350 

© 2024 The Author(s). Published by College of Education for Pure Science (Ibn Al-Haitham), 

University of Baghdad. This is an open-access article distributed under the terms of the Creative Commons 

Attribution 4.0 International License 

 

 

Ibn Al-Haitham Journal for Pure and Applied Sciences 

Journal homepage: jih.uobaghdad.edu.iq 

PISSN: 1609-4042, EISSN: 2521-3407 

 IHJPAS. 2024, 37(4) 

 

Spin Characters' Decomposition Matrices of 𝐒𝟐𝟕, 𝐒𝟐𝟖modulo, 𝐩 =13 

 

Ahmed H. Jassim1,*  and  Saeed A. Taban2  
1,2 Department of Mathematics, College of Science, Basrah  University, Basrah, Iraq. 

*Corresponding Author. 

 

  Received: 23 March 2023                       Accepted: 12 June 2023                 Published: 20 October 2024 

 doi.org/10.30526/37.4.3360 

 
 Abstract   

        In this study, when the field characteristic is 13, we calculate decomposition matrices for 

the spin characters  S27 and S28  which are broken down into blocks, where the decomposition 

matrices are connected between irreducible spin characters and irreducible modular spin 

characters. The technique used in this study is (r, r̅)-inducing, which produces projective 

characters for symmetric group S27 by projecting S26's character, and symmetric group S28 by 

projecting S27's character. We can find it by fixing all bar divisions, finding all irreducible spin 

characters for S27 (S28), p = 13, and all irreducible modular spin characters for S27 (S28), p =

13. In order to explore irreducible modular spin characteristics, general correlations and 

theorems will be discovered as a result of this research.  

Keywords: Decomposition Matrix, Irreducible Modular Spin Character, projective character. 

 

1. Introduction 

Symmetric group Sn has a representation group S̅nwith a central Z = {−1,1} such that 

S̅n/Z ≅  Sn. The representations which do not have Z in their kernel are called the spin 

representations of Sn for more information, see [1-3]. The spin characters of the spin 

representations of Sn are labelled by the distinct parts of the partitions of n and denoted by 〈α〉. 

In fact, if  α = (α1, α2, . . . , αm) is partition of n and n − m is even, then there is one irreducible 

spin character denoted by 〈α〉∗ which is self-associate(double), and if n − m is odd, then there 

are two associate spin characters denoted by 〈α〉 and 〈α〉′see [4-6]. The number of rows and 

columns of decomposition matrix corresponds to the number of projective characters and (p, α)-

regular classes, respectively [3]. In this study we found the decomposition matrices of spin 

characters for S27 and S28 modulo p = 13. The distribution of the spin characters into p-blocks 

is accomplished using the (r, r̅)-inducing (restricting) approach [7,8]. Numerous people conduct 

research on this subject, have contributed to this field of study [9-22]. Before we declare any 

results, let's define certain notations and terminologies. "p.s." is the principal spin character 

("p.i.s." indecomposable), "m.s." is means modular spin character ("i.m.s." irreducible), "di" is 

p.i.s. of Sn, "Di" is p.i.s. of Sn−1, and "〈⬚〉no" is the number of i.m.s. 

https://creativecommons.org/licenses/by/4.0/
https://creativecommons.org/licenses/by/4.0/
https://orcid.org/0000-0001-5581-8326
mailto:ahmed_h.jassim@uobasrah.edu.iq
https://orcid.org/0000-0003-2133-6817
mailto:saeed.taban@uobasrah.edu.iq


IHJPAS. 2024, 37(4)  

 

351 

 

1. Preliminaries 

      For the study, some important conclusions were needed. 

Theorem 2.1. Degree of the spin character 〈𝛼1, … , 𝛼m〉 = 2[(𝑛−𝑚)/2] 𝑛!

∏𝑚
𝑖=1 𝛼𝑖!

∏1≤𝑖<𝑗≤𝑚
(𝛼𝑖−𝛼𝑗)

(𝛼𝑖+𝛼𝑗)
 [1]. 

Theorem 2.2. Given that 𝑏 is the number of 𝑝-conjugate characters to the irreducible ordinary 

character 𝜒 of G and that 𝐵 is an ablock of defect one, then:   

a. ∃𝑁 ∈ ℤ+ such that the irreducible ordinary characters lying in the block 𝐵 can be 

partitioned into two disjoint classes:𝐵1={𝜒 ∈ 𝐵   |𝑏 deg 𝑥 ≡ 𝑁𝑚𝑜𝑑𝑝𝑎}, 𝐵2={𝜒 ∈
𝐵  |𝑏 deg 𝑥 ≡ −𝑁𝑚𝑜𝑑𝑝𝑎} 

b. The block 𝐵's decomposition matrix has coefficients that are either 1 or 0 [23]. 

Theorem 2.3. Let𝐺 be a group of order |𝐺| = 𝑚𝑜𝑝𝑎, where (𝑝, 𝑚𝑜) = 0. If 𝑐 is a principal 

character of sub group 𝐻 of 𝐺, then deg 𝑐 ≡ 0 𝑚𝑜𝑑𝑝𝑎 [24,25]. 

Theorem 2.4. Let 𝑝 be odd then 

1. If 𝑛 be even, 𝑝 ∤ 𝑛, then 〈𝑛〉 = 𝜑〈𝑛〉 and 〈𝑛〉′ = 𝜑〈𝑛〉′ are distinct irreducible modular spin 

characters. 

2. If 𝑛 is odd, 𝑝 ∤ 𝑛 or 𝑝 ∤ (𝑛 − 1), then 〈𝑛 − 1,1〉 and 〈𝑛 − 1,1〉′ are distinct irreducible 

modular spin characters of degree 2[(𝑛−3)/2] × (𝑛 − 2) which are denoted by 𝜑〈𝑛 − 1,1〉 
and 𝜑〈𝑛 − 1,1〉′respectively [2]. 

 

2. Decomposition matrix for �̅�𝟐𝟕 

      The decomposition matrix for 𝑆2̅7 of degree (288,253), and it is decomposed in to blocks of 

character it consists of 69 blocks which 𝐵1of defect two, 𝐵2 , 𝐵3, . . . , 𝐵16 are defect one, and the 

remaining blocks are defect zero, decomposition matrix is equal to 𝐵1⨁𝐵2⨁. . . ⨁𝐵69. 

 

Lemma 3.1. Decomposition matrix for the block 𝑩𝟏 of type double as shown in the Table 1. 

 

Table 1. Block 𝐵1 

Spin characters Decomposition matrix 

〈27〉∗ 1                          
〈26,1〉 1 1                         

〈24,2,1〉∗  1 1                        
〈23,3,1〉∗   1 1                       
〈22,4,1〉∗    1 1                      
〈21,5,1〉∗     1 1                     
〈20,6,1〉∗      1                     
〈19,7,1〉∗      1 1                    
〈18,8,1〉∗     1 1 1 1                   
〈17,9,1〉∗    1 1   1 1                  

〈16,10,1〉∗   1 1     1 1                 
〈15,11,1〉∗  1 1       1 1                

〈14,13〉 1 1          1               
〈14,12,1〉∗ 2 1         1 2 2              
〈14,11,2〉∗          1 1 1 2 1             
〈14,10,3〉∗         1 1    1 1            
〈14,9,4〉∗        1 1      1 1           
〈14,8,5〉∗       1 1        1 1          
〈14,7,6〉∗       1          1          

〈13,11,2,1〉            1 1 1    1         
〈13,10,3,1〉             1 1 1   1 1        
〈13,9,4,1〉               1 1   1 1       
〈13,8,5,1〉                1 1   1 1      
〈13,7,6,1〉                 1    1      

〈11,10,3,2,1〉∗             1      1   1     



IHJPAS. 2024, 37(4)  

 

352 

 

〈11,9,4,2,1〉∗                  1 1 1  1 1    
〈11,8,5,2,1〉∗                    1 1  1 1   
〈11,7,6,2,1〉∗                     1   1   
〈10,9,4,3,1〉∗                  1     1  1  
〈10,8,5,3,1〉∗                      1 1 1 1 1 
〈10,7,6,3,1〉∗                        1  1 
〈9,8,5,4,1〉∗                      1    1 
〈9,7,6,4,1〉∗                         1 1 
〈8,7,6,5,1〉∗                         1  

 𝑑1 𝑑2 𝑑3 𝑑4 𝑑5 𝑑6 𝑑7 𝑑8 𝑑9 𝑑10 𝑑11 𝑑12 𝑑13 𝑑14 𝑑15 𝑑16 𝑑17 𝑑18 𝑑19 𝑑20 𝑑21 𝑑22 𝑑23 𝑑24 𝑑25 𝑑26 

 

Proof: By using (0,1)-inducing of p.i.s. method on 𝐷1 in 𝑆27  we have  

𝐷1 ↑(0,1) 𝑆27 = 〈26〉 + 〈26〉′ + 〈25,1〉∗ + 2〈14,12〉∗ + 〈13,1,2,1〉 + 〈13,12,1〉′ ↑(0,1) 𝑆27 

 = 2〈27〉∗ + 2〈26,1〉 + 2〈26,1〉′ + 2〈14,13〉 + 2〈14,13〉′ + 4〈14,12,1〉∗ 

 = 2𝑑1 

similarly, using (r, �̅�)-inducing of p.i.s. 𝐷2, 𝐷3, 𝐷4, 𝐷5, 𝐷6, 𝐷46, 𝐷40, 𝐷34, 𝐷28, 𝐷52, 𝐷12, 𝐷13, 

𝐷29,  𝐷35, 𝐷41, 𝐷47, 𝐷18, 𝐷19, . . . , 𝐷26 of 𝑆26 to 𝑆27 we get on 𝑑2, 𝑑3, … , 𝑑26 respectively, and on 

(13, 𝛼)-regular classes we have   

1. 〈26,1〉 = 〈26,1〉′ 

2. 〈14,13〉 = 〈14,13〉′ 

3. 〈13,11,2,1〉 = 〈13,11,2,1〉′ 

4. 〈13,10,3,1〉 = 〈13,10,3,1〉′ 

5. 〈13,9,4,1〉 = 〈13,9,4,1〉′ 

6. 〈13,8,5,1〉 = 〈13,8,5,1〉′ 

7. 〈13,7,6,1〉 = 〈13,7,6,1〉′ 

8. 〈21,5,1〉∗ = 〈20,6,1〉∗ + 〈22,4,1〉∗ − 〈23,3,1〉∗ + 〈24,2,1〉∗ − 〈26,1〉 + 〈27〉∗ 
9. 〈14,8,5〉∗ = 〈14,7,6〉∗ + 〈14,9,4〉∗ − 〈14,10,3〉∗ + 〈14,11,2〉∗ − 〈14,12,1〉∗ + 〈14,13〉 + 〈27〉∗ 

10. 〈13,8,5,1〉 = 〈13,7,6,1〉 + 〈13,9,4,1〉 − 〈13,10,3,1〉 + 〈13,11,2,1〉 − 〈14,13〉 + 〈26,1〉 
11. 〈11,8,5,2,1〉∗ = 〈11,7,6,2,1〉∗ + 〈11,9,4,2,1〉∗ − 〈11,10,3,2,1〉∗ − 〈13,11,2,1〉 +

〈14,11,2〉∗ − 〈15,11,1〉∗ + 〈24,2,1〉∗ 

Table 2. Block 𝐵2, 𝐵3 

Block Spin characters Decomposition matrix 

𝐵2 

〈25,2〉 1                        

〈25,2〉′  1                       
〈15,12〉 1  1                      
〈15,12〉′  1  1                     

〈13,12,2〉∗   1 1 1 1                   

〈12,10,3,2〉     1  1                  
〈12,10,3,2〉′      1  1                 

〈12,9,4,2〉       1  1                

〈12,9,4,2〉′        1  1               
〈12,8,5,2〉         1  1              

〈12,8,5,2〉′          1  1             

〈12,7,6,2〉           1              
〈12,7,6,2〉′            1             

𝐵3 

〈24,3〉             1            
〈24,3〉′              1           
〈16,11〉             1  1          
〈16,11〉′              1  1         

〈13,11,3〉∗               1 1 1 1       
〈12,11,3,1〉                 1  1      
〈12,11,3,1〉′                  1  1     
〈11,9,4,3〉                   1  1    



IHJPAS. 2024, 37(4)  

 

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12. 〈10,8,5,3,1〉∗ = 〈10,7,6,3,1〉∗ + 〈10,9,4,3,1〉∗ + 〈11,10,3,2,1〉∗ − 〈13,10,3,1〉 +
〈14,10,3〉∗ − 〈16,10,1〉∗ + 〈23,3,1〉∗ 

13. 〈9,8,5,4,1〉∗ = 〈10,8,5,3,1〉∗ − 〈8,7,6,5,1〉∗ − 〈11,8,5,2,1〉∗ + 〈13,8,5,1〉 − 〈14,8,5〉∗ +
〈18,8,1〉∗ − 〈21,5,1〉∗ 

14. 〈10,9,4,3,1〉∗ = 〈8,7,6,5,1〉∗ + 〈11,9,4,2,1〉∗ − 〈11,10,3,2,1〉∗ − 〈13,9,4,1〉 +
〈13,10,3,1〉 + 〈14,9,4〉∗ − 〈14,10,3〉∗ + 〈16,10,1〉∗ − 〈17,9,1〉∗ + 〈22,4,1〉∗ − 〈23,3,1〉∗ 

15. 〈10,7,6,3,1〉∗ = 〈9,7,6,4,1〉∗ − 〈8,7,6,5,1〉∗ + 〈11,7,6,2,1〉∗ − 〈13,7,6,1〉 − 〈14,9,4〉∗ +
〈19,7,1〉∗ − 〈20,6,1〉∗ 

then the matrix contains at most 41 columns since the number of the i.m.s. is equal or less than the 

number of the spin characters, but Table 1 contains at most 26 columns since there are 15 equations 

corresponding the spin characters of 𝑆27 in 𝑩𝟏, and because 𝑑𝑖 − 𝑑𝑗  is not p.s. to 𝑆27 ∀1 ≤ 𝑖 <

𝑗 ≤ 26, and 𝑑1, 𝑑2, ⋯ , 𝑑26 are linearly independent, then we get Table 1. 

 

Lemma 3.2. The blocks 𝑩𝟐, 𝑩𝟑 of type associate as shown in the Table 2. 

 

 

 

 

 

 

Proof:. By using (r, �̅�)-inducing of p.i.s. 𝐷3, 𝐷10, 𝐷14, 𝐷15, 𝐷16, 𝐷17, 𝐷2, 𝐷9, 𝐷12, 𝐷168, 𝐷169, 𝐷20, 

𝐷21 of 𝑆26 to 𝑆27we get on 𝑘1, 𝑘2, . . . , 𝑘9, 𝑑45,𝑑46, 𝑘10, 𝑘11respectively. Since 〈25,2〉 ≠ 〈25,2〉′ are 

distinct irreducible modular spin characters then 𝑘1 must split to 𝑑27, 𝑑28, also since 𝑩𝟐 of defect 

one then from (theorem 2.2) 𝑘2,𝑘3 must splits to 𝑑30, 𝑑31 and 𝑑32, 𝑑33, respectively. Since 

〈12,9,4,2〉 ≠ 〈12,9,4,2〉′ so 𝑘4 or 𝑘5 is split. If 𝑘4 is split to 𝑑35, 𝑑36,but 〈12,8,5〉 ≠ 〈12,8,5〉′ then 

𝑘5 split to, 𝑑37, 𝑑38. If 𝑘5 is split and from  (13, 𝛼)-regular classes, 

〈12,9,4,2〉 + 〈12,7,6,2〉 − 〈12,8,5,2〉 ≠ 〈12,9,4,2〉′ + 〈12,7,6,2〉′ − 〈12,8,5,2〉′ (1) 

then 𝑘4 must split, so in both cases we get 𝑘4 and 𝑘5  are splits. Since〈12,7,6,2〉 ≠ 〈12,7,6,2〉′ then 

𝑘6 must split to 𝑑37, 𝑑38. For 𝑩𝟑 since 〈24,3〉 ≠ 〈24,3〉′ then 𝑘7 must split to 𝑑39,  𝑑40, also since 

𝑩𝟑 of defect one then 𝑘8,𝑘9 must splits to 𝑑41, 𝑑42 and 𝑑43, 𝑑44, respectively. Since 〈11,8,5,3〉 ≠

〈11,8,5,3〉′ so 𝑘10 or 𝑘11 is split. If 𝑘9 is split to 𝑑47, 𝑑48, but 〈11,7,6,3〉 ≠ 〈11,7,6,3〉′ then 𝑘10 split 

to, 𝑑49, 𝑑50. If 𝑘10 is split , from (13, 𝛼)-regular classes, 

〈11,8,5,3〉 − 〈11,7,6,3〉 ≠ 〈11,8,5,3〉′ − 〈11,7,6,3〉′ (2) 

then 𝑘9 must split,so in both cases we get 𝑘9 and 𝑘10 are splits,then we get Table 2. 

Lemma 3.3. The blocks 𝑩𝟒, 𝑩𝟓 of type associate as shown in the Table 3. 

 

 

 

 

〈11,9,4,3〉′                    1  1   
〈11,8,5,3〉                     1  1  
〈11,8,5,3〉′                      1  1 
〈11,7,6,3〉                       1  
〈11,7,6,3〉′                        1 

  𝑑27 𝑑28 𝑑29 𝑑30 𝑑31 𝑑32 𝑑33 𝑑34 𝑑35 𝑑36 𝑑37 𝑑38 𝑑39 𝑑40 𝑑41 𝑑42 𝑑43 𝑑44 𝑑45 𝑑46 𝑑47 𝑑48 𝑑49 𝑑50 



IHJPAS. 2024, 37(4)  

 

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Table 3. Blocks𝐵4, 𝐵5 

Block Spin characters Decomposition matrix 

𝐵4 

〈23,4〉 1                        

〈23,4〉′  1                       

〈17,10〉 1  1                      
〈17,10〉′  1  1                     

〈13,10,4〉∗   1 1 1 1                   

〈12,10,4,1〉     1  1                  
〈12,10,4,1〉′      1  1                 

〈11,10,4,2〉       1  1                

〈11,10,4,2〉′        1  1               
〈10,8,5,4, 〉         1  1              
〈10,8,5,4〉′          1  1             

〈10,7,6,4〉           1              
〈10,7,6,4〉′            1             

𝐵5 

〈22,5〉             1            
〈22,5〉′              1           
〈18,9〉             1  1          
〈18,9〉′              1  1         

〈13,9,5〉∗               1 1 1 1       
〈12,9,5,1〉                 1  1      
〈12,9,5,1〉′                  1  1     
〈11,9,5,2〉                   1  1    
〈11,9,5,2〉′                    1  1   
〈10,9,5,3〉                     1  1  
〈10,9,5,3〉′                      1  1 
〈9,7,6,5〉                       1  
〈9,7,6,5〉′                        1 

  𝑑51 𝑑52 𝑑53 𝑑54 𝑑55 𝑑56 𝑑57 𝑑58 𝑑59 𝑑60 𝑑61 𝑑62 𝑑63 𝑑64 𝑑65 𝑑66 𝑑67 𝑑68 𝑑69 𝑑70 𝑑71 𝑑72 𝑑73 𝑑74 

 

Proof:. By using (r, �̅�)-inducing of  p.i.s.𝐷3, 𝐷8, 𝐷170, 𝐷171, 𝐷13, 𝐷23, 𝐷24,𝐷4, 𝐷7, 𝐷172, 𝐷173, 𝐷19, 

𝐷22, 𝐷26 of 𝑆26 to 𝑆27we get on 𝑘1, 𝑘2, 𝑑55,𝑑56, 𝑘3, 𝑘4, 𝑘5,𝑘6, 𝑘7, 𝑑67,𝑑68, 𝑘8, 𝑘9, 𝑘10 respectively. 

Since 〈23,4〉 ≠ 〈23,4〉′ then 𝑘1 must split to 𝑑51, 𝑑52, also since 𝑩𝟒 of defect one then 𝑘2 must 

split to 𝑑53, 𝑑54. 〈11,10,4,2〉 ≠ 〈11,10,4,2〉′ so 𝑘3 or 𝑘4 is split. If 𝑘3 is split to 𝑑57, 𝑑58, but 

〈10,8,5,4〉 ≠ 〈10,8,5,4〉′ then 𝑘4 split to, 𝑑59, 𝑑60. If 𝑘4 is split and from (13, 𝛼)-regular classes, 

〈11,10,4,2〉 + 〈10,7,6,4〉 − 〈10,8,5,4〉 ≠ 〈11,10,4,2〉′ + 〈10,7,6,4〉′ − 〈10,8,5,4〉′ (3) 

then 𝑘3 must split, so in both cases we get 𝑘3 , 𝑘4  are splits. Since〈10,7,6,4〉 ≠ 〈10,7,6,4〉′ then 𝑘5 

split to 𝑑61, 𝑑62. In 𝑩𝟓 〈22,5〉 ≠ 〈22,5〉′ then 𝑘6 must split to 𝑑63, 𝑑64, also 𝑩𝟓 of defect one then 

𝑘7 must split to 𝑑65, 𝑑66. Since 〈11,9,5,2〉 ≠ 〈11,9,5,2〉′ so 𝑘8 or 𝑘9 is split. If 𝑘8  is split to 𝑑69, 

𝑑70, but 〈10,9,5,3〉 ≠ 〈10,9,5,3〉′ then 𝑘9 split to, 𝑑71, 𝑑72. If 𝑘9 is split and   

〈11,9,5,2〉 + 〈9,7,6,5〉 − 〈10,9,5,3〉 ≠ 〈11,9,5,2〉′ + 〈9,7,6,5〉′ − 〈10,9,5,3〉′ (4) 

then 𝑘8 must split,so in both cases we get 𝑘8 and 𝑘9  are splits. Finally. Since〈9,7,6,5〉 ≠ 〈9,7,6,5〉′ 

then 𝑘10 must split to 𝑑73, 𝑑74, so we get Table 3. 

Lemma 3.4. Blocks 𝑩𝟔, 𝑩𝟖 of type double andblock 𝑩𝟕 of type associateas given in Table 4. 

 

 

 

 

 

 

 



IHJPAS. 2024, 37(4)  

 

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Table 4. Blocks 𝐵6, 𝐵7, 𝐵8 

Block Spin characters Decomposition matrix 

𝐵6 

〈22,3,2〉∗ 1                        
〈16,9,2〉∗ 1 1                       
〈15,9,3〉∗  1 1                      
〈13,9,3,2〉   1 1                     

〈12,9,3,2,1〉∗    1 1                    
〈9,8,5,3,2〉∗     1 1                   
〈9,7,6,3,2〉∗      1                   

𝐵7 

〈21,6〉       1                  
〈21,6〉′        1                 
〈19,8〉       1  1                
〈19,8〉′        1  1               

〈13,8,6〉∗         1 1 1 1             
〈12,8,6,1〉           1  1            
〈12,8,6,1〉′            1  1           
〈11,8,6,2〉             1  1          
〈11,8,6,2〉′              1  1         
〈10,8,6,3〉               1  1        
〈10,8,6,3〉′                1  1       
〈9,8,6,4〉                 1        
〈9,8,6,4〉′                  1       

𝐵8 

〈21,4,2〉∗                   1      
〈17,8,2〉∗                   1 1     
〈15,8,4〉∗                    1 1    
〈13,8,4,2〉                     1 1   

〈12,8,4,2,1〉∗                      1 1  
〈10,8,4,3,2〉∗                       1 1 
〈8,7,6,4,2〉∗                        1 

  𝑑75 𝑑76 𝑑77 𝑑78 𝑑79 𝑑80 𝑑81 𝑑82 𝑑83 𝑑84 𝑑85 𝑑86 𝑑87 𝑑88 𝑑89 𝑑90 𝑑91 𝑑92 𝑑93 𝑑94 𝑑95 𝑑96 𝑑97 𝑑98 

 

Proof: Since   

• degree {〈16,9,2〉∗, 〈13,9,3,2〉 + 〈13,9,3,2〉′, 〈9,8,5,3,2〉∗} ≡ 156 mod 132 

• degree {〈22,3,2〉∗, 〈15,9,3〉∗, 〈12,9,3,2,1〉∗, 〈9,7,6,3,2〉∗} ≡ −156 mod 132,  

• degree {〈21,4,2〉∗, 〈15,8,5〉∗, 〈12,8,4,2,1〉∗, 〈8,7,6,4,2〉∗} ≡ 91 mod 132 

• degree {〈17,8,2〉∗, 〈13,8,4,2〉 + 〈13,8,4,2〉′, 〈10,8,4,3,2〉∗} ≡ −91 mod 132, 

and by(2,12)-inducing of p.i.s. 𝐷39, 𝐷41, 𝐷43, 𝐷45, 𝐷47, 𝐷49,𝐷51, 𝐷53, 𝐷55, 𝐷57, 𝐷59, 𝐷61 of  𝑆26 to  

S27, and on (13, 𝛼)-regular classes we have:   

1. 〈13,9,3,2〉 = 〈13,9,3,2〉′ 
2. 〈12,9,3,2,1〉∗ = 〈9,8,5,3,2〉∗ − 〈9,7,6,3,2〉∗ + 〈13,9,3,2〉 − 〈15,9,3〉∗ + 〈16,9,2〉∗ − 〈22,3,2〉∗ 

3. 〈13,8,4,2〉 = 〈13,8,4,2〉′ 
4. 〈12,8,4,2,1〉∗ = 〈10,8,4,3,2〉∗ − 〈8,7,6,4,2〉∗ + 〈13,8,4,2〉 − 〈15,8,4〉∗ + 〈17,8,2〉∗ − 〈21,4,2〉∗ 

then each blocks 𝑩𝟔, 𝑩𝟖 contains at most 6 columns, so we get 𝑩𝟔, 𝑩𝟖. To find block 𝑩𝟕 by using 

(r, �̅�)-inducing of p.i.s. 𝐷5, 𝐷8, 𝐷175, 𝐷176, 𝐷20, 𝐷23, 𝐷22 of 𝑆26 to 𝑆27 get on 𝑘1, 𝑘2, 𝑑85,𝑑86, 𝑘3, 

𝑘4, 𝑘5. Since 〈21,6〉 ≠ 〈21,6〉′ then 𝑘1 split to 𝑑81, 𝑑82, also since 𝑩𝟕 of defect one then 𝑘2 split to 

𝑑83, 𝑑84. Since 〈11,8,6,2〉 ≠ 〈11,8,6,2〉′ so 𝑘3 or 𝑘4 is split. If 𝑘3 is split to 𝑑87, 𝑑88, but 

〈10,8,6,3〉 ≠ 〈10,8,6,3〉′ then 𝑘4  split to, 𝑑89, 𝑑90. If 𝑘4 is split and 

〈11,8,6,2〉 + 〈9,8,6,4〉 − 〈10,8,6,3〉 ≠ 〈11,8,6,2〉′ + 〈9,8,6,4〉′ − 〈10,8,6,3〉′ (5) 

then 𝑘3  split, so in both cases we get 𝑘3 and 𝑘4  are splits. Finally. Since〈9,8,6,4〉 ≠ 〈9,8,6,4〉′ then 

𝑘5 must split to 𝑑91, 𝑑92, then we get Table 4. 

 

Lemma 3.5. Block 𝑩𝟗 of type associate, and 𝑩𝟏𝟎, 𝑩𝟏𝟏 of type double as shown in the Table 5. 



IHJPAS. 2024, 37(4)  

 

356 

 

Table 5. Blocks 𝐵9, 𝐵10, 𝐵11 

Block Spin characters Decomposition matrix 

𝐵9 

〈21,3,2,1〉 1                        

〈21,3,2,1〉′  1                       

〈16,8,2,1〉 1  1                      
〈16,8,2,1〉′  1  1                     

〈15,8,3,1〉   1  1                    

〈15,8,3,1〉′    1  1                   
〈14,8,3,2〉     1  1                  

〈14,8,3,2〉′      1  1                 

〈13,8,3,2,1〉∗       1 1 1 1               
〈9,8,4,3,2,1〉         1  1              
〈9,8,4,3,2,1〉′          1  1             

〈8,7,6,3,2,1〉           1              
〈8,7,6,3,2,1〉′            1             

𝐵10 

〈20,5,2〉∗             1            

〈18,7,2〉∗             1 1           

〈15,7,5〉∗              1 1          
〈13,7,5,2〉               1 1         

〈12,7,5,2,1〉∗                1 1        

〈10,7,5,3,2〉∗                 1 1       

〈9,7,5,4,2〉∗                  1       

𝐵11 

〈20,4,3〉∗                   1      
〈17,7,3〉∗                   1 1     
〈16,7,4〉∗                    1 1    
〈13,7,4,3〉                     1 1   

〈12,7,4,3,1〉∗                      1 1  
〈11,7,4,3,2〉∗                       1 1 
〈8,7,5,4,3〉∗                        1 

  

𝑑
9

9
 

𝑑
1

0
0
 

𝑑
1

0
1
 

𝑑
1

0
2
 

𝑑
1

0
3
 

𝑑
1

0
4
 

𝑑
1

0
5
 

𝑑
1

0
6
 

𝑑
1

0
7
 

𝑑
1

0
8
 

𝑑
1

0
9
 

𝑑
1

1
0
 

𝑑
1

1
1
 

𝑑
1

1
2
 

𝑑
1

1
3
 

𝑑
1

1
4
 

𝑑
1

1
5
 

𝑑
1

1
6
 

𝑑
1

1
7
 

𝑑
1

1
8
 

𝑑
1

1
9
 

𝑑
1

2
0
 

𝑑
1

2
1
 

𝑑
1

2
2
 

 

Proof: To find 𝑩𝟗 using (r, �̅�)-inducing of  p.i.s. 𝐷63, 𝐷64, 𝐷65, 𝐷66, 𝐷101, 𝐷69, 𝐷71, 𝐷72, 𝐷73, 𝐷74 

of 𝑆26 to 𝑆27we get on 𝑑99, 𝑑100, 𝑑101, 𝑑102, 𝑘1, 𝑘2, 𝑑107,𝑑108,𝑑109,𝑑110 respectively. Since 

〈14,8,3,2〉 ≠ 〈14,8,3,2〉′ then 𝑘1 must split to 𝑑103, 𝑑104, also since 𝑩𝟗 of defect one then 𝑘2 must 

split to 𝑑105, 𝑑106,then we get block 𝑩𝟗.To find 𝑩𝟏𝟎 and 𝑩𝟏𝟏 since  

• degree {〈20,5,2〉∗, 〈15,7,5〉∗, 〈12,7,5,2,1〉∗, 〈9,7,5,4,2〉∗} ≡ 143 mod 132 

• degree {〈18,7,2〉∗, 〈13,7,5,2〉 + 〈13,7,5,2〉′, 〈10,7,5,3,2〉∗} ≡ −143 mod 132,  

• degree {〈17,7,3〉∗, 〈13,7,4,3〉 + 〈13,7,4,3〉′, 〈11,7,4,3,2〉∗} ≡ 143 mod 132 

• degree {〈20,4,3〉∗, 〈16,7,4〉∗, 〈12,7,4,3,1〉∗, 〈8,7,5,4,3〉∗} ≡ −143 mod 132,  

by inducing of p.i.s. 𝐷75, 𝐷77, …, 𝐷97 of  𝑆26 to  S27, and    

1. 〈13,7,5,2〉 = 〈13,7,5,2〉′ 
2. 〈12,7,5,2,1〉∗ = 〈10,7,5,3,2〉∗ − 〈9,7,5,4,2〉∗ + 〈13,7,5,2〉 − 〈15,7,5〉∗ + 〈18,7,2〉∗ − 〈20,5,2〉∗ 

3. 〈13,7,4,3〉 = 〈13,7,4,3〉′ 
4. 〈12,7,4,3,1〉∗ = 〈11,7,4,3,2〉∗ − 〈8,7,5,4,3〉∗ + 〈13,7,4,3〉 − 〈16,7,4〉∗ + 〈17,7,3〉∗ − 〈20,4,3〉∗ 

then ech blocks 𝑩𝟏𝟎, 𝑩𝟏𝟏 contains at most 6 columns, so we get Table 5. 

 

Lemma 3.6. Block 𝑩𝟏𝟐 of type associateand and 𝑩𝟏𝟑 of type double as given in the Table 6. 

 

 

 

 

 



IHJPAS. 2024, 37(4)  

 

357 

 

Table 6. Blocks 𝐵12, 𝐵13 

Block Spin characters Decomposition matrix 

𝐵12 

〈20,4,2,1〉 1                  
〈20,4,2,1〉′  1                 
〈17,7,2,1〉 1  1                
〈17,7,2,1〉′  1  1               
〈15,7,4,1〉   1  1              
〈15,7,4,1〉′    1  1             
〈14,7,4,2〉     1  1            
〈14,7,4,2〉′      1  1           

〈13,7,4,2,1〉∗       1 1 1 1         
〈10,7,4,3,2,1〉         1  1        
〈10,7,4,3,2,1〉′          1  1       
〈8,7,5,4,2,1〉           1        
〈8,7,5,4,2,1〉′            1       

𝐵13 

〈19,5,3〉∗             1      
〈18,6,3〉∗             1 1     
〈16,6,5〉∗              1 1    
〈13,6,5,3〉               1 1   

〈12,6,5,3,1〉∗                1 1  
〈11,6,5,3,2〉∗                 1 1 
〈9,6,5,4,3〉∗                  1 

  𝑑123 𝑑124 𝑑125 𝑑126 𝑑127 𝑑128 𝑑129 𝑑130 𝑑131 𝑑132 𝑑133 𝑑134 𝑑135 𝑑136 𝑑137 𝑑138 𝑑139 𝑑140 

 

Proof:. By using inducing of  p.i.s. 𝐷87, 𝐷88, 𝐷89, 𝐷90, 𝐷101, 𝐷93, 𝐷95, 𝐷96, 𝐷97, 𝐷98 of 𝑆26 to 𝑆27we 

get on 𝑑123, 𝑑124, 𝑑125, 𝑑126, 𝑘1, 𝑘2, 𝑑131,𝑑132,𝑑133,𝑑134 respectively. Since 〈14,7,4,2〉 ≠

〈14,7,4,2〉′𝑘1 must split to 𝑑127, 𝑑128, also since 𝑩𝟏𝟐 of defect one then 𝑘2 must split to 

𝑑129, 𝑑130.To find block 𝑩𝟏𝟑, Since   

• degree {〈18,6,3〉∗, 〈13,6,5,3〉 + 〈13,6,5,3〉′, 〈11,6,5,3,2〉∗} ≡ 156 mod 132 

• degree {〈19,5,3〉∗, 〈16,6,5〉∗, 〈12,6,5,3,1〉∗, 〈9,6,5,4,3〉∗} ≡ −156 mod 132,  

by inducing of p.i.s.𝐷105, 𝐷107, 𝐷109, 𝐷111, 𝐷113, 𝐷115 of  𝑆26 to  S27, and  

1. 〈13,6,5,3〉 = 〈13,6,5,3〉′ 
2. 〈12,6,5,3,1〉∗ = 〈11,6,5,3,2〉∗ − 〈9,6,5,4,3〉∗ + 〈13,6,5,3〉 − 〈16,6,5〉∗ + 〈18,6,3〉∗ − 〈19,5,3〉∗ 

then the block 𝑩𝟏𝟑 contains at most 6 columns , then we get Table 6. 

 

Lemma 3.7. Blocks 𝑩𝟏𝟒, 𝑩𝟏𝟓 of type associate as shown in the Table 7. 

Table 7. Blocks 𝐵14, 𝐵15 

Block Spin Character Block 

𝐵14 

〈19,5,2,1〉 1                        
〈19,5,2,1〉′  1                       
〈18,6,2,1〉 1  1                      
〈18,6,2,1〉′  1  1                     
〈15,6,5,1〉   1  1                    
〈15,6,5,1〉′    1  1                   
〈14,6,5,2〉     1  1                  
〈14,6,5,2〉′      1  1                 

〈13,6,5,2,1〉∗       1 1 1 1               
〈10,6,5,3,2,1〉         1  1              
〈10,6,5,3,2,1〉′          1  1             
〈9,6,5,4,2,1〉           1              
〈9,6,5,4,2,1〉′            1             

𝐵15 

〈19,4,3,1〉             1            
〈19,4,3,1〉′              1           
〈17,6,3,1〉             1  1          
〈17,6,3,1〉′              1  1         



IHJPAS. 2024, 37(4)  

 

358 

 

〈16,6,4,1〉               1  1        
〈16,6,4,1〉′                1  1       
〈14,6,4,3〉                 1  1      
〈14,6,4,3〉′                  1  1     

〈13,6,4,3,1〉∗                   1 1 1 1   
〈11,6,4,3,2,1〉                     1  1  
〈11,6,4,3,2,1〉′                      1  1 
〈8,6,5,4,3,1〉                       1  
〈8,6,5,4,3,1〉′                        1 

 
 

𝑑
1

4
1
 

𝑑
1

4
2
 

𝑑
1

4
3
 

𝑑
1

4
4
 

𝑑
1

4
5
 

𝑑
1

4
6
 

𝑑
1

4
7
 

𝑑
1

4
8
 

𝑑
1

4
9
 

𝑑
1

5
0
 

𝑑
1

5
1
 

𝑑
1

5
2
 

𝑑
1

5
3
 

𝑑
1

5
4
 

𝑑
1

5
5
 

𝑑
1

5
6
 

𝑑
1

5
7
 

𝑑
1

5
8
 

𝑑
1

5
9
 

𝑑
1

6
0
 

𝑑
1

6
1
 

𝑑
1

6
2
 

𝑑
1

6
3
 

𝑑
1

6
4
 

 

Proof: Using inducing of p.i.s. 𝐷105, 𝐷106, 𝐷107, 𝐷108, 𝐷131, 𝐷111, 𝐷113, 𝐷114, 𝐷115, 𝐷116,𝐷117, 

𝐷118, 𝐷119, 𝐷120, 𝐷131, 𝐷123, 𝐷125, 𝐷126, 𝐷127, 𝐷128 of 𝑆26 to 𝑆27we get on 𝑑141, 𝑑142, 𝑑143, 𝑑144, 

𝑘1, 𝑘2, 𝑑149,𝑑150,𝑑151,𝑑152,𝑑153, 𝑑154, 𝑑155, 𝑑156, 𝑘3, 𝑘4, 𝑑161, 𝑑162, 𝑑163, 𝑑164. Since 

〈14,6,5,2〉 ≠ 〈14,6,5,2〉′ then 𝑘1 split to 𝑑145, 𝑑146, also 𝑩𝟏𝟒 of defect one 𝑘2 must split to 

𝑑147, 𝑑148. For the block 𝑩𝟏𝟓, 〈14,6,4,3〉 ≠ 〈14,6,4,3〉′ then 𝑘3 must split to 𝑑155, 𝑑156, also since 

𝑩𝟏𝟓 of defect one then 𝑘4 must split to 𝑑157, 𝑑158, then we get Table 7. 

 

Lemma 3.8. The block 𝐵16 of type associate as shown in the Table 8. 

Table 8. Block 𝐵16 

spin characters decomposition matrix 

〈18,4,3,2〉 1            
〈18,4,3,2〉′  1           
〈17,5, 3,2 〉 1  1          
〈 17 ,5,3,2〉′  1  1         
〈16,5,4,2〉   1  1        
〈16,5,4,2〉′    1  1       
〈15,5,4,3〉     1  1      
〈15,5,4,3〉′      1  1     

〈13,5,4,3,2〉∗       1 1 1 1   
〈12,5,4,3,2,1〉         1  1  
〈12,5,4,3,2,1〉′          1  1 
〈7,6,5,4,3,2〉           1  
〈7,6,5,4,3,2〉′            1 

 𝑑165 𝑑166 𝑑167 𝑑168 𝑑169 𝑑170 𝑑171 𝑑172 𝑑173 𝑑174 𝑑175 𝑑176 

 

Proof:We find the required matrix by using inducing of p.i.s. 𝐷135, 𝐷136, 𝐷137, 𝐷138, 𝐷180, 𝐷181, 

𝐷139, 𝐷140 get on 𝑘1, 𝑘2, 𝑘3, 𝑘4, 𝑑173, 𝑑174, 𝑘5. Since 〈18,4,3,2〉 ≠ 〈18,4,3,2〉′ then 𝑘1 split to 

𝑑165, 𝑑166. Since 〈16,5,4,2〉 ≠ 〈16,5,4,2〉′ so 𝑘2 or 𝑘3 is split. If 𝑘2 is split to 𝑑167, 𝑑168, but 

〈15,5,4,3〉 ≠ 〈15,5,4,3〉′ then 𝑘3 split to, 𝑑169,  𝑑170. If 𝑘3 is split, and  

〈16,5,4〉 + 〈13,5,4,3,2〉∗ − 〈15,5,4,3〉 ≠ 〈16,5,4〉′ + 〈13,5,4,3,2〉∗ − 〈15,5,4,3〉′ (6) 

then 𝑘2  split, so in both cases we get 𝑘2 and 𝑘3  are splits, also 𝑩𝟏𝟔 of defect one then 𝑘4 split to 

𝑑171, 𝑑172. Finally. 〈7,6,5,4,3,2〉 ≠ 〈7,6,5,4,3,2〉′ then 𝑘5 split to 𝑑175, 𝑑176, we get Table 8. 

 

3.Decomposition matrix for �̅�𝟐𝟖 

     Decomposition matrix for 𝑆2̅8 of degree (334,295), and it is decomposed in to blocks of 

character it consists of 69 blocks which 𝐵1of defect two, 𝐵2 , 𝐵3, . . . , 𝐵21 are defect one, and the 

remaining blocks are defects zero, decomposition matrix is equal to 𝐵1⨁𝐵2⨁. . . ⨁𝐵69 

 

Lemma 4.1.The blocks 𝑩𝟐, 𝑩𝟑, 𝑩𝟒 of type double as shown in Table 9.  



IHJPAS. 2024, 37(4)  

 

359 

 

 Table 9. Blocks 𝐵2, 𝐵3,𝐵4 

Block spin characters decomposition matrix 

𝐵2 

〈27,1〉∗ 1                  
〈14,13,1〉 1 1                 

〈14,11,2,1〉∗  1 1                
〈14,10,3,1〉∗   1 1               
〈14,9,4,1〉∗    1 1              
〈14,8,5,1〉∗     1 1             
〈14,7,6,1〉∗      1             

𝐵3 

〈25,3〉∗       1            
〈16,12〉∗       1 1           
〈13,12,3〉        1 1          

〈12,11,3,2〉∗         1 1         
〈12,9,4,3〉∗          1 1        
〈12,8,5,3〉∗           1 1       
〈12,7,6,3〉∗            1       

𝐵4 

〈24,4〉∗             1      
〈17,11〉∗             1 1     
〈13,11,4〉              1 1    

〈12,11,4,1〉∗               1 1   
〈11,10,4,3〉∗                1 1  
〈11,8,5,4〉∗                 1 1 
〈11,7,6,4〉∗                  1 

  𝑑55 𝑑56 𝑑57 𝑑58 𝑑59 𝑑60 𝑑61 𝑑62 𝑑63 𝑑64 𝑑65 𝑑66 𝑑67 𝑑68 𝑑69 𝑑70 𝑑71 𝑑72 

 

 

Proof: Since   

• degree {〈14,13,1〉 + 〈14,13,1〉′, 〈14,10,3,1〉∗, 〈14,8,5,1〉∗} ≡ 117 mod 132, 

• degree {〈27,1〉∗, 〈14,11,2,1〉∗, 〈14,9,4,1〉∗, 〈14,7,6,1〉∗} ≡ −117 mod 132, 

• degree {〈16,12〉∗, 〈12,11,3,2〉∗, 〈12,8,5,3〉∗} ≡ 91 mod 132, 

• degree {〈25,3〉∗, 〈13,12,3〉 + 〈13,12,3〉′, 〈12,9,4,3〉∗, 〈12,7,6,3〉∗} ≡ −91 mod 132,  

• degree {〈17,11〉∗, 〈12,11,4,1〉∗, 〈11,8,5,4〉∗} ≡ 156 mod 132, 

• degree {〈24,4〉∗, 〈13,11,4〉 + 〈13,11,4〉′, 〈11,10,4,3〉∗, 〈11,7,6,4〉∗} ≡ −156 mod 132,  

used inducing of p.i.s.𝐷2, 𝐷11, 𝐷10, 𝐷9, 𝐷8,  𝐷7, 𝐷27, 𝐷29, 𝐷31, 𝐷33, 𝐷35, 𝐷37, 𝐷39, 𝐷41, 𝐷43, 𝐷45, 

𝐷47, 𝐷49 of  𝑆27 to  S28, and   

1. 〈14,13,1〉 = 〈14,13,1〉′ 
2. 〈14,11,2,1〉∗ = 〈14,10,3,1〉∗ − 〈14,9,4,1〉∗ + 〈14,13,1〉 − 〈27,1〉∗ + 〈14,8,5,1〉∗ − 〈14,7,6,1〉∗ 

3. 〈13,12,3〉 = 〈13,12,3〉′ 
4. 〈12,11,3,2〉∗ = 〈12,9,4,3〉∗ − 〈12,8,5,3〉∗ + 〈13,12,3〉 − 〈16,12〉∗ + 〈25,3〉∗ + 〈12,7,6,3〉∗ 

5. 〈13,11,4〉 = 〈13,11,4〉′ 
6. 〈12,11,4,1〉∗ = 〈11,10,4,3〉∗ − 〈11,8,5,4〉∗ + 〈13,11,4〉 − 〈17,11〉∗ + 〈24,4〉∗ + 〈11,7,6,4〉∗ 

so each of these blocks contains 6 columns, so we get Table 9 

 

Lemma 4.2. The block 𝑩𝟓 of type associate and 𝑩𝟔 of type doubleas shown in the Tables 10. 

Table 10. Blocks 𝐵5, 𝐵6 

Block Spin characters Decomposition matrix 

𝐵5 

〈24,3,1〉 1                  
〈24,3,1〉′  1                 
〈16,11,1〉 1  1                
〈16,11,1〉′  1  1               
〈14,11,3〉   1  1              
〈14,11,3〉′    1  1             

〈13,11,3,1〉∗     1 1 1 1           



IHJPAS. 2024, 37(4)  

 

360 

 

〈11,9,4,3,1〉       1  1          
〈11,9,4,3,1〉′        1  1         
〈11,8,5,3,1〉         1  1        
〈11,8,5,3,1〉′          1  1       
〈11,7,6,3,1〉           1        
〈11,7,6,3,1〉′            1       

𝐵6 

〈23,5〉∗             1      
〈18,10〉∗             1 1     
〈13,10,5〉              1 1    

〈12,10,5,1〉∗               1 1   
〈11,10,5,2〉∗                1 1  
〈10,9,5,4〉∗                 1 1 
〈10,7,6,5〉∗                  1 

  𝑑73 𝑑74 𝑑75 𝑑76 𝑑77 𝑑78 𝑑79 𝑑80 𝑑81 𝑑82 𝑑83 𝑑84 𝑑85 𝑑86 𝑑87 𝑑88 𝑑89 𝑑90 

 

Proof: By using inducing ofp.i.s. 𝐷39, 𝐷40, 𝐷11, 𝐷12, 𝐷45, 𝐷46, … , 𝐷51, 𝐷53, 𝐷55, 𝐷57, 𝐷59, 𝐷61 of 

𝑆27 to 𝑆28we get on 𝑑73, 𝑑74, 𝑘1, 𝑘2, 𝑑79, 𝑑80, . . . , 𝑑90. Since 〈14,11,3〉 ≠ 〈14,11,3〉′ then 𝑘1 split 

to 𝑑75, 𝑑76, also since 𝑩𝟓 of defect one then 𝑘2 split to 𝑑77, 𝑑78. To find block 𝑩𝟔, since  

• degree {〈23,5〉∗, 〈13,10,5〉 + 〈13,10,5〉′, 〈11,10,5,2〉∗, 〈10,7,6,5〉∗} ≡ 117 mod 132 

• degree {〈18,10〉∗, 〈12,10,5,1〉∗, 〈10,9,5,4〉∗} ≡ −117 mod 132,  

and on (13, 𝛼)-regular classes we have:   

1. 〈13,10,5〉 = 〈13,10,5〉′ 
2. 〈12,10,5,1〉∗ = 〈11,10,5,2〉∗ + 〈13,10,5〉 − 〈18,10〉∗ + 〈23,5〉∗ − 〈10,9,5,4〉∗ + 〈10,7,6,5〉∗ 

then the block contains at most 6 columns , so we get Table 10 

 

Lemma 4.3. The block 𝑩𝟕 of type associate and 𝑩𝟖 of type double as shown in the Table 11. 

Table 11. Blocks 𝐵7, 𝐵8 

Block Spin characters Decomposition matrix 

𝐵7 

〈23,4,1〉 1                  
〈23,4,1〉′  1                 

〈17,10,1〉 1  1                

〈17,10,1〉′  1  1               
〈14,10,4〉   1  1              

〈14,10,4〉′    1  1             

〈13,10,4,1〉∗     1 1 1 1           
〈11,10,4,2,1〉       1  1          
〈11,10,4,2,1〉′        1  1         

〈10,8,5,4,1〉         1  1        

〈10,8,5,4,1〉′          1  1       
〈10,7,6,4,1〉           1        

〈10,7,6,4,1〉′            1       

𝐵8 

〈22,6〉∗             1      
〈19,9〉∗             1 1     
〈13,9,6〉              1 1    

〈12,9,6,1〉∗               1 1   
〈11,9,6,2〉∗                1 1  
〈10,9,6,3〉∗                 1 1 
〈9,8,6,5〉∗                  1 

  𝑑91 𝑑92 𝑑93 𝑑94 𝑑95 𝑑96 𝑑97 𝑑98 𝑑99 𝑑100 𝑑101 𝑑102 𝑑103 𝑑104 𝑑105 𝑑106 𝑑107 𝑑108 

 

Proof: By inducing of  p.i.s. 𝐷51, 𝐷52, 𝐷8, 𝐷14, 𝐷57, 𝐷58, …, 𝐷62, 𝐷63, 𝐷65, 𝐷67, 𝐷69, 𝐷71, 𝐷73 of 

𝑆27 to 𝑆28we get on 𝑑91, 𝑑92, 𝑘1, 𝑘2, 𝑑97, 𝑑98, …, 𝑑108. Since 〈14,10,4〉 ≠ 〈14,10,4〉′ then 𝑘1 split 

to 𝑑93, 𝑑94, also 𝑩𝟕 of defect one then 𝑘2 split to 𝑑95, 𝑑96.To find the 𝑩𝟖 since   



IHJPAS. 2024, 37(4)  

 

361 

 

• degree {〈22,6〉∗, 〈13,9,6〉 + 〈13,9,6〉′, 〈11,9,6,2〉∗, 〈9,8,6,5〉∗} ≡ 117 mod 132, 

• degree {〈19,9〉∗, 〈12,9,6,1〉∗, 〈10,9,6,3〉∗} ≡ −117 mod 132,  

and on (13, 𝛼)-regular classes we have:   

1. 〈13,9,6〉 = 〈13,9,6〉′ 

2. 〈12,9,6,1〉∗ = 〈11,9,6,2〉∗ + 〈13,9,6〉 − 〈19,9〉∗ + 〈22,6〉∗ − 〈10,9,6,3〉∗ + 〈9,8,6,5〉∗ 

then the block 𝑩𝟖 contains at most 6 columns, so we get Table 11 

 

Lemma 4.4. Block 𝑩𝟗 is associate and 𝑩𝟏𝟎, 𝑩𝟏𝟏 are double as shown in the Table 12. 

Table 12. Blocks 𝐵9, 𝐵10, 𝐵11 

Block Spin characters Decomposition matrix 

𝐵9 

〈22,5,1〉 1                        

〈22,5,1〉′  1                       

〈18,9,1〉 1  1                      
〈18,9,1〉′  1  1                     

〈14,9,5〉   1  1                    

〈14,9,5〉′    1  1                   
〈13,9,5,1〉∗     1 1 1 1                 

〈11,9,5,2,1〉       1  1                

〈11,9,5,2,1〉′        1  1               
〈10,9,5,3,1〉         1  1              
〈10,9,5,3,1〉′          1  1             

〈9,7,6,5,1〉           1              
〈9,7,6,5,1〉′            1             

𝐵10 

〈22,3,2,1〉∗             1            

〈16,9,2,1〉∗             1 1           

〈15,9,3,1〉∗              1 1          
〈14,9,3,2〉∗               1 1         
〈13,9,3,2,1〉                1 1        

〈9,8,5,3,2,1〉∗                 1 1       

〈9,7,6,3,2,1〉∗                  1       

𝐵11 

〈21,7〉∗                   1      
〈20,8〉∗                   1 1     
〈13,8,7〉                    1 1    

〈12,8,7,1〉∗                     1 1   
〈11,8,7,2〉∗                      1 1  
〈10,8,7,3〉∗                       1 1 
〈9,8,7,4〉∗                        1 

 
 

𝑑
1

0
9
 

𝑑
1

1
0
 

𝑑
1

1
1
 

𝑑
1

1
2
 

𝑑
1

1
3
 

𝑑
1

1
4
 

𝑑
1

1
5
 

𝑑
1

1
6
 

𝑑
1

1
7
 

𝑑
1

1
8
 

𝑑
1

1
9
 

𝑑
1

2
0
 

𝑑
1

2
1
 

𝑑
1

2
2
 

𝑑
1

2
3
 

𝑑
1

2
4
 

𝑑
1

2
5
 

𝑑
1

2
6
 

𝑑
1

2
7
 

𝑑
1

2
8
 

𝑑
1

2
9
 

𝑑
1

3
0
 

𝑑
1

3
1
 

𝑑
1

3
2
 

 

Proof: Using inducing of  p.i.s. 𝐷63, 𝐷64, 𝐷7, 𝐷15, 𝐷69, 𝐷70, … , 𝐷74,  𝐷99, 𝐷101, 𝐷103, 𝐷105, 𝐷107, 

𝐷109, 𝐷81, 𝐷83, 𝐷85, 𝐷87, 𝐷89, 𝐷91 of 𝑆27 to 𝑆28we get on 𝑑109, 𝑑110, 𝑘1, 𝑘2, 𝑑115, 𝑑116 , …, 𝑑132. 

Since 〈14,9,5〉 ≠ 〈14,9,5〉′ then 𝑘1 split to 𝑑111, 𝑑112, also 𝑩𝟗 of defect 𝑘2 split to 𝑑113, 𝑑114. To 

find blocks 𝑩𝟏𝟎 , 𝑩𝟏𝟏, since   

• degree {〈16,9,2,1〉∗, 〈14,9,3,2〉∗, 〈9,8,5,3,2,1〉∗} ≡ 143 mod 132, 

• degree{〈22,3,2,1〉∗, 〈15,9,3,1〉∗, 〈13,9,3,2,1〉 + 〈13,9,3,2,1〉′, 〈9,7,6,3,2,1〉∗} ≡ −143 mod 

132,  

• degree {〈20,8〉∗, 〈12,8,7,1〉∗, 〈10,8,7,3〉∗} ≡ 143 mod 132, 

• degree {〈21,7〉∗, 〈13,8,7〉 + 〈13,8,7〉′, 〈11,8,7,2〉∗, 〈9,8,7,4〉∗} ≡ −143 mod 132,  

and on (13, 𝛼)-regular classes we have:   

1. 〈13,9,3,2,1〉 = 〈13,9,3,2,1〉′ 



IHJPAS. 2024, 37(4)  

 

362 

 

2. 〈9,8,5,3,2,1〉∗ = 〈9,7,6,3,2,1〉∗ + 〈13,9,3,2,1〉 − 〈14,9,3,2〉∗ + 〈15,9,3,1〉∗ − 〈16,9,2,1〉∗ +
〈22,3,2,1〉∗ 

3. 〈13,8,7〉 = 〈13,8,7〉′ 

4. 〈12,8,7,1〉∗ = 〈11,8,7,2〉∗ + 〈13,8,7〉 − 〈20,8〉∗ + 〈21,7〉∗ − 〈10,8,7,3〉∗ + 〈9,8,7,4〉∗ 

so each of these blocks contains 6 columns, thenwe get Table 12 

 

Lemma 4.5. The blocks 𝑩𝟏𝟐, 𝑩𝟏𝟑 are associate as shown in the Table 13. 

Table 13. Blocks 𝐵12, 𝐵13 

Block Spin characters Decomposition matrix 

𝐵12 

〈21,6,1〉 1                        

〈21,6,1〉′  1                       
〈19,8,1〉 1  1                      

〈19,8,1〉′  1  1                     

〈14,8,6〉   1  1                    
〈14,8,6〉′    1  1                   

〈13,8,6,1〉∗     1 1 1 1                 

〈11,8,6,2,1〉       1  1                
〈11,8,6,2,1〉′        1  1               
〈10,8,6,3,1〉         1  1              

〈10,8,6,3,1〉′          1  1             
〈9,8,6,4,1〉           1              
〈9,8,6,4,1〉′            1             

𝐵13 

〈21,4,3〉             1            
〈21,4,3〉′              1           
〈17,8,3〉             1  1          
〈17,8,3〉′              1  1         
〈16,8,4〉               1  1        
〈16,8,4〉′                1  1       

〈13,8,4,3〉∗                 1 1 1 1     
〈12,8,4,3,1〉                   1  1    
〈12,8,4,3,1〉′                    1  1   
〈11,8,4,3,2〉                     1  1  
〈11,8,4,3,2〉′                      1  1 
〈8,7,6,4,3〉                       1  
〈8,7,6,4,3〉′                        1 

  

𝑑
1

3
3
 

𝑑
1

3
4
 

𝑑
1

3
5
 

𝑑
1

3
6
 

𝑑
1

3
7
 

𝑑
1

3
8
 

𝑑
1

3
9
 

𝑑
1

4
0
 

𝑑
1

4
1
 

𝑑
1

4
2
 

𝑑
1

4
3
 

𝑑
1

4
4
 

𝑑
1

4
5
 

𝑑
1

4
6
 

𝑑
1

4
7
 

𝑑
1

4
8
 

𝑑
1

4
9
 

𝑑
1

5
0
 

𝑑
1

5
1
 

𝑑
1

5
2
 

𝑑
1

5
3
 

𝑑
1

5
4
 

𝑑
1

5
5
 

𝑑
1

5
6
 

 

Proof: By inducing of  p.i.s. 𝐷81, 𝐷82, 𝐷8, 𝐷16, 𝐷87, 𝐷88, … , 𝐷92, 𝐷117, 𝐷118, 𝐷119, 𝐷217, 𝐷218, 𝐷221, 

𝐷222 of 𝑆27 to 𝑆28we get on 𝑑133, 𝑑134, 𝑘1, 𝑘2, 𝑑139, 𝑑140, …, 𝑑144, 𝑘3, 𝑘4, 𝑘5, 𝑑151, 𝑑152, 𝑘6, 𝑘7. 

Since 〈14,8,6〉 ≠ 〈14,8,6〉′ then 𝑘1 split to 𝑑135, 𝑑136, also 𝑩𝟏𝟐 of defect one then 𝑘2 split to 

𝑑137, 𝑑138. To find block 𝑩𝟏𝟐, 〈17,8,3〉 ≠ 〈17,8,3〉′ so 𝑘3 or 𝑘4 is split. If 𝑘3 is split to 𝑑145, 𝑑146, 

but 〈16,8,4〉 ≠ 〈16,8,4〉′ then 𝑘4 split to, 𝑑147, 𝑑148. If 𝑘4 is split and 

〈16,8,4〉 + 〈17,8,3〉 − 〈21,4,3〉 ≠ 〈16,8,4〉′ + 〈17,8,3〉′ − 〈21,4,3〉′ (7) 

then 𝑘3  split, so, in both cases we get 𝑘3 and 𝑘4  are splits, also 𝑩𝟏𝟑 of defect one then 𝑘5 split to 

𝑑149, 𝑑150. Since 〈11,8,4,3,2〉 ≠ 〈11,8,4,3,2〉′ so 𝑘6 or 𝑘7 is split. If 𝑘6 is split to 𝑑153, 𝑑154, but 
〈8,7,6,4,3〉 ≠ 〈8 7 6 4 3〉′ then 𝑘7 split to, 𝑑155, 𝑑156. If 𝑘7 is split, and 

〈11,8,4,3,2〉 − 〈8,7,6,4,3〉 ≠ 〈11,8,4,3,2〉′ − 〈8,7,6,4,3〉′ (8) 

then 𝑘6  split, so in both cases we get 𝑘6 and 𝑘7  are splits, then we get Table 13. 

 

Lemma 4.6.  Blocks 𝑩𝟏𝟒, 𝑩𝟏𝟔 of type double and 𝑩𝟏𝟓 is associate as shown in Table 14.  

 

 



IHJPAS. 2024, 37(4)  

 

363 

 

Table 14. Blocks 𝐵14, 𝐵15,𝐵16 

Block Spin characters Decomposition matrix 

𝐵14 

〈21,4,2,1〉∗ 1                        
〈17,8,2,1〉∗ 1 1                       
〈15,8,4,1〉∗  1 1                      
〈14,8,4,2〉∗   1 1                     
〈13,8,4,2,1〉    1 1                    

〈10,8,4,3,2,1〉∗     1 1                   
〈8,7,6,4,2,1〉∗      1                   

𝐵15 

〈20,5,3〉       1                  
〈20,5,3〉′        1                 
〈18,7,3〉       1  1                
〈18,7,3〉′        1  1               
〈16,7,5〉         1  1              
〈16,7,5〉′          1  1             

〈13,7,5,3〉∗           1 1 1 1           
〈12,7,5,3,1〉             1  1          
〈12,7,5,3,1〉′              1  1         
〈11,7,5,3,2〉               1  1        
〈11,7,5,3,2〉′                1  1       
〈9,7,5,4,3〉                 1        
〈9,7,5,4,3〉′                  1       

𝐵16 

〈20,5,2,1〉∗                   1      
〈18,7,2,1〉∗                   1 1     
〈15,7,5,1〉∗                    1 1    
〈14,7,5,2〉∗                     1 1   
〈13,7,5,2,1〉                      1 1  

〈10,7,5,3,2,1〉∗                       1 1 
〈9,7,5,4,2,1〉∗                        1 

  

𝑑
1

5
7
 

𝑑
1

5
8
 

𝑑
1

5
9
 

𝑑
1

6
0
 

𝑑
1

6
1
 

𝑑
1

6
2
 

𝑑
1

6
3
 

𝑑
1

6
4
 

𝑑
1

6
5
 

𝑑
1

6
6
 

𝑑
1

6
7
 

𝑑
1

6
8
 

𝑑
1

6
9
 

𝑑
1

7
0
 

𝑑
1

7
1
 

𝑑
1

7
2
 

𝑑
1

7
3
 

𝑑
1

7
4
 

𝑑
1

7
5
 

𝑑
1

7
6
 

𝑑
1

7
7
 

𝑑
1

7
8
 

𝑑
1

7
9
 

𝑑
1

8
0
 

 

Proof: By inducing of p.i.s.𝐷99, 𝐷101, 𝐷103, 𝐷105, 𝐷107, 𝐷109, 𝐷111, 𝐷112, 𝐷113, 𝐷219, 𝐷220, 𝐷115, 

𝐷116, 𝐷123, 𝐷125, 𝐷127, 𝐷129, 𝐷131, 𝐷133 of 𝑆27 to 𝑆28we get on 𝑑157,𝑑158,…,𝑑162, 𝑘1,  𝑘2, 𝑘3,𝑑169, 

𝑑170, 𝑘4, 𝑘5, 𝑑175, 𝑑176,…,𝑑180. To find blocks 𝑩𝟏𝟒 , 𝑩𝟏𝟔, since   

• degree {〈21,4,2,1〉∗, 〈15,8,4,1〉∗, 〈13,8,4,2,1〉 + 〈13,8,4,2,1〉′, 〈8,7,6,4,2,1〉∗} ≡ 156 mod 132, 

• degree {〈17,8,2,1〉∗, 〈14,8,4,2〉∗, 〈10,8,4,3,2,1〉∗} ≡ −156 mod 132,  

• degree {〈20,5,2,1〉∗, 〈15,7,5,1〉∗, 〈13,7,5,2,1〉 + 〈13,7,5,2,1〉′, 〈9,7,5,4,2,1〉∗} ≡ 104 mod 132, 

• degree {〈18,7,2,1〉∗, 〈14,7,5,2〉∗, 〈10,7,5,3,2,1〉∗} ≡ −104 mod 132,  

and on (13, 𝛼)-regular classes:   

1. 〈13,8,4,2,1〉 = 〈13,8,4,2,1〉′ 
2. 〈14,8,4,2〉∗ = 〈15,8,4,1〉∗ + 〈13,8,4,2,1〉 − 〈17,8,2,1〉∗ + 〈21,4,2,1〉∗ − 〈10,8,4,3,2,1〉∗ +

〈8,7,6,4,2,1〉∗ 

3. 〈13,7,5,2,1〉 = 〈13,7,5,2,1〉′ 
4. 〈14,7,5,2〉∗ = 〈15,7,5,1〉∗ + 〈13,7,5,2,1〉 − 〈10,7,5,3,2,1〉∗ + 〈9,7,5,4,2,1〉∗ − 〈18,7,2,1〉∗ +

〈20,5,2,1〉∗ 

so each blocks contains 6 columns. In 𝑩𝟏𝟓, 〈18,7,3〉 ≠ 〈18,7,3〉′ so 𝑘1 or 𝑘2 is split. If 𝑘1 is split 

to 𝑑163, 𝑑164, but 〈16,7,5〉 ≠ 〈16,7,5〉′ then 𝑘2 split to, 𝑑165, 𝑑166. If 𝑘2 is split and 

〈16,7,5〉 − 〈18,7,3〉 + 〈20,5,3〉 ≠ 〈16,7,5〉′ − 〈18,7,3〉′ + 〈20,5,3〉′ (9) 

then 𝑘1  split, so, in both cases we get 𝑘1 and 𝑘2  are splits, also 𝑩𝟏5 of defect one then 𝑘3 split to 

𝑑167, 𝑑168. Since 〈11,8,4,3,2〉 ≠ 〈11,8,4,3,2〉′ so 𝑘4 or 𝑘5 is split. If 𝑘4 is split to 𝑑171,𝑑172, but 

〈9,7,5,4,3〉 ≠ 〈9,7,5,4,3〉′ then 𝑘5 split to, 𝑑173, 𝑑174. If 𝑘5 is split, and  



IHJPAS. 2024, 37(4)  

 

364 

 

〈11,7,5,3,2〉 − 〈9,7,5,4,3〉 ≠ 〈11,7,5,3,2〉′ − 〈9,7,5,4,3〉′ (10) 

then 𝑘4  split to 𝑑171, 𝑑172, so, in both cases we get 𝑘4 , 𝑘5  are splits, then we get Table 14.  

 

Lemma 4.7 Blocks 𝑩𝟏𝟕 , 𝑩𝟏𝟗 are double and 𝑩𝟏𝟖 of type associate as shown in Table 15.  

Table 15. Blocks 𝐵17, 𝐵18,𝐵19 

Block Spin characters Decomposition matrix 

𝐵17 

〈20,4,3,1〉∗ 1                        
〈17,7,3,1〉∗ 1 1                       
〈16,7,4,1〉∗  1 1                      
〈14,7,4,3〉∗   1 1                     
〈13,7,4,3,1〉    1 1                    

〈11,7,4,3,2,1〉∗     1 1                   
〈8,7,5,4,3,1〉∗      1                   

𝐵18 

〈19,5,4〉       1                  
〈19,5,4〉′        1                 
〈18,6,4〉       1  1                
〈18,6,4〉′        1  1               
〈17,6,5〉         1  1              
〈17,6,5〉′          1  1             

〈13,6,5,4〉∗           1 1 1 1           
〈12,6,5,4,1〉             1  1          
〈12,6,5,4,1〉′              1  1         
〈11,6,5,4,2〉               1  1        
〈11,6,5,4,2〉′                1  1       
〈10,6,5,4,3〉                 1        
〈10,6,5,4,3〉′                  1       

𝐵19 

〈19,5,3,1〉∗                   1      
〈18,6,3,1〉∗                   1 1     
〈16,6,5,1〉∗                    1 1    
〈14,6,5,3〉∗                     1 1   
〈13,6,5,3,1〉                      1 1  

〈11,6,5,3,2,1〉∗                       1 1 
〈9,6,5,4,3,1〉∗                        1 

  𝑑
1

8
1
 

𝑑
1

8
2
 

𝑑
1

8
3
 

𝑑
1

8
4
 

𝑑
1

8
5
 

𝑑
1

8
6
 

𝑑
1

8
7
 

𝑑
1

8
8
 

𝑑
1

8
9
 

𝑑
1

9
0
 

𝑑
1

9
1
 

𝑑
1

9
2
 

𝑑
1

9
3
 

𝑑
1

9
4
 

𝑑
1

9
5
 

𝑑
1

9
6
 

𝑑
1

9
7
 

𝑑
1

9
8
 

𝑑
1

9
9
 

𝑑
2

0
0
 

𝑑
2

0
1
 

𝑑
2

0
2
 

𝑑
2

0
3
 

𝑑
2

0
4
 

 

Proof: By inducing of p.i.s.𝐷153, 𝐷155, 𝐷157, 𝐷159, 𝐷161, 𝐷163,𝐷135, 𝐷136, 𝐷137, 𝐷221, 𝐷222, 

𝐷139, 𝐷140, 𝐷141, 𝐷143, 𝐷145, 𝐷147, 𝐷149, 𝐷151 of 𝑆27 to 𝑆28we get on𝑑181, 𝑑182,…,𝑑186,𝑘1, 𝑘2, 𝑘3 

,𝑑193, 𝑑194, 𝑘4,𝑘5, 𝑑199, 𝑑200,…,𝑑204. To find blocks 𝑩𝟏𝟕, 𝑩𝟏𝟗 since   

• degree {〈20,4,3,1〉∗, 〈16,7,4,1〉∗, 〈13,7,4,3,1〉 + 〈13,7,4,3,1〉′, 〈8,7,5,4,3,1〉∗} ≡ 130 mod 132, 

• degree {〈17,7,3,1〉∗, 〈14,7,4,3〉∗, 〈11,7,4,3,2,1〉∗} ≡ −130 mod 132,  

• degree {〈19,5,3,1〉∗, 〈16,6,5,1〉∗, 〈13,6,5,3,1〉 + 〈13,6,5,3,1〉′, 〈9,6,5,4,3,1〉∗} ≡ 143 mod 132, 

• degree {〈18,6,3,1〉∗, 〈14,6,5,3〉∗, 〈11,6,5,3,2,1〉∗} ≡ −143 mod 132,  

and on (13, 𝛼)-regular classes:   

1. 〈13,7,4,3,1〉 = 〈13,7,4,3,1〉′ 

2. 〈14,7,4,3〉∗ = 〈16,7,4,1〉∗ + 〈13,7,4,3,1〉 − 〈11,7,4,3,2,1〉∗ + 〈8,7,5,4,3,1〉∗ −
〈17,7,3,1〉∗ + 〈20,4,3,1〉∗ 

3. 〈13,6,5,3,1〉 = 〈13,6,5,3,1〉′ 

4. 〈14,6,5,3〉∗ = 〈16,6,5,1〉∗ + 〈13,6,5,3,1〉 − 〈11,6,5,3,2,1〉∗ + 〈9,6,5,4,3,1〉∗ −
〈18,6,3,1〉∗ + 〈19,5,3,1〉∗ 

so each blocks contains 6 columns.To find blocks 𝑩𝟏𝟖, 〈18,6,4〉 ≠ 〈18,6,4〉′ so 𝑘1 or 𝑘2 is split. If 

𝑘1 is split to 𝑑187, 𝑑188, but 〈17,6,5〉 ≠ 〈1,6,5〉′ then 𝑘2 split to, 𝑑189, 𝑑190. If 𝑘2 is split, and  



IHJPAS. 2024, 37(4)  

 

365 

 

〈17,6,5〉 − 〈18,6,4〉 + 〈19,5,4〉 ≠ 〈17,6,5〉′ − 〈18,6,4〉′ + 〈19,5,4〉′ (11) 

then 𝑘1  split, so, in both cases we get 𝑘1 and 𝑘2  are splits, also 𝑩𝟏𝟖 of defect one then 𝑘3 split to 

𝑑191, 𝑑192. Since 〈11,6,5,4,2〉 ≠ 〈11,6,5,4,2〉′ so 𝑘4 or 𝑘5 is split. If 𝑘4 is split to 𝑑195,𝑑196, but 
〈10,6,5,4,3〉 ≠ 〈10,6,5,4,3〉′ then 𝑘5 split to, 𝑑197, 𝑑198. If 𝑘5 is split and  

〈11,6,5,4,2〉 − 〈10,6,5,4,3〉 ≠ 〈11,6,5,4,2〉′ − 〈10,6,5,4,3〉′ (12) 

then 𝑘4  split, so, in both cases we get 𝑘4 and 𝑘5  are splits, then we get Table 15. 

 

Lemma 4.8.  Block 𝑩𝟐𝟎is a double and 𝑩𝟐𝟏is associateas given in Table 16.  

Table 16. Blocks 𝐵20, 𝐵21 

Block Spin characters Decomposition matrix 

𝐵20 

〈19,4,3,2〉∗ 1                  
〈17,6,3,2〉∗ 1 1                 
〈16,6,4,2〉∗  1 1                
〈15,6,4,3〉∗   1 1               
〈13,6,4,3,2〉    1 1              

〈12,6,4,3,2,1〉∗     1 1             
〈8,6,5,4,3,2〉∗      1             

𝐵21 

〈18,4,3,2,1〉       1            
〈18,4,3,2,1〉′        1           
〈17,5,3,2,1〉       1  1          
〈17,5,3,2,1〉′        1  1         
〈16,5,4,2,1〉         1  1        
〈16,5,4,2,1〉′          1  1       
〈15,5,4,3,1〉           1  1      
〈15,5,4,3,1〉′            1  1     
〈14,5,4,3,2〉             1  1    
〈14,5,4,3,2〉′              1  1   

〈13,5,4,3,2,1〉∗               1 1 1 1 
〈7,6,5,4,3,2,1〉                 1  
〈7,6,5,4,3,2,1〉′                  1 

  𝑑205 𝑑206 𝑑207 𝑑208 𝑑209 𝑑210 𝑑211 𝑑212 𝑑213 𝑑214 𝑑215 𝑑216 𝑑217 𝑑218 𝑑219 𝑑220 𝑑221 𝑑222 

 

Proof: By inducing of p.i.s. 𝐷153, 𝐷155, 𝐷157, 𝐷159, 𝐷161, 𝐷163, 𝐷165, 𝐷166, …, 𝐷170, 𝐷212, 

𝐷173, 𝐷175, 𝐷176 , we get on 𝑑211, 𝑑212, 𝑑213, 𝑑214, 𝑑215, 𝑑216,  𝑘1, 𝑘2, 𝑑221, 𝑑222. Since  

• degree {〈19,4,3,2〉∗, 〈16,6,4,2〉∗, 〈13,6,4,3,2〉 + 〈13,6,4,3,2〉′, 〈8,6,5,4,3,2〉∗} ≡ 117 mod 132 

• degree {〈17,6,3,2〉∗, 〈15,6,4,3〉∗, 〈12,6,4,3,2,1〉∗} ≡ −117 mod 132,  

and on (13, 𝛼)-regular classes we: 

1. 〈13,6,4,3,2〉 = 〈13,6,4,3,2〉′ 
2. 〈15,6,4,3〉∗ = 〈16,6,4,2〉∗ + 〈13,6,4,3,2〉 − 〈12,6,4,3,2,1〉∗ + 〈8,6,5,4,3,2〉∗ − 〈17,6,3,2〉∗ +

〈19,4,3,2〉∗ 

then the block 𝑩𝟐𝟎 contains at most 6 columns. To find the 𝑩𝟐𝟏, 〈14,5,4,3,2〉 ≠ 〈14,5,4,3,2〉′, so 

𝑘1 divided to 𝑑217, 𝑑218 or there are two columns: 

𝜑1 = 𝑎1〈15,5,4,3,1〉 + 𝑎2〈14,5,4,3,2〉 + 𝑎3〈13,5,4,3,2,1〉∗ + 𝑎4〈7,6,5,4,3,2,1〉,  

𝜑2 = 𝑎1〈15,5,4,3,1〉′ + 𝑎2〈14,5,4,3,2〉′ + 𝑎3〈13,5,4,3,2,1〉∗ + 𝑎4〈7,6,5,4,3,2,1〉′,  

to describe columns, since 𝑩𝟐𝟏 of defect one then 𝑎1, 𝑎2, 𝑎3, 𝑎4 ∈ {0,1}, but 〈7,6,5,4,3,2,1〉 ↓ 𝑆27 

has only one of i.m.s. and from table has only one of i.m.s. then 𝑎4 = 0, so that degree 𝜑1, 𝜑2 ≡

0 mod 73 (theorem 2.3) only when  𝜑1 + 𝜑2 = 𝑑217 + 𝑑218, then 𝑘1 = 𝑑217 + 𝑑218, also 𝑩𝟐𝟏 of 

defect one then 𝑘2 split to 𝑑219, 𝑑220, from abve we get Table 16. 

Lemma 4.9. Decomposition matrix for the block 𝐵1 of type double as shown in the Tables 17. 



IHJPAS. 2024, 37(4)  

 

366 

 

Table 17. Block 𝐵1 

spin 

character

s 

spin characters 

〈28〉 1                                                      
〈28〉′  1                                                     

〈26,2〉∗ 1 1 1 1                                                   

〈25,2,1〉   1  1                                                  

〈25,2,1〉′    1  1                                                 
〈23,3,2〉     1  1                                                

〈23,3,2〉′      1  1                                               

〈22,4,2〉       1  1                                              
〈22,4,2〉′        1  1                                             

〈21,5,2〉         1  1                                            

〈21,5,2〉′          1  1                                           
〈20,6,2〉           1  1                                          
〈20,6,2〉′            1  1                                         

〈19,7,2〉           1  1  1                                        

〈19,7,2〉′            1  1  1                                       
〈18,8,2〉         1  1    1  1                                      

〈18,8,2〉′          1  1    1  1                                     

〈17,9,2〉       1  1        1  1                                    
〈17,9,2〉′        1  1        1  1                                   

〈16,10,2〉     1  1            1  1                                  

〈16,10,2〉′      1  1            1  1                                 
〈15,13〉∗ 1 1 1 1                   1 1                               
〈15,12,1〉 1  1  1                  1  1                              

〈15,12,1〉′  1  1  1                  1  1                             
〈15,11,2〉     1                1    1  1                            
〈15,11,2〉′      1                1    1  1                           

〈15,10,3〉                   1  1      1  1                          

〈15,10,3〉′                    1  1      1  1                         
〈15,9,4〉                 1  1          1  1                        

〈15,9,4〉′                  1  1          1  1                       

〈15,8,5〉               1  1              1  1                      
〈15,8,5〉′                1  1              1  1                     

〈15,7,6〉               1                  1                      

〈15,7,6〉′                1                  1                     
〈14,12,2〉 1 1                     1 1 1  1        1 1                   
〈14,12,2〉′ 1 1                     1 1  1  1       1 1                   

〈13,12,2,1〉∗                       1 1   1 1       1 1 1 1                 

〈13,10,3,2〉∗                           1 1 1 1     1 1 1 1 1 1               
〈13,9,4,2〉∗                             1 1 1 1       1 1 1 1             

〈13,8,5,2〉∗                               1 1 1 1       1 1 1 1           

〈13,7,6,2〉∗                                 1 1         1 1           
〈12,10,3,2,1〉                                   1    1      1          

〈12,10,3,2,1〉′                                    1    1      1         

〈12,9,4,2,1〉                                     1  1  1    1  1        
〈12,9,4,2,1〉′                                      1  1  1    1  1       
〈12,8,5,2,1〉                                         1  1    1  1      

〈12,8,5,2,1〉′                                          1  1    1  1     

〈12,7,6,2,1〉                                           1      1      
〈12,7,6,2,1〉′                                            1      1     

〈10,9,4,3,2〉                                     1          1    1    

〈10,9,4,3,2〉′                                      1          1    1   
〈10,8,5,3,2〉                                             1  1  1  1  1  

〈10,8,5,3,2〉′                                              1  1  1  1  1 

〈10,7,6,3,2〉                                                 1    1  
〈10,7,6,3,2〉′                                                  1    1 
〈9,8,5,4,2〉                                             1        1  



IHJPAS. 2024, 37(4)  

 

367 

 

〈9,8,5,4,2〉′                                              1        1 

〈9,7,6,4,2〉                                                   1  1  

〈9,7,6,4,2〉′                                                    1  1 
〈8,7,6,5,2〉                                                   1    

〈8,7,6,5,2〉′                                                    1   

 𝑑
1
 

𝑑
2
 

𝑑
3
 

𝑑
4
 

𝑑
5
 

𝑑
6
 

𝑑
7
 

𝑑
8
 

𝑑
9
 

𝑑
1

0
 

𝑑
1

1
 

𝑑
1

2
 

𝑑
1

3
 

𝑑
1

4
 

𝑑
1

5
 

𝑑
1

6
 

𝑑
1

7
 

𝑑
1

8
 

𝑑
1

9
 

𝑑
2

0
 

𝑑
2

1
 

𝑑
2

2
 

𝑑
2

3
 

𝑑
2

4
 

𝑑
2

5
 

𝑑
2

6
 

𝑑
2

7
 

𝑑
2

8
 

𝑑
2

9
 

𝑑
3

0
 

𝑑
3

1
 

𝑑
3

2
 

𝑑
3

3
 

𝑑
3

4
 

𝑑
3

5
 

𝑑
3

6
 

𝑑
3

7
 

𝑑
3

8
 

𝑑
3

9
 

𝑑
4

0
 

𝑑
4

1
 

𝑑
4

2
 

𝑑
4

3
 

𝑑
4

4
 

𝑑
4

5
 

𝑑
4

6
 

𝑑
4

7
 

𝑑
4

8
 

𝑑
4

9
 

𝑑
5

0
 

𝑑
5

1
 

𝑑
5

2
 

𝑑
5

3
 

𝑑
5

4
 

 

Proof:  By inducing of p.i.s. 𝐷1, 𝐷27, 𝐷28,  𝐷3, 𝐷4, 𝐷5,  𝐷6,    𝐷179, 𝐷7, 𝐷8, 𝐷9, 𝐷193, 𝐷29, 𝐷30, 

𝐷11, 𝐷14, 𝐷15, 𝐷16, 𝐷17, 𝐷31, 𝐷32, 𝐷18, 𝐷33, 𝐷34, … , 𝐷38, 𝐷22, 𝐷23, . . . , 𝐷26 of  𝑆27 to 𝑆28. All i.m.s. 

are associated in block 𝐵1, since 〈28〉 ≠ 〈28〉′, according to (theorem 2.4) 〈28〉, 〈28〉′ have the 

same multiplicity, hence 𝑘1 = 𝑑1 + 𝑑2. Since〈23,3,2〉 ≠ 〈23,3,2〉′ so 𝑘2 or 𝑘3 is split. If 𝑘2 is split 

to 𝑑5, 𝑑6, but 〈22,4,2〉 ≠ 〈22,4,2〉′ then 𝑘3 split to, 𝑑7, 𝑑8. If 𝑘3 is split, and 

〈23,3,2〉 + 〈21,5,2〉 − 〈22,4,2〉 ≠ 〈23,3,2〉′ + 〈21,5,2〉′ − 〈22,4,2〉′ (13) 

then 𝑘2 , so in both cases we get 𝑘2 , 𝑘3  splits. Since〈21,5,2〉 ≠ 〈21,5,2〉′ so 𝑘4 or 𝑘5 is split. If 

𝑘4  is split to 𝑑9, 𝑑10, but 〈20,6,2〉 ≠ 〈20,6,2〉′ then 𝑘5 split to, 𝑑11, 𝑑12. If 𝑘5 is split, and  

21,5,2+19,7,2−〈20,6,2〉 ≠ 〈21,5,2〉′ + 〈19,7,2〉′ − 〈20,6,2〉′ (14) 

then 𝑘4 is split, so we get 𝑘4 , 𝑘5   splits. Since〈19,7,2〉 ≠ 〈19,7,2〉′ so 𝑘6 or 𝑘7 is split. If 𝑘7  is 

split to 𝑑15, 𝑑16, but 〈20,6,2〉 ≠ 〈20,6,2〉′ then 𝑘6 split to, 𝑑13, 𝑑14. If 𝑘6 is split, and  

〈19,7,2〉 − 〈20,6,2〉 ≠ 〈19,7,2〉′ − 〈20,6,2〉′ (15) 

then 𝑘7 is split, so we get 𝑘6 and 𝑘7  are splits.Since〈17,9,2〉 ≠ 〈17,9,2〉′ so 𝑘8 or 𝑘9 is split. If 𝑘8  

is split to 𝑑17, 𝑑18, but 〈16,10,2〉 ≠ 〈16,10,2〉′ then 𝑘9 split to, 𝑑19, 𝑑20. If 𝑘9 is split, and  

〈17,9,2〉 − 〈16,10,2〉 ≠ 〈17,9,2〉′ − 〈16,10,2〉′ (16) 

then 𝑘8 is split, then 𝑘8 , 𝑘9  splits. Since 〈16,10,2〉 ≠ 〈16,10,2〉′. then 𝑘10 = 𝑑21 + 𝑑22 has been 

divided or has two columns 𝜑1, 𝜑2, to explain these columns, since 〈16,10,2〉 ↓ 𝑆27 =

〈15,10,2〉∗1
+ 〈16,9,2〉∗2

+ 〈16,10,1〉∗4
 has 7 of i.m.s. we have 𝑎1 ∈ {0,1,2,3}.  In the same way 

we 𝑎24 ∈ {0,1}, 𝑎2, 𝑎5, 𝑎8, 𝑎10, 𝑎19 ∈ {0,1,2}, 𝑎14, 𝑎18,  𝑎21,  𝑎22,  𝑎23 ∈ {0,1,2,3}, 𝑎6, 𝑎7,

𝑎11,   𝑎15 ∈ {0,1, … ,4} 𝑎12, 𝑎13, 𝑎16, 𝑎17, 𝑎20 ∈ {0,1, … ,6},  𝑎4 ∈ {0,1, … ,7}, 𝑎3 ∈ {0,1, … ,8},

𝑎9 ∈ {0,1, … ,10}. Let 𝑎1 ∈ {1,2,3} (if𝑎1 = 0 contradiction). Since 〈16,10,2〉 ↓ 𝑆27 ∩ 〈15,13〉∗ ↓

𝑆23 has no i.m.s so 𝑎2 = 0, the same way we get 𝑎7, 𝑎8, 𝑎10, 𝑎11, … , 𝑎24 are equal to zero, and 

since inducing m.s. is m.s. then we get:  

(〈17,8,2〉∗ − 〈15,8,4〉∗ + 〈13,8,4,2〉) ↑(5,9) 𝑆28 ℎ𝑒𝑛𝑐𝑒 𝑎6 = 0 (17) 

(〈13,12,2〉∗ − 〈15,12〉 + 〈25,2〉) ↑(0,1) 𝑆28 ℎ𝑒𝑛𝑐𝑒 𝑎3 = 0 (18) 

Therefore, we only obtain degree 𝜑1, 𝜑2 ≡ 0 𝑚𝑜𝑑132 when 𝜑1 + 𝜑2 = 𝑚(𝑑21 + 𝑑22), 𝑚 ∈

{1,2}, which is basically the partition of 𝑘10 into 𝑑21, 𝑑22. 〈15,10,3〉 ≠ 〈15,10,3〉′ so 𝑘12 or 𝑘13 is 

split. If 𝑘12 split to 𝑑27, 𝑑28, but 〈15,9,4〉 ≠ 〈15,9,4〉′ then 𝑘13 split to, 𝑑29, 𝑑30. If 𝑘13  split, and  

〈15,8,5〉 + 〈15,10,3〉 + 〈17,9,2〉 + 〈21,5,2〉 + 〈23,3,2〉 − 〈15,9,4〉 − 〈16,10,2〉 −
〈18,8,2〉 − 〈22,4,2〉 ≠ 〈15,8,5〉′ + 〈15,10,3〉′ + 〈17,9,2〉′ + 〈21,5,2〉′ + 〈23,3,2〉′ −
〈15,9,4〉′ − 〈16,10,2〉′ − 〈18,8,2〉′ − 〈22,4,2〉′ 

 

 

(19) 

so that 𝑘12 is split, then 𝑘12 , 𝑘13  splits. Since〈15,8,5〉 ≠ 〈15,8,5〉′ so 𝑘14 or 𝑘15 is split. If 𝑘14 is 

split to 𝑑31, 𝑑32, but 〈15,7,6〉 ≠ 〈15,7,6〉′ then 𝑘15 split to, 𝑑33, 𝑑34. If 𝑘15 is split,  and  

〈15,8,5〉 − 〈18,8,2〉 − 〈20,6,2〉 − 〈15,7,6〉 + 〈19,7,2〉 + 〈21,5,2〉 ≠ 
〈15,8,5〉′ − 〈18,8,2〉′ − 〈20,6,2〉′ − 〈15,7,6〉′ + 〈19,7,2〉′ + 〈21,5,2〉′ 

 

(20) 



IHJPAS. 2024, 37(4)  

 

368 

 

then 𝑘14 split, so 𝑘14,𝑘15  splits. Since〈10,9,4,3,2〉 ≠ 〈10,9,4,3,2〉′ so 𝑘18 or 𝑘20 is split. If 𝑘18 is 

split to 𝑑47, 𝑑48, but 〈8,7,6,5,2〉 ≠ 〈8,7,6,5,2〉′ then 𝑘20 split to, 𝑑51, 𝑑52. If 𝑘20  is split, and  

〈10,9,4,3,2〉 − 〈8,7,6,5,2〉 ≠ 〈10,9,4,3,2〉′ − 〈8,7,6,5,2〉′ (21) 

then 𝑘18 is split, so 𝑘18, 𝑘20  splits. Since〈10,7,6,3,2〉 ≠ 〈10,7,6,3,2〉′ so 𝑘19 or 𝑘21 is split. If 𝑘19 

is split 𝑑49, 𝑑50, but 〈9,8,5,4,2〉 ≠ 〈9,8,5,4,2〉′ then 𝑘21 split, 𝑑53, 𝑑54. If 𝑘21 , is split and  

〈10,8,5,3,2〉 − 〈10,7,6,3,2〉 ≠ 〈10,8,5,3,2〉′ − 〈8,7,6,5,2〉′ (22) 

then 𝑘19 is split, so 𝑘19, 𝑘21  splits. Since〈12,9,4,2,1〉 ≠ 〈12,9,4,2,1〉′ so 𝑘16 or 𝑘17 is split. If 𝑘16 

split 𝑑37, 𝑑38, but 〈10,8,5,3〉 ≠ 〈10,8,5,3〉′ then 𝑘17 split 𝑑45, 𝑑46. If 𝑘17  split, and  

〈10,9,4,3,2〉 + 〈10,7,6,3,2〉 − 〈9,7,6,4,2〉 − 〈10,8,5,3,2〉 + 〈9,8,5,4,2〉 + 〈8,7,6,5,2〉 ≠ 
〈10,9,4,3,2〉′ + 〈10,7,6,3,2〉′ − 〈9,7,6,4,2〉′ − 〈10,8,5,3,2〉′ − 〈9,8,5,4,2〉′ + 〈8,7,6,5,2〉′ 

 

(23) 

then 𝑘19 is split, so 𝑘19, 𝑘21 splits. Since 〈15,11,2〉 ≠ 〈15,11,2〉′ on (13, 𝛼)-regular classes and we 

have 294 columns in the decomposition matrix, then  𝑘11  must be split to 𝑑25, 𝑑26.  

 

Conclusions 

      There is no prescribed method to find irreducible modular spin properties when the field 

property is primary, especially when the investigation concerns the same field with a group change. 

As a result, we had to conduct a series of studies to collect enough information to find new 

properties and theorems, including decomposition matrices that establish a connection between 

irreducible spin characteristics and irreducible modular spin characteristics. This opens the way for 

a comprehensive investigation that first examines the properties of irreducible modular spins before 

classifying entities. 

 

Acknowledgment 

         The authors are very grateful to the executive manager and editorial board members of the 

Ibn AL-Haitham Journal of Pure and Applied Sciences. 

 

Conflict of Interest  

         The authors declare that they have no conflicts of interest. 

 

Funding 

        There is no funding for the article. 
 

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