows that there are three choices for [ϕ]. Moreover each of those [ϕ] is adjacent with exactly one vertex of type [13 , 22 ], which in turn is adjacent with three vertices of type [1, 2, 4] and these elements are not adjacent to any other c0 (A7 )T4 = vertex. In particular we have kC4 (T4 ) = 1 and thus, by (2.4), e S µ[14 ,3] [A7 ] = 35. Since 4i=1 T (Ci ) = T (A7 ), the procedure 2.3 closes giving e c0 (A7 ) = 421 and c0 (T (A7 )) = 4. Moreover c0 (O(A7 )) = 3 with the three connected components of O0 (A7 ) having as vertex sets {2, 3, 4, 6}, {5} and {7}. 4. The degrees 8 and 9 In this section, we determine e c0 (An ) = c0 (An ), c0 (T (An )) and c0 (O(An )) for n ∈ {8, 9}. We start recalling en elementary fact about edges in a quotient power graph. Lemma 4.1. ( [2, Remark 6]) For each ψ, ϕ ∈ An , such that [ψ] 6= [ϕ], {[ψ], [ϕ]} is an edge in Pe0 (An ) if and only if {ψ, ϕ} is an edge in P0 (An ).

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Lemma 4.2. Let n ≥ 8. Then: i) the vertices of type [1n−4 , 22 ] belong to the same connected component Ωn of Pe0 (An ); ii) T (Ωn ) ⊇ {[1n−4 , 22 ], [1n−3 , 3], [1n−7 , 22 , 3]}; iii) for every T ∈ T (Ωn ), Ωn contains all the vertices of [An ] of type T and T (Ωn ) = VC(T ) . Proof. i) Let [π1 ] and [π2 ] be distinct vertices in [An ] of type [1n−4 , 22 ]. Then π1 , π2 ∈ An are distinct and share at most one transposition. Let π1 = (a b)(c d) for suitable distinct a, b, c, d ∈ N. We analyze the two possibilities: 1) π1 and π2 share a transposition; 2) π1 and π2 share no transposition. 1) In this case we have π2 = (a b)(e f ), for suitable distinct e, f ∈ N and {c, d} = 6 {e, f }. If |{c, d} ∩ {e, f }| = 1, we may assume that e = c, that is, π2 = (a b)(c f ). Since n ≥ 8, there exist x, y, z ∈ N \ {a, b, c, e, f } and we have the following path between [π1 ] and [π2 ]: [π1 ], [(a b)(c d)(x y z)], [(x y z)], [(a b)(c f )(x y z)], [π2 ]. If |{c, d} ∩ {e, f }| = 0, let π = (a b)(c e). By the previous case, there is a path between [π1 ] and [π] as well as a path between [π2 ] and [π] and so also a path between [π1 ] and [π2 ]. 2) We distinguish the three possible subcases: a) |Mπ1 ∩ Mπ2 | = 2. Then π2 = (a e)(c f ), for suitable e, f ∈ N, such that {a, e, c, f } ∩ {a, b, c, d} = {a, c}. Let π = (a b)(c f ). Then, by 1), there exists a path between [π1 ] and [π] and a path between [π2 ] and [π]. So there exists a path also between [π1 ] and [π2 ]. b) |Mπ1 ∩ Mπ2 | = 1. Then π2 = (a e)(f g), for suitable e, f, g ∈ N such that {a, e, f, g} ∩ {a, b, c, d} = {a}. Now consider π = (a b)(f g) and argue as in a). c) |Mπ1 ∩ Mπ2 | = 0. Then π2 = (e f )(g h) for suitable e, f, g, h ∈ N such that {e, f, g, h} ∩ {a, b, c, d} = ∅. Now consider π = (a b)(e f ) and argue as in a). This shows that all the vertices of Pe0 (An ) of type [1n−4 , 22 ] lie in the same connected component Ωn . ii) Collecting the types met in the paths used for i), we get immediately T (Ωn ) ⊇ {[1n−4 , 22 ], [1n−3 , 3], [1n−7 , 22 , 3]}.

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iii) Let T ∈ T (Ωn ). The fact that T (Ωn ) = VC(T ) is an application of Lemma 2.1. The fact that Ωn contains all the vertices of [An ] of type T is instead a consequence of i) and of [2, Corollary 5.14]. Lemma 4.3. Let n ∈ {8, 9}. Then: i) the vertices of type [1n−8 , 24 ] belong to the same connected component Λn of Pe0 (An ); ii) T (Λn ) = {[1n−8 , 24 ], [1n−8 , 2, 6], [1n−6 , 32 ], [1n−8 , 42 ]}; iii) Λn 6= Ωn . Proof. Let first n = 8. Let [π1 ] and [π2 ] be distinct elements in [A8 ] of type [24 ], with π1 = (a b)(c d)(e f )(g h), where {a, b, c, d, e, f, g, h} = N. Since π1 , π2 are distinct, they share at most two transpositions. We analyze the three possibilities: 1) π1 and π2 share one transposition; 2) π1 and π2 share two transposition; 3) π1 and π2 share no transposition. 1) In this case, since the 2-cycles in which π1 splits commute and also the entries in each cycle commute, we can assume that π2 = (a b)(c e)(h f )(g d) and look to the path : [π1 ], [(a b)(c e g d f h)], [(c g f )(e d h)], [(c h g e f d)(a b)], [π2 ]. 2) Here, for the same considerations as in 1), we reduce to π2 = (a b)(c d)(e g)(f h). Consider π = (a b)(c e)(d h)(f g). Then by 1), there exists a path between [π1 ] and [π] and a path between [π2 ] and [π]. Thus there exists also a path between [π1 ] and [π2 ]. 3) Here we reduce to the two cases π2 = (a c)(b d)(e g)(f h) or π2 = (a c)(b g)(e d)(f h). In the first case let π = (a b)(c d)(e g)(f h). Then by 2), there exists a path between [π1 ] and [π] and a path between [π2 ] and [π]. In the second case let π = (a b)(c g)(e d)(f h). Then by 1), there exists a path between [π1 ] and [π] and by 2), there exists a path between [π2 ] and [π]. In both cases we then get a path between [π1 ] and [π2 ]. This shows that all vertices of type [24 ] in [A8 ] are in the same connected component Λ8 of Pe0 (A8 ). Collecting the types we met along the considered paths we get also T (Λ8 ) ⊇ {[24 ], [2, 6], [12 , 32 ]}. It is also clear that [42 ] ∈ T (Λ8 ) because [(1 2 3 4)(5 6 7 8)]2 = (1 3)(2 4)(5 7)(6 8). To show that T (Λ8 ) = {[24 ], [2, 6], [12 , 32 ], [42 ]} it is enough to observe

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that the set of types {[24 ], [2, 6], [12 , 32 ], [42 ]} is closed by proper powers within the types admissible for A8 . Finally, since the type [14 , 22 ] is admissible for Ω8 but not for Λ8 we deduce that Λ8 6= Ω8 . Next let n = 9 and let [π1 ] and [π2 ] be distinct elements in [A9 ] of type [1, 24 ]. If π1 , π2 have the same fixed point we are inside a copy of A8 and so there is a path between [π1 ] and [π2 ] in Pe0 (A8 ), which is also a path in Pe0 (A9 ). If π1 , π2 have a different fixed point, without loss of generality we may assume that π1 fixes 9 and π2 fixes 8. Now consider ϕ1 = (1 2)(3 4)(5 6)(7 8) which fixes 9 and ϕ2 = (1 2)(3 4)(5 6)(7 9) which fixes 8. We have the following path between [ϕ1 ] and [ϕ2 ] : [ϕ1 ], [(1 3 5 2 4 6)(7 8)], [(1 5 4)(3 2 6)], [(1 3 5 2 4 6)(7 9)], [ϕ2 ]. Since πi and ϕi have the same fixed point, there is a path between [πi ] and [ϕi ], for all i = 1, 2 and thus also a path between [π1 ] and [π2 ]. This shows that all vertices of type [1, 24 ] in [A9 ] are in the same connected component Λ9 of Pe0 (A9 ). Collecting the types met along the paths we get T (Λ9 ) ⊇ {[1, 24 ], [1, 2, 6], [13 , 32 ]}. It is also clear that [1, 42 ] ∈ T (Λ9 ) because [(1 2 3 4)(5 6 7 8)]2 = (1 3)(2 4)(5 7)(6 8). Arguing as for n = 8, we then get T (Λ9 ) = {[1, 24 ], [1, 2, 6], [13 , 32 ], [1, 42 ]}. Finally, since the type [15 , 22 ] is admissible for Ω9 but not for Λ9 we deduce that Λ9 6= Ω9 . Theorem 4.4. The values of c0 (An ),c0 (T (An )) and c0 (O(An )) for n ∈ {8, 9} are given by the following table. Table 2. c0 (An ) and c0 (T (An )) for n = 8, 9. n 8 9 c0 (An ) 962 5442 c0 (T (An )) 3 4 c0 (O(An )) 2 2 Proof. Let n = 8. By Lemma 3.1, the prime 7 is isolated in O0 (A8 ). It follows that also T1 = [1, 7] is isolated in Pe0 (A8 ) and thus e c0 (A8 )T1 = e 960. By Lemma 4.2, all the vertices of P0 (A8 ) of type T2 = [14 , 22 ] are in Ω8 , so that e c0 (A8 )T2 = 1, and T (Ω8 ) ⊇ {[14 , 22 ], [15 , 3], [1, 22 , 3]}.

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Moreover looking at the types admissible for A8 which can have as proper power one of the types in {[14 , 22 ], [15 , 3], [1, 22 , 3]}, we get immediately that T (Ω8 ) = {[14 , 22 ], [15 , 3], [1, 22 , 3], [3, 5], [13 , 5], [12 , 2, 4]}. Thus oe(Ω8 ) = {2, 3, 4, 5, 6, 15} is included in the connected component of O0 (A8 ) containing 2. By Lemma 4.3, all the vertices of type T3 = [24 ] are in Λ8 and T (Λ8 ) = {[24 ], [2, 6], [12 , 32 ], [42 ]},

oe(Λ8 ) = {2, 3, 4, 6}

so that e c0 (A8 )T3 = 1 and the procedure 2.3 closes giving c0 (A8 ) = 962. The discussion above shows also that c0 (T (A8 ) = 3 and c0 (O(A8 )) = 2. Let n = 9. By Lemma 3.1, the prime 7 is isolated in O0 (A9 ). Thus, for T1 = [12 , 7] we have e c0 (A9 )T1 = 4320. By Lemma 4.2, all the vertices 5 2 of type T2 = [1 , 2 ] are in Ω9 and T (Ω9 ) ⊇ {[15 , 22 ], [16 , 3], [12 , 22 , 3]}. In particular e c0 (A9 )T2 = 1. Moreover, looking at the types admissible for A9 which can have as proper power one of the types in {[15 , 22 ], [16 , 3], [12 , 22 , 3]}, we get that T (Ω9 ) = {[15 , 22 ], [16 , 3], [12 , 22 , 3], [1, 3, 5], [14 , 5], [13 , 2, 4]} and so oe(Ω9 ) = {2, 3, 4, 5, 6, 15}. By Lemma 4.3, all the vertices of type T3 = [1, 24 ] are in Λ9 and T (Λ9 ) = {[1, 24 ], [1, 2, 6], [13 , 32 ], [1, 42 ]},

oe(Λ9 ) = {2, 3, 4, 6}

so that e c0 (A9 )T3 = 1. Finally consider the type T4 = [33 ]. Let ϕ ∈ A9 with Tϕ = T4 and let C4 be the connected component of Pe0 (A9 ) containing [ϕ]. We show that kC4 (T4 ) = 1. Note, first of all, that T4 has no proper power and that it is the proper power only of [9]. Moreover, by [2, Lemma 5.2], the type [9] is not a proper power and its only proper power has exponent a ∈ N, with 1 < a < 9 and not coprime to 9, so that a = 3. Thus, by Lemma 2.1, we have VC(T4 ) = {T4 , [9]} = T (C4 ) and so also oe(C4 ) = {3, 9}. We show that [ϕ] is the only vertex of type T4 in C4 , showing that there exists no path between [ϕ] and some [σ] ∈ [A9 ] such that [σ] 6= [ϕ] and Tσ = T4 . Let γ be a path with end vertex [ϕ]. Since the only types admissible for C4 are T4 and [9] and in a quotient power graph there exists no edge incident to vertices having the same type, we have that the first edge of γ is {[ϕ], [ψ]}, for some ψ ∈ A9 such that Tψ = [9].

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Thus, by Lemma 4.1, {ϕ, ψ} is an edge in P0 (A9 ), which necessarily gives ψ 3 = ϕ. Pick one of those [ψ] and consider the edges of type {[ψ], [δ]} in Pe0 (A9 ). We necessarily have Tδ = T4 . Thus, again by Lemma 4.1, ψ and δ are, in a suitable order, one the power of the other. It follows that δ = ψ 3 = ϕ. This means that γ is a path of length 1 in which one end is not of type T4 . Then e c0 (A9 )T4 = µT4 [A9 ] = 1120. Since all the types admissible for A9 have appeared, the procedure 2.3 stops giving e c0 (A9 ) = 5442. The discussion above shows also that c0 (T (A9 )) = 4 and that c0 (O(A9 )) = 2. 5. Higher degrees In this section, we determine e c0 (An ), c0 (T (An )) and c0 (O(An )) for n ≥ 10. We start showing that the fact that the vertices of even order are distributed in more than one connected components is a peculiarity of n ∈ {8, 9}. Recall that Ωn is defined by Lemma 4.2. Lemma 5.1. For n ≥ 10, the vertices in [An ]0 of even order are contained in Ωn . Proof. By Lemma 4.2, all the vertices of type [1n−4 , 22 ] are contained in Ωn of Pe0 (An ). Let [π] ∈ [An ]0 with o(π) = 2k, for a positive integer k. Then o(π k ) = 2, so that π k is the product of s cycles of length two, for some s ∈ N even. If s = 2, then [π k ] ∈ Ωn and so [π] ∈ Ωn . If s ≥ 4 then π k = (a b)(c d)(e f )(g h)σ, for suitable distinct a, b, c, d, e, f, g, h ∈ N and σ ∈ An , with σ = id or the product of s − 4 cycles of order two. Let ϕ = (a c e b d f )(g h)σ. Then ϕ ∈ An and ϕ3 = π k . Due to n ≥ 10, there exist distinct elements i, j ∈ N \ {a, b, c, d, e, f, g, h} and we have the following path: [π], [π k ], [ϕ], [(a e d)(c b f )], [(a e d)(c b f )(g h)(i j)], [(g h)(i j)]. Since [(g h)(i j)] ∈ Ωn , we get also [π] ∈ Ωn .

Theorem 5.2. c0 (A10 ) = 29345, c0 (T (A10 )) = 3 and c0 (O(A10 )) = 1. Proof. By Lemma 5.1, the vertices of even order in [A10 ]0 are contained in Ω10 . Thus, for T1 = [16 , 22 ], we have e c0 (A10 )T1 = 1. Moreover from the paths: [(1 2 3)(4 5 6)], [(1 2 3)(4 5 6)(7 8)(9 10)],

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and [(1 2 3 4 5 6 7)], [(1 2 3 4 5 6 7)(8 9 10)], [(8 9 10)], [(8 9 10)(1 2 3 4 5)], [(1 2 3 4 5)], [(1 2 3 4 5)(6 7)(8 9)] we deduce that the types [14 , 32 ], [13 , 7], [3, 7], [17 , 3], [12 , 3, 5] and [15 , 5] are also admissible for Ω10 . Since [10] ∈ / T (A10 ), the vertices of type T2 = [52 ] ∈ T (A10 ) are instead isolated and thus e c0 (A10 )T2 = µ[52 ] [A10 ] = e 18144. The same argument used for P0 (A9 ) shows also that for T3 = [1, 33 ] we have e c0 (A10 )T3 = µ[1,33 ] [A10 ] = 11200. Since all the types adc0 (A10 ) = missible for A10 have appeared, the procedure 2.3 stops giving e 29345 and c0 (T (A10 )) = 3. Moreover, since oe(Ω10 ) contains all the prime less than 10, we deduce also that c0 (O(A10 )) = 1. Lemma 5.3. Let n ≥ 11. Then the vertices of order 3 in [An ] are contained in Ωn . Proof. Let n ≥ 11 and [π] ∈ [An ], with o(π) = 3. Then π is the product of s ≥ 1 cycles of length 3. To show that [π] ∈ VΩn , we show that whatever s is, there exists ϕ ∈ An such that ϕ2 = π with o(ϕ) even and apply Lemma 5.1. If s = 1 then, since n ≥ 11, there exist a, b, c, d ∈ N \ Mπ and we take ϕ = π 2 (a b)(c d). If s ≥ 2, then π = (a b c)(d e f )σ, for suitable a, b, c, d, e, f ∈ N and σ ∈ An , with σ 3 = id and we take ϕ = (a d b e c f )σ 2 . Lemma 5.4. Let n ≥ 11 and p a prime number such that 5 ≤ p ≤ n − 3 and p 6= n2 , n−1 2 . Then the vertices of order p in [An ] are contained in Ωn . Proof. Let [π] ∈ [An ], with o(π) = p. Then Tπ ∈ T , where T = {Ts = [1n−sp , ps ] : s ∈ N, s ≤ n/p}. By Lemma 4.2 iii) it is enough to show that T ⊆ T (Ωn ). Recall now that, by Lemma 2.1, T (Ωn ) is the set of vertices of a connected component of P0 (T (An )). Thus, by Lemma 5.1 and Lemma 5.3, it is enough to prove that for all s ∈ N, with s ≤ n/p, there exist T ∗ ∈ T (An ) with o(T ∗ ) even or o(T ∗ ) = 3 and a path in P0 (T (An )) between T ∗ and Ts . If sp ≤ n − 3, consider T = [ps , 3, 1n−sp−3 ]. Since T 3 = Ts , we take T ∗ = T p = [3, 1n−3 ]. If sp ≥ n − 2, being p ≤ n − 3, we have s ≥ 2. Let first s = 2. Then n − 2 ≤ 2p ≤ n and, by p 6= n2 , n−1 2 , we get 2p = n − 2. So we can consider T ∗ = [2p, 2], because o(T ∗ ) is even and (T ∗ )2 = T2 .

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Next let s = 3 and consider T = [3p, 1n−3p ]; being p ≥ 5, we have T 3 = T3 while o(T p ) = 3 and we take T ∗ = T p . Finally, for s ≥ 4, we take T ∗ = [(2p)2 , ps−4 , 1n−sp ] after having observed that (T ∗ )2 = Ts and that o(T ∗ ) is even. Lemma 5.5. Let n ≥ 11 with n = 2p or n = 2p + 1, for some prime p. Let [π] ∈ [An ], with o(π) = p. i) If Tπ = [1n−2p , p2 ], then [π] is an isolated vertex of Pe0 (An ); ii) if Tπ = [1n−p , p], then [π] ∈ Ωn ; iii) p ∈ oe(Ωn ).

Proof. If Tπ = [1n−2p , p2 ], being n − 2p ≤ 1, Tπ admits no proper power and is not a proper power of a type admissible for An . Thus, by Lemma 2.2, [π] is isolated in Pe0 (An ). Next let Tπ = [1n−p , p]. By n ≥ 11, we get n − p ≥ p ≥ 5 and thus, the type T = [1n−p−3 , 3, p] ∈ T (An ) satisfies o(T p ) = 3, while T 3 = Tπ . So, by Lemma 5.3, T p ∈ T (Ωn ) and, by Lemma 2.1, Tπ ∈ T (Ωn ). Finally, Lemma 4.2 iii), guarantees [π] ∈ Ωn . Lemma 5.6. For n ≥ 11, all the vertices in [An ] apart those of order a prime p such that p ≥ n − 2 or p = n2 , n−1 2 are contained in Ωn .

Proof. Let [π] ∈ [An ], with o(π) not a prime p such that p ≥ n − 2 or p = n2 , n−1 2 . By Lemma 5.1, we can assume that o(π) is odd. Let first o(π) = p, for some prime p. If p = 3, then Lemma 5.3 applies. If p ≥ 5, then, being also p ≤ n − 3, Lemma 5.4 applies. Assume now that o(π) is composite. If o(π) = 3k for some k ∈ N, k ≥ 2, then we k−1 k−1 have o(π 3 ) = 3 and, by Lemma 5.3, we get [π 3 ] ∈ Ωn , so that also [π] ∈ Ωn . Finally let o(π) = tpq, with p ≥ 3, q ≥ 5 prime numbers and t an odd positive integer. Since o(π t ) = pq, then Tπ contains a term equal to pq or two terms equal to p and q respectively. In any case, being n ≥ 11, this implies o(π tp ) = q ≤ n − 3 so that, by Lemma 5.4, [π tp ] ∈ Ωn and hence also [π] ∈ Ωn . We are ready to clarify the crucial role played by the set of integers A = P ∪ (P + 1) ∪ (P + 2) ∪ (2P ) ∪ (2P + 1) in our connectivity problem. Theorem 5.7. Let n ≥ 11 and Γ0 ∈ {Pe0 (An ), P0 (T (An )), O0 (An )}. Then: i) all the connected components of Γ0 up to one, the main component, are isolated vertices;

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ii) the main component of Pe0 (An ) is Ωn ; the main component of P0 (T (An )) is the subgraph induced by T (Ωn ); the main component of O0 (An ) is the subgraph induced by oe(Ωn ); iii) the values of c0 (An )and c0 (T (An )) and c0 (O(An )) are given by the following table. Table 3. c0 (An ), c0 (T (An )) and c0 (O(An )) for n ≥ 11. c0 (An ) c0 (T (An )) c0 (O(An )) 3 2 + 4n(n−2)(n−4)! + 1 n−1 4n(n−2)(n−4)! (n − 2)! + 3 2 +1 n−1 n(n−1)(n−4)! (n − 2)! + 3 3 +1 2 n(n−1)(n−4)! 2 2 +1 2 4n(n−2)(n−4)! 2 1 + 1 n−1 (n − 2)! + 1 2 2 n(n − 3)! + 1 2 2 4(n−1)(n−3)! 3 2 + n(n − 3)! + 1 n 4(n−1)(n−3)! 2 1 +1 n 1 1 1

n(n−1)(n−4)! 2

n ≥ 11 n − 2, n−1 /P 2 ∈ P, n ∈ n−1 n, 2 ∈ P, n − 2 ∈ /P n−1 n, n − 2 ∈ P, 2 ∈ /P n−1 n − 2 ∈ P, n, 2 ∈ /P n−1 ∈ P, n, n − 2 ∈ /P 2 n−1 /P n ∈ P, n − 2, 2 ∈ n − 1 ∈ P, n2 ∈ /P n − 1, n2 ∈ P n /P 2 ∈ P, n − 1 ∈ n∈ /A

Proof. i)-iii) Let n ≥ 11 be fixed and recall that c0 (An ) = e c0 (An ). First of all, observe that P ∩ O0 (An ) = {p ∈ P : p ≤ n}. It follows that we can reformulate Lemma 5.6 saying that all the vertices in [An ]0 of order not belonging to n n−1 B(n) = P ∩ {n, n − 1, n − 2, , } 2 2 are in Ωn . In particular, oe(Ωn ) ⊇ O0 (An )\B(n). We refer to the numbers in B(n) as to the critical orders for n. By Lemma 2.1 ii), there exists a unique connected component Θn of O0 (An ) such that VΘn ⊇ oe(Ωn ). If n ∈ / A, then B(n) = ∅ and thus all the vertices of [An ]0 are in Ωn . c0 (An ) = c0 (T (An )) = c0 (O(An )) = 1. Next let Thus, by Remark 1, e n ∈ A. We examine all the possible positions of n with respect to the sets P, P + 1, P + 2, 2P, 2P + 1 whose union is A. Since 2P, P + 1 are subsets of the even positive integers, while P, P + 2, 2P + 1 are subsets of the odd positive integers, we have ten cases to deal with. For all of them we need to understand the number of connected components of

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Pe(An ) and P0 (T (An )) containing vertices of order belonging to B(n), and the number of connected components of O0 (An ), different from Θn , containing a vertex belonging to B(n). / P , so Let n ∈ 2P \ (P + 1). Then n is even, n2 = p ∈ P and n − 1 ∈ that the only critical order is p. But, by Lemma 5.5, p ∈ oe(Ωn ) and thus VΘn = O0 (An ) and c0 (O(An )) = 1. To examine the graph Pe0 (An ), let [π] ∈ [An ] with o(π) = p. Then Tπ ∈ {[p2 ], [1p , p]} and, by Lemma 5.5 the vertices of type [1p , p] are in Ωn , while those of type T1 = [p2 ] are isolated. Thus e c0 (An )T1 = µ[p2 ] [An ] and hence e c0 (An ) = 4(n−1)(n−3)! + 1, n while c0 (T (An )) = 2. Let n ∈ 2P ∩ (P + 1). Then n is even, n2 = p ∈ P and n − 1 = q ∈ P , so that the critical orders are p and q. By Lemmas 5.5 and 3.1, we have that p ∈ oe(Ωn ) ⊆ VΘn while q is isolated in O0 (An ). Thus c0 (O(An )) = 2 and, by Lemma 2.2, VΘn = O0 (An ) \ {q} = oe(Ωn ). To examine the graph Pe0 (An ), let first [π] ∈ [An ] with o(π) = p. As in the previous case, Tπ ∈ {T1 = [p2 ], T = [1p , p]} and T ∈ T (Ωn ) while e c0 (An )T1 = µ[p2 ] [An ]. Moreover each [π] ∈ [An ] with o(π) = q is isolated and Tπ = [1, q]. So, for T2 = [1, q], we get e c0 (An )T2 = n(n − 3)!. It 4(n−1)(n−3)! follows that e c0 (An ) = + n(n − 3)! + 1 and c0 (T (An )) = 3. n Let n ∈ (P + 1) \ 2P . Here n is even, n2 ∈ / P and n − 1 = p ∈ P , so that the only critical order is p. By Lemma 3.1, p is isolated in O0 (An ) and so c0 (O(An )) = 2. Thus, by Lemma 2.2, we also get VΘn = O0 (An ) \ {p} = oe(Ωn ). Moreover, each [π] ∈ [An ] with o(π) = p is isolated in Pe0 (An ). Thus, for T1 = [1, p], we get e c0 (An )T1 = n(n − 3)! and therefore e c0 (An ) = n(n − 3)! + 1 and c0 (T (An )) = 2. Let n ∈ P \[(P +2)∪(2P +1)]. Here the only critical order is n, which by Lemma 3.1 is isolated, so that c0 (O(An )) = 2. Then, by Lemma 2.2, VΘn = O0 (An ) \ {n} = oe(Ωn ). For T1 = [n], we get e c0 (An )T1 = (n − 2)!, so that e c0 (An ) = (n − 2)! + 1 and c0 (T (An )) = 2. Let n ∈ (2P +1)\[P ∪(P +2)]. Here the only critical order is p = n−1 2 , which by Lemma 5.5, belongs to oe(Ωn ). Thus c0 (O(An )) = 1 and VΘn = O0 (An ) = oe(Ωn ). Moreover, for T1 = [1, p2 ], we get e c0 (An )T1 = µT1 [An ]. + 1 and c (T (A )) = 2. Thus e c0 (An ) = (n−1)4n! 2 (n−3) 0 n Let n ∈ (P + 2) \ [P ∪ (2P + 1)]. Here the only critical order is n − 2 and, by Lemmas 3.1 and 2.2, we get c0 (O(An )) = 2, VΘn = oe(Ωn ), + 1 and c0 (T (An )) = 2. e c0 (An ) = n(n−1)(n−4)! 2 Let n ∈ [P ∩ (P + 2)] \ (2P + 1). Here the critical orders are n, n − 2, c0 (An ) = (n − 2)! + both isolated by Lemma 3.1. Thus c0 (O(An )) = 3, e

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n(n−1)(n−4)! + 2 VΘn = oe(Ωn ).

1 and c0 (T (An )) = 3. Applying Lemma 2.2 we also get

Let n ∈ [P ∩ (2P + 1)] \ (P + 2). The critical orders are n, n−1 2 . By Lemmas 3.1, 2.2 and 5.5, with the usual arguments, we get c0 (O(An )) = + 1 and c0 (T (An )) = 3. 2, VΘn = oe(Ωn ), e c0 (An ) = (n − 2)!+ 4n(n−2)(n−4)! n−1 Let n ∈ [(P + 2) ∩ (2P + 1)] \ P . The critical orders are n − 2 and n−1 n−1 e(Ωn ). Thus 2 . By Lemmas 3.1 and 5.5, n is isolated while 2 ∈ o 4n(n−2)(n−4)! n(n−1)(n−4)! + +1 c0 (O(An )) = 2, VΘn = oe(Ωn ), e c0 (An ) = 2 n−1 and c0 (T (An )) = 3. Finally we show that P ∩ (P + 2) ∩ (2P + 1) = ∅, proving that if n ∈ P ∩ (P + 2), then n−1 / P. Let n = p ∈ P and n − 2 = p − 2 = 2 ∈ q ∈ P . Then, due to n ≥ 11, we have p, q ≥ 11. Since 3 divides n−1 n(n − 1)(n − 2) = p(p − 1)q, we deduce that 3 also divides p−1 2 = 2 . n−1 The possibility 3 = n−1 / P. 2 is excluded by n ≥ 11 and thus 2 ∈ Our case by case analysis have shown also that VΘn = oe(Ωn ) and that Ωn is the only connected component of Pe0 (An ), not reduced to an isolated vertex. Thus, by Lemmas 2.1 and 2.2, we also have that T (Ωn ) is the only connected component of P0 (T (An )) which is not reduced to an isolated vertex. We emphasize that the remarkable fact that all the connected components of Γ0 up to one are isolated vertices is a peculiarity of n ≥ 11. Namely, looking to the details given for the degrees 3 ≤ n ≤ 10, it is immediately controlled that a lot of exceptions arise for the various graphs under consideration. We give now a more concise overview on the number of connected components of the graph O0 (An ). Corollary 5.8. The values of c0 (O(An )) for n ≥ 4, with n 6= 6, are given by the following table. Table 4. c0 (O(An )), for n ≥ 4, n 6= 6. c0 (O(An )) n ≥ 4, n 6= 6 1 n∈ / P ∪ (P + 1) ∪ (P + 2) 2 n ∈ [P \ (P + 2)] ∪ [(P + 2) \ P ] ∪ (P + 1) 3 n ∈ P ∩ (P + 2) Proof. A check on the Tables 1, 2 and 3.

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Corollary 5.9. O0 (An ) is connected if and only if n = 3 or n, n − 1, n − 2 are not prime. The maximum number of connected components of O0 (An ) is 3 and it is realized if and only if n is a prime admitting a twin prime. Proof. It follows from Theorem 3.2 and Corollary 5.8.

Corollary 5.10. P0 (An ) is connected if and only if P0 (T (An )) is connected, that is, if and only if n = 3 or n ∈ / A. Proof. It is immediate by checking that c0 (T (An )) = 1 if and only if c0 (An ) = 1, by Theorems 3.2 , 4.4, 5.2 and 5.7 and noting that the natural numbers n such that 4 ≤ n ≤ 10 are contained in A. Corollary 5.11. The minimum n ∈ N such that P (An ) is 2-connected and An is non-abelian, is n = 16. There exists infinite n such that P (An ) is 2-connected. Proof. By Corollary 5.10 we need to show that N \ A contains infinite elements. Let n = 4k2 , where k ≥ 2 is a positive integer. Since n is even and n ≥ 16, then we have n ∈ / P ∪ (P + 2) ∪ (2P + 1). Moreover n2 = 2k2 is not a prime so that n ∈ / 2P. Finally, n − 1 = (2k − 1)(2k + 1), so that n∈ / P + 1. References [1] J. Abawajy, A. Kelarev and M. Chowdhury, Power Graphs: A Survey, Electronic Journal of Graph Theory and Applications 1 (2) (2013), 125-147. [2] D. Bubboloni, M. A. Iranmanesh, S. M. Shaker, Quotient graphs for group power graphs (2014), preprint. [3] G. R. Pourgholi, H. Yousefi-Azari and A. R. Ashrafi, The Undirected Power Graph of a Finite Group, Bull. Malays. Math. Sci. Soc. (to appear)

Daniela Bubboloni Department of Economics and Management, University of Firenze, via delle Pandette 9(D-6), 50127, Firenze, Italy Email: [email protected] Mohammad A. Iranmanesh Department of Mathematics, Yazd University, 89195-741, Yazd, Iran Email: [email protected]

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Seyed M. Shaker Department of Mathematics, Yazd University, 89195-741, Yazd, Iran Email: seyed− [email protected]
