The rainbow k-connectivity of the non-commutative graph of a finite group

The non-commuting graph Γ( G ) of a non-abelian group G is defined as follows. The vertex set V (Γ( G )) of ℾ( G ) is G \ Z ( G ) where Z ( G ) denotes the center of G and two vertices x and y are adjacent if and only if xy ≠ yx . We prove that the rainbow k -connectivity of Γ( G ) is equal to ⌈k/2⌉ + 2, for 3 ≤ k ≤ | Z ( G )|.


Introduction
Let G be a group and Z(G) be the center of G. The non-commuting graph (G) associated to G is the graph with vertex set G \ Z(G) and such that two vertices x and y are adjacent whenever xy 6 = yx. The non-commuting graph of a group was first considered by Paul Erdös in 1975, [6]. Subsequently, it was strongly developed in [1].
Let be a connected graph with the vertex set V ( ) and the edge set E( ). Define a coloring ' : E( ) ! {1, 2, . . . , t}, t 2 N, where adjacent edges may be colored the same. Given an edge coloring of , a path P is rainbow if no two edges of P are colored the same. An edge-colored graph is rainbow connected if every pair of vertices of are connected by a rainbow. The rainbow connection number rc 1 ( ) of is defined to be the minimum integer t such that there exists an edge-coloring of with t colors that makes rainbow connected.
From a generalization given by Chartrand, Johns, McKeon and Zhang in 2009 [2], an edgecolored graph is called rainbow k-connected if any two distinct vertices of are connected by at least k internally disjoint rainbow paths. The rainbow k-connectivity of , denoted by rc k ( ), is the minimum number of colors required to color the edges of to make it rainbow k-connected, and ' is called a rainbow k-coloring of . We usually denote rc 1 ( ) by rc( ).
The importance of rainbow connection number emerge from applications to the secure transfer of classified information between agencies [2]. Recently, Septyanto in [8], showed another form to see the application.
The commutator of an ordered pair g 1 , g 2 of elements of G is the element [g 1 , g 2 ] = g 1 1 g 1 2 g 1 g 2 2 G G is abelian if and only if [g 1 , g 2 ] = 1 Let G(V, E), and let a = (e 1 , ..., e j ) be a path with e i 2 E. Then l(a) := j is called the length of a.
We denote by P (x, y) the set of all x, y paths in G. Then d(x, y) := min{l(a)|a 2 P (x, y)} is called the distance from x to y.
We call diam(G) := max{d(x, y)|x, y 2 G} the diameter of G. The length of a shortest cycle of G is called the girth of G.
When a pair of vertices g i , g j are joined, we denoted by g i ⇠ g j . In otherwise we denoted by g i ⌧ g j .

About edge-connectivity
We need to find k-rainbow paths between any two vertices for (G), with k 3. We may ask for the maximum number of paths from v 1 to v 2 vertices, no two of which have an edge in common (such paths are called edge-disjoint paths). As a consequence of Menger's theorem about max-flow and min-cut, Witney [10] presented that a graph is k-connected if and only if any two vertices are connected by k internally disjoint paths. With Whitney's result we can answer how many edge-disjoint paths are connecting a given pair of vertices on (G).
Definition 3.1. The edge-connectivity is the minimum size of a subset C ⇢ E(G) for which G C is not connected for a graph G. The edge-connectivity of G is denoted by (G). If (G) k then G es called k-edge connected.
The next theorem is a result implied by Menger's theorem. This form can be found in [7, Chapter 15]. As we can obtain the rainbow-connectivity number of (G) and this graph is connected by blocks with s = |Z(G)| as size of each block, we have that the graph (G) is s-edge-connected and there exist s edge-disjoint paths in (G). Then, our problem now is coloring the s edge-disjoint paths of (G). Remark 3.3. By 1.1 we note that there exist two cases that we need analyze, for g i , g j , g k , g l 2 S M (G) and z r , z t , z w , z p 2 Z(G). The first case is when g i z r ⇠ g j z t which give us a bipartite complete graph in (G). The second case is when we have g i z r ⇠ g j z t ⇠ g k z w , but g i z r ⌧ g k z w . Remark 3.4. We note that (G) s. Then, if we want a path between end vertices g i z r and g j z t , without loss of generality we start with g i z r , necessarily, from 3.2, the edges g i z r ⇠ g j z t b with t b 2 {1, ..., s}, are in the set of edge-disjoint paths. The same happens for the edges g i z ra ⇠ g j z t with r a 2 {1, ..., s} because we have s disjoint paths, therefore we need all outedge from g i z r , and all in-edge to g j z t , thus all our edge-disjoint paths have the following form: (g i z r , g j z t b , ..., g i z ra , g j z t ), with t a , r b 2 {1, ..., s}.

Rainbow k-connectivity
, then the set of edges is given by www.ejgta.org The rainbow k-connectivity of the non-commutative graph of a finite group | Luis A. Dupont et al.
. The coloring given by: For an easier study of this kind of graph we use a table called rainbow table, whose entries (r a , t b ) are the color from edge (g i z ra , g j z t b ). This table is the following form: The (n + 2)-color in the table is given by white space.

Case when
, then the set of edges is given by www.ejgta.org The rainbow k-connectivity of the non-commutative graph of a finite group | Luis A. Dupont et al.
. The coloring given by: and (n + 2)-color with white spaces.

How to build the rainbow table
Example 5.1. We give the case when s = 6 and g 1 ⇠ g 2 in S M (G) with the coloring assigned before. Without loss of generality suppose that ({g 1 z p , g 2 z p }) = 1, then the rainbow table is given by: We can see that there is not exist a rainbow k-connectivity with 4 colors. To give s edge-disjoint paths with ends vertices g 1 z 2 and g 2 z 4 , the first path cross above g 2 z 1 , then we start the path with g 1 z 2 4 ⇠ g 2 z 1 . Now, we need move from g 2 z 1 but our only options are g 2 z 1 1 ⇠ g 1 z 1 , g 2 z 1 3 ⇠ g 1 z 5 and g 2 z 1 2 ⇠ g 1 z 6 and these edges can not arrive to g 2 z 4 because all the in-edge repeat color 4. For this reason we need to ensure that there exist enough in-edge that cover complete the out-edge in the set edges with majority color. For the existence of all edge-disjoint paths for any vertex we need to add one color more, and the table is given by 2 6 6 6 6 6 6 4 Example 5.2. We will do an example step-by-step about how we found all the edge-disjoint paths with our table. Let g 1 ⇠ g 2 in S M (G) and |Z(G)| = 4. Then, we will build our rainbow table with 3 colors the following form.
From this table we can found rc 3 ( (G)) = 3 for any vertices. For example, for end vertices g 1 z 3 , g 2 z 4 g 2 z 1 If we note, we can not find 4 edge-disjoint paths with 3 colors, because g 1 z 1 to g 2 z 1 passes through g 2 z 3 , the paths are the followings: Then, we need add another color, then the table is 4 colors the following form: Then, with all this 4 colors we found all 4 edge-disjoint paths from g 1 z 1 to g 2 z 1 , and they are the followings: The coloring given before can not help us to find all the disjoint-edge paths for the case when g i ⇠ g j ⇠ g l but g i ⌧ g l in S M (G) , for example, the rainbow table for this case is the next But, we can see that for go from g i z 1 to g l z 2 we have same colors then, we need to do paths with length at least 4 like the following picture: The coloring given for this specifical case is the following: The rainbow tables for each case are the following: www.ejgta.org The rainbow k-connectivity of the non-commutative graph of a finite group | Luis A. Dupont et al.

4
g i z 1 g i z 2 g i z 3 g l z 1 g l z 2 g l z 3 Theorem 6.1. Let G be a non-abelian group with |Z(G)| = 3 and (G) be the non-commutative graph associated to G, then rc 3 ( (G)) = 4.
Proof. Let the set of edges be the following form: {e 2 E( (G))|g j z ja ⇠ g l z ja such that ({g j , g l }) = 2 for g j , g l 2 S M (S) and j a = 1, 3} And the coloring is given by The following are all the 3 edge-disjoint paths for each pair of vertices when ({g j , g l }) = 2 All the edge-disjoint paths when ({g i , g j }) = 2, ({g j , g l }) = 2 and g i ⇠ g j ⇠ g l but g i ⌧ g l All the edge-disjoint paths when ({g i , g j }) = 1 Theorem 6.2. Let G be a finite non-abelian group. Then rc k ( (G))  ⌃ k Proof. We will proof that 4 is a coloring works for our graph.
1. Case g i ⇠ g j Let g i z ia , g j z j b be the end vertices. We want to find the edge-disjoint paths between them. Let 4.1 the rainbow table assigned for this case. From 4.1 it is evident that the first path is given by g i z ia Let j 1 be the column assigned to the row i a such that (i a , j 1 ) = f 1 then, we remove the entries with color f 1 to the column g j z j 1 and, the same happen to column g j z j b . Remark 6.3. When we say remove the entry we say that entry is not consider to form the rainbow path.
Thus, the path for this case is with (i a 1 , j 1 ) 6 = f 1 6 = (i a 1 , j b ) the colors assigned to remaining entries and g j z j 1 , g i z ia 1 the respective vertices from remaining entries.
Let (i a , j 2 ) be the entry with j 2 6 = j 1 , such that (i a , j 2 ) = f 2 then, we remove the entries with same color as f 2 in column g j z j 2 . We can not use the entry (g i z a 1 , g j z j b ) because is an edge for 1, moreover we remove all the entries with same color as f 2 in column g j z j b . Thus, the path is the following: with (i a 2 , j 2 ), (i a 2 , j b ) the colors assigned to remaining entries and g j z j 2 , g i z ia 2 the respective vertices from remaining entries.
· · · f · · · . . . . . . Under the conditions stated above we apply the same to all the colors assigned to i a -raw. We take edges from remaining entries to form the rest paths with the same method. Let j 0 1 such that f 0 = (i a , j 0 1 ) from j b -column we remove the row with entry same color like f 0 . The new path is the following: Take (i a , j 0 1 ), (i a 0 1 , j 0 1 ) as remaining entries from all the entries do not removed before with a dofferent color as f 0 . Remark 6.4. Suppose that we can coloring with only ⌅ k 2 ⇧ + 1 colors. Let g i z im any start vertex, then there exists a pair of vertices g j z jn , g j z j n 0 such that {(a ir , b jn )|(a ir , b jn ) color 6 = ( ⌅ k 2 ⇧ + 1) color} identify with {(a ir , b j n 0 )|(a ir , b jn ) color = the last color}, therefore is impossible to built k paths between any end vertices g i z im , g j z jn passes through g j z j n 0 , just like 5.1.

2.
Case: g i ⇠ g j ⇠ g l with g i ⌧ g l in S M (G) . (a) Repetition of different color to the last color Case: repetition of one color between columns. Suppose that f is the repeated color between the columns assigned to the end vertices g i z ia and g l z l b i.e. f = (j c , i a ) = (j c , l b ) in the rainbow table, for some c = {1, ..., |Z(G)|}, with l b 2 g l Z and i a 2 g i Z.
Suppose that f is in the path passes through g j z jc , thus for do the rainbow path we need Thus, there are cases when we have not free columns for do the rainbow paths. The same happens for case 5: Therefore, we can not form k rainbow paths with ⌃ k 2 ⌥ + 1 different colors. Theorem 1.3 Let G be a finite non-abelian group. Then rc k ( (G)) = ⌃ k Proof. From 6.2, 6.5 and 6.6.
Given the structure of (G), it could be considered a generalization of study in [5] to find the Harary index of (G).
By 3.2 we know that we can found 5 edge-disjoint paths for any pair of vertices then, without loss of generality we give the 5 edge-disjoint paths for end vertices x, ax 2 2 S M (G) . By 1.3 we know that we need b 5 2 c + 2 -color. The rainbow table is given below Then, the 5 edge-disjoin paths are given by: 5 If we note, we can not find 5 edge-disjoint paths with only 4 colors, for example, for the end vertices x 4 b 4 and x 3 b 2 we have the following paths: Start with color 1 Start with color 2 Start with color 3 Start with color x 4 b 4 4 ⇠ a 3 b 2 Start with color 4 from