| The authors of [ABCCHM] studied the normalized integral table algebra (A, B) generated by a faithful non-real element of degree 3, where they assumed that B contains only one element of degree 1, i.e., L1(B)={1}, and does not contain any element of degree 2, i.e., L2{B)=φ. So it is a valuable topic to study the normalized integral table algebra (A, B) generated by a faithful non-real el-ement of degree 3 with L1(B)= 1 and L2{B)≠φ, this is what we study in this thesis。The whole thesis consists of six chapters. The contents of Chapter 1 to 3 are background, preliminaries and the goals of this thesis and line of research, the new results we come to are exhibited in Chapters 4,5 and 6.In Chapter 4, first we give the table algebra generated by a faithful element of degree 2 and without non-trivial element of degree 1. Secondly, we give the relations between Irr(SL(2,5)) and products by the non-real generater b3 and some other element.Proposition 4.1.1 Let (A, B) be a NITA generated by an element b2 E B of degree 2 and L1(B)= 1. Then (A, B)-x (Ch(SL(2,5)),Irr(SL(2,5))), B has exactly two elements of degree 2, both are real.Theorem 4.2.1 Let (A,B) be a NITA with a non-real faithful element b3∈B of degree 3 such that Bb3= B, and c2, c*2 be as in Proposition 4.1.1. Assume that L1{B)= 1, and L2(B)={c2,c2*}. Thus BC2(?)X Irr(SL(2,5)), Bc2*(?)XIrr(SL(2,5)). Then Irr(b3b3)\{1} (?)Bb3\Bc2. Furthermore, b3c2= x6∈B and b3c2*= yb∈B.Corollary 4.2.3 Let (A,B) be a NITA with a non-real faithful element b3 ∈ B of degree 3 such that Bb3= B, and c2, c2* be as in Proposition 4.1.1. Assume that L1(B)= 1, and L2(B)={c2,c*2}. If a ∈Irr(SL(2,5)), then(1) ab3∈B\Irr(SL(2,5)) and Irr(b23) (?)B\Irr(SL(2,5)).(2) For b∈B\Irr(SL(2,5)), it follows that Supp{ab}∩Irr(SL(2,5))=φ.Lemma 4.3.1 Let (A,B) be a NITA. Assume L1(B)=1, L2(B) ≠φ. Then L2(B)= {c1, c2,......,cn,c1*,c2*,......,cn*},where Bc1(?)X Bc2(?)X..........X Bcn(?)X(Ch(SL(2,5)),Irr(SL(2, 5))) and ci*∈Bci 1≤(?)i≤n. Also Bci is table subset of B, 1≤(?)≤n.Theorem 4.3.3 Let G be a perfect finite group i.e. (G=G’) with a faith-ful irreducible character of degree 3. Then G has no irreducible character degree 2.In Chapter 5, we try to classify normalized integral table algebra generated by a faithful element of degree 3 with trivial base element of degree 1 and contains exactly two base elements of degree 2 appearing in (Ch(SL(2, 5), Irr(SL(2, 5)). At first we give the following theorem:Theorem 5.1.1 Let (A,B) be a NITA with a non-real element b3∈B of degree 3 such that Bb3 =B. Assume that L1 (B) = 1, L2(t3) = {c2, c2*}. Then Bc2 = Bc2*(?)X Irr(SL(2, 5)) and b3b3 has one of the following possibilities.HP 1 b3b3 = 1+2d4, where d4∈B is real.HP 2:b3b3 = 1+d4 + d4, where d4∈B.HP 3:b3b3 = 1+d3 + d5, where d3, d5∈B are real.HP 4: b3b3 = 1+d4+e4, where d4, e4∈B are real and d4≠e4.HP 5:b3b3=1+b8, where b8 b8∈B is real.We continue in the sections 2, 3, 4, 5 to investigate the NITAs satisfying HP1, HP2, HP3, HP4 respectively, and come to the following theorems.Theorem 5.2.1 There exists no NITA satisfying HP 1.Theorem 5.3.1 There exists no NITA satisfying HP 2.Theorem 5.4.1 There exists no NITA satisfying HP 3.Theorem 5.5.1 If(A, B) is a NITA satisfying HP 4, i. e., b3b3 = 1+d4+e4, where d4,e4∈B are real. Then b32=α3+β3+γ3, where α3,β3,γ3∈B \ {b3}. Moreoverα3β3= d4+r5, where r5∈L+(B),α3γ3= d4’+r5’, where r5’∈L+(B),β3β3= d4"+r5", where r5"∈L+(B), where d4,d4’,d4" is equal to either d4 or e4, and r5, r5’ and r5" can not be reducible or irreducible individually.Further if (b3d4, b3d4) = 3, then(1) d42 maybe one of the following forms:(ⅰ) d42 = 1+d4 +e4+m7, where m7∈L+(B)(ⅱ) d42 = 1+2de+m7, where m7∈ L+(B)(ⅲ) d42 = 1+2e4+m7, where m7∈ L+(B)(2) e24 maybe one of the following forms:(ⅰ) e42 = 1+d4 +e4+n7, where n7∈L+(B)(ⅱ)e42 = 1+2d4+n7, where n7∈L+(B)(ⅲ)e42 =1+2e4+n7, where n7∈L+(B)(3) m7 and n7 can not be reducible or irreducible individually.Lemma 5.5.2 Theorem 5.5.1 holds while (b3d4, b3d4)=4.In Chapter 6, we discuss NITAs satisfying HP-5 in detail upon four SUB-HP cases as described in the following Lemma 6.1.1. We elaborate and solve SUB-HP 1 and SUB-HP 2 and confirm that there exists no such NITA. The remaining SUB-HP 3 and SUB-HP 4 are still open problems.Lemma 6.1.1 Let (A, B) a NITA generated by a faithful non-real element b3∈ B of degree 3. Assume that L1(B) = 1, L2(B) = {c2, c2*}. If(A,B) satisfying HPS, then one of the following holds.S UB-HP 1. b32 =b3 + b6, b6∈ B is real of degree 6.SUB-HP 2. b32= b3+b6, b6∈B is non-real of degree 6.SUB-HP 3. b32 = c3+b6, c3, b6∈B and c3≠ b3, b3.SUB-HP 4. b32 = c4+c5, c4, c5∈ B.Theorem 6.2.1 There exists no NITA satisfying HP5 and SUB-HP1.Theorem 6.3.1 If (A, t3) is a NITA satisfying HP5 and SUB-HP 2. Then (b3bs, b3bs)=3.As a Corollary of the proof of Theorem 6.3.1, we come to the following the-orem, which is a correction of the gap for the case (b3b8, b3b8)=4 in [ABCCHM].Theorem 6.3.2 Let (A, B) be a NITA generated by a faithful non-real ele-ment b3∈B of degree 3. Assume that L1(B) = 1, L2(B) =φ. Then there exists no NITA such that b363=1+b8, where b8∈B a real dement of degree 8, and b32 = b3 + b6, b6∈B is non-real of degree 6, and (b3b8, b3b8) = 4. |
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