# Preliminary Classification of Sub-case (3)

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## Abstract

In this chapter the case (3) is studied in more details. It is shown that if the element *b* _{10} appearing in the product *b* _{3} *b* _{8} is non-real, then *b* _{3} *b* _{10} contains two constituents one of which is a non-real element of degree 15.

## 4.1 Introduction

In Chap. 4 we shall freely use the definitions and notation used in Chap. 2. The Main Theorem 2 of Chap. 3 left two unsolved problems for the complete classification of the Normalized Integral Table Algebra (NITA) (*A*,*B*) generated by a faithful non-real element of degree 3 with *L* _{1}(*B*)=1 and *L* _{2}(*B*)=∅. In this chapter we solve the second problem where (*b* _{3} *b* _{10},*b* _{3} *b* _{10})≠2, and prove that the NITA that satisfies case (2) of the Main Theorem 1 of Chap. 2 where (*b* _{3} *b* _{10},*b* _{3} *b* _{10})≠2 does not exist. Consequently, we can state the Main Theorem 3 of this chapter as follows.

### Main Theorem 3

*Let*(

*A*,

*B*)

*be a NITA generated by a non*-

*real element*

*b*

_{3}∈

*B*

*of degree*3

*and without non*-

*identity basis element of degree*1

*and*2.

*Then*\(b_{3} \bar{b}_{3} = 1 + b_{8}\),

*b*

_{8}∈

*B*,

*and one of the following holds*:

- (1)
*There exists a real element**b*_{6}∈*B**such that*\(b_{3}^{2} =\bar{b}_{3} + b_{6}\)*and*(*A*,*B*)≅_{ x }(*CH*(*PSL*(2,7)),*Irr*(*PSL*(2,7))). - (2)
*There exist**b*_{6},*b*_{10},*b*_{15}∈*B*,*where**b*_{6}*is non*-*real*,*such that*\(b_{3}^{2} = \bar{b}_{3} + b_{6}\), \(\bar{b}_{3} b_{6} = b_{3} + b_{15}\),*b*_{3}*b*_{6}=*b*_{8}+*b*_{10},*and*(*b*_{3}*b*_{8},*b*_{3}*b*_{8})=3.*Moreover*,*if**b*_{10}*is real then*(*A*,*B*)≅_{ x }(*CH*(3⋅*A*_{6}),*Irr*(3⋅*A*_{6}))*of dimension*17,*and if**b*_{10}*is non*-*real then*(*b*_{3}*b*_{10},*b*_{3}*b*_{10})=2*and**b*_{15}*is a non*-*real element*. - (3)
*There exist**c*_{3},*b*_{6}∈*B*,*c*_{3}≠*b*_{3}*or*\(\bar{b}_{3}\),*such that*\(b_{3}^{2} = c_{3} + b_{6}\)*and either*(*b*_{3}*b*_{8},*b*_{3}*b*_{8})=3*or*4.*If*(*b*_{3}*b*_{8},*b*_{3}*b*_{8})=3*and**c*_{3}*is non*-*real*,*then*(*A*,*B*)≅_{ x }(*A*(3⋅*A*_{6}⋅2),*B*_{32})*of dimension*32. (*See Theorem*2.9*of Chap*. 2*for the definition of this specific NITA*.)*If*(*b*_{3}*b*_{8},*b*_{3}*b*_{8})=3*and**c*_{3}*is real*,*then*(*A*,*B*)≅_{ x }(*A*(7⋅5⋅10),*B*_{22})*of dimension*22. (*See Theorem*2.10*of Chap*. 2*for the definition of this specific NITA*.)

In the above Main Theorem 3 we still have 2 open problems in the cases (2) and (3). In case (2) we must classify the NITA such that *b* _{10} is non-real and (*b* _{3} *b* _{10},*b* _{3} *b* _{10})=2, and in case (3) we must classify the NITA such that (*b* _{3} *b* _{8},*b* _{3} *b* _{8})=4. In Chap. 5, the open case (2) will be solved.

Let us emphasize that the NITA’s of dimension 7 and 17 in the cases (1) and (2) of the Main Theorem 2 are strictly isomorphic to the NITA’s induced from finite groups *G* via the basis of the irreducible characters of *G*. However, the NITA’s of dimensions 22 and 32 are not induced from finite groups as described in Chap. 2.

The Main Theorem 3 follows from the Main Theorem 2 in Chap. 3 and the next theorem.

### Theorem 4.1

*Let*(

*A*,

*B*)

*be a NITA generated by a non*-

*real element*

*b*

_{3}∈

*B*

*of degree*3

*and without non*-

*identity basis element of degree*1

*or*2.

*Then*\(b_{3}\bar{b}_{3}=1+b_{8}\),

*b*

_{8}∈

*B*.

*Assume that*

*where*

*b*

_{6},

*b*

_{10}

*are non*-

*real elements in*

*B*

*and*

*b*

_{15}∈

*B*.

*Then*

*b*

_{15}

*is a non*-

*real element and*(

*b*

_{3}

*b*

_{10},

*b*

_{3}

*b*

_{10})=2.

The rest of this chapter is devoted to proving the above theorem.

## 4.2 Preliminary Results

*A*,

*B*) is a NITA generated by a non-real element

*b*

_{3}∈

*B*of degree 3 and without non-identity basis element of degree 1 and 2 such that

*R*

_{15}must all be ≥5.

*b*

_{15}is non-real is simple.

*b*

_{15}is a real element and we shall derive a contradiction. By (4.3) we have that (

*b*

_{6},

*b*

_{3}

*b*

_{15})=1 and together with (4.12), we obtain that \((\bar{b}_{6},b_{3}b_{15})=1\) which implies that (

*b*

_{15},

*b*

_{3}

*b*

_{6})=1, and we have a contradiction to (4.4). Therefore

The remainder of this chapter will consist of the proof that (*b* _{3} *b* _{10},*b* _{3} *b* _{10})=2.

*b*

_{3}

*b*

_{10},

*b*

_{3}

*b*

_{10})≥2. By (4.4) we obtain that \((\bar{b}_{6},b_{3}b_{10})=0\). Now by (4.3), (4.4) and (4.7) we obtain that \((b_{3}b_{10},b_{3}b_{10})=(b_{6}^{2},b_{3}b_{10})=(\bar{b}_{3}b_{6},\bar{b}_{6}b_{10})=(b_{3},\bar{b}_{6}b_{10})+(b_{15},\bar{b}_{6}b_{10})=1+(b_{15},\bar{b}_{6}b_{10})\). Since \((b_{3},\bar{b}_{6}b_{10})=1\), we have that \((b_{15},\bar{b}_{6}b_{10})\leq 3\) which implies that (

*b*

_{3}

*b*

_{10},

*b*

_{3}

*b*

_{10})≤4. Therefore

Assume henceforth that *b* _{15}=*R* _{15} and we shall derive a contradiction. Then by (4.8) we obtain that *b* _{3} *b* _{10}=2*b* _{15} which implies that \((\bar{b}_{10},b_{3}\bar{b}_{15})=2\), and by (4.12) we obtain that \((\bar{b}_{10},b_{6}\bar{b}_{6})=1\) which implies that \((b_{10},b_{6}\bar{b}_{6})=1\). Now (4.12) implies that \((\bar{b}_{15},\bar{b}_{3}b_{10})=(b_{10},b_{3}\bar{b}_{15})=1\). By (4.4) we obtain that \((b_{6},\bar{b}_{3}b_{10})=1\) and since (*b* _{3} *b* _{10},*b* _{3} *b* _{10})=4, we obtain that \(\bar{b}_{3}b_{10}=b_{6}+\bar{b}_{15}+\alpha+\beta\) where *α*,*β*∈*B*, |*α*|+|*β*|=9 and |*α*|,|*β*|≥5, which is a contradiction to our assumption that *b* _{15}=*R* _{15}.

Splitting to five cases

Case 1 | |

Case 2 | |

Case 3 | |

Case 4 | |

Case 5 | |

In the rest of this chapter, we derive a contradiction for Cases 1–4 in the above table. In Case 5, by (4.8) we then have that (*b* _{3} *b* _{10},*b* _{3} *b* _{10})=(*b* _{15},*b* _{15})+(*R* _{15},*R* _{15})=2. Then we will have proven Theorem 1.1.

## 4.3 Case *R* _{15}=*x* _{5}+*x* _{10}

*R*

_{15}=

*x*

_{5}+

*x*

_{10}. Then by (4.7), (4.8) and (4.13), we have that

*b*

_{3}

*x*

_{5},

*b*

_{3}

*x*

_{5})=2, by (4.23) we have that \((b_{3}x_{5},\bar{b}_{3}y_{5})=1\). Hence

*b*

_{3}

*b*

_{10},

*b*

_{3}

*b*

_{10})=3 we have a contradiction.

*b*

_{3}

*x*

_{5}=

*c*

_{15}, which is a contradiction to (4.19). Now (4.29) implies that

*b*

_{8}

*x*

_{5},

*b*

_{3}

*y*

_{5})≥2. In addition, (4.5), (4.19) and (4.31) imply that \((b_{8}x_{5},b_{3}y_{5})=(\bar{b}_{3}b_{8},y_{5}\bar{x}_{5})=(\bar{b}_{3},y_{5}\bar{x}_{5})+(b_{6},y_{5}\bar{x}_{5})+(\bar{b}_{15},y_{5}\bar{x}_{5})=1+(b_{6},y_{5}\bar{x}_{5})\). Hence

*x*

_{5},

*b*

_{3}

*y*

_{5})=1. Now (4.30) implies that

*x*=

*z*

_{5}and

*y*=

*z*

_{10}, which imply that

**Case 1**

**Case 2**

*z*

_{10},

*b*

_{3}

*b*

_{15})=1, by (4.3) we have that (

*b*

_{6},

*b*

_{3}

*b*

_{15})=1 and together with (4.10), we obtain that

*b*

_{10},

*b*

_{3}

*z*

_{10})=1, by (4.39) we obtain that (

*y*

_{5},

*b*

_{3}

*z*

_{10})=1, and by (4.40) and (4.41) we obtain that (

*b*

_{3}

*z*

_{10},

*b*

_{3}

*z*

_{10})=3. Thus

*y*

_{5},

*b*

_{3}

*z*

_{5})=1, and by (4.40) and (4.41) we obtain that (

*b*

_{3}

*z*

_{5},

*b*

_{3}

*z*

_{5})=2 which implies that

## 4.4 Case *R* _{15}=*x* _{6}+*x* _{9}

*R*

_{15}=

*x*

_{6}+

*x*

_{9}. Then by (4.7), (4.8) and (4.13), we have that

Assume henceforth that \(\bar{b}_{3}x_{6}=b_{10}+2b_{4}\) and we shall derive a contradiction. Then *b* _{3} *b* _{4}=2*x* _{6} which implies by (4.2) that \(4=(b_{3}b_{4},b_{3}b_{4})=(b_{3}\bar{b}_{3},b_{4}\bar{b}_{4})=1+(b_{8},b_{4}\bar{b}_{4})\). Thus \((b_{8},b_{4}\bar{b}_{4})=3\), and we have a contradiction.

*b*

_{3}

*x*

_{6},

*b*

_{3}

*x*

_{6})=2, so we can write that

*s*|,|

*t*|≥5.

Assume henceforth that *s*=*s* _{5} and we shall derive a contradiction. Then by (4.68) we obtain that \(b_{3}\bar{b}_{15}=b_{8}+\bar{b}_{10}+s_{5}+t_{22}\) and *b* _{3} *s* _{5}=*b* _{15}. Hence (4.2) and \(\bar{b}_{3}(b_{3}s_{5})=(b_{3}\bar{b}_{3})s_{5}\) imply that *b* _{8} *s* _{5}=*b* _{8}+*b* _{10}+*t* _{22}. Since *b* _{8} and *s* _{5} are real elements and *b* _{10} is non-real, we have a contradiction.

*s*=

*s*

_{6}and we shall derive a contradiction. Then by (4.68) we obtain that \(b_{3}\bar{b}_{15}=b_{8}+\bar{b}_{10}+s_{6}+t_{21}\) and

*b*

_{3}

*s*

_{6}=

*b*

_{15}+

*c*

_{3}where

*c*

_{3}∈

*B*. Now \(\bar{b}_{3}(b_{3}s_{6})=(b_{3}\bar{b}_{3})s_{6}\) implies that \(b_{8}s_{6}=b_{8}+b_{10}+t_{21}+\bar{b}_{3}c_{3}\). Since

*b*

_{8}and

*s*

_{6}are reals and

*b*

_{10}is non-real, we obtain that \((\bar{b}_{10},\bar{b}_{3}c_{3})=1\) and we have a contradiction, which implies that

*z*

_{7}∈

*B*. Now (4.65) implies that \((\bar{b}_{6}x_{6},\bar{b}_{6}x_{6})\leq 3\) and together with (4.71) we obtain that \((\bar{b}_{6}x_{6},\bar{b}_{6}x_{6})=3\), and by (4.61), (4.67) and \((b_{10},x_{6}\bar{x}_{6})=1\) we obtain that \(\bar{b}_{6}x_{6}=b_{6}+c+d\), which is a contradiction. Thus (4.67) implies that

*b*

_{3}

*b*

_{10},

*b*

_{3}

*b*

_{10})=3 imply that \(\bar{b}_{3}b_{10}=b_{6}+c+d\) which is a contradiction, and we obtain that \((\bar{b}_{3}y_{8},b_{3}x_{6})=2\). Now we have that either \((c,\bar{b}_{3}y_{8})=2\) or \((c,\bar{b}_{3}y_{8})=(d,\bar{b}_{3}y_{8})=1\), which implies that (

*b*

_{3}

*y*

_{8},

*b*

_{3}

*y*

_{8})≥3. By (4.11), (4.66) and (4.72), we obtain that \((b_{8}x_{6},b_{8}x_{6})=(b_{8}^{2},x_{6}\bar{x}_{6})=1+(b_{6}\bar{b}_{6},x_{6}\bar{x}_{6})\). Now (4.71) implies that (

*b*

_{8}

*x*

_{6},

*b*

_{8}

*x*

_{6})≤6 together with (4.63) and since (

*b*

_{3}

*y*

_{8},

*b*

_{3}

*y*

_{8})≥3, we have that

*b*

_{3}

*y*

_{8},

*b*

_{3}

*y*

_{8})≤4. If \((c,\bar{b}_{3}y_{8})=2\) then \(\bar{b}_{3}y_{8}=2c_{12}\) and \((\bar{b}_{6}x_{6},\bar{b}_{6}x_{6})=5\). Now (4.64) implies that \(\bar{b}_{3}b_{10}=b_{6}+d_{6}+f_{18}\) where

*f*

_{18}∈

*B*and \(\bar{b}_{6}x_{6}=b_{6}+c_{12}+f_{18}\), which is a contradiction.

*b*

_{3}

*y*

_{8},

*b*

_{3}

*y*

_{8})=3 which implies by (4.75) and (4.81) that 2≤(

*b*

_{3}

*c*,

*b*

_{3}

*c*)≤4. By (4.77) and (4.78), we obtain that (

*b*

_{10},

*b*

_{3}

*c*)=(

*y*

_{8},

*b*

_{3}

*c*)=1 which implies that (

*b*

_{3}

*c*,

*b*

_{3}

*c*)≠2.

*b*

_{3}

*y*

_{8},

*b*

_{3}

*y*

_{8})=3 and by (4.62), we obtain that \((y_{8},\allowbreak\bar{b}_{3}x_{6})=\nobreak 1\) which implies that

*b*

_{3}

*c*,

*b*

_{3}

*c*)=3. Then by (4.61) and (4.80) we obtain that

*c*|≥9, and by (4.84) we obtain that \(g+\bar{b}_{3}d=x_{6}+x_{9}+\alpha+\beta\). If

*g*=

*x*

_{9}then by (4.85) we obtain that |

*c*|=10, and if

*g*≠

*x*

_{9}then we obtain that either

*g*=

*α*or

*g*=

*β*. If

*α*=3=

*α*

_{3}, then (4.5) and (4.83) imply that

*α*=

*b*

_{3},

*y*

_{8}=

*b*

_{8}and \(x_{6}=\bar{b}_{6}\) which implies by (4.57) that \(b_{3}b_{10}=b_{15}+\bar{b}_{6}+x_{9}\). Thus \((\bar{b}_{10},b_{3}b_{6})=1\), and we have a contradiction to (4.4). In the same way, we can show that |

*β*|≠3 which implies by (4.83) that

*g*|≤12 and |

*c*|≤11. Therefore

*b*

_{3}

*c*,

*b*

_{3}

*c*)=4. Then by (4.62) and by (4.68), we obtain that

*g*+

*h*=

*α*+

*β*which implies by (4.83) and (4.88) that |

*c*|=13. If \((x_{9},\bar{b}_{3}d)=0\) then we can assume that

*g*+

*h*=

*x*

_{9}+

*α*. Now from 4≤|

*α*|≤14 and by (4.88) we obtain that

Splitting to seven cases

Case 1 | | | | ( |

Case 2 | | | | ( |

Case 3 | | | | ( |

Case 4 | | | | ( |

Case 5 | | | | ( |

Case 6 | | | | ( |

Case 7 | ( |

*c*|≤11.

### 4.4.1 Case *c*=*c* _{9}

*c*=

*c*

_{9}. Then (4.77), (4.78) and (4.91) imply that

*b*

_{3}

*c*

_{9}=

*y*

_{8}+

*b*

_{10}+

*q*

_{9}where

*q*

_{9}∈

*B*which implies that

*c*

_{3}∈

*B*,

*b*

_{3}

*c*

_{3}=

*h*

_{9}and (

*c*

_{3},

*b*

_{3}

*g*

_{6})=1. Since

*b*

_{3}

*c*

_{3}=

*h*

_{9}we obtain that (

*b*

_{3}

*c*

_{3},

*b*

_{3}

*c*

_{3})=1, which is a contradiction to (

*c*

_{3},

*b*

_{3}

*g*

_{6})=1.

### 4.4.2 Case *c*=*c* _{10}

*c*=

*c*

_{10}. Then (4.77), (4.78) and (4.91) imply that

*q*

_{12}is a real element. By (4.62), (4.77), (4.83), (4.93) and \(b_{3}(\bar{b}_{3}y_{8})=\bar{b}_{3}(b_{3}y_{8})\) we obtain that \(q_{12}+b_{3}d_{8}+b_{3}e_{6}=\bar{b}_{3}\alpha+\bar{b}_{3}\beta\). Now we can assume that \((q_{12},\bar{b}_{3}\alpha)=1\) and by (4.83) we obtain that \((y_{8},\bar{b}_{3}\alpha)=1\) which implies that |

*α*|≥8, and since

*q*

_{12}is a real element we obtain that

Assume henceforth that \(\alpha=\bar{c}_{10}\) and we shall derive a contradiction. Then by (4.81), (4.83) and (4.94) we obtain that \((\bar{c}_{10},\bar{b}_{3}d_{8})=1\), and we have a contradiction to (4.94). By (4.86) we obtain that *α*≠*b* _{15}. Now by (4.93), (4.95) and (4.96) we obtain that \((c_{10},\bar{b}_{3}q_{12})=1\), \((\bar{\alpha},\bar{b}_{3}q_{12})=1\) and \((\bar{b}_{15},\bar{b}_{3} q_{12})=1\) which imply that either |*α*|≤7 or |*α*|=11. Now |*α*|≥8 implies that *α*=*α* _{11} and by (4.83) we obtain that *β*=*β* _{7}.

*q*

_{15}∈

*B*. Hence (4.2), (4.93) and \(\bar{b}_{3}(b_{3}q_{12})=(b_{3}\bar{b}_{3})q_{12}\) imply that

*Σ*

_{13}∈ℕ

*B*and

*b*

_{3}

*d*

_{8}=

*y*

_{8}+

*Σ*

_{13}+

*ν*

_{3}where

*ν*

_{3}∈

*B*which implies that \((\nu_{3},\allowbreak\bar{b}_{3}\beta_{7})=\nobreak 1\) and we have a contradiction. Now we have that \((\bar{b}_{3}\alpha_{11},b_{3}e_{6})\geq 2\).

If *b* _{3} *e* _{6}=*y* _{8}+*r* _{10} where *r* _{10}∈*B*, then \(\bar{b}_{3}\alpha_{11}=c_{3}+y_{8}+r_{10}+q_{12}\) and we have a contradiction.

If *b* _{3} *e* _{6}=*y* _{8}+*r* _{4}+*r* _{6} where *r* _{4},*r* _{6}∈*B*, then \((r_{6},\bar{b}_{3}\alpha_{11})=1\) and (4.98) implies that *r* _{6} is a real element, and we have that \((\bar{e}_{6},b_{3}r_{6})=(\alpha_{11},b_{3}r_{6})=1\), which is a contradiction.

If *b* _{3} *e* _{6}=*y* _{8}+*r* _{5}+*γ* _{5} where *r* _{5},*γ* _{5}∈*B* then we have that \((\bar{e}_{6},b_{3}r_{5})=(\alpha_{11},b_{3}r_{5})=1\) and we have a contradiction.

### 4.4.3 Case *c*=*c* _{11}

*c*=

*c*

_{11}. Then (4.77), (4.78) and (4.91) imply that

*q*

_{15}is a real element, and we obtain that

*β*=

*g*

_{12}.

*q*

_{15}is a real element, (4.101) implies that

*q*

_{15}is a real element, we have that

*c*|≤14.

### 4.4.4 Case *c*=*c* _{12}

*c*=

*c*

_{12}. Then by (4.61), (4.80), (4.81) and (4.111), we obtain that

*b*

_{3}

*g*

_{6},

*b*

_{3}

*g*

_{6})=3 and we shall derive a contradiction. Then (4.112) implies that

*b*

_{3}

*g*

_{6}=

*c*

_{12}+

*v*

_{3}+

*w*

_{3}and \(\bar{b}_{3}g_{6}=y_{8}+\nu+\mu\) where

*v*

_{3},

*w*

_{3},

*ν*,

*μ*∈

*B*. Now by (4.112), (4.113) and \(\bar{b}_{3}(b_{3}g_{6})=b_{3}(\bar{b}_{3}g_{6})\), we obtain that \(x_{9}+b_{15}+\bar{b}_{3}v_{3}+\bar{b}_{3}w_{3}=\beta_{12}+b_{3}\nu+b_{3}\mu\), which is a contradiction. Now we have that (

*b*

_{3}

*g*

_{6},

*b*

_{3}

*g*

_{6})=2, by (4.112) and (4.113) we obtain that

*v*

_{10}and

*v*

_{17}are real elements. By (4.77), (4.78) and (4.111) we obtain that

*q*≠

*y*

_{8},

*b*

_{10}, and by (4.118) we obtain that |

*q*|≠17. Thus

*q*≠

*v*

_{17}which implies that

*q*=

*v*

_{10}. Now since

*v*

_{10}is a real element, (4.118) implies \((\bar{c}_{12},b_{3}v_{10})=1\) and by (4.115) we obtain that (

*g*

_{6},

*b*

_{3}

*v*

_{10})=1, which is a contradiction to (4.116).

### 4.4.5 Case *c*=*c* _{13}

*c*=

*c*

_{13}. Then by (4.77) we obtain that

*b*

_{3}

*d*

_{5},

*b*

_{3}

*d*

_{5})=2 and together with (4.119), we obtain that

*r*

_{9}+

*b*

_{3}

*z*

_{7}=

*b*

_{6}

*d*

_{5}. By (4.65), (4.69) and (4.70), we obtain that (

*b*

_{6}

*d*

_{5},

*b*

_{6}

*d*

_{5})≤3 which implies that

*4.124*) we obtain that \((y_{8},\bar{b}_{3}z_{9})=(y_{8},\bar{b}_{3}r_{9})=1\). Now we can assume by (4.127) that

*u*|≤12.

Assume henceforth that \((u,b_{3}\bar{b}_{15})=1\) and we shall derive a contradiction. Then by (4.12) we obtain that *u* is a real element. By (4.130) we obtain that (*z* _{9},*b* _{3} *u*)=1, by (4.126) we obtain that \((\bar{c}_{13},b_{3}u)=1\), and by the assumption we obtain that (*b* _{15},*b* _{3} *u*)=1 which implies that |*u*|≥14, and we have a contradiction to |*u*|≤12.

*q*is a real element. By (4.130) we obtain that (

*r*

_{9},

*b*

_{3}

*q*)=1, by (4.126) we obtain that \((\bar{c}_{13},b_{3}q)=1\), and together with (

*b*

_{15},

*b*

_{3}

*q*)=1 we obtain that |

*q*|≥14.

*u*|≥6. Now (4.129) implies that |

*q*|≤15, and we obtain that

*r*

_{9},

*b*

_{3}

*λ*)=1 and 4≤

*λ*≤5, we obtain that (

*x*

_{9},

*b*

_{3}

*λ*)=0 which implies that either

*x*

_{9}=

*z*

_{9}or (

*x*

_{9},

*b*

_{3}

*q*)=1. If

*x*

_{9}=

*z*

_{9}then by (4.124), (4.128) and (4.57), we obtain that \((y_{8},\bar{b}_{3}x_{9})=(z_{7},\bar{b}_{3}x_{9})=(b_{10},\bar{b}_{3}x_{9})=1\) and we have a contradiction.

Now we have that (*x* _{9},*b* _{3} *q*)=1. By (4.126), (4.130) and (4.131) we obtain that \((b_{15},b_{3}q)=(\bar{c}_{13},b_{3}q)=(r_{9},b_{3}q)=1\), and we have a contradiction to (*x* _{9},*b* _{3} *q*)=1.

*b*

_{6}

*d*

_{5},

*b*

_{6}

*d*

_{5})≤3 which implies that

*x*

_{9}=

*α*. By (4.57), (4.136) and (4.137) we obtain that \((z_{7},\bar{b}_{3}x_{9})=(b_{10},\bar{b}_{3}x_{9})=(y_{8},\bar{b}_{3}x_{9})=1\), which is a contradiction.

### 4.4.6 Case *c*=*c* _{14}

*c*=

*c*

_{14}. Then by (4.77) we obtain that

*q*|+|

*u*|=24, and by (4.142) we obtain that \((y_{8},\allowbreak \bar{b}_{3}y_{12})=\nobreak 1\).

Assume henceforth that \((u,\bar{b}_{3}y_{12})=(q,\bar{b}_{3}y_{12})=1\) and we shall derive a contradiction. Then \(\bar{b}_{3}y_{12}=q+u+y_{8}+z_{4}\) where *z* _{4}∈*B* which implies that *b* _{3} *z* _{4}=*y* _{12} and by (4.143) and (4.145) we obtain that (*z* _{4},*b* _{3} *e* _{6})=1, which is a contradiction to *b* _{3} *z* _{4}=*y* _{12}.

### 4.4.7 Case (*b* _{3} *x* _{6},*b* _{3} *x* _{6})=3

*b*

_{3}

*x*

_{6},

*b*

_{3}

*x*

_{6})=3 Then

*c*|≥7, and by (4.147) we obtain that |

*c*|≤12. Now we have that

*v*=

*v*

_{3}and we shall derive a contradiction. Then by (4.148) we obtain that

*α*≠

*b*

_{6}and |

*α*|≤6, so (4.151) and (4.152) imply that

*α*≠

*c*and

*α*≠

*f*. Now we have a contradiction to (4.152) and (4.155).

*f*≤17, and together with (

*b*

_{3}

*v*

_{4},

*b*

_{3}

*v*

_{4})=2 we obtain that \((f,\bar{b}_{3}v_{4})=0\). Now we have that

*c*|≤9, and together with (4.151) we obtain that

*c*=

*c*

_{7}then by (4.147) and (4.153) we obtain that \(\bar{b}_{3}c_{7}=b_{15}+x_{6}\), and by (4.152) and (4.159) we obtain that

*b*

_{3}

*c*

_{7}=

*v*

_{4}+

*b*

_{10}, which is a contradiction.

If *c*=*c* _{8} then by (4.147) and (4.153), we obtain that \(\bar{b}_{3}c_{8}=b_{15}+x_{6}+r_{3}\) where *r* _{3}∈*B*. Now we have that \(r_{3}=\bar{b}_{3}\) and *b* _{8}=*c* _{8}, since *b* _{15} is a non-real element we have a contradiction to (4.5).

*c*=

*c*

_{9}then by (4.147) and (4.153), we obtain that

*v*

_{4},

*b*

_{3}

*m*

_{3})=1 implies that (

*w*

_{4},

*b*

_{3}

*m*

_{3})=0, so (

*w*

_{4},

*Σ*

_{13})=1. By (4.161) we obtain that

*b*

_{3}

*w*

_{4}=

*x*

_{6}+

*l*

_{3}+

*n*

_{3}, which is a contradiction to (

*b*

_{3}

*w*

_{4},

*b*

_{3}

*w*

_{4})≤2.

## 4.5 Case *R* _{15}=*x* _{7}+*x* _{8}

*R*

_{15}=

*x*

_{7}+

*x*

_{8}. Then by (4.7) and (4.8) we have that

*s*=

*s*

_{21}and we shall derive a contradiction. Then by (4.165)

*t*=

*t*

_{6}and

*b*

_{3}

*t*

_{6}=

*b*

_{15}+

*c*

_{3}where

*c*

_{3}∈

*B*. Now \(\bar{b}_{3}(b_{3}t_{6})=(b_{3}\bar{b}_{3})t_{6}\) implies that \(b_{8}t_{6}=b_{8}+b_{10}+s_{21}+\bar{b}_{3}c_{3}\). Since

*b*

_{8}and

*t*

_{6}are real elements and

*b*

_{10}is a non-real element, we obtain that \((\bar{b}_{10},\bar{b}_{3}c_{3})=1\) and we have a contradiction. Now we have that

*b*

_{3}

*x*

_{7},

*b*

_{3}

*x*

_{7})=2. Then we obtain that

*b*

_{15},

*b*

_{3}

*y*

_{11})=1 and we shall derive a contradiction. Then we have that \((b_{15},\bar{b}_{3}d)=1\). By (4.171) we obtain that

*e*≠

*d*, by (4.169) and (4.3) we obtain that

*b*

_{6}≠

*d*and by (4.173) and (4.169) we obtain that \(d\neq\bar{b}_{15}\), which is a contradiction to (4.172). Now we have that

*y*

_{11}is a real element which implies that (

*b*

_{15},

*b*

_{3}

*y*

_{11})=1, and we have a contradiction to (4.175). Now we have that

*u*∈

*B*. Then by (4.174) we obtain that |

*u*|≥5.

*u*=

*u*

_{5}and we shall derive a contradiction. Then by (4.174) we obtain that \(\bar{b}_{3}u_{5}=y_{11}+y_{4}\) where

*y*

_{4}∈

*B*and

*b*

_{3}

*u*

_{5}=

*c*+

*q*where

*q*∈

*B*. Now by \(b_{3}(\bar{b}_{3}u_{5})=\bar{b}_{3}(b_{3}u_{5})\) we obtain that \(\bar{b}_{3}c+\bar{b}_{3}q=b_{3}y_{11}+b_{3}y_{4}\). Now by (4.172) we obtain that \((b_{15},\bar{b}_{3}c)=1\) and by (4.175) we obtain that (

*b*

_{15},

*b*

_{3}

*y*

_{11})=0 which implies that (

*b*

_{15},

*b*

_{3}

*y*

_{4})=1, and we have a contradiction. Hence

*c*=

*c*

_{9}and we shall derive a contradiction. Then by (4.169) and (4.172) we obtain that \(\bar{b}_{3}c_{9}=x_{7}+b_{15}+u_{5}\) where

*u*

_{5}∈

*B*, which is a contradiction to (4.178). Now (4.173) implies that

*d*≠

*b*

_{6}and by (4.171) we obtain that

*d*≠

*e*. Now (4.177) implies that \((d,\bar{b}_{3}y_{11})=1\) and by (4.180) we obtain that

*b*

_{10}is a non-real element we obtain that \(e\neq\bar{x}_{7}\), \(e\neq\bar{x}_{8}\) and \(e\neq\bar{b}_{15}\). Thus \((\bar{b}_{10},b_{3}e)=(\bar{b}_{3}\bar{b}_{10},e)=0\) and \((b_{8},b_{3}e)=(\bar{b}_{3}b_{8},e)=\nobreak 0\). Now by (4.165), (4.183) and (4.186) we can assume that (

*s*,

*C*)=1 which implies that

*C*|=3|

*e*|−21 which implies that

*e*|.

Assume henceforth that *e*=*e* _{11} and we shall derive a contradiction. Then by (4.189) we obtain that 11≤|*s*| and by (4.183) we obtain that |*C*|=12 which implies by ( *4.188* ) that |*s*|≤12. Hence 11≤|*s*|≤12 which implies by (4.164) that 15≤|*t*|≤16, and we have a contradiction to (4.181).

*e*=

*e*

_{14}and we shall derive a contradiction. Then by (4.189) we obtain that 12≤|

*s*| and by (4.183) we obtain that |

*C*|=21 which implies by (4.188) that |

*s*|≤21. Therefore

*s*|≤15, and by (4.164) we obtain that 12≤|

*t*|≤15, which is a contradiction to (4.181).

*e*=

*e*

_{12}and we shall derive a contradiction. Then by (4.189) we obtain that 9≤|

*s*| and by (4.188) we obtain that |

*s*|≤15. Hence 9≤|

*s*|≤15. Now (4.164) and (4.181) imply that 9≤|

*s*|≤10. By (4.189) it is impossible that

*s*=

*s*

_{10}. Therefore

*s*=

*s*

_{9}and \(\bar{b}_{3}s_{9}=e_{12}+\bar{b}_{15}\) and by (4.183) we obtain that

*b*

_{8}and

*s*

_{9}are real elements we obtain that

*y*

_{11}is a real element. By (4.172) we obtain that

*c*=

*c*

_{12}and by (4.169)

*d*=

*d*

_{9}. Now (4.170) and (4.182) imply that

*w*

_{5},

*b*

_{3}

*w*

_{6})=1 we obtain that (

*e*

_{12},

*b*

_{3}

*w*

_{6})=0 which implies that \((e_{12},b_{3}\bar{d}_{9})=1\). By (4.169) we obtain that \((\bar{x}_{7},b_{3}\bar{d}_{9})=1\) and together with \((w_{5},b_{3}\bar{d}_{9})=1\), we obtain that

*e*=

*e*

_{13}which implies by (4.188) that |

*s*|≤18, and by (4.189) we obtain that 11≤|

*s*|. Therefore 11≤|

*s*|≤18. Now (4.181) and (4.164) imply that 17≤|

*s*|≤18. By (4.183) and (4.187) we obtain that |

*s*|≠17. Thus

*s*=

*s*

_{18}and by (4.165) we obtain that

*t*=

*t*

_{9}and

*b*

_{8}and

*t*

_{9}are real elements and

*b*

_{10}is a non-real element, we obtain that \((\bar{b}_{10},b_{3}E_{12})=1\), which is a contradiction to (4.171).

*y*

_{11},

*b*

_{3}

*c*)=1 and together with (4.171) we obtain that

*c*=

*c*

_{10}then by (4.197) we obtain that

*H*=

*H*

_{8}∈

*B*and (

*c*

_{10},

*b*

_{3}

*H*

_{8})=1, which is a contradiction to (4.198), and if

*c*=

*c*

_{17}then (4.169) implies that

*d*=

*d*

_{4}, which is a contradiction to (4.196). Now (4.179) implies that

Splitting to seven cases

Case 1 | | | |

Case 2 | | | |

Case 3 | | | |

Case 4 | | | |

Case 5 | | | |

Case 6 | | | |

Case 7 | ( |

### 4.5.1 Case *c*=*c* _{11}

*c*=

*c*

_{11}. Then (4.178) and (4.197) imply that

*y*

_{11}is a real element and we shall derive a contradiction. Then by (4.203) and (4.204) we obtain that \(c_{11}=\bar{f}_{11}\) and

*y*

_{11}is a non-real element. By (4.200) and (4.201) we obtain that

*b*

_{8}and

*g*

_{12}are real elements and

*y*

_{11}is a non-real element, we obtain that \((\bar{y}_{11},b_{3}\varSigma _{10})=1\). Now (4.209) implies that

*Σ*

_{10}∈

*B*and by (4.203) we obtain that

### 4.5.2 Case *c*=*c* _{12}

*c*=

*c*

_{12}. Then (4.195) and (4.197) imply that

*y*

_{11}is a real element and we shall derive a contradiction. Then by (4.170) and (4.196) we obtain that

*y*

_{11}is a non-real element. By (4.198) we obtain that

*b*

_{8},

*F*

_{15})=(

*b*

_{8},

*b*

_{3}

*c*

_{12})=0, and by (4.162) we obtain that \((\bar{b}_{10},F_{15})=(\bar{b}_{10},b_{3}c_{12})=0\). Hence by (4.165) we can assume that

*s*|≤15, and by (4.164)

*s*is a real element. Now we have that

*s*|. Hence

*s*=

*s*

_{15}and we shall derive a contradiction. Then by (4.215) and (4.218) we obtain that

*b*

_{8}and

*s*

_{15}are real elements and

*y*

_{11}is a non-real element, we obtain that

*f*

_{14},

*b*

_{3}

*y*

_{11})=1, and together with (4.170) we obtain that

*α*in

*B*such that \(\alpha\in \mathit{Irr}(\bar{ \varSigma }_{18})\cap \mathit{Irr}(S_{12})\) which implies that 10≤|

*α*|. Hence

*S*

_{12}=

*α*

_{12}. Now we have that

*s*|≠15, so (4.215), (4.219) and (4.221) imply that

*b*

_{8}and

*s*are real elements and

*y*

_{11}is a non-real element, we obtain that \(\bar{y}_{11}\in \mathit{Irr}(b_{3}T)\), which is a contradiction to (4.228).

### 4.5.3 Case *c*=*c* _{13}

*c*=

*c*

_{13}. Then (4.195) and (4.197) imply that

*y*

_{11}is a real element and we shall derive a contradiction. Then by (4.170) and (4.196) we obtain that

*y*

_{11}is a non-real element. By (4.198) we obtain that

*s*=

*s*

_{18}and we shall derive a contradiction. Then (4.230) and (4.232) imply that

*b*

_{8}and

*s*

_{18}are real elements and

*y*

_{11}is a non-real element, we obtain that

*s*|≠18 which implies by (4.230) and (4.233) that 11≤|

*s*|≤12. By (4.165), (4.230) and (4.232) we obtain that

*b*

_{8}and

*s*are real elements and

*y*

_{11}is a non-real element, we obtain that \(s\neq\bar{y}_{11}\); and by (4.232) and 11≤|

*s*|≤12, we have that \(\bar{y}_{11}\notin F_{18}\). Hence \(\bar{y}_{11}\in b_{3}T\), which is a contradiction to (4.245).

### 4.5.4 Case *c*=*c* _{14}

*c*=

*c*

_{14}. Then (4.195) and (4.197) imply that

*b*

_{8}∉

*Irr*(

*b*

_{3}

*c*

_{14}) and \(\bar{b}_{10}\notin \mathit{Irr}(b_{3}c_{14})\), and by (4.165) and (4.247) we can assume that

*y*

_{11}is a real element and we shall derive a contradiction. Then by (4.170) and (4.196) we obtain that either

*b*

_{3}

*c*

_{14},

*b*

_{3}

*c*

_{14})=4. Therefore

*Σ*

_{12}=

*h*

_{12}where

*h*

_{12}∈

*B*and

*H*

_{20}=

*x*

_{8}+

*h*

_{12}. Therefore

*g*≠

*t*. By (4.165) and (4.250) we obtain that 12≤|

*t*|≤15. Now (4.171) implies that

*g*is a real element and we shall derive a contradiction. By (4.254) and (4.258) we obtain that \((\bar{c}_{14},b_{3}g)=(x_{8},b_{3}g)\) which implies that 9≤|

*g*|. By (4.254) we obtain that

*g*=

*g*

_{9}. Now we have that

*h*

_{12},

*b*

_{3}

*k*

_{5})=1, which is a contradiction to (4.258).

*g*is a non-real element. By (4.2), (4.256), (4.170), (4.254) and \((b_{3}\bar{b}_{3})y_{11}=b_{3}(\bar{b}_{3}y_{11})\), we obtain that

*b*

_{8}and

*y*

_{11}are real elements and

*g*is a non-real element, we obtain that \((h_{12},b_{3}g)=(\bar{g},b_{3}\bar{h}_{12})=1\) and by (4.258) we obtain that (

*x*

_{8},

*b*

_{3}

*g*)=1 which implies that 8≤|

*g*|. By (4.254) we obtain that either

*g*=

*g*

_{8}or

*g*=

*g*

_{9}.

*g*=

*g*

_{9}and we shall derive a contradiction. Then by (4.258) and (4.257) we obtain that

*g*

_{9},

*b*

_{3}

*h*

_{5})=1, which is a contradiction to (4.260).

*g*=

*g*

_{8}. By (4.258) we obtain that (

*x*

_{8},

*b*

_{3}

*g*

_{8})=1, and since (

*h*

_{12},

*b*

_{3}

*g*

_{8})=1 we obtain that

*γ*

_{4}∈

*B*. Hence

*b*

_{3}

*γ*

_{4}=

*h*

_{12}which implies that (

*b*

_{3}

*γ*

_{4},

*b*

_{3}

*γ*

_{4})=1. On the other hand, we have that either \((m,\bar{b}_{3}\gamma_{4})=1\) or \((n,\bar{b}_{3}\gamma_{4})=1\) which implies that (

*b*

_{3}

*γ*

_{4},

*b*

_{4}

*γ*

_{4})≠1, and we have a contradiction.

*y*

_{11}is a non-real element. By (4.2), (4.165), (4.247), (4.251) and \(b_{3}(\bar{b}_{3}s)=(b_{3}\bar{b}_{3})s\) we obtain that

*b*

_{8}and

*s*are real elements and

*y*

_{11}is a non-real element we obtain that

*s*=

*s*

_{12}then by (4.251) we obtain that

*s*=

*s*

_{13}then by (4.251), we obtain that

*c*

_{14}∈

*Irr*(

*b*

_{3}

*Q*

_{16}).

- Case 1:
*x*_{7}∈*Irr*(*Q*_{16}). - Case 2:
*g*_{10}∈*Irr*(*Q*_{16}). - Case 3:
\(\bar{f}_{10}\in \mathit{Irr}(Q_{16})\).

**Case 1** By (4.169) and (4.268) we obtain that \((d_{7},\bar{b}_{3}h_{6})=1\) and by (4.263) we obtain that \((f_{10},\bar{b}_{3}h_{6})=1\), which is a contradiction.

**Case 2**By (4.267) we obtain that

*g*

_{10}∈

*Irr*(

*Q*

_{16}) we obtain by (4.268) and (4.269) that \(\beta_{6}=\bar{g}_{6}\). Now by (4.269) we obtain that \((\bar{g}_{10},b_{3}g_{6})=1\), and by (4.266) we obtain that (

*d*

_{7},

*b*

_{3}

*g*

_{6})=1, which is a contradiction.

**Case 3**By (4.267) we obtain that

*s*=

*s*

_{14}then by (4.251) we obtain that

*f*∈

*B*in

*Irr*(

*T*

_{13}) such that 10≤|

*f*| which implies that

*f*=

*T*

_{13}. Hence

*α*

_{7}∈

*Irr*(

*G*

_{12}) then there exists an element

*α*

_{5}∈

*B*such that

*α*

_{5}∈

*Irr*(

*G*

_{12}) which implies that \((\alpha_{5},b_{3}\bar{f}_{13})=1\), and we have a contradiction. Hence

*α*

_{7}=

*β*

_{7}. By (4.273) we obtain that 3≤(

*b*

_{3}

*y*

_{11},

*b*

_{3}

*y*

_{11}). Now (4.196) and (4.274) imply that

*s*=

*s*

_{15}then by (4.251) we obtain that

*f*∈

*B*in

*Irr*(

*T*

_{16}) such that 10≤|

*f*| which implies that

*f*=

*T*

_{16}. Hence

*f*

_{4}∈

*B*, and we have a contradiction since \((f_{4},\bar{b}_{3}c_{14})=1\).

### 4.5.5 Case *c*=*c* _{15}

*c*=

*c*

_{15}. Then (4.195) and (4.197) imply that

*y*

_{11}is a real element and we shall derive a contradiction. Then by (4.170) and (4.196) we obtain that

*d*

_{6},

*b*

_{3}

*h*

_{5})=1, which is a contradiction.

*y*

_{11}is a non-real element. By (4.198) we obtain that

*y*

_{11},

*b*

_{3}

*c*

_{15})=1, and by (4.165) and since

*y*

_{11}is a non-real element we obtain that \((y_{11},b_{3}\bar{b}_{15})=0\) which implies that \(c_{15}\neq\bar{b}_{15}\). Now by (4.5) and (4.162) we obtain that

*b*

_{8}∉

*Irr*(

*b*

_{3}

*c*

_{15}) and \(\bar{b}_{10}\notin \mathit{Irr}(b_{3}c_{15})\) and by (4.165) (4.280) we can assume that

*b*

_{8}and

*s*are real elements and

*y*

_{11}is a non-real element we obtain that either \(\bar{y}_{11}\in \mathit{Irr}(S)\) or \(\bar{y}_{11}\in \mathit{Irr}(b_{3}T)\). By (4.174) we obtain that

*4.169*), we obtain that

*α*

_{3},

*b*

_{3}

*y*

_{11})=1, and we have a contradiction. Hence \((x_{8},\allowbreak\bar{b}_{3}c_{15})=\nobreak 1\) which implies that

Now we have that \(\bar{y}_{11}\in b_{3}T\) which implies that 5≤|*T*| and by (4.285) we obtain that 12≤|*s*|.

*b*

_{3}

*x*

_{8},

*b*

_{3}

*x*

_{8})=2 and we shall derive a contradiction. Then by (4.162) we obtain that

*s*| we obtain by (4.287) that |

*S*|≤12 which implies that

*α*

_{14}∈

*Irr*(

*b*

_{3}

*e*

_{9}) and by (4.171) we obtain that (

*b*

_{10},

*b*

_{3}

*e*

_{9})=1, which is a contradiction.

*b*

_{3}

*x*

_{8},

*b*

_{3}

*x*

_{8})=3. Then by (4.162) we obtain that

*b*

_{8}≠

*x*

_{8}, and by (4.288) we obtain that

*α*|, and by (4.289) we obtain that |

*α*|≤10. Therefore

*α*≠

*s*,

*α*≠

*y*

_{11}and

*s*≠

*y*

_{11}. Since (

*b*

_{3}

*x*

_{8},

*b*

_{3}

*x*

_{8})=3 we obtain by (4.289) that

*α*≠

*b*

_{10}. Now by (4.292) we obtain that (

*b*

_{3}

*c*

_{15},

*b*

_{3}

*c*

_{15})=4. By (4.279) and (4.288) we obtain that

*v*

_{4}∉

*Irr*(

*G*

_{12}) and since 14≤|

*s*| we have that \(v_{4}\notin \mathit{Irr}(\bar{b}_{3}s)\). Therefore \((v_{4},\bar{b}_{3}\alpha)=1\) which implies that |

*α*|≤9 and by (4.292) we obtain that \((c_{15},\bar{b}_{3}\alpha)=1\) which implies that 8≤|

*α*|. Thus

*s*|≤16. By (4.162), (4.289), (4.290), (4.293) and \(b_{3}(\bar{b}_{3}x_{8})=\bar{b}_{3}(b_{3}x_{8})\) we obtain that

*x*

_{8},

*b*

_{3}

*α*)=(

*x*

_{8},

*b*

_{3}

*β*)=1 which implies by (4.296) that (

*f*

_{15},

*b*

_{3}

*β*)=0 and (

*f*

_{15},

*b*

_{3}

*α*)=1. So 9≤|

*α*| and by (4.295), we obtain that

*α*=

*α*

_{9}which implies by (4.292) that

*s*=

*s*

_{15}. By (4.285) we obtain that

*b*

_{3}

*f*

_{15}=

*G*

_{12}+

*T*

_{15}+

*c*

_{15}+

*h*

_{3}where

*h*

_{3}∈

*B*, which is impossible, or

*v*

_{5}∈

*Irr*(

*G*

_{12}).

*v*

_{5}∈

*Irr*(

*G*

_{12}). By (4.196) we obtain that

*m*

_{3},

*b*

_{3}

*m*

_{7})=1, and we have a contradiction.

### 4.5.6 Case *c*=*c* _{16}

*c*=

*c*

_{16}. Then (4.195) and (4.197) imply that

*y*

_{11}is a real element and we shall derive a contradiction. Then by (4.170) and (4.196) we obtain that

*y*

_{11}is a non-real element. By (4.198) we obtain that

*b*

_{8}∉

*Irr*(

*b*

_{3}

*c*

_{16}) and \(\bar{b}_{10}\notin \mathit{Irr}(b_{3}c_{16})\) and by (4.298) we can assume that

- Case 1:
*S*=*t*and \(\varSigma _{14}=\varSigma ^{*}_{14}\). - Case 2:
\(S=\varSigma ^{*}_{14}\) and

*Σ*_{14}=*t*_{14}. - Case 3:
\(\varSigma ^{*}_{14}=\alpha+\beta\) where

*α*,*β*∈*B*,*S*=*α*and*Σ*_{14}=*β*+*t*.

**Case 1**By (4.304) we obtain that

*t*|. By (4.305) we obtain that

*s*∈

*Irr*(

*b*

_{3}

*T*) which implies that 8≤|

*T*| and 13≤|

*s*|. In the same way we can show that

*t*|. Now by (4.165) we can assume that

*s*=

*s*

_{13}and

*t*=

*t*

_{14}. Now (4.302) and (4.309) imply that

**Case 2**By (4.304) we obtain that

*x*

_{8},

*Σ*

_{18})=1 which implies that

*v*

_{8},

*b*

_{3}

*γ*

_{8})=1 which implies that

*g*

_{8}=

*x*

_{8}. By (4.318) we obtain that \((y_{11},\bar{b}_{3}x_{8})=(x_{8},b_{3}y_{11})=1\), and by (4.5) we obtain that

*b*

_{8}≠

*x*

_{8}. Now by (

*4.162*) we obtain that \((b_{10},\bar{b}_{3}x_{8})=1\), which is a contradiction.

**Case 3**By (4.306) we obtain that

*t*| and |

*β*|≤5. By (4.5) and (4.319) we obtain that

*b*

_{8}≠

*e*

_{8}. Therefore |

*β*|≠3 which implies that 4≤|

*β*|≤5.

*β*=

*β*

_{4}and we shall derive a contradiction. Then (4.319) implies that

*t*

_{10}=

*α*

_{10}. By (4.165) we obtain that (

*b*

_{15},

*b*

_{3}

*t*

_{10})=1 and by (4.304) we obtain that \((\bar{c}_{16},b_{3}t_{10})=1\), which is a contradiction.

*β*=

*β*

_{5}, and by (4.165) and (4.319) we obtain that

*t*=

*t*

_{9},

*s*=

*s*

_{18}and

*α*=

*α*

_{9}. By (4.165), (4.172) and (4.319) we obtain that

## 4.6 Case (*b* _{3} *x* _{7},*b* _{3} *x* _{7})=3

*b*

_{3}

*x*

_{7},

*b*

_{3}

*x*

_{7})=3 we obtain that

*c*,

*b*

_{3}

*b*

_{15})=1. By (4.3) we obtain that (

*b*

_{6},

*b*

_{3}

*b*

_{15})=1, by (4.10) we obtain that \((\bar{b}_{15},b_{3}b_{15})=1\) and (

*b*

_{3}

*b*

_{15},

*b*

_{3}

*b*

_{15})=4. Now since \(c\neq b_{6},\bar{b}_{15}\) we obtain that

*c*|≤13 and \((\bar{b}_{3}c,x_{7})=1\) and by (4.323) we obtain that \((\bar{b}_{3}c,b_{15})=1\) which implies that

*α*|≤9, by (4.4) we obtain that

*α*≠

*b*

_{6}and by (4.326) we obtain that

*f*≠

*α*. Now we have by (4.324) that

*c*=

*α*, by (4.325) we obtain that

*c*=

*c*

_{9}and by (4.327) we obtain that

*β*=

*β*

_{3}. By (4.328) we obtain that \(b_{3}c_{9}+b_{3}\beta_{3}=b_{10}+z_{4}+w_{7}+\bar{b}_{3}g_{5}\) which implies that (

*w*

_{7},

*b*

_{3}

*c*

_{9})=1. By (4.327) we obtain that (

*z*

_{4},

*b*

_{3}

*c*

_{9})=1 and together with (4.329) we obtain that 4≤(

*b*

_{3}

*c*

_{9},

*b*

_{3}

*c*

_{9}). By (4.320) and (4.323) we obtain that (

*b*

_{3}

*c*

_{9},

*b*

_{3}

*c*

_{9})≤3, and we have a contradiction.

*b*

_{3}

*z*

_{5},

*b*

_{3}

*z*

_{5})≤3.

*b*

_{3}

*z*

_{5},

*b*

_{3}

*z*

_{5})=3 and we shall derive a contradiction. Then by (4.325), (4.326) and (4.332) we obtain that

*b*

_{3}

*c*

_{9}=

*b*

_{10}+

*z*

_{5}+

*w*

_{6}+

*v*

_{6}where

*v*

_{6}∈

*B*. By (4.320) and (4.323) we obtain that (

*b*

_{3}

*c*

_{9},

*b*

_{3}

*c*

_{9})≤,3 and we have a contradiction. Now (4.331) implies that

*c*=

*c*

_{13}and we shall derive a contradiction. Then by (4.324) we obtain that

*f*=

*f*

_{11}and (4.332) implies that

*q*

_{5},

*b*

_{3}

*w*

_{6})=(

*u*

_{5},

*b*

_{3}

*w*

_{6})=1, and by (4.330) we obtain that (

*x*

_{7},

*b*

_{3}

*w*

_{6})=1, which is a contradiction. Now we have by (4.325) that 9≤|

*c*|≤12 and by (4.324) we obtain that 12≤|

*f*|≤15.

*f*=

*f*

_{12}and

*c*=

*c*

_{12}. By (4.1), (4.320), (4.330) and \(b_{3}(b_{3}x_{7})=b^{2}_{3}x_{7}\) we obtain that

*b*

_{3}

*c*

_{12}+

*b*

_{3}

*d*+

*b*

_{3}

*e*=

*b*

_{10}+

*z*

_{5}+

*w*

_{6}+

*b*

_{6}

*x*

_{7}. By (4.320) and since

*c*=

*c*

_{12}we obtain that |

*d*|,|

*e*|≠3,12 which implies by (4.334) that (

*z*

_{5},

*b*

_{3}

*d*)=(

*z*

_{5},

*b*

_{3}

*e*)=0. Hence (

*z*

_{5},

*b*

_{3}

*c*

_{12})=1 which implies that

*c*

_{12}=

*f*

_{12}and since (

*b*

_{3}

*b*

_{10},

*b*

_{3}

*b*

_{10})=3 we have a contradiction to (4.324).

*δ*|≠15 which implies that \(\delta\neq \bar{b}_{15}\). Therefore \(\bar{b}_{15}\in \mathit{Irr}(\bar{b}_{3}D)\). Since 9≤|

*c*|≤12 we obtain by (4.5), (4.162) and (4.336) that

*b*

_{8}∉

*Irr*(

*D*) and \(\bar{b}_{10}\notin \mathit{Irr}(D)\). Therefore by (4.165) we can assume that

*s*∈

*Irr*(

*D*). Now (4.336) implies that

*b*

_{8}and

*s*are real elements and

*z*

_{5}is a non-real element, we obtain that either \(\bar{z}_{5}\in \mathit{Irr}(E)\) or \(\bar{z}_{5}\in \mathit{Irr}(b_{3}F)\). If \(\bar{z}_{5}\in \mathit{Irr}(E)\) then by (4.338) we obtain that \((\bar{c},b_{3}z_{5})=1\), and since 9≤|

*c*|≤12 we have a contradiction to (4.333). Therefore \(\bar{z}_{5}\in \mathit{Irr}(b_{3}F)\). Now by (4.333) we obtain that either \(\bar{x}_{7}\in \mathit{Irr}(F)\) or \(\bar{g}_{8}\in \mathit{Irr}(F)\). If \(\bar{x}_{7}\in \mathit{Irr}(F)\) then \((s,\bar{b}_{3}x_{7})=(\bar{x}_{7},\bar{b}_{3}s)=1\). Since

*s*is a real element and

*b*

_{10},

*z*

_{5}are non-real elements we obtain that

*s*≠

*b*

_{10},

*z*

_{5}, so (4.330) implies that

*s*=

*w*

_{6}, which is a contradiction to (4.165) and (4.330). Now we have that

*c*=

*c*

_{9}and we shall derive a contradiction. Then by (4.337) we obtain that

*b*

_{3}

*u*

_{4}=

*s*

_{12}. Now by (4.320), (4.323), (4.324), (4.335) and \(b_{3}(\bar{b}_{3}c_{9})=\bar{b}_{3}(b_{3}c_{9})\) we obtain that \(d+e+b_{3}u_{5}=\delta_{6}+c_{9}+\bar{g}_{8}+u_{4}\). Hence (

*b*

_{3}

*u*

_{4},

*b*

_{3}

*u*

_{4})≠1, and we have a contradiction.

*c*=

*c*

_{10}and we shall derive a contradiction. Then by (4.337) we obtain that

*d*|+|

*e*|=11, and now we can assume that

*d*=

*δ*

_{5},

*e*=

*e*

_{6}and

*e*

_{6}∈

*Irr*(

*G*

_{12}) which implies that

*c*=

*c*

_{11}and we shall derive a contradiction. Then by (4.335) we obtain that

*δ*=

*δ*

_{4}and (

*z*

_{5},

*b*

_{3}

*δ*

_{4})=1 which implies that (

*w*

_{6},

*b*

_{3}

*δ*

_{4})=0. By (4.330), (4.333), (4.335), (4.340), (4.338) and \(b_{3}(\bar{b}_{3}z_{5})=\bar{b}_{3}(b_{3}z_{5})\), we obtain that

*w*

_{6}+

*z*

_{5}+

*H*=

*E*+

*b*

_{3}

*δ*

_{4}. Now we have that

*w*

_{6}∈

*Irr*(

*E*) which implies that (

*w*

_{6},

*b*

_{3}

*c*

_{11})=1, and by (4.338) we obtain that

*b*

_{3}

*c*

_{11},

*b*

_{3}

*c*

_{11}). Now (4.337) implies that

*c*

_{11},

*b*

_{3}

*u*

_{5})=(

*c*

_{11},

*b*

_{3}

*u*

_{6})=1 we obtain that \((\bar{g}_{8},b_{3}u_{5})=(\bar{g}_{8},\allowbreak b_{3}u_{6})=\nobreak 0\). Therefore we have that either \(d=\bar{g}_{8}\) or \(e=\bar{g}_{8}\), but by (4.169) we obtain that |

*d*|+|

*e*|=10 and we have a contradiction.

*c*=

*c*

_{12}which implies by (4.335) that

*δ*=

*δ*

_{3}and (

*z*

_{5},

*b*

_{3}

*δ*

_{3})=1 which implies that (

*w*

_{6},

*b*

_{3}

*δ*

_{3})=0. By (4.330), (4.333), (4.335), (4.340), (4.338) and \(b_{3}(\bar{b}_{3}z_{5})=\bar{b}_{3}(b_{3}z_{5})\), we obtain that

*w*

_{6}+

*z*

_{5}+

*H*=

*E*+

*b*

_{3}

*δ*

_{3}. Now we have that

*w*

_{6}∈

*Irr*(

*E*) which implies that (

*w*

_{6},

*b*

_{3}

*c*

_{12})=1, and by (4.338) we obtain that

*s*|≥15. If |

*s*|=15 then By (4.320), (4.323), (4.324), (4.335) and \(b_{3}(\bar{b}_{3}c_{12})=\bar{b}_{3}(b_{3}c_{12})\) we obtain that \(d+e+\bar{b}_{15}+b_{3}C_{14}=\delta_{3}+\bar{b}_{3}s_{15}+\bar{b}_{3}w_{6}\). By (4.320) we obtain that

*d*,

*e*≠

*δ*

_{3}and since

*c*=

*c*

_{12}we obtain that \((d,\bar{b}_{3}s_{15})=(e,\bar{b}_{3}s_{15})=0\). Now we have that \((d,\bar{b}_{3}w_{6})=(e,\bar{b}_{3}w_{6})=1\), which is a contradiction to \((c_{12},\bar{b}_{3}w_{6})=1\). If |

*s*|≠15 then |

*s*|≤11, which is a contradiction to (4.339).

## 4.7 Case *b* _{3} *b* _{10}=*b* _{15}+*x* _{5}+*y* _{5}+*z* _{5}

*b*

_{3}

*b*

_{10}=

*b*

_{15}+

*x*

_{5}+

*y*

_{5}+

*z*

_{5}we obtain that

*b*

_{3}

*x*

_{5},

*b*

_{3}

*y*

_{5}) which implies that

*d*=

*e*and we shall derive a contradiction. Then (

*b*

_{3}

*x*

_{5},

*b*

_{3}

*y*

_{5})=2 and by (4.341) and (4.342), we obtain that

*x*

_{5},

*b*

_{3}

*u*

_{5})=(

*y*

_{5},

*b*

_{3}

*u*

_{5})=1. Now we have that

*c*|≥10 and by (4.343) and (4.347) we obtain that

*c*=

*c*

_{10}and

*d*=

*d*

_{5}. Since (

*c*

_{10},

*b*

_{3}

*b*

_{15})=1 we obtain by (4.10) that \((c_{10},\bar{b}_{3}b_{10})=1\) which implies by (4.4) that

*d*

_{5},

*b*

_{3}

*b*

_{15})=1, and we have a contradiction.

*b*

_{3}

*u*

_{5},

*b*

_{3}

*u*

_{5})=3 which implies that

*b*

_{10},

*b*

_{3}

*γ*)=(

*b*

_{10},

*b*

_{3}

*δ*)=1. By (4.350) and (4.4) we obtain that

*γ*,

*δ*≠

*b*

_{6}and |

*γ*|+|

*δ*|=10 which implies that

*γ*,

*δ*≠

*c*

_{10}. Now (4.348) implies that

*γ*+

*δ*=

*α*+

*β*and |

*α*|+|

*β*|=14, which is a contradiction.

*b*

_{3}

*z*

_{5}=

*d*+

*e*or

*b*

_{3}

*z*

_{5}=

*c*+

*f*where

*f*∈

*B*. If

*b*

_{3}

*z*

_{5}=

*d*+

*e*then 2|

*c*|=15, so

*c*,

*b*

_{3}

*b*

_{15})=0 and we shall derive a contradiction. Then (

*d*,

*b*

_{3}

*b*

_{15})=(

*e*,

*b*

_{3}

*b*

_{15})=(

*f*,

*b*

_{3}

*b*

_{15})=1. In addition, we have that |

*d*|=|

*e*|=|

*f*| which implies by (4.4) and (4.10) that

*c*=

*c*

_{7}and by (4.354) we obtain that \((c_{7},\allowbreak \bar{b}_{3}u_{5})=\nobreak1\). By (4.4), (4.341), (4.342), (4.351) and \(b_{3}(\bar{b}_{3}b_{10}=\bar{b}_{3}(b_{3}b_{10})\), we obtain that \(b_{8}+b_{3}d_{8}+b_{3}e_{8}+b_{3}f_{8}=\bar{b}_{3}b_{15}+2b_{10}+u_{5}+v_{5}+w_{5}\) which implies that \((\bar{b}_{3}u_{5},\allowbreak\bar{b}_{3}u_{5})=\nobreak 2\). By (4.353) and since (

*c*,

*b*

_{3}

*b*

_{15})=0 we obtain that \((b_{15},\bar{b}_{3}d_{8})=1\) and together with (4.343) we obtain that

*g*

_{4},

*b*

_{3}

*u*

_{5})=1. Now by (4.341) and \((\bar{b}_{3}u_{5},\bar{b}_{3}u_{5})=2\) we obtain that

*b*

_{3}

*u*

_{5}=

*g*

_{4}+

*x*

_{5}, which is a contradiction.

*b*

_{3}

*x*

_{5}=

*c*+

*d*we obtain that either

*c*=

*c*

_{10}or

*c*=

*c*

_{12}. By (4.341) we obtain that (

*x*

_{5},

*b*

_{3}

*b*

_{10})=(

*x*

_{5},

*b*

_{3}

*u*

_{5})=1 which implies that

*c*=

*c*

_{12}and we shall derive a contradiction. Then by (4.356) we obtain that

*α*=

*α*

_{6},

*β*=

*β*

_{6}and

*b*

_{10},

*b*

_{3}

*α*

_{6})=1 and by (4.359) we obtain that (

*b*

_{3}

*α*

_{6},

*b*

_{3}

*α*

_{6})=2 which implies that \(0=(u_{5},b_{3}\alpha_{6})=(\alpha_{6},\bar{b}_{3}u_{5})\). In the same way, we can show that \((\beta_{6},\bar{b}_{3}u_{5})=0\) which implies by (4.358) and (4.360) that \(\bar{b}_{3}u_{5}=d_{3}+c_{12}\). Now we have that (

*b*

_{3}

*u*

_{5},

*b*

_{3}

*u*

_{5})=2 and by (4.341) we obtain that

*c*=

*c*

_{10}which implies that

*α*,

*β*≠

*d*

_{5}which implies that

*b*

_{3}

*b*

_{15},

*b*

_{3}

*b*

_{15})=5 we have a contradiction. Now (4.358) implies that (

*u*

_{5},

*b*

_{3}

*c*

_{10})=1. In the same way, we can show that (

*v*

_{5},

*b*

_{3}

*c*

_{10})=(

*w*

_{5},

*b*

_{3}

*c*

_{10})=1 and by (4.355) we obtain that (

*b*

_{10},

*b*

_{3}

*c*

_{10})=1 which implies that (

*b*

_{3}

*c*

_{10},

*b*

_{3}

*c*

_{10})≠4, and we have a contradiction to (4.361).