Cyclic Permutation Groups that are Automorphism Groups of Graphs

In this paper we establish conditions for a permutation group generated by a single permutation to be an automorphism group of a graph. This solves the so called concrete version of König’s problem for the case of cyclic groups. We establish also similar conditions for the symmetry groups of other related structures: digraphs, supergraphs, and boolean functions.


Introduction
Frucht's theorem, conjectured by Dénes König states that every abstract finite group is isomorphic to the automorphism group of a graph [30]. On the other hand it is known that not every permutation group is an automorphism group of a graph. For example, there is no graph on n vertices whose automorphism group is the cyclic group C n generated by an n-element cycle. The problem asking which permutation groups can be represented as automorphism groups of graphs is known as the concrete version of König's problem [29].
This problem turned out much harder and was studied first for regular permutation groups as the problem of Graphical Regular Representation. There were many partial results (see for instance [15][16][17][22][23][24][26][27][28]) until the full characterization was obtained by Godsil [6] in 1979. In [2], Babai uses the result of Godsil to prove a similar characterization in the case of directed graphs.
In [20,21], Mohanty et al., consider permutation groups generated by a single permutation (they call them cyclic permutation groups) whose order is a prime or a Mariusz.Grech@math.uni.wroc.pl 1 Institute of Mathematics, University of Wroclaw, pl.Grunwaldzki 2, 50-384 Wrocław, Poland power of a prime. In [21,Theorem 3], they described those cyclic permutation groups of prime power order greater than 5 that are automorphism groups of graphs. However, the results contained some gaps. The authors made a false claim that there are no such groups of prime order 3 or 5. Also the proof of the main result contained a gap. All this has been corrected in [9]. Our aim is to generalize these results to cyclic permutation groups of arbitrary order.
When comparing the results in [7,8,10,14,25] one may observe that usually formulations of theorems concerning graphical representability are more natural and nicer when the problems are considered for edge-colored graphs rather than for simple graphs. In [13] we provide a relatively simple characterization of those cyclic permutation groups that are automorphism groups of edge-colored graphs. We also prove that each such permutation group is an automorphism group of a 3-colored graph.
In fact, the problem for edge-colored graphs has been considered already by H. Wielandt in [29]. Permutation groups that are automorphism groups of edge-colored digraphs were called 2-closed, and those that are automorphism groups of colored graphs were named 2 * -closed. In [18], A. Kisielewicz introduced the so-called graphical complexity of permutation groups. By G R(k) we denote the class of automorphism groups of k-colored graphs, by which we mean the graphs whose edges are colored with at most k colors. By G R we denote the union of all classes G R(k), which is the class of 2 * -closed groups. Moreover, we put G R * (k) = G R(k)\G R(k − 1), and for a permutation group A, we say that A has a graphical complexity k if A ∈ G R * (k). Then, G R (2) is the class of automorphism groups of simple graphs.
In this paper we fully characterize those cyclic permutation groups that are automorphism groups of simple graphs. In the last section we consider the same problem for other structures: digraphs, supergraphs, and boolean functions.
In Sect. 2, we recall some definitions concerning edge-colored graphs and permutation groups. We recall two results from [10], and prove their generalizations we need in the sequel. In Sect. 3, we recall results concerning cyclic permutation groups of prime order. In Sect. 4, we present some minimal (in a sense) permutation groups that do not belong to G R (2), while in Sect. 5, we present minimal cyclic permutation groups which do belong to G R (2). These will be used in the proof of the main results. In Sect. 6, we prove another auxiliary result which we call the extension lemma. The main results of the paper are given in Sect. 7. They include the results of [13]. Our approach is a little different than that in [13], and therefore we obtain, by the way, another proof of the results of [13]. The last section presents the corresponding results for digraphs, supergraphs, and boolean functions.

Definitions and Basic Facts
We assume that the reader has the basic knowledge in the areas of graphs and permutation groups. In fact, the terminology is standard and the reader is referred to [1,30]. The permutation groups are considered up to permutation isomorphism.
We need to refer to some results on the automorphism groups of k-colored graphs, so we recall here related terminology. A k-colored graph (or more precisely k-edgecolored graph) is a pair G = (V , E), where V is the set of vertices, and E is an edge-color function from the set P 2 (V ) of two elements subsets of V into the set of colors {0, . . . , k − 1} (in other words, G is a complete simple graph with each edge colored by one of k colors). In some situations it is helpful to treat the edges colored 0 as missing. In particular, the 2-colored graph can be treated as a usual graph. Also, if no confusion can arise, we omit the adjective "colored". By a (sub)graph of G spanned on a subset W ⊆ V we mean G = (W , E ) with E ({v, w}) = E({v, w}), for all v, w ∈ W .
Direct sum. For permutation groups (A, V ), (B, W ), by a direct sum of A and B we mean a permutation group (A ⊕ B, V ∪ W ) with the action given by Parallel product. For a permutation group (A, V ), the parallel product A \ \n is a permutation group (A, V × {1, . . . , n}) with the following natural action.
Thus, A \ \n ≈ A × I n . Now, we recall two theorems which are proved in [10] and will be used later.
Later on, we will need one more lemma.

Lemma 2.3
Let k ≥ 1 and B / ∈ G R(k) be a permutation group such that for every k-colored graph G, with the property B ⊆ Aut(G), there is a permutation f ∈ Aut(G)\B preserves all the orbits of B. Then, B ⊕ C / ∈ G R(k) for every permutation group C.
Proof Let B = (B, V ) and G be a k-colored graph such that B ⊕ C ⊆ Aut(G ). Then, obviously, the graph G, spanned on the set V , has the mentioned property. Let f ∈ Aut(G)\B be a permutation which preserves all the orbits of B. By f we denote a permutation which acts as f on V and fixes all other vertices of G . Obviously, We have to show that the colors of the edges of the graph G are preserved by f . If an edge e is contained in the graph G, then f (e) = f (e) and E( f (e)) = E( f (e)) = E(e) as required. If neither of the ends of e belongs to V , then f (e) = e and the statement is still true. The only nontrivial case is the edges of the form e = {v, w}, , as required. This shows that f ∈ Aut(G ) and completes the proof of lemma. This is well known (see [10], for instance) that I 1 ∈ G R(0), I 2 / ∈ G R, I n ∈ G R * (3), for n ∈ {3, 4, 5}, and I n ∈ G R (2), otherwise. This completes the case of permutation group of order one (which, as we see, is not quite trivial). In future consideration, we assume that the order of a cyclic permutation group is at least two.

Earlier Results
In this section we recall the results completing the description of the graphical complexity of cyclic permutation groups of prime order started in [20,21]. Those groups have the form C p \ \r ⊕ I q . The result from [20] can be written as follows.
A situation is a little different when p = 2.

Permutation Groups Outside GR(2)
In this section, we show a few one-generated permutation groups that have a small number of orbits, and are not automorphism groups of a 2-colored graph. In the general case, for every one-generated permutation group (A, V ), there is a subset W ⊆ V such that A restricted to W is one of the permutation groups of this form.
As it was mentioned before, for n > 2 and for every k-colored graph G, if Aut(G) ⊇ C n , then Aut(G) ⊇ D n . Hence, C n / ∈ G R, for n > 2. We prove the similar statement for some other cases.  (2). (a) all orbits are colored in one color, (b) one orbit is colored in some color and two orbits in another.
In the case (a), the group of automorphisms of the graph contains D |O 1 | ⊕ D |O 2 | . In the case (b), we may exchange the names of the vertices (in a cyclic way) such that a reflection will be an automorphism of the graph. For x = 4 and x = 5 we have four kinds of coloring: (a) all orbits are colored in one color, (b) one orbit is colored in different color than the rest of the orbits, (c) the orbits which contain the edges {v 0 , w 0 } and {v 0 , w 2 } are colored in one color and the rest of the orbits are colored in the second color, (d) the orbits which contain the edges {v 0 , w 0 } and {v 0 , w 1 } are colored in one color and the rest of the orbits are colored in the second color.
In the case (a), we have the same situation as in the case (a) for x = 3. For x = 5, in the remaining cases, we have the same situation as in the case (b) for x = 3. The same is true in cases (b) and (c) for x = 4. In the case (d) for x = 4, the situation is a little different. There is no automorphism that acts as a fixing point reflection on every orbit but still there is an automorphism that acts as a reflection on every orbit. After exchanging the names of the vertices (in a cyclic way) the permutation is an automorphism of the graph. This permutation fixes two points in the orbit O 1 ; v 0 and v |O 1 |/2 but fixes no point in the orbit O 2 . This is clear that we may also find and automorphism of the graph that fixes two point in the orbit O 2 and fixes no point in the orbit O 1 . Since in every case, we have an automorphism of a graph that does not belong to A, we have A / ∈ G R (2).
We note that in the case, where gcd(|O 1 |, |O 2 |) ∈ {3, 5}, and for every graph G, if Aut(G) ⊇ A, then Aut(G) contains a permutation that acts as a fixing point reflection on each orbit. Observe also that if |O i | is divided by 2 for some i ∈ {1, 2}, then there is an automorphism f 3 of a graph that acts as no fixing point reflection of the orbit O i and as a fixing point reflection on the other orbit. Using Theorem 4.1, and observations from the proof, we prove the following three theorems.
We consider the action of A on the set of edges. There are few types of orbits in this action. We color these orbits to obtain a graph such that This shows that if gcd(|O 1 |, |O 2 |) ∈ {3, 5} and gcd(|O 2 |, |O 3 |) ∈ {3, 5}, then, after exchanging the names of the vertices, the permutation belongs to Aut(G)\A. If gcd(|O 1 |, |O 2 |) ∈ 4 and gcd(|O 2 |, |O 3 |) ∈ {3, 5}, then, after exchanging the names of the vertices, the permutation belongs to Aut(G)\A. This completes the proof.
All the possibilities that occur in Theorem 4.3 are similar. Therefore, we may assume gcd(|O 1 |, |O 2 |) = 4, gcd(|O 2 |, |O 3 |) = 3, and gcd(|O 3 |, |O 4 |) = 5 in this case. Any coloring of the graph G, such that Aut(G) ⊇ A, permits to an action of a permutation that, after exchanging the names of the vertices, is equal either to None of these permutations belong to A. This completes the proof. A little more complicated proof is in the situation, where A has three orbits such that 3 divides |O 1 | and |O 2 | but not |O 3 |, 5 divides |O 2 | and |O 3 | but not |O 1 |, and 2 divides |O 1 | and |O 3 | but not |O 2 |. However, the statement is the same. There are only two such orbits. Those orbits are preserved by σ . Hence, the group of automorphisms of the graph is not equal to A. Moreover, it contains a permutation that acts as a fixing point reflection on every of the three orbits. The remaining case is where 4 divides both |O 1 | and |O 3 |. We consider the restriction of A to the set O 2 ∪ O 3 (we denote it B = τ ). By Theorem 4.1, for every graph G such that Aut(G) ⊇ B, there exists a permutation σ that acts as a fixing point reflection on both orbits O 2 and O 3 . We consider the action of the group A on the set of the edges.

Theorem 4.5 Let A be an one-generated permutation group with three orbits O
Then, A has three orbits which are contained in R and four orbits which are contained in S. The only three nontrivial colorings of those orbits (up to symmetries) are when we color one orbit which is contained in R in color 1 and other two in color 0 and we color either one or two orbits which are contained in T in color 1 and other in color 0 (as in Theorem 4.1 in case x = 4). It is easy to verify that in all those three cases there is an automorphism of a graph which acts as reflection on O 1 and as τ n σ on the set O 2 ∪ O 3 , for some n. Hence, the group of automorphisms of the graph is not equal to A.

The First Step of Induction
In this section we study one-generated permutation groups (A, V ) ∈ G R(k) such that whenever we remove one of the orbits (say O), then a restriction of A to the set V \O does not belong to G R(k).
In [21], it is proved the following. At first we generalize this theorem and prove the following.

Lemma 5.2 Let A be a one-generated permutation group with two orbits O
This is easy to verify that A preserves the colors of the edges of C(x y, xz) and therefore, A ⊆ Aut(C(x y, xz)). We prove the opposite inclusion. The 1-degree of a vertex which belongs to O 1 is equal to 3z + 2. The 1-degree of a vertex which belongs to O 2 is equal to 3y + 2. Since y = z, every automorphism of C(x y, xz) preserves the partition of V on O 1 and O 2 . Consequently, Aut(C(x y, xz)) ⊆ D xy ⊕ D xz .
We show that reflections are forbidden. Since A is transitive on O 1 and on O 2 , and moreover, A ⊆ Aut(C(x y, xz)), it is enough to exclude one reflection on O 1 and one reflection on O 2 . We show that an element ab ∈ D xy ⊕ D xz , where b ∈ D xz and does not belong to Aut(C(x y, xz)). For v ∈ V , by N (v), we denote the set of 1-neighbors of v in the opposite orbit. Then, . Since x > 5, this intersection is empty. In the similar way, one may exclude a reflection on O 2 . Hence, Aut(C(x y, xz)) ⊆ C xy ⊕ C xz .
We show that if ab ∈ Aut(C(x y, xz)), where a = id and b ∈ C xz , then b = σ xl 2 for some l. We know that ab fixes v 0 , v 1 and v 3 . Hence, the image of w 0 has to belong to the intersection In similar way one may show that if ab ∈ Aut(C(x y, xz)), where b = id and a ∈ C xy , then a = σ xl 1 for some l. Since A ⊆ Aut(C(x y, xz)), this implies the inclusion Aut(C(x y, xz)) ⊆ A.
Again, it is easy to verify that A preserves the colors of the edges of C(x, x). Hence, we have A ⊆ Aut(C(x, x)). We prove the opposite inclusion. The 1-degree of a vertex which belongs to O 1 is equal to 5. The 1-degree of a vertex which belongs to O 2 is equal to 3. Hence, every automorphism of C(x, x) preserves the partition of the set V on the orbits O 1 and O 2 . The graph spanned on O 1 is isomorphic with a x-cycle. This implies that Aut(C(x, x)), restricted to the set O 1 , is contained in D x . Let a ∈ Aut(C(x, x)) be a permutation such that a(v 0 ) = (v 0 ). We consider the possibilities on a(w 0 ). We obtain that a( Assume first that a does not act trivially on O 1 . Then, for i > 1, we is empty. Consequently, if a ∈ Aut(C(x, x)) and it fixes v 0 , then it fixes every vertex which belongs to O 1 Hence, a fixes w 0 . Since C(x, x) is preserved by σ , we have immediately that a fixes every vertex in O 2 . Hence a is a trivial permutation. Consequently, Aut(C(x, x)) = A. Now, we consider the cases where gcd(|O 1 |, |O 2 |) ∈ {3, 4, 5}. We prove the following two lemmas.

Lemma 5.3 Let A be a one-generated permutation group with two orbits O
Proof In the case y = z = 1, this is a consequence of Lemma 3.3 in [9]. We have also defined the graph C(x, x) there. In the other cases, we use the same symbols as in the proof of Lemma 5.2. We define C(x y, xz) as follows.
From the definition of C(x y, xz), we have immediately that A preserves the colors of the edges of C(x y, xz). Hence, A ⊆ Aut(C(x y, xz)). We prove that Aut(C(x y, xz)) preserves the sets O 1 and O 2 . Indeed, the 1-degree of a vertex, which belongs to O 1 is equal to z + 2. The 1-degree of a vertex, which belongs to O 2 is equal to y + 2. These numbers are different.
The remaining part of the proof is similar to the appropriate part of the proof of Lemma 5.2. By N 1 (v), we denote the set of 1-neighbors of the vertex v in the set O 2 , and by N 2 (v), the set of 2-neighbors of the vertex v in the set O 2 . We show that permutations that acts as reflections on some of the sets O 1 or O 2 are forbidden. We take a ∈ Aut(C(x y, xz)) that fixes v 0 . Observe that We have to show that if a ∈ Aut(C(x y, xz)) fixes v 0 , then a acts as σ xl 2 on O 2 for some l. We have N 1 (v 0 ) = {w xl ; l ∈ {0, . . . , z − 1}}. Therefore, a(w 0 ) = w xl for some l. Since the subgraph of (C(x y, xz)) spanned on O 2 is a |O 2 |-cycle, we have that a acts as σ xl 2 on O 2 . Again the role of O 1 and O 2 are symmetric, therefore, the same is true for O 1 (a permutation that fixes w 0 acts as σ xl 1 on O 1 for some l. Thus, A = Aut(C(x y, xz)). 3 , where y, z, t 2 , t 3 are positive integers such that gcd(y, zt 2 ) = 1, gcd(y, zt 3 ) = 1 and gcd(t 2 , t 3 ) = 1. We may assume that either t 2 = t 3 = 1 or t 2 < t 3 . By changing the names of orbits we may exclude the situation when y = 1, z = 1, t 2 = 1 and t 3

Lemma 5.4 Let A be a one-generated permutation group with three orbits O
From the definition of G, we have immediately Aut(G) ⊆ A. We prove the opposite inclusion. At the beginning, we show that Aut(G) preserves the partition of V on the sets O 1 , O 2 and O 3 . We count the 1-degrees of the elements which belong to these orbits. By d i , we denote here the 1-degree of the vertices that belong to O i . If y = z = t 1 = t 2 = 1, then d 1 = 5, d 2 = 3 and d 3 = 2. This implies that the partition is preserved by Aut(G). Similarly, if y = 3, z = 2, t 2 = 1, t 3 = 2, then we have Otherwise, d 2 = 2+2y+t 3 , d 3 = 2+y+t 2 . Moreover, if neither y = (2z−1)t 2 +zt 3 nor y = zt 1 + (z−1) 2 t 2 , then d 1 is equal to D 1 = 2 + 2zt 2 + zt 3 and d 1 is equal to y − 2 + 2zt 2 + zt 3 , otherwise. Since t 2 ≤ t 3 , we have d 2 > d 3 . If d 1 = D 1 , then obviously, Then, zt 2 + zt 3 2 = 4. This is possible only if either y = 3, z = 2, t 2 = 1 and t 3 = 2 or y = 1, z = 1, t 2 = 1 and t 3 = 6. The former case was considered earlier. The other case, we have excluded. Now, let d 1 = d 3 . Then, (2z − 1)t 2 + t 3 = 4. This is possible only if y = 1, z = 1, t 2 = 1, t 3 = 3. This case was also excluded. As a consequence we have that, in every case, Aut(G) preserves the partition of the set V on the sets O 1 , O 2 and O 3 .
We show that Aut(G) = A. When G = C(3x, 2x, 4x), one may check it directly. However, the fact that, when A has the orbits of cardinality 3x, 2x, 4x respectively, is immediate consequence of Lemmas 5.2 and 6.1. In proofs of those two lemmas, we do not use Lemma 5.4. We consider the case when G = C(x, x, x). This is easy to verify that A ⊆ Aut (C(x, x, x)). We have to prove the opposite inclusion. The graph spanned on O 1 is a |O 1 |-cycle. Moreover, every vertex, which belongs to O 3 has exactly one 1-neighbor in the set O 1 and every vertex, which belongs to O 2 has exactly one 1-neighbor in the set O 3 . This implies that Aut (C(x, x, x)) is equal either to A or to D x \ \3 . We have to exclude the second case. Assume that a ∈ Aut (C(x, x, x)) fixes v 1 0 . Then, a fixes v 2 0 . The vertex v 2 0 has only one 1-neighbor that belong O 1 and is not v 1 0 . This is the vertex v 1 x−1 . This implies that a fixes v 1 x−1 . Hence, a = id and A = Aut (C(x, x, x)).
We consider the remaining cases. The graphs spanned on the sets O i , i ∈ {1, 2, 3}, are either |O i |-cycles or their complements. This implies that Aut(G) ⊆ D |O 1 | ⊕D |O 2 | ⊕ D |O 3 | . We have to exclude the reflections and unwanted elements of Assume that a ∈ Aut(G) fixes v 1 0 . We have In the remaining part of the paper, we will use not only the statements of Lemmas 5.2, 5.3 and 5.4 but also the constructions of the graphs C(n, m) and C(n, m, k).
On the end of this section, we write one more theorem.

Extension Lemmas
We prove here the extension lemmas that we will use in a proof of general case.
. Moreover, we denote x i = gcd(c 0 , c i ). We assume that, for i < j, we have x i ≤ x j . This implies that if x i = 2, for some i, then x 1 = 2 and c 0 = 2.
Let G = (V \O, E) be a k-colored graph such that Aut(G) = C. We construct a finite number of graphs G We define E 0 = E * * , for s = 1 and for c 0 / ∈ {2, 4}. For s = 1 and c 0 ∈ {2, 4}, we define Assume now that we have constructed a graph G k , for k ≥ 0. Since O is the orbit in V , every element of O has the same 1-degree (say m k ) in G k . By P k we denote the set of those elements of V \O that have 1-degree (in G k ) which is equal to m k . Obviously, P k is a union of orbits of the group C.
We will construct a graph G k+1 in the situation when either |P k | > |O| or O i ⊆ P k , for some i ∈ {1, . . . , s}. In the other cases G k is the final step of our construction.
If |P k | > |O| and no set O i is a subset of P k , then we put The 1-degree of the elements which belong to O ∪ P k is increased. The 1-degree of remaining elements does not change. Obviously, after this change, the 1-degree of the elements which belong to O is larger than of the 1-degree of those elements which belong to P k .
Let i ∈ {1, . . . , s} be such that O i ⊆ P k . We consider three situations. Case 1. s = 1 and c 1 = c 0 . Then, C = D ⊕C c 1 . By Theorem 2.1, C ∈ G R(k), implies c 1 = 2. Then, by assumption c 0 = 2. Consequently, C = D ⊕ C 2 \ \2 ∈ G R(k). Case 2. Either s = 1, c 1 = c 0 and c 0 > 2 or s > 1 and x 1 = 2. We put Since x 1 = 2, in every case, the 1-degree of the elements that belong to O is increased. The 1-degree of the elements that belong O i is increased at most as 1-degree of the elements that belong to O. Remaining elements have the same 1-degree as before. In particular, 1-degree of the elements which belong to O is larger than the 1-degree of those elements which belong to P k . Case 3. Either s = 1, c 0 = 2 and c 1 > 2 or s > 1 and otherwise.
If i = s, then since c 0 / ∈ {3, 5}, the 1-degree of the elements which belong to O is increased. Remaining elements have the same 1-degree as before. If i < s, then the 1-degree of the elements which belong to O is increased. The 1-degree of the elements which belong to O i+1 is increased at most as 1-degree of the elements which belong to O. Remaining elements have the same 1-degree as before. In both cases, the 1-degree of the elements which belong to O is larger than the 1-degree of those elements which belong to P k .
Since the group C has a finite number of orbits, the procedure will finish after a finite number of steps. If k is the final step, then |P k | ≤ |O| and there is no i ∈ {1, . . . , s} such that O i ⊆ P k . We show that Aut(G k ) = B.
First, we prove that every a ∈ Aut(G k ) stabilizes O. By the construction of the graph G k , if there is a vertex v ∈ V \O with the same 1-degree as the elements which belong to O, then |O| / ∈ {2, 4}. This implies that the graph spanned on O is 1-connected. We know that every vertex v ∈ V \O with the same 1-degree as the elements of O is a 0-neighbor of every vertex that belong to O. Moreover, the number of those elements is not greater than |O|. Hence, if some vertex that belongs to O is moved out of O, then it is true for all elements which belong to O. This implies that the graph spanned on the set a(O) has to be either O-cycle or its complement. Since a(O) consists of at least two orbits of B, this is impossible. Consequently, the set O is stabilized by every a ∈ Aut(G k ).
The graph spanned on the set V \O is equal to G. Hence, the group Aut(G k ), restricted to V \O, is equal to C. The graph spanned on the set O is equal either to |O|-cycle or to the complement of |O|-cycle. Hence, the group Aut(G k ), restricted to O, is equal either to C c 0 or to D c 0 . For c 0 > 2, there is an orbit O such that gcd(|O|, |O |) > 2. In similar way as in proof of Lemma 5.2, a reflection on O implies a reflection on O . Hence, the second case is excluded. This is immediately from the definition of G * that τ preserves the colors of G * . Moreover, this is clear that the changes which are done in the graphs G * * , G 0 , . . . , G k do not break it down. Consequently, every b ∈ B preserves the colors of the edges of G k . Hence, B ⊆ Aut(G k ).
Let S be the set of elements f ∈ Aut(G k ) which acts on the set V \O as τ . Since B ⊆ Aut(G k ), such elements exist (τ is one of them). Since f acts on V \O as τ , the group generated by f , restricted to the set V \O, is equal to C. This implies that |Aut(G k )| = |C||S|. We check the possible actions of f on the set O.  Proof Obviously, k ≥ 2. Let C be a restriction of the group B to the set Since k ≥ 2, without loss of generality we may assume that, for a vertex v ∈ O, the 1-degree of the vertex v (in a graph G 1 ) is greater than 0. Let (x y, xz). We define a graph G = (V , E) as follows. otherwise.

We show that Aut(G) = B. The inclusion B ⊆ Aut(G) is an immediate consequence of the definition of the graph G.
We prove the opposite inclusion. We show that Aut(G) preserves the set O 1 ∪ O 2 . We count the 1-degree of the vertices which belong to V . If v ∈ O 2 , then according to the constructions of C(x y, xz) in Lemmas 5.2 and 5.3, d 1 (v) ≤ 3y + 2 for x > 5, and In the case when x > 6, we have 3y + 2 < x y/2. If x = 5, then y + 2 < x y/2. In these two cases, the only vertices which have the same 1-degree as the vertices which belong to O 2 , could be those which belong to O 1   We have: the set O 1 ∪ O 2 is preserved by Aut(G), the graph spanned on O 1 ∪O 2 is equal to G 2 and the graph spanned on V \(O 1 ∪ O 2 ) is equal to G 1 . Hence, Aut(G) ⊆ C ⊕ Aut (C(x y, xz)). Moreover, since for every orbit O / ∈ {O 1 , O 2 }, we have gcd(|O |, |O i |) ≤ 2 for i ∈ {1, 2}, and gcd(|O|, |O 1 |) = 2, the group B has index 2 in C ⊕ Aut (C(x y, xz)). Since the vertices v 0 and v 1 have the different sets of 1-neighbors in V \(O 1 ∪ O 2 ), the group Aut(G) has index at least 2 in the group C ⊕ Aut (C(x y, xz)). Consequently, since B ⊆ Aut(B), we have B = Aut(G).
We consider the case |O 1 | = 3x, |O 2 | = 2x, |O 3 | = 4x. In proof of Lemma 5.4, we have remained the proof that Aut (C(3x, 2x, 4x)) = A to the reader. In this case we prove that B ∈ G R(k) in another way than in other cases. By Lemma 6.2, the group B restricted to the set V \O 1 belongs to G R(k). Consequently, by Lemma 6.1, Further, we consider those cases for which we have proved in Lemma 5.4 (without using this lemma) that Aut(C(x y, xzt 2 , We show that Aut(G) = B. The inclusion B ⊆ Aut(G) is an immediate consequence of the definition of the graph G. We prove the opposite inclusion. We show that Aut(G) preserves the set O 1 ∪ O 2 ∪O 3 . We count the 1-degree of the vertices which belong to V . If v ∈ O 3 , then according to the constructions of C(x y, xzt 2 , xzt 3 The number y + t 2 + 2 is the smallest of them. The only vertices, which have the same 1-degree as the vertices that belong to O 3 could be those that belong to O 1 . If the situation take place, then obviously O 1 ∪ O 3 is preserved by Aut(G). If x y = xzt 3 , then for v ∈ O 1 , we have d 1 (v) > y + zt 2 + 2, and this is not this case. Otherwise, the vertices which belong to O 1 • there is a graph G such that Aut(G) = C and its subgraph spanned on the orbit O is colored by two colors.
Proof Let |O 1 | = 4x, |O 2 | = 4y, |O| = 2z. We may assume that either , E) be a graph as in the assumption. Let t be the number of 1-neighbors of u 0 in O. Since |O| is even, we may assume (eventually exchanging the colors of G) that z and t have a different parity. We define the graphs G j = (V , E j ) inductively.
we denote here the number of 1-neighbors of v is the graph G i . Obviously, d G 0 (v 0 ) > d G 0 (w 0 ). We will construct the graph G j+1 in the case when there exists a vertex v / . Assume that we have constructed a graph G j and there exists such a vertex v . Then, v belongs to the orbit O / ∈ {O 1 , O 2 , O}. We construct a graph G j+1 by modifying the graph G j .
We do not change the 1-degree of the vertices which belong to O 2 and we increase the 1-degree of v . This is easy to see that in every graph G j , we have . Hence,the procedure will finish after a finite number of steps. Let G k be the last graph in our construction. We show that Aut(G k ) = B. The inclusion B ⊆ Aut(G k ) is an immediate consequence of the definition.
We prove the opposite inclusion. First, we show that O 1 and O 2 are preserved by Aut(G k ). We have Let N C be the set of cyclic permutation groups consisting of C n , n ≥ 3, groups with two orbits O 1 and O 2 such that gcd(|O 1 |, |O 2 |) ∈ {3, 5}, groups with three orbits O 1 , O 2 , O 3 such that gcd(|O 1 |, |O 2 |) = 3, gcd(|O 2 |, |O 3 |) = 5, gcd(|O 1 |, |O 2 |) ≤ 2. By Theorems 4.1, 4.2, and 4.5, if A ∈ N C and G is a graph such that A ⊆ Aut(G), then there exists a permutation σ ∈ Aut(G) such that σ acts as a fixing point reflection on each of orbits of A. Now, we prove a negative extension lemma.
, α(w i j ) = w i j .) Obviously, α / ∈ B, which completes the proof in this case. We may use at most two colors to color these edges. Hence, exactly the same proof as above works here.

Main Results
In this section, we prove the theorems that characterize the graphical complexity of one-generated permutation groups. First we give an alternate proof of the result of [13]. The "if" part, we prove by induction on the number of orbits. By Lemma 2.3, we may restrict our proof to the case where there is no orbit of cardinality one. If there are the only two orbits, then the statement holds by Lemmas 5.2 and 5.3. Now, we assume that the statement holds in all the cases where there are less than k orbits, k > 2. We prove that this implies that the statement holds for k orbits. The remaining cases are when s(O) > 1. In those cases the orbits of B satisfy the conditions and we have the same as in the first case.
Since in this paper, there is no place where we have used more than three colors (the step of induction preserves the number of colors), we have

Corollary 7.2 Let A be a one-generated permutation group. Then, A ∈ G R(3) if and only if A ∈ G R.
When we want to describe one-generated permutation groups that belong to G R(2), the theorem becomes more complicated. We may give a number of conditions that each orbit has to satisfy. However, it will be clearer if we write it in another way.
Let A be a one-generated permutation group of order n. We introduce now a graph Graph(A) with loops which gives an information how the "prime powers parts" of A are joined.
• The vertices of Graph(A) are those primes that divide n.  4 , this reflection will be eliminated. Hence, also in this situation, A H 1 ∈ G R(2) and we can continue the procedure. Case 2. There is no orbit of A with cardinality larger than 4, divided by 2 but not by 4. Then, there is an orbit of cardinality 2. We add the orbit of cardinality two as the first element of the list of the orbits which belong to O(H 1 ). We may use A 1 H 1 = C 2 as in the first step of induction. Then, we use Lemma 6.4 to show that A 3 H 1 ∈ G R(2) and continue the procedure. Since there are only two orbits of even cardinality greater then 2, i. e., O 1 and O 2 , the conditions in Lemma 6.1 are satisfied and the procedure works.
This completes the proof of the theorem.

Other Structures
In this section, we deal with the same problem for other structures, digraphs, supergraphs and boolean functions.

Digraphs
We start with the digraphs. A digraph G is a pair (V , E), where V is a set of the vertices of G and E ⊆ (V ×V )\{ (v, v) : v ∈ V } is a set of directed edges of G (without loops). A permutation σ of the set V belongs to Aut(G) if, for every pair (v, w), we have (v, w) ∈ E if and only if (σ (v), σ (w)) ∈ E. For a digraph G = (V , E) and a vertex v ∈ V , we define d f (v) to be the number of these vertex w ∈ V that (w, v) ∈ E. We say that a permutation group A belongs to the class DG R(2) if there exists a digraph G such that Aut(G) = A. In [18], A. Kisielewicz stayed the following.
Theorem 8.1 [18] Every one generated permutation group belong to the class DG R (2).
In [18], Kisielewicz has proved this only for one example. He has written that it can be easily generalized for the general case. Since it may be not so obvious how to do it, we write a more detailed proof. Let G 1 = (V \O, E 1 ) be a digraph such that Aut(G 1 ) = B, and, for every v ∈ V \O, we have d f (v) ≥ 1. We define a digraph G = (V , E) as follows. (v, w) ∈ E if and only if one of the following holds.
• {v, w} ⊆ (V \O) and (v, w) ∈ E 1 , Obviously, d f (v) = 1, for every v ∈ O and d f (v) > 1, otherwise. This implies that the set O is stabilized by Aut(G). The rest of proof is the same as in the proof of Lemma 6.1.
We continue the proof of the theorem. If A is of order two, then, by Theorem 7.3, A ∈ G R(2) ⊆ DG R (2). Let A be a one-generated group of order greater than two. There exists an orbit O of cardinality n > 2. Then, the group A restricted to O is equal to C n . There exists a graph G = (O, E) such that d f (v) = 1 for every v ∈ O. Consequently, using repetitively Lemma 8.2, we have A ∈ DG R(2).

Supergraphs
The supergraphs is another graphical structure. It was introduced in [18] by A. Kisielewicz. This is an extension of a graph. A supergraph G is a pair (V , E), where V is a set of vertices of G. The set of the edges is defined inductively. Every vertex is an edge of order 0. Let E i be the set of the edges of order i. If k > 0, then every edge of order k is a pair {v, w}, v = w, such that v ∈ E j , w ∈ E l and j < k, l < k. Then, E = E i . We say that G is of order k, if E = E k and E = E k−1 . A permutation σ of V belongs to Aut(G), if σ preserves the structure of G. We say that A ∈ SG R(k) if there exists a supergraph G of order at most k such that Aut(G) = A. In [18], we can find the following.

Theorem 8.3 [18] G R(2) = SG R(1) and DG R(2) ⊆ SG R(2).
An immediate consequence of Theorems 8.1 and 8.3 is: Corollary 8.4 Every one-generated permutation group belongs to SG R (2). Moreover, a one-generated permutation group belongs to SG R (1) if and only if it belongs to G R(2). By BG R we denote BG R(k). The boolean function can be identifying with ndimension k-colored simplex, i. e., where every subsimplex is colored one of k-colors. In this sense boolean functions are graphical structures, which is one of the natural generalizations of edge-colored graphs. The theorem stated in [4] and repeated in [19], we can write as follows.

Boolean Functions
Theorem 8.5 [4] A one-generated permutation group A belongs to BG R if and only if whenever there exists an orbit O such that |O| ∈ {3, 4, 5}, then there exists an orbit O such that gcd(|O|, |O |) > 2. Moreover, if A ∈ BG R, then A ∈ BG R (2).
In [4], the proof is very complicated. In [19], it is much simpler. However, the proof in [19] contains a gap. At the end of the proof, it is used an extension theorem ( [19,Theorem 4.4]) without checking the assumptions. In my opinion, the assumptions were forgotten. Obviously, one can prove that they are satisfied. However, the proof of those conditions is as hard as the proof of whole the theorem. The extension lemma that should be used there is a stronger version of [19,Theorem 4.4] but in less general case.

Lemma 8.6 Let (A, V ) be a one-generated permutation group. Let W be a proper subset of V preserved by A such that A restricted to W belongs to BG R(2) and A restricted to V \W belongs to BG R(2). Then, A ∈ BG R(2).
The proof is similar to the proof of Lemma 6.1 and the proof of Lemma 8.2. This is as hard as the proof of Lemma 8.2, easier than the proof of Lemma 6.1 and definitely easier than the proof of [19,Theorem 4.4]. We leave it to the reader.
Open Access This article is distributed under the terms of the Creative Commons Attribution 4.0 International License (http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made.