Genus of commuting graphs of some classes of finite rings
aa r X i v : . [ m a t h . R A ] J a n Genus of commuting graphs of some classes offinite rings
Walaa Nabil Taha Fasfous and Rajat Kanti Nath ∗ Department of Mathematical Sciences, Tezpur University,Napaam-784028, Sonitpur, Assam, IndiaEmail Addresses: [email protected] (W.N.T. Fasfous),[email protected] (R.K. Nath)
Abstract:
In this paper, we compute the genus of commuting graphs of non-commutative rings of order p , p , p q and p q , where p and q are prime integers.We also characterize those finite rings such that their commuting graphs are planar ortoroidal. Key words:
Commuting graph, Genus, planar graph, toroidal graph.
The smallest non-negative integer n such that a graph G can be embedded on thesurface obtained by attaching n handles to a sphere is called the genus of G . We write γ ( G ) to denote the genus of a graph G . It is worth mentioning that γ ( K n ) = (cid:24) ( n − n − (cid:25) if n ≥ . (1.1)Also, if G = m ⊔ i =1 K n i then by [3, Corollary 2] we have γ ( G ) = m X i =1 γ ( K n i ) . (1.2)A graph G is called planar or toroidal if γ ( G ) = 0 or 1, respectively. The commutinggraph Γ c ( R ), of a finite non-commutative ring R with center Z ( R ), is an undirectedgraph whose vertex set is R \ Z ( R ) and two distinct vertices are adjacent if theycommute. The concept of Γ c ( R ) was introduced by Akbari, Ghandehari, Hadianand Mohammadian [2] in 2004. However, they studied Γ c ( R ) for semisimple rings.In [1, 7] Abdollahi and Mohammadian respectively, considered commuting graphs ofsome matrix rings and Omidi and Vatandoost, in [8], initiated the study of commuting ∗ Corresponding author raphs of any finite non-commutative rings. It is noteworthy that Γ c ( R ) is less exploredcompared to the study of commuting graphs of finite groups which was introduced byBrauer and Fowler [4] in 1955. In [5], Dutta, Fasfous and Nath have computed genusof commuting graphs of some classes of finite rings and characterized non-commutativerings of order p and p (for any prime) with unity such that their commuting graphsare planar or toroidal. In this paper, we consider non-commutative rings of order p , p , p q and p q , where p and q are two primes, and compute the genera of theircommuting graphs. We also characterize those rings such that their commuting graphsare planar or toroidal. It is worth mentioning that the structures of Γ c ( R ) for the abovementioned classes of rings have been described in [9, 10] and their spectral aspects havebeen explored in [6]. We begin with the following result.
Theorem 2.1.
Let R be a non-commutative ring with unity and | R | = p . (a) Let | Z ( R ) | = p .(i) If p = 2 then Γ c ( R ) is planar, toroidal or γ (Γ c ( R )) = 2 .(ii) If p ≥ then γ (Γ c ( R )) = ( p + p + 1) l ( p − p − p − p − m or l l ( p − p − p − p − m + l l ( p − p − p − p − m for some positive integers l , l such that l + l ( p + 1) = p + p + 1 ; and hence Γ c ( R ) is neitherplanar nor toroidal. (b) Let | Z ( R ) | = p .(i) Γ c ( R ) is planar if and only if p = 2 .(ii) If p ≥ then γ (Γ c ( R )) = ( p + 1) l ( p − p − p − p − m ; and hence Γ c ( R ) is neither planar nor toroidal.Proof. (a) By [10, Theorem 2.5] we have Γ c ( R ) = ( p + p + 1) K p − p or l K p − p ⊔ l K p − p , where l + l ( p + 1) = p + p + 1. Case 1: Γ c ( R ) = ( p + p + 1) K p − p By (1.2) we have γ (Γ c ( R )) = ( p + p + 1) γ ( K p − p ). If p = 2 then p − p = 2 andso γ (Γ c ( R )) = 7 γ ( K ) = 0. Therefore, Γ c ( R ) is planar. If p ≥ p − p ≥
6. By(1.1) we have γ (Γ c ( R )) = ( p + p + 1) (cid:24) ( p − p − p − p − (cid:25) . Since p ≥ ( p − p − p − p − ≥ and so γ (Γ c ( R )) ≥
13. Thus Γ c ( R ) isneither planar nor toroidal. Case 2: Γ c ( R ) = l K p − p ⊔ l K p − p By (1.2) we have γ (Γ c ( R )) = l γ ( K p − p ) ⊔ l γ ( K p − p ). If p = 2 then p − p = 2and p − p = 6. Also, l + 3 l = 7 which gives l = 4 and l = 1 or l = 1 and l = 2.Therefore, γ (Γ c ( R )) = 4 γ ( K ) + γ ( K ) = 1 or γ (Γ c ( R )) = γ ( K ) + 2 γ ( K ) = 2. That s, Γ c ( R ) is toroidal or γ (Γ c ( R )) = 2. If p ≥ p − p ≥ p − p ≥
24. By(1.1) we have γ (Γ c ( R )) = l (cid:24) ( p − p − p − p − (cid:25) + l (cid:24) ( p − p − p − p − (cid:25) . Note that ( p − p − p − p − ≥ and ( p − p − p − p − ≥
35. Therefore, γ (Γ c ( R )) ≥ l + 35 l >
36. That is, Γ c ( R ) is neither planar nor toroidal. This completes the proofof part (a).(b) By [10, Theorem 2.5] we have Γ c ( R ) = ( p + 1) K p − p . Therefore, using (1.2)we get γ (Γ c ( R )) = ( p +1) γ ( K p − p ). If p = 2 then p − p = 4. Therefore, γ (Γ c ( R )) =3 γ ( K ) = 0. That is, Γ c ( R ) is planar. If p ≥ p − p ≥
18 and so by (1.1) wehave γ (Γ c ( R )) = ( p + 1) (cid:24) ( p − p − p − p − (cid:25) . Note that ( p − p − p − p − ≥ and so γ (Γ c ( R )) ≥
72. Thus Γ c ( R ) is neitherplanar nor toroidal. This completes the proof of part (b). Theorem 2.2.
Let R be a non-commutative ring with unity where | R | = p and Z ( R ) not a field. (a) Let | Z ( R ) | = p .(i) If p = 2 then Γ c ( R ) is planar, toroidal or γ (Γ c ( R )) = 2 .(ii) If p ≥ then γ (Γ c ( R )) = ( p + p + 1) l ( p − p − p − p − m or l l ( p − p − p − p − m + l l ( p − p − p − p − m for some positive integers l , l such that l + l ( p +1) = p + p +1 ; and hence Γ c ( R ) is neither planarnor toroidal. (b) Let | Z ( R ) | = p . Then γ (Γ c ( R )) = ( p + 1) l ( p − p − p − p − m ; and hence Γ c ( R ) is neither planar nor toroidal.Proof. (a) By [10, Theorem 2.5] we have Γ c ( R ) = ( p + p + 1) K p − p or l K p − p ⊔ l K p − p , where l + l ( p + 1) = p + p + 1. Case 1: Γ c ( R ) = ( p + p + 1) K p − p By (1.2) we have γ (Γ c ( R )) = ( p + p + 1) γ ( K p − p ). If p = 2 then p − p = 4 andso γ (Γ c ( R )) = 7 γ ( K ) = 0. Therefore, Γ c ( R ) is planar. If p ≥ p − p ≥ γ (Γ c ( R )) = ( p + p + 1) (cid:24) ( p − p − p − p − (cid:25) . Since p ≥ ( p − p − p − p − ≥ and so γ (Γ c ( R )) ≥ c ( R ) isneither planar nor toroidal. Case 2: Γ c ( R ) = l K p − p ⊔ l K p − p By (1.2) we have γ (Γ c ( R )) = l γ ( K p − p ) ⊔ l γ ( K p − p ). If p = 2 then p − p = 4and p − p = 6. Also, l + 3 l = 7 which gives l = 4 and l = 1 or l = 1 and l = 2.Therefore, γ (Γ c ( R )) = 4 γ ( K ) + γ ( K ) = 1 or γ (Γ c ( R )) = γ ( K ) + 2 γ ( K ) = 2. That s, Γ c ( R ) is toroidal or γ (Γ c ( R )) = 2. If p ≥ p − p ≥
18 and p − p ≥
24. By(1.1) we have γ (Γ c ( R )) = l (cid:24) ( p − p − p − p − (cid:25) + l (cid:24) ( p − p − p − p − (cid:25) . Note that ( p − p − p − p − ≥ and ( p − p − p − p − ≥
35. Therefore, γ (Γ c ( R )) ≥ l + 35 l >
53. That is, Γ c ( R ) is neither planar nor toroidal. This completes theproof of part (a).(b) By [10, Theorem 2.5] we have Γ c ( R ) = ( p + 1) K p − p . By (1.2) we have γ (Γ c ( R )) = ( p + 1) γ ( K p − p ). If p ≥ p − p ≥ γ (Γ c ( R )) = ( p + 1) (cid:24) ( p − p − p − p − (cid:25) . Note that ( p − p − p − p − ≥ and so γ (Γ c ( R )) ≥
6. Thus Γ c ( R ) is neither planarnor toroidal. This completes the proof of part (b). Theorem 2.3.
Let R be a non-commutative ring where | R | = p q and Z ( R ) = { } . (a) Let t ∈ { p, q, p , pq } and ( t − | ( p q − .(i) Γ c ( R ) is planar if t = p = q = 2 ; or t = p = 4 and q ≥ ; or t = p = 2 , , and q ≥ ; or t = q = 2 , , and p ≥ .(ii) If t = p ≥ and q ≥ ; or t = q ≥ and p ≥ ; or p ≥ , q ≥ and t = p or pq then γ (Γ c ( R )) = p q − t − l ( t − t − m ; and hence Γ c ( R ) isneither planar nor toroidal. (b) Let l ( p −
1) + l ( q −
1) + l ( p −
1) + l ( pq −
1) = p q − for some positiveintegers l , l , l and l .(i) If p = 2 = q then Γ c ( R ) is planar.(ii) If p = 2 and q = 3 then γ (Γ c ( R )) = l ; and hence Γ c ( R ) is not planar buttoroidal if and only if l = 1 .(iii) If p = 2 and q ≥ then γ (Γ c ( R )) = l l ( q − q − m + l l (2 q − q − m .(iv) If q = 2 and p = 3 then γ (Γ c ( R )) = 2 l + l .(v) If q = 2 and p ≥ then γ (Γ c ( R )) = l l ( p − p − m + l l ( p − p − m + l l (2 p − p − m .(vi) If p = 3 = q then γ (Γ c ( R )) = 2( l + l ) .(vii) If p = 3 and q ≥ then γ (Γ c ( R )) = l l ( q − q − m +2 l + l l (3 q − q − m .(viii) If p ≥ and q = 3 then γ (Γ c ( R )) = l l ( p − p − m + l l ( p − p − m + l l (3 p − p − m .(ix) If p ≥ and q ≥ then γ (Γ c ( R )) = l l ( p − p − m + l l ( q − q − m + l l ( p − p − m + l l ( pq − pq − m . t follows that Γ c ( R ) is neither planar nor toroidal in all the cases ( iii ) − ( ix ) .Proof. (a) By [9, Theorem 2.9] we have Γ c ( R ) = p q − t − K t − . By (1.2) we have γ (Γ c ( R )) = p q − t − γ ( K t − ). Case 1: p = 2 = q In this case t = 2, since ( t − | ( p q −
1) = 7. Therefore, γ (Γ c ( R )) = 7 γ ( K ) = 0.That is, Γ c ( R ) is planar. Case 2: p = 2 and q ≥ q − ∤ (4 q −
1) and (2 q − ∤ (4 q − t = q and t = pq = 2 q .If t = p = 2 or t = p = 4 and q ≥ t − γ (Γ c ( R )) = 0. Thatis, Γ c ( R ) is planar. Case 3: p ≥ q = 2We have ( p − ∤ (2 p − p − ∤ (2 p −
1) and (2 p − ∤ (2 p − t = p , p and 2 p = pq . If t = q = 2 then t − γ (Γ c ( R )) = 0. That is,Γ c ( R ) is planar. Case 4: p ≥ q ≥ t = p = 3 or 5 then t − γ (Γ c ( R )) = 0. That is, Γ c ( R ) is planar.If t = p ≥ t − ≥
6. Therefore, by (1.1) we have γ (Γ c ( R )) = p q − t − l ( t − t − m . Since ( t − t − ≥ and p q − t − > γ (Γ c ( R )) >
2. That is, Γ c ( R ) is neitherplanar nor toroidal. If t = p then t − ≥
8. Therefore, by (1.1) we have γ (Γ c ( R )) = p q − t − l ( t − t − m . Since ( t − t − ≥ and p q − t − > γ (Γ c ( R )) >
4. That is, Γ c ( R ) is neitherplanar nor toroidal. If t = q = 3 or 5 then t − γ (Γ c ( R )) = 0. Thatis, Γ c ( R ) is planar. If t = q ≥ t − ≥
6. Therefore, by (1.1) we have γ (Γ c ( R )) = p q − t − l ( t − t − m . Since ( t − t − ≥ and p q − t − > γ (Γ c ( R )) >
2. That is, Γ c ( R ) is neitherplanar nor toroidal. If t = pq then t − ≥
8. Therefore, by (1.1) we have γ (Γ c ( R )) = p q − t − l ( t − t − m . Since ( t − t − ≥ and p q − t − > γ (Γ c ( R )) >
4. That is, Γ c ( R ) is neitherplanar nor toroidal.(b) By [9, Theorem 2.9] we have Γ c ( R ) = l K p − ⊔ l K q − ⊔ l K p − ⊔ l K pq − .By (1.2) we have γ (Γ c ( R )) = l γ ( K p − ) + l γ ( K q − ) + l γ ( K p − ) + l γ ( K pq − ). Case 1: p = 2 = q In this case γ (Γ c ( R )) = l γ ( K ) + l γ ( K ) + l γ ( K ) + l γ ( K ) = 0. Therefore,Γ c ( R ) is planar. Case 2: p = 2 and q ≥ γ (Γ c ( R )) = l γ ( K q − ) + l γ ( K q − ). If q = 3 then γ (Γ c ( R )) = l γ ( K ) + l γ ( K ) = l . Therefore, Γ c ( R ) is not planar since l = 0 and Γ c ( R ) istoroidal if l = 1. If q ≥ q − ≥ q − ≥
9. Therefore, by (1.1) we have γ (Γ c ( R )) = l l ( q − q − m + l l (2 q − q − m . ince ( q − q − ≥ (2 q − q − ≥ we have γ (Γ c ( R )) ≥ l ≥
3. That is, Γ c ( R )is neither planar nor toroidal. Case 3: q = 2 and p ≥ γ (Γ c ( R )) = l γ ( K p − ) + l γ ( K p − ) + l γ ( K p − ). If p = 3 then γ (Γ c ( R )) = l γ ( K ) + l γ ( K ) + l γ ( K ) = 2 l + l ≥
3. Therefore, Γ c ( R ) is neitherplanar nor toroidal. If p ≥ p − ≥ p − ≥
24 and 2 p − ≥
9. Therefore,by (1.1) we have γ (Γ c ( R )) = l l ( p − p − m + l l ( p − p − m + l l (2 p − p − m .Since ( p − p − ≥ ( p − p − ≥
35 and (2 p − p − ≥ we have γ (Γ c ( R )) ≥ l + 3 l ≥
38. That is, Γ c ( R ) is neither planar nor toroidal. Case 4: p ≥ q ≥ p = 3 = q then γ (Γ c ( R )) = l γ ( K ) + l γ ( K ) = 2( l + l ) ≥
4. Therefore, Γ c ( R )is neither planar nor toroidal. If p = 3 and q ≥ γ (Γ c ( R )) = l γ ( K ) + l γ ( K q − ) + l γ ( K ) + l γ ( K q − )= l l ( q − q − m + 2 l + l l (3 q − q − m . Since ( q − q − ≥ (3 q − q − ≥ we have γ (Γ c ( R )) ≥ l + 10 l ≥
12. Thatis, Γ c ( R ) is neither planar nor toroidal. If q = 3 and p ≥ γ (Γ c ( R )) = l γ ( K p − ) + l γ ( K ) + l γ ( K p − ) + l γ ( K p − )= l l ( p − p − m + l l ( p − p − m + l l (3 p − p − m . Since ( p − p − ≥ ( p − p − ≥
35 and (3 q − q − ≥ we have γ (Γ c ( R )) ≥ l + 10 l ≥
45. That is, Γ c ( R ) is neither planar nor toroidal. If p ≥ q ≥ γ (Γ c ( R )) = l l ( p − p − m + l l ( q − q − m + l l ( p − p − m + l l ( pq − pq − m . Since ( p − p − ≥ ( q − q − ≥ ( p − p − ≥
35 and ( pq − pq − ≥
35 we have γ (Γ c ( R )) ≥ l + 35 l ≥
70. That is, Γ c ( R ) is neither planar nor toroidal. Theorem 2.4.
Let R be a non-commutative ring with unity where | R | = p q and | Z ( R ) | = pq . (a) If p = 2 = q then Γ c ( R ) is planar. (b) If p = 2 and q ≥ then γ (Γ c ( R )) = 3 l (2 q − q − m . (c) If q = 2 and p ≥ then γ (Γ c ( R )) = ( p + 1) l (2 p − p − p − p − m . (d) If p ≥ and q ≥ then γ (Γ c ( R )) = ( p + 1) l ( p q − pq − p q − pq − m .It follows that Γ c ( R ) is neither planar nor toroidal in all the cases ( b ) − ( d ) . roof. By [9, Theorem 2.12] we have Γ c ( R ) = ( p + 1) K p q − pq . By (1.2) we have γ (Γ c ( R )) = ( p + 1) γ ( K p q − pq ). Case 1: p = 2 = q In this case γ (Γ c ( R )) = 3 γ ( K ) = 0. Therefore, Γ c ( R ) is planar. Case 2: p = 2 and q ≥ γ (Γ c ( R )) = 3 γ ( K q ) and 2 q ≥
6. Therefore, by (1.1) we have γ (Γ c ( R )) = 3 l (2 q − q − m .Since (2 q − q − ≥ we have γ (Γ c ( R )) ≥
3. That is, Γ c ( R ) is neither planar nortoroidal. Case 3: q = 2 and p ≥ γ (Γ c ( R )) = ( p + 1) γ ( K p − p ) and 2 p − p ≥
12. Therefore,by (1.1) we have γ (Γ c ( R )) = ( p + 1) l (2 p − p − p − p − m .Since (2 p − p − p − p − ≥ p + 1 ≥ γ (Γ c ( R )) ≥
24. That is, Γ c ( R )is neither planar nor toroidal. Case 4: p ≥ q ≥ p q − pq ≥
18. Therefore, by (1.1) we have γ (Γ c ( R )) = ( p + 1) l ( p q − pq − p q − pq − m .Since ( p q − pq − p q − pq − ≥ and p + 1 ≥ γ (Γ c ( R )) ≥
72. That is, Γ c ( R )is neither planar nor toroidal. Theorem 2.5.
Let R be a non-commutative ring where | R | = p q and | Z ( R ) | = p . (a) Let ( q − | ( pq − .(i) If p = 2 = q then Γ c ( R ) is planar.(ii) If q = 2 and p ≥ then γ (Γ c ( R )) = 2 p − l ( p − p − m .(iii) If p ≥ and q ≥ then γ (Γ c ( R )) = pq − q − l ( p q − p − p q − p − m .It follows that Γ c ( R ) is neither planar nor toroidal in the cases ( ii ) and ( iii ) . (b) Let ( p − | ( pq − .(i) If p = 2 and q ≥ then Γ c ( R ) is planar.(ii) If p ≥ and q ≥ then γ (Γ c ( R )) = pq − p − l ( p − p − p − p − m ; and hence Γ c ( R ) is neither planar nor toroidal. (c) Let l ( p −
1) + l ( q −
1) = pq − , where l and l are positive integers.(i) If p = 2 = q then Γ c ( R ) is planar.(ii) If p = 2 and q ≥ then γ (Γ c ( R )) = l l (4 q − q − m .(iii) If q = 2 and p ≥ then γ (Γ c ( R )) = l l ( p − p − p − p − m + l l ( p − p − m . iv) If p ≥ and q ≥ then γ (Γ c ( R )) = l l ( p − p − p − p − m + l l ( p q − p − p q − p − m .It follows that Γ c ( R ) is neither planar nor toroidal in all the cases ( ii ) − ( iv ) .Proof. (a) By [9, Theorem 2.12] we have Γ c ( R ) = pq − q − K p q − p . By (1.2) we have γ (Γ c ( R )) = pq − q − γ ( K p q − p ). We shall complete the proof by considering the followingcases. Note that the case p = 2 and q ≥ q − ∤ (2 q − Case 1: p = 2 = q In this case we have Γ c ( R ) = 3 K and so γ (Γ c ( R )) = 0. That is, Γ c ( R ) is planar. Case 2: q = 2 and p ≥ γ (Γ c ( R )) = (2 p − γ ( K p ) where p ≥
9. Therefore, by (1.1) we have γ (Γ c ( R )) = (2 p − l ( p − p − m .Since ( p − p − ≥ and 2 p − ≥ γ (Γ c ( R )) >
15. That is, Γ c ( R ) isneither planar nor toroidal. Case 3: p ≥ q ≥ p q − p ≥
18. Therefore, by (1.1) we have γ (Γ c ( R )) = pq − q − l ( p q − p − p q − p − m .Since ( p q − p − p q − p − ≥ and pq − q − > γ (Γ c ( R )) >
36. That is, Γ c ( R )is neither planar nor toroidal.(b) By [9, Theorem 2.12] we have Γ c ( R ) = pq − p − K p − p . By (1.2) we have γ (Γ c ( R )) = pq − p − γ ( K p − p ). We shall complete the proof by considering the followingcases. Note that the case q = 2 and p ≥ p − ∤ (2 p − Case 1: p = 2 and q ≥ γ (Γ c ( R )) = (2 q − γ ( K ) = 0. That is, Γ c ( R ) is planar. Case 2: p ≥ q ≥ p − p ≥
18. Therefore, by (1.1) we have γ (Γ c ( R )) = pq − p − l ( p − p − p − p − m .Since ( p − p − p − p − ≥ and pq − p − > γ (Γ c ( R )) >
36. That is, Γ c ( R )is neither planar nor toroidal.(c) By [9, Theorem 2.12] we have Γ c ( R ) = l K p − p ⊔ l K p q − p . By (1.2) we have γ (Γ c ( R )) = l γ ( K p − p ) + l γ ( K p q − p ). Case 1: p = 2 = q In this case γ (Γ c ( R )) = l γ ( K ) + l γ ( K ) = 0. Therefore, Γ c ( R ) is planar. Case 2: p = 2 and q ≥ γ (Γ c ( R )) = l γ ( K ) + l γ ( K q − ) = l γ ( K q − ). We have 4 q − ≥ γ (Γ c ( R )) = l l (4 q − q − m . ince (4 q − q − ≥ we have γ (Γ c ( R )) ≥ l ≥
2. That is, Γ c ( R ) is neither planarnor toroidal. Case 3: q = 2 and p ≥ γ (Γ c ( R )) = l γ ( K p − p ) + l γ ( K p ). Since p − p ≥
18 and p ≥ γ (Γ c ( R )) = l l ( p − p − p − p − m + l l ( p − p − m .Since ( p − p − p − p − ≥ and ( p − p − ≥ we have γ (Γ c ( R )) ≥ l + 3 l ≥
21. That is, Γ c ( R ) is neither planar nor toroidal. Case 4: p ≥ q ≥ p − p ≥
18 and p q − p ≥
18. Therefore, by (1.1) we have γ (Γ c ( R )) = l l ( p − p − p − p − m + l l ( p q − p − p q − p − m . Since ( p − p − p − p − ≥ and ( p q − p − p q − p − ≥ we have γ (Γ c ( R )) ≥ l + 18 l ≥
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