Beiträge zur Algebra und Geometrie Contributions to Algebra and Geometry Volume 41 (2000), No. 1, 33-47.

Towards a Classification of Homogeneous Integral Table Algebras of Degree Five

M. R. Darafsheh A. Rahnamai Barghi

Center for Theoretical Physics and Mathematics, AEOI, Tehran, Iran Department of Mathematics, University of Tehran, Tehran, Iran Institute for Advanced Studies in Basic Sciences, Gava Zang, Zanjan, Iran Institute for Studies in Theoretical Physics and Mathematics, Tehran, Iran

Abstract. In this paper we classify some of the homogeneous integral table algebras of degree 5 containing a faithful element of width 3 such that in a given basis of the algebra the basis elements are standard. This work is towards classifying homogeneous integral table algebras of degree 5. MSC 1991: 20C05, 20C99, 16P10

1. Introduction

The notation of a table algebra in its new formulation is defined in [2]. Most of related notions and properties of C-algebras and table algebras are defined and proved in [4]. Of particular importance is the classification of homogeneous integral table algebras of small degrees. The homogeneous integral table algebras of degree 1 are (ℂG, G) where G is a finite abelian group [6]. The homogeneous integral table algebras of degree 2 which contain a faithful element are classified in [5]. Also the homogeneous integral table algebras of degree 3 which contain a faithful real element are classified in [7]. In [3] one can also find the classification of homogeneous integral table algebras of degree 4 which contain a faithful element. In [1] there exists an open problem asking for the classification of homogeneous standard integral table algebras of degree 5 with a faithful element.

The main purpose of this paper is to state classification theorems about some of the homogeneous standard integral table algebras of degree 5. This work solves part of the stated open problem.

0138-4821/93 $ 2.50 \begin{matrix}c \\ ◯ \end{matrix} 2000 Heldermann Verlag

In what follows we remind some basic definitions from [2] and [4]. Note that throughout this paper ℂ denotes the field of complex numbers and ℝ^{+} the positive reals.

Definition 1.1. Let B = {b_{1}, b_{2}, …, b_{n}} be a basis of a finite-dimensional associative and commutative algebra A over the complex field ℂ with identity element b_{1} = 1_{A}. Then (A, B) is called a table algebra (and B is said to be a table basis) if and only if the following conditions hold:

(i) For all i, j, m, b_{i}b_{j} = ∑_{m=1}^{n} λ_{ijm}b_{m} with λ_{ijm} ∈ ℝ^{+} ∪ {0}.

(ii) There is an algebra automorphism (denoted by ^{−}) of A whose order divides 2, such that b_{i} ∈ B implies that \overline{b}_{i} ∈ B (then \overline{i} is defined by b\overline{i} = \overline{b}_{i}, and b_{i} ∈ B is called real if \overline{b_{i}} = b or equivalently \overline{i} = i).

(iii) For all i, j, λ_{ij1} ≠ 0 if and only if j = \overline{i}.

By [1, Lemma 2.9], there is an algebra homomorphism f : A ⟶ ℂ such that f(b_{i}) = f(\overline{b_{i}}) ∈ ℝ^{+} for all i and such a map f is uniquely determined. The positive real numbers f(b_{i}), 1 ≤ i ≤ n, are called the degrees of (A, B).

In this paper when we refer to a table algebra (A, B) we assume f : A ⟶ ℂ is the unique linear character associated to (A, B).

Definition 1.2. An integral table algebra (abbreviated ITA) is a table algebra (A, B) such that all the structure constants λ_{ijm} are non-negative integers and all the degrees f(b_{i}) are rational integers.

For example any finite group G yields two examples of ITA’s; Z(ℂG,Cla(G)), the center of the group algebra of a finite group G with the table algebra basis the set of sums Ĉ of G-conjugacy classes C of G, with automorphism ^{−} extended linearly from inversion in G, and with degrees f(Ĉ) = |C| for all Ĉ ∈ Cla(G), and (Ch(G), Irr(G)), the ring of complex valued class functions on G, with table basis the set of irreducible characters of G, with automorphism ^{−} extended linearly from complex conjugation of characters, and with degrees f(χ) = χ(1) for all χ ∈ Irr(G).

Definition 1.3. [2] Let (A, B) be a table algebra. For a ∈ A we define

Supp_{B}(a) = {b_{i} | a = ∑_{i=1}^{n} λ_{i}b_{i} where b_{i} ∈ B and λ_{i} ≠ 0}.

Let (A, B) be a table algebra. A nonempty subset N ⊆ B is called a table subset (or a C-subset) of B iff Supp_{B}(b_{i}b_{j}) ⊆ N for all b_{i}, b_{j} ∈ N.

Any table subset is stable under ^{−} and contains 1_{A} by [1, Proposition 2.7]. For any c ∈ B, the set B_{c} defined by

B_{c} = ⋃_{n=1}^{∞} Supp(c^{n})

is easily seen to be a table subset of B, called the table subset generated by c. Any element c of B is called faithful iff B_{c} = B.

Definition 1.4. [5, Definition 1.3] A table algebra (A, B) is called homogeneous (of degree λ) iff |B| ≥ 2 and, for some fixed λ ∈ ℝ^{+}, f(b) = λ for all b ∈ B − {1}.

For any table algebra (A, B), there is a positive definite Hermitian form ( , ) on A with the following properties [4, Proposition 2.4]: B is an orthogonal set with respect to ( , ); for all b ∈ ℝB (the real span of B ) and all a, c ∈ A,

(ab, c) = (a, \overline{b}c);

and for all b_{i} ∈ B, and λ_{i\overline{i}1} the constants in Definition 1.1,

(b_{i}, b_{i}) = λ_{i\overline{i}1}.

It follows that, for all a, b, c ∈ B,

c ∈ Supp_{B}(ab) ⟺ a ∈ Supp_{B}(\overline{b}c).

Definition 1.5. Let (A, B) be a table algebra and b ∈ B. Then b is called standard iff (b, b) = f(b). Furthermore, B is called standard iff every element of B is standard.

Any table algebra may be rescaled (replacing each table basis element by a positive scalar multiple) to one which is homogeneous, and any ITA can be rescaled to a homogeneous ITA ([6, Theorem 1]). Therefore a classification theorem for all homogeneous integral table algebras (HITA) is an impossible mission.

Definition 1.6. [5] Two table algebras (A, B) and (C, D) are called isomorphic (denoted by B ≅ D) if there exists an algebra isomorphism ϕ : A ⟶ C such that ϕ(B) is a rescaling of D; and the algebras are called exactly isomorphic (denoted by B ≅_{x} D) if ϕ(B) = D. So B ≅_{x} D means that B and D yield the same structure constants.

2. Main results

If (A, B) is a homogeneous standard integral table algebra degree of 5 which contains a faithful element of width 2, then [6, Theorem 3] implies the complete classification of (A, B). The main purpose of this section is to state a classification theorem about homogeneous standard ITA of degree 5 with the additional hypothesis that the basis B contains a faithful element of width 3. The notion of width is defined in [6] as follows:

Definition 2.1. Let (A, B) be a table algebra and b ∈ B. The width of b is defined to be |Supp_{B}(b\overline{b})|.

It is easy to check that if B contains an element b of width 3 then one of the following possibilities hold: If b is a real element, i.e., b = \overline{b}, then

(i)  b^{2} = 5.1 + 3b + d, d = \overline{d},
(ii) b^{2} = 5.1 + 3c + b, c = \overline{c},
(iii) b^{2} = 5.1 + 3c + d, c = \overline{c}, d = \overline{d},
(iv) b^{2} = 5.1 + 2c + 2\overline{c},
(v) b^{2} = 5.1 + 2c + 2d, c = \overline{c}, d = \overline{d},
(vi) b^{2} = 5.1 + 2b + 2d, d = \overline{d},

and if b is a nonreal element, then

(vii) b\overline{b} = 5.1 + 2c + 2d, c = \overline{c}, d = \overline{d},
(viii) b\overline{b} = 5.1 + 2b + 2\overline{b},
(ix) b\overline{b} = 5.1 + 2c + 2\overline{c},
(x) b\overline{b} = 5.1 + 3c + d, c = \overline{c}, d = \overline{d},

for some distinct elements b, c, d of B.

As far as one can see considering all the above cases is a very difficult task and we classify the cases (i),(ii),(iii) and (x) in this paper. Therefore in all of our statement we assume that (A, B) is a HITA of degree 5 where B is standard, i.e., f(b) = (b, b)=5 for all b ∈ B − {1}.

Proposition 2.2. Let (A, B) be a homogeneous standard integral table algebra of degree 5. If B contains a real faithful element b such that b^{2} = 5.1 + 3b + d then (A, B) is one of the following:

B = {1, b, c, d}, b^{2} = 5.1 + 3b + d, c^{2} = 5.1 + 2b + 2d,
d^{2} = 5.1 + 3c + b, bc = 3d + 2c, dc = 3b + 2c;

or

B = {1, b, d}, b^{2} = 5.1 + 3b + d, d^{2} = 5.1 + 4b, bd = b + 4d.

Proof. Let b ∈ B such that b^{2} = 5.1 + 3b + d for some d ∈ B and d = \overline{d}, d ≠ b. Since f(bd) = 25 and (bd, b) = (b^{2}, d) = 5 we have one of the following cases:

bd = b + e_{1} + \overline{e_{1}} + e_{2} + \overline{e_{2}}, b, e_{1}, e_{2} are distinct elements of B,  (1)
bd = b + 3e_{1} + e_{2}, e_{1} = \overline{e_{1}}, e_{2} = \overline{e_{2}}, b, e_{1}, e_{2} are distinct elements of B,  (2)
bd = b + 2e + 2\overline{e}, b, e are distinct elements of B,  (3)
bd = b + 4e, e = \overline{e}, b, e are distinct elements of B,  (4)
bd = b + e_{1} + e_{2} + e_{3} + e_{4}, b, e_{1}, e_{2}, e_{3}, e_{4} are distinct real elements of B,  (5)
bd = b + e_{1} + e_{2} + e_{3} + \overline{e_{3}}, e_{1} = \overline{e_{1}}, e_{2} = \overline{e_{2}}, b, e_{1}, e_{2}, e_{3} are distinct elements of B, (6)
bd = b + 2e_{1} + e_{2} + e_{3}, b, e_{1}, e_{2}, e_{3} are distinct real elements of B,  (7)
bd = b + 2e_{1} + e_{2} + \overline{e_{2}}, e_{1} = \overline{e_{1}}, b, e_{1}, e_{2} are distinct elements of B,  (8)
bd = b + 2e_{1} + 2e_{2}, b, e_{1}, e_{2} are distinct real elements of B.  (9)

If (1) holds then by the equality b(bd) = b^{2}d we obtain

d^{2} + 4d + 3e_{1} + 3\overline{e_{1}} + 3e_{2} + 3\overline{e_{2}} = 5.1 + be_{1} + b\overline{e_{1}} + be_{2} + b\overline{e_{2}}.  (∗)

Now we claim that e_{1} does not appear in the decomposition of be_{1}. If (be_{1}, e_{1}) ≠ 0 then it is easy that (be_{1}, e_{1}) = 5, 10 or 15. We deal with these cases separately. First we assume (be_{1}, e_{1}) = 5 and consider the following cases:

case (i) be_{1} = d + e_{1} + x_{1} + x_{2} + x_{3}, where x_{1}, x_{2}, x_{3} are distinct real elements of B,

case (ii) be_{1} = d + e_{1} + 2x_{1} + x_{2}, where x_{1}, x_{2} are distinct real elements of B,

case (iii) be_{1} = d + e_{1} + 3x, where x is real elements of B.

In the case (i) we have:

25 = (be_{1}, be_{1}) = (b^{2}, e_{1}\overline{e_{1}}) = (5.1 + 3b + d, 5.1 + b + ⋯) ≥ 40,

which is a contradiction. In the case (ii) we have

35 = (be_{1}, be_{1}) = (b^{2}, e_{1}\overline{e_{1}}) = (5.1 + 3b + d, 5.1 + b + ⋯) ≥ 40,

again a contradiction. Let case (iii) hold, and e_{1}\overline{e_{1}} = 5.1 + αb + βd + ⋯ for some non-negative integers α, β. Hence

55 = (be_{1}, be_{1}) = (b^{2}, e_{1}\overline{e_{1}}) = 25 + 15α + 5β.

Thus 3α + β = 6. This implies that α = 1, β = 3 or α = 2, β = 0.

If α = 1, β = 3 then e_{1}\overline{e_{1}} = 5.1 + 3d + b. So

(de_{1}, e_{1}) = (e_{1}\overline{e_{1}}, d) = 15.

This shows that e_{1} appears thrice in the decomposition of de_{1}. On the other hand, d and b don’t appear in the decompositon of d^{2} by (∗). This implies that (d^{2}, e_{1}\overline{e_{1}}) = 25. Therefore by (de_{1}, de_{1}) ≥ 45, we have a contradiction.

If α = 2 and β = 0, then e_{1}\overline{e_{1}} = 5.1 + 2b + · · · . Hence (be_{1}, e_{1}) = (b, e_{1}\overline{e_{1}}) = 10. This implies that e_{1} does appear twice in the decomposition of be_{1}, that is a contradiction. This proves that e_{1} does not appear once in the decomposition of be_{1}.

Now if e_{1} appears twice in the decomposition of be_{1}, then we have be_{1} = d + 2e_{1} + · · · . Hence (be_{1}, be_{1}) ≤ 45 and

(b^{2}, e_{1}\overline{e_{1}}) = (5.1 + 3b + d, 5.1 + 2b + ⋯) ≥ 55.

Since (be_{1}, be_{1}) = (b^{2}, e_{1}\overline{e_{1}}), we obtain a contradiction. Now if e_{1} appears thrice in the decomposition of be_{1}, then be_{1} = d + 3e_{1} + · · · . Hence (be_{1}, be_{1}) = 55 and

(b^{2}, e_{1}\overline{e_{1}}) = (5.1 + 3b + d, 5.1 + 3b + ⋯) ≥ 70.

Since (be_{1}, be_{1}) = (b^{2}, e_{1}\overline{e_{1}}) we have a contradiction. Therefore e_{1} does not appear in the decomposition of be_{1}. Similarly we can prove that e_{2} does not appear in the decomposition of be_{2}.

Now we can write be_{1} = d + αe_{2} + β\overline{e_{2}} + γ\overline{e_{1}} + ⋯ for some non-negative integers α, β, γ. Hence

(bd, be_{1}) = (b + e_{1} + \overline{e_{1}} + e_{2} + \overline{e_{2}}, d + αe_{2} + β\overline{e_{2}} + γ\overline{e_{1}} + ⋯) = 5(α + β + γ).

Also, (b^{2}, de_{1}) = (5.1 + 3b + d, b + δd + ⋯) = 15 + 5δ. Since (be_{1}, bd) = (b^{2}, de_{1}), we have α + β + γ = 3 + δ. Now we show that every elements \overline{e_{1}}, e_{2} and \overline{e_{2}} appear at least once in the decomposition of be_{1}.

Otherwise, if be_{1} = d + 2e_{2} + ⋯, then be_{2} = d + 2e_{1} + ⋯ and \overline{e_{1}}e_{2} = 2b + · · · . Since (be_{1}, be_{2}) = (d + 2e_{2} + ⋯, d + 2e_{1} + ⋯) ≤ 25 and (b^{2}, \overline{e_{1}}e_{2}) = (5.1 + 3b + d, 2b + ⋯) ≥ 30, we have a contradiction, by the equality (be_{1}, be_{2}) = (b^{2}, \overline{e_{1}}e_{2}).

If be_{1} = d + 2\overline{e_{1}} + ⋯ then b\overline{e_{1}} = d + 2e_{1} + ⋯ and e_{1}^{2} = 2b + · · ·. Since

(be_{1}, b\overline{e_{1}}) = (d + 2\overline{e_{1}} + ⋯, d + 2e_{1} + ⋯) ≤ 25,

and (b^{2}, e_{1}^{2}) = (5.1 + 3b + d, 2b + ⋯) ≥ 30, by the equality (be_{1}, b\overline{e_{1}}) = (b^{2}, e_{1}^{2}) we have a contradiction.

If be_{1} = d + 2\overline{e_{2}} + ⋯ then b\overline{e_{2}} = d + 2e_{1} + ⋯, and e_{1}e_{2} = 2b + · · · . Since (be_{1}, b\overline{e_{2}}) = (d + 2\overline{e_{2}} + ⋯, d + 2e_{1} + ⋯) ≤ 25 and (b^{2}, e_{1}e_{2}) = (5.1 + 3b + d, 2b + ⋯) ≥ 30, by the equality (b^{2}, e_{1}e_{2}) = (be_{1}, b\overline{e_{2}}) we obtain a contradiction.

Hence we must have the following possibilities by (∗)

be_{1} = d + \overline{e_{1}} + e_{2} + \overline{e_{2}} + x,
be_{2} = d + \overline{e_{1}} + e_{1} + \overline{e_{2}} + y,
b\overline{e_{1}} = d + \overline{e_{2}} + e_{2} + e_{1} + \overline{x},
b\overline{e_{2}} = d + \overline{e_{1}} + e_{2} + e_{1} + \overline{y},

for some x, y ∈ B − \{1\} such that x ∉ \{d, \overline{e_{1}}, e_{2}, \overline{e_{2}}\} and y ∉ \{d, e_{1}, \overline{e_{1}}, \overline{e_{2}}\}. Now by the above and equality (∗) we obtain d^{2} = 5.1 + x + \overline{x} + y + \overline{y}. Since (b^{2}, d^{2}) = (bd, bd) = 25, we have Supp_{B}(b^{2}) ∩ Supp_{B}(d^{2}) = {1}. It follows that x ≠ b, d. Now, by the equality b(be_{1}) = b^{2}e_{1} we have

e_{1} + 3x + de_{1} = \overline{x} + \overline{y} + y + b + bx.

If x = \overline{x} = y, then d^{2} = 5.1+4x and e_{1} +de_{1} = b+bx. Hence bx = e_{1} +e_{2} +\overline{e_{1}} +\overline{e_{2}} +k and de_{1} = b+\overline{e_{1}}+\overline{e_{2}}+e_{2}+k where k = \overline{k}, k ∈ B−\{1, b, d, x, e_{1}, e_{2}, \overline{e_{1}}, \overline{e_{2}}\}. Since b^{2}x = b(bx), we can see that 4d+bk = x+3k+dx. It follows that there is a basis element s such that bk = s+x+3k and dx = s + 4d. If s ≠ k then (bk, bk) ≤ 65, (b^{2}, k^{2}) = (5.1 + 3b + d, 5.1 + 3b + ⋯) ≥ 70. Since (bk, bk) = (b^{2}, k^{2}), we get a contradiction. If s = k, then bk = x + 4k and dx = 4d + k. Let x appear α times in x^{2}. Since (dx, dx) = 85, (d^{2}, x^{2}) = (5.1 + 4x, 5.1 + αx + ⋯), we get that x^{2} = 5.1 + 3x + l for some l ∈ B. But d(dx) = 20.1 + 16x + dk, d^{2}x = 5x + 4x^{2}, so we have dk = x + 4l. Hence (dk, dk) = 85, (d^{2}, k^{2}) = 25, because b ≠ x. Since (dk, dk) = (d^{2}, k^{2}), we obtain a contradiction.

If x = \overline{x}, x ≠ y, \overline{y}, then e_{1} + 2x + de_{1} = b + y + \overline{y} + bx. Hence bx = e_{1} + \overline{e_{1}} + αx + ⋯, where α ≥ 2 is an integer number. If α = 2, then (bx, bx) = 35 and (b^{2}, x^{2}) ≥ 55. Since (bx, bx) = (b^{2}, x^{2}), we get a contradiction. If α = 3, then (bx, bx) = 55 and (b^{2}, x^{2}) ≥ 70. Since (bx, bx) = (b^{2}, x^{2}), we have a contradiction.

If x = y, x ≠ \overline{x}, y ≠ \overline{y}, then e_{1} +2x+de_{1} = 2\overline{x}+b+bx implies that bx = e_{1} +e_{2} +βx+⋯, where β ≥ 2. We obtain a contradiction by the equality (bx, bx) = (b^{2}, x^{2}).

If x = \overline{y}, x ≠ \overline{x}, y ≠ \overline{y}, then e_{1}+2x+de_{1} = 2y+b+bx. Hence we have bx = e_{1}+\overline{e_{2}}+γx+⋯, for some γ ≥ 2. It is easy to check that this case cannot hold.

If x is different from \overline{x}, \overline{y}, x, then e_{1} + 3x + de_{1} = \overline{x} + \overline{y} + y + b + bx. Hence bx = e_{1} + 4x or bx = e_{1} + 3x + z. If bx = e_{1} + 4x, then de_{1} = \overline{x} + \overline{y} + y + b + x. Since b(bx) = b^{2}x, we can see that dx = e_{1} + e_{2} + \overline{e_{1}} + \overline{e_{2}} + d. Also, since d(bx) = 4dx + de_{1}, (bd)x = d(bx), we have

4x + e_{1} + e_{1}x + e_{2}x + \overline{e_{1}}x + \overline{e_{2}}x = b + y + x + \overline{y} + 4e_{1} + 4e_{2} + 4\overline{e_{1}} + 4\overline{e_{2}} + 4d.

Since the intersection parameters of table algebras are non-negative, x appears at least 4 times on the left-hand side in the above equality. But x appears exactly once on the righthand side in the above equality, which is a contradiction. If bx = 3x + e_{1} + z, for some z ∈ B − \{b, y, \overline{x}, \overline{y}\}. This implies that b appears thrice in the decompositoion of x\overline{x}. Also we have

(bx, bx) = (e_{1} + 3x + z, e_{1} + 3x + z) = 55, (b^{2}, x\overline{x}) = (5.1 + 3b + d, 5.1 + 3b + ⋯) ≥ 70.

But since (b^{2}, x\overline{x}) = (bx, bx) we get a contradiction. This proves that case (1) cannot occur.

If (2) holds then bd = b+3e_{1}+e_{2} where e_{1} = \overline{e_{1}} and e_{2} = \overline{e_{2}}. Since (b, de_{1}) = (e_{1}, bd) = 15, b appears thrice in the decomposition of de_{1}. Let d^{2} = 5.1 + αb + βd + ⋯, for some nonnegative integers α, β. Hence (b^{2}, d^{2}) = 25 + 15α + 5β. By the equality (b^{2}, d^{2}) = (bd, bd) = (b + 3e_{1} + e_{2}, b + 3e_{1} + e_{2}) = 55, we have 3α + β = 6. This implies that α = 2, β = 0 or α = 1, β = 3.

If α = 2, β = 0 then d^{2} = 5.1 + 2b + · · · . Thus 10 = (d^{2}, b) = (d, bd) that is d appears twice in the decomposition of bd, a contradiction. If α = 1, β = 3 then d^{2} = 5.1+3d+b. Since 5 = (d^{2}, b) = (bd, d) we have e_{2} = d, i.e., bd = 3e_{1} +b+d. By the equality b(bd) = b^{2}d we have be_{1} = 3d + 2e_{1} and by the equality (bd)d = bd^{2} we obtain that de_{1} = 3b + 2e_{1}. This shows that b appears twice in the decomposition of e_{1}^{2} and so does d. Therefore e_{1}^{2} = 5.1 + 2b + 2d, and B_{b} = {1, b, d, e_{1}}. Since b is a faithful element we obtain that B = {1, b, d, e_{1}}.

If (3) holds then bd = b + 2e + 2\overline{e} such that b ≠ e. We have (bd, bd) = 45. Let d^{2} = 5.1 + αb + βd + ⋯, for some non-negative integers α, β. Since (b^{2}, d^{2}) = (5.1 + 3b + d, 5.1 + αb + βd + ⋯) = 25 + 15α + 5β and (bd, bd) = (b^{2}, d^{2}), we have 3α + β = 4. This implies that α = 1, β = 1 or α = 0, β = 4. If α = 1, β = 1 then d^{2} = 5.1 + b + d + · · · . Thus (bd, d) = (d^{2}, b) = 5 which implies that d appears once in the decomposition of bd which is a contradiction to the form of bd. If α = 0, β = 4 then d^{2} = 5.1 + 4d. Thus bd^{2} = 5b + 4bd and (bd)d = bd + 2de + 2d\overline{e}, therefore by the equality bd^{2} = (bd)d we obtain

5b + 3bd = 2de + 2d\overline{e}.

Hence by the above equality there are two cases as follow:

de = 2b + 3e, d\overline{e} = 2b + 3\overline{e} (∗∗) de = 2b + 3\overline{e}, d\overline{e} = 2b + 3e (∗∗∗)

If (∗∗) holds then (de, de) = 65 and (e\overline{e}, d) = (d\overline{e}, \overline{e}) = 15 implies that d appears thrice in the decomposition of e\overline{e}. Since 65 = (de, de) = (d^{2}, e\overline{e}) and

(d^{2}, e\overline{e}) = (5.1 + 4d, 5.1 + 3d + ⋯) = 80,

we have a contradiction.

If (∗∗∗) holds then (d^{2}, e^{2}) = (de, d\overline{e}) = 20 and (e^{2}, d) = (d\overline{e}, e) = 15 implies that d appears thrice in the decomposition of e^{2}. Since (e^{2}, d^{2}) = 60 we have a contradiction. This proves that case (3) cannot occur.

If (4) holds then bd = b + 4e, such that e is a real element. Thus (bd, bd) = 85. On the other hand, let d^{2} = 5.1 + αb + βd + ⋯, for some non-negative integers α, β. Then (b^{2}, d^{2}) = (5.1 + 3b + d, 5.1 + αb + βd + ⋯) = 25 + 5(3α + β). Since f(d^{2}) = 25 and (bd, bd) = (b^{2}, d^{2}) we have

3α + β = 12, α + β ≤ 4.

Therefore α = 4, β = 0, i.e., d^{2} = 5.1 + 4b. Moreover, (bd, d) = (d^{2}, b) = 20. Hence d appears 4 times in the decomposition of bd, i.e., bd = b + 4d. This shows that B_{b} = {1, b, d}. Since b is faithful, B = {1, b, d}.

If (5) holds, then by the equality b(bd) = b^{2}d we obtain

5.1 + d + be_{1} + be_{2} + be_{3} + be_{4} = 5d + 3e_{1} + 3e_{2} + 3e_{3} + 3e_{4} + d^{2}.

Since (b^{2}, d^{2}) = (bd, bd) = 25, we have Supp_{B}(d^{2})∩Supp_{B}(b^{2}) = {1}. Now we claim that e_{1} does not appear in the decomposition of be_{1}. If (be_{1}, e_{1}) ≠ 0, then (be_{1}, e_{1}) = 5, 10, 15 or 20. We assume (be_{1}, e_{1}) = 5. Hence e_{2}, e_{3} or e_{4} may appear in the decomposition of be_{1}. For example, let e_{2} appear α times in be_{1} for some α ∈ ℤ^{+}. Thus be_{1} = d+e_{1} +αe_{2} +· · · . If α = 1, then 35 ≥ (be_{1}, be_{1}) = (b^{2}, e_{1}^{2}) = (5.1 + 3b + d, 5.1 + b + ⋯) ≥ 40 which is a contradiction. If α = 2, then be_{1} = d+e_{1}+2e_{2}+⋯ and 35 ≥ (be_{1}, be_{1}) = (b^{2}, e_{1}^{2}) = (5.1+3b+d, 5.1+b+⋯) ≥ 40 which is a contradiction. If α = 3, then be_{1} = d + e_{1} + 3e_{2} and be_{2} = d + 3e_{1} + · · ·. Thus 35 ≥ (be_{1}, be_{2}) = (b^{2}, e_{1}e_{2}) ≥ 45, which is a contradiction. Hence, we have (be_{1}, e_{1}) ≠ 5. By the same way the other cases are proved and so e_{1} cannot occur in the decomposition of be_{1}. Similarly, we can prove that e_{i} cannot occur in the decomposition of be_{i}, for i = 2, 3, 4. Now, it is easy to check that

be_{1} = d + e_{2} + e_{3} + e_{4} + x,
be_{2} = d + e_{1} + e_{3} + e_{4} + y,
be_{3} = d + e_{1} + e_{2} + e_{4} + z,
be_{4} = d + e_{1} + e_{2} + e_{3} + w,

where d^{2} = 5.1 + x + y + z + w. Since b(be_{1}) = bd + be_{2} + be_{3} + be_{4} + bx, b^{2}e_{1} = 5e_{1} + 3be_{1} + de_{1} and b(be_{1}) = b^{2}e_{1}, we have e_{1}+3x+de_{1} = b+y+z+w+bx. The obvious analogue of the proof of case (1) shows that bx = e_{1} + 3x + k and de_{1} = b + y + z + w + k where k ∈ B − {1, e_{1}, x}. Hence x^{2} = 5.1 + 3b + ⋯ which implies that

55 = (bx, bx) = (b^{2}, x^{2}) = (5.1 + 3b + d, 5.1 + 3b + ⋯) ≥ 70,

that is contradiction. Therefore the case (5) cannot hold.

If (6) holds, then we can prove that

be_{1} = d + e_{2} + e_{3} + \overline{e_{3}} + x, x = \overline{x}
be_{2} = d + e_{1} + e_{3} + \overline{e_{3}} + y, y = \overline{y}

be_{3} = d + e_{1} + e_{2} + \overline{e_{3}} + z,
be_{4} = d + e_{1} + e_{2} + e_{3} + \overline{z},

where x, y, z are elements (not necessarily distinct) of B − \{1\} and d^{2} = 5.1 + x + y + z + \overline{z}. Since b(be_{1}) = b^{2}e_{1}, b(be_{1}) = bd + be_{1} + be_{2} + be_{3} + b\overline{e_{3}} + bx and b^{2}e_{1} = 5e_{1} + 3be_{1} + de_{1}, we have

e_{1} + 3x + de_{1} = b + y + z + \overline{z} + bx

The obvious analogue of the proof of case (1) shows bx = e_{1}+3x+k and de_{1} = b+y+z+\overline{z}+k such that k ≠ e_{1}, x. It follows that 55 = (bx, bx) = (b^{2}, x^{2}) = (5.1+3b+d, 5.1+3b+⋯) ≥ 70 which is a contradiction. Therefore the case (6) cannot occur.

Now, we consider the case (7). Let b, d appear α, β times in d^{2} respectively. Since 35 = (bd, bd) = (b^{2}, d^{2}), we obtain α = 0, β = 2, and so d^{2} = 5.1 + 2d + x + y, for some (not necessarily distinct) elements x, y ∈ B − {1, d}. Since b(bd) = b^{2} + 2be_{1} + be_{2} + be_{3}, b^{2}d = 5d + 3bd + d^{2} and b(bd) = b^{2}d^{,} we obtain

2be_{1} + be_{2} + be_{3} = 6e_{1} + 3e_{2} + 3e_{3} + 6d + x + y.  (I)

The obvious analogue of (5) shows that e_{1} does not appear in the decomposition of be_{1}. Now, from (I) e_{1} appears at least thrice in be_{2}. In fact, if e_{1} appears 4 times in be_{2}, then we have be_{2} = d + 4e_{1}. Hence we have (be_{1}, e_{2}) = (be_{2}, e_{1}) = 20, (be_{1}, d) = 10. It follows that |Supp_{B}(be_{1})| ≥ 6 which is a contradiction. Thus e_{1} appears thrice in the decomposition of be_{2}. It implies that be_{2} = d + 3e_{1} + ⋯ and be_{3} = d + 3e_{1} + · · · . Hence e_{2}, e_{3} both appear thrice in be_{1} which is a contradiction. Therefore the case (7) cannot occur.

We can observe exactly in the same way that the cases (8) and (9) cannot occur. Now the proof of proposition is complete. □

Proposition 2.3. Let (A, B) be a homogeneous standard integral table algebra of degree 5 such that B contains a real faithful element b so that b^{2} = 5.1 + 3c + b. Then B = {1, b, c} and

c^{2} = 5.1 + 2b + 2c, bc = 3b + 2c, b^{2} = 5.1 + 3c + b.

Proof. Let b ∈ B such that b^{2} = 5.1 + 3c + b and b, c are two real elements. Since (bc, b) = (b^{2}, c) = 15, b appears thrice in the decomposition of bc. Thus bc = 3b + e_{1} + e_{2}, for some e_{1}, e_{2} ∈ B − {1, b}. Let c^{2} = 5.1 + γc + δb + ⋯, for some non-negative integers δ, γ. Hence we have δ + γ ≤ 4 by the equality f(c^{2}) = 25. Moreover,

(b^{2}, c^{2}) = (5.1 + 3c + b, 5.1 + γc + δb + ⋯) = 25 + 15γ + 5δ.  (1)

If e_{1} = e_{2} = d then bc = 3b+2d and so (bc, bc) = 65. Since (bc, bc) = (c^{2}, b^{2}), we have 3γ+δ = 8 and therefore γ = 2 = δ. This shows that c^{2} = 5.1 + 2c + 2b and so (bc, c) = (c^{2}, b) = 10. Hence c appears twice in the decomposition of bc. Thus bc = 3b+2c. Therefore B_{b} = {1, b, c}. Since b is faithful, we obtain B = {1, b, c}.

If e_{1} ≠ e_{2}, then by (1)

(bc, bc) = (b^{2}, c^{2}) = (5.1 + 3c + b, 5.1 + γc + δb + ⋯).

Hence 25 + 15γ + 5δ = 55. This shows that γ = 1, δ = 3 or γ = 2, δ = 0. If γ = 2, δ = 0 then c^{2} = 5.1 + 2c + x + y for some x, y ∈ B − {1, b, c} and bc = 3b + e_{1} + e_{2}.

First we assume that x ≠ y. Since (bc)c = bc^{2} we have

be_{1} + be_{2} = e_{1} + e_{2} + 2c + 3x + 3y.  (2)

Hence we have

(be_{1}, c) = (bc, e_{1}) = 5 and (be_{2}, c) = (bc, e_{2}) = 5.  (3)

This implies that c appears once in the decomposition of be_{1} and also be_{2}. Moreover from (2) we know that only one of e_{1} or e_{2} appears in be_{2}. Let e_{1}\overline{e_{2}} = α′c + β′b ⋯ for some positive integers α′, β′. Thus

(be_{1}, be_{2}) = (b^{2}, e_{1}\overline{e_{2}}) = 5(3α′ + β′).  (4)

Now from (2) we must have one of the following cases:

(∗) \{ \begin{matrix}be_{1} = c + e_{1} + αx + βy, (I) \\ be_{2} = c + e_{2} + αy + βx. (II) \end{matrix}

(∗∗) \{ \begin{matrix}be_{1} = c + e_{2} + αx + βy, (I) \\ be_{2} = c + e_{1} + αy + βx, (II) \end{matrix}

such that α + β = 3.

Moreover, from (4) and the above eqalities we have

3α′ + β′ = 1 + 2αβ.  (5)

Assume that x, y ∉ {e_{1}, e_{2}}. If the case (∗) holds, then from (5) we have one of the following cases:

α = 0, β = 3 and α′ = 0, β′ = 1, (i)
α = 1, β = 2 and α′ = 1, β′ = 2 or α′ = 0, β′ = 5, (ii)
α = 2, β = 1 and α′ = 1, β′ = 2 or α′ = 0, β′ = 5, (iii)
α = 3, β = 0 and α′ = 0, β′ = 1. (iv)

If the case (i) holds, then e_{1}\overline{e_{2}} = b + · · · . So (be_{2}, e_{1}) = (e_{1}\overline{e_{2}}, b) = 5. But this shows that e_{1} appears in the decomposition of be_{2}, a contradiction .

If the case (ii) holds, then be_{1} = c + e_{1} + 2y + x, be_{2} = c + e_{2} + 2x + y and e_{1}\overline{e_{2}} = c + 2b + · · · . So (be_{2}, e_{1}) = (e_{1}\overline{e_{2}}, b) = 10. Thus e_{1} appears twice in the decomposition of be_{2}, a contradiction. Also if (ii) holds for case α′ = 0, β′ = 5 then e_{1}\overline{e_{2}} = 5b. This shows that (be_{2}, e_{1}) = (b, e_{1}\overline{e_{2}}) = 25, that is, e_{1} appears 5 times in the decomposition of be_{2}, a contradiction.

If (iii) holds, for case α′ = 1, β′ = 2 then

e_{1}\overline{e_{2}} = c + 2b + ⋯, be_{1} = c + e_{1} + 2x + y, be_{2} = c + e_{2} + x + 2y.

Hence (be_{2}, e_{1}) = (e_{1}\overline{e_{2}}, b) = 10. This shows that e_{1} appears in the decomposition of be_{2}, a contradiction. Also, if (iii) holds for case α′ = 0, β′ = 5, then e_{1}\overline{e_{2}} = 5b. This shows that (be_{2}, e_{1}) = (b, e_{1}\overline{e_{2}}) = 25, that is, e_{1} appears 5 times in be_{2}, a contradiction.

If the case (iv) holds, it is easy to check that e_{1} appears in the decomposition of be_{2} that is a contradiction. This proves that the case (∗) cannot occur.

Now we assume that the case (∗∗) holds. Since (e_{1}\overline{e_{2}}, b) = (be_{1}, e_{2}) = 5, b appears once in e_{1}\overline{e_{2}}. Assume that c appears α′ times in e_{1}\overline{e_{2}}. Thus e_{1}\overline{e_{2}} = b + α′c + ⋯ and so (b^{2}, e_{1}\overline{e_{2}}) = (5.1 + 3c + b, b + α′c + ⋯) = 5 + 15α′. Since

(b^{2}, e_{1}\overline{e_{2}}) = (be_{1}, be_{2}) = 5 + 10αβ,

we have 2αβ = 3α′. Hence one of the following cases hold:

α = 0, β = 3 and α′ = 0, (v) α = 3, β = 0 and α′ = 0. (vi)

If the case (v) holds then be_{1} = e_{2} + c + 3y and be_{2} = e_{2} + c + 3x by (∗∗). Since b(be_{1}) = b^{2}e_{1} we have

by + b + x = e_{1}c + y + e_{1}.  (6)

Obviously by = 3e_{1} + y + z and e_{1}c = b + x + z + 2e_{1} for some z ∈ B − {1, e_{1}}, by (6) and the decomposition of be_{1}. Since (\overline{y}e_{1}, b) = (by, e_{1}) = 15, b appears thrice in the decomposition of \overline{y}e_{1}. Therefore (e_{1}\overline{y}, bc) ≥ 45. Also (by, e_{1}c) = (3e_{1} + y + z, b + 2e_{1} + x + z) ≤ 40. Since (\overline{y}e_{1}, bc) = (by, e_{1}c) we have a contradiction.

If the case (vi) holds, then be_{1} = e_{2} + c + 3x and be_{2} = e_{1} + c + 3y. Since b(be_{1}) = b^{2}e_{1} we have bx + b + y = e_{1} + x + e_{1}c. Hence

bx = x + z + 3e_{1}, e_{1}c = b + 2e_{1} + y + z,

for some z ∈ B − {1, e_{1}}. Thus (bx, e_{1}c) = (x + z + 3e_{1}, b + 2e_{1} + y + z) ≤ 40. On the other hand, (bc, \overline{e_{1}}x) = (bx, e_{1}c) leading to contradiction.

If e_{1} = x then c^{2} = 5.1 + 2c + e_{1} + y and bc = 3b + e_{1} + e_{2}. Since b(bc) = b^{2}c, we have

be_{1} + be_{2} = 2c + 4e_{1} + 3y + e_{2}.  (7)

Hence

be_{1} = c + e_{2} + αe_{1} + βy and be_{2} = c + α′e_{1} + β′y,  (8)

such that α′ + α = 4, β′ + β = 3, or we must have

be_{2} = c + e_{2} + α′e_{1} + β′y and be_{1} = c + αe_{1} + βy,  (9)

such that α′ + α = 4, β′ + β = 3.

If (8) holds, then we have

be_{1} = c + 3e_{1} + e_{2} and be_{2} = c + e_{1} + 3y.  (8.1)

If (9) holds, then we have

be_{1} = c + 4e_{1} and be_{2} = c + e_{2} + 3y.  (9.1)

First assume that (8.1) holds. Then

(be_{1}, be_{1}) = (c + 3e_{1} + e_{2}, c + 3e_{1} + e_{2}) = 55.

Also, (b^{2}, e_{1}\overline{e_{1}}) = (5.1 + 3c + b, 5.1 + 3b + λc + ⋯) = 40 + 15λ. Since (b^{2}, e_{1}\overline{e_{1}}) = (be_{1}, be_{1}), we must have λ = 1, that is, e_{1}\overline{e_{1}} = 5.1 + 3b + c. By the equality b(be_{1}) = b^{2}e_{1} we have e_{1}c = b + c + e_{1} + e_{2} + y. Moreover

(c^{2}, e_{1}\overline{e_{1}}) = (5.1 + 2c + e_{1} + y, 5.1 + 3b + c) = 35,

and that (e_{1}c, e_{1}c) = 25. Since (e_{1}c, e_{1}c) = (c^{2}, e_{1}\overline{e_{1}}) we have a contradiction.

Next if the case (9.1) holds then (be_{1}, be_{1}) = 85. Moreover, (b^{2}, e_{1}\overline{e_{1}}) = (5.1 + 3c + b, 5.1 + 4b) = 45. Now by the equality (b^{2}, e_{1}\overline{e_{1}}) = (be_{1}, be_{1}) we have a contradiction.

Similary if e_{1} = y we will obtain a contradiction and so this cases cannot occure.

Now if {e_{1}, e_{2}} = {x, y} then without loss of generality we can assume that x = e_{1} and y = e_{2}. Hence c^{2} = 5.1 + 2c + e_{1} + e_{2} and bc = 3b + e_{1} + e_{2}. By the equality b(bc) = b^{2}c we must have

be_{1} + be_{2} = 2c + 4e_{1} + 4e_{2}.

Hence we have one of the following cases:

be_{1} = c + e_{1} + 3e_{2},  be_{2} = c + 3e_{1} + e_{2}  (2.1)

be_{1} = be_{2} = c + 2e_{1} + 2e_{2}, (2.2)

be_{1} = c + 3e_{1} + e_{2},  be_{2} = c + 3e_{2} + e_{1}.  (2.3)

If the case (2.1) holds then by the equality (be_{1}, be_{2}) = (b^{2}, e_{1}\overline{e_{1}}), since (be_{1}, be_{2}) = 35 and (b^{2}, e_{2}\overline{e_{1}}) = (5.1 + 3b + c, 3b + ⋯) ≥ 45, we have a contradiction. If the case (2.2) holds, then (be_{1}, be_{1}) = (c+2e_{1} +2e_{2}, c+2e_{1} +2e_{2}) = 45, and (b^{2}, e_{1}\overline{e_{1}}) = (5.1+3b+c, 5.1+2b+…) ≥ 55. Since (be_{1}, be_{1}) = (b^{2}, e_{1}\overline{e_{1}}) we have a contradiction.

By similar considerations we see that the case (2.3) cannot hold. Therefore we have proved that {x, y} ∩ {e_{1}, e_{2}} = ∅.

Now we assume that x = y. Then c^{2} = 5.1+2c+2x, b^{2} = 5.1+3c+b and bc = 3b+e_{1} +e_{2}. By the equality b^{2}c = b(bc) we have

be_{1} = c + e_{2} + 3x and be_{2} = c + e_{1} + 3x.

Hence 15 = (be_{1}, x) = (bx, e_{1}) and 15 = (be_{2}, x) = (bx, e_{2}). This shows that e_{1} and e_{2} appear thrice in bx. But f(bx) = 25 and f(3e_{1} + 3e_{2}) = 30, that is a contradiction. Similarly if γ = 1, δ = 3 we will obtain a contradiction. The proposition is proved now. □

Proposition 2.4. There is no homogeneous standard integral table algebra of degree 5 such that its table basis contains a real faithful element, say b, where

b^{2} = 5.1 + 3c + d, d ≠ c, d = \overline{d}, c = \overline{c}.

Proof. Let b ∈ B such that b^{2} = 5.1 + 3c + d, and c, d be two distinct real elements of B. Since 15 = (b^{2}, c) = (bc, b) and f(bc) = 25, one of the following cases holds:

bc = 3b + 2e, e ∈ B − \{b\}, e = \overline{e},  (1)
bc = 3b + e_{1} + e_{2}, e_{1}, e_{2} ∈ B − {b}, e_{1} ≠ e_{2}.  (2)

If the case (1) holds, then bc = 3b + 2e, e = \overline{e}. Hence

(bc, bc) = (3b + 2e, 3b + 2e) = 65.

Let c^{2} = 5.1 + αc + βd ⋯ for some non-negative integers α, β. Since f(c^{2}) = 25, we have α + β ≤ 4. Moreover

(b^{2}, c^{2}) = (5.1 + 3c + d, 5.1 + αc + βd + ⋯) = 25 + 15α + 5β.

By the equality (b^{2}, c^{2}) = (bc, bc) we have 3α + β = 8 and α + β ≤ 4. This shows that α = 2 = β. Hence c^{2} = 5.1 + 2c + 2d. Since 10 = (c^{2}, d) = (dc, c), c appears twice in the decomposition of dc. Now by the equality b(bc) = b^{2}c we must have

2be = 2c + 3d + dc (1.1)

Obviously, from (1.1) d must appear in the decomposition of dc. Let d appear λ times in dc, so dc = 2c + λd + · · · . By (1.1) we have be = 2c + γd + ⋯ such that 2γ = 3 + λ. Hence λ = 1, γ = 2 or λ = 3, γ = 3.

If λ = 1, γ = 2 then dc = 2c + d + 2x and be = 2c + 2d + x for some x ∈ B − {c, d} by (1.1). Hence

(dc, dc) = (2c + d + 2x, 2c + d + 2x) = 45.

Let d^{2} = 5.1 + δd + c + ⋯, for some non-negative integers δ. We have

(d^{2}, c^{2}) = (5.1 + c + δd + ⋯, 5.1 + 2c + 2d) = 25 + 10(1 + δ).

Since (dc, dc) = (d^{2}, c^{2}), we have δ = 1 and so d^{2} = 5.1+c+d+· · · . Since (bd, bd) = (b^{2}, d^{2}) = 45, f(bd) = 25, 10 = (be, d) = (e, bd) and 5 = (b^{2}, d) = (bd, b), we have bd = b + 2e + 2e_{1} for some e_{1} ∈ B − {1, b, e}. Hence d^{2} = 5.1 + c + d + v + w for some v, w ∈ B − {1, c, d}. Since b(bd) = b^{2}d we must have be_{1} = 2x+2d+v and d^{2} = 5.1+c+d+2v. Let ec = 2b+δc+εe_{1}+⋯ for some non-negative integers δ, ε by the equality bc = 2c+2d+x. Hence (bd, ec) = (b+2e+ 2e_{1}, 2b+δe+εe_{1}+⋯) = 10+10(δ+ε). Since (bd, ec) = (dc, be) = (2c+2x+d, 2c+2d+x) = 40, we have δ + ε = 3. Moreover,

40 = (c^{2}, be) = (bc, ec) = (3b + 2e, 2b + δe + εe_{1} + ⋯).

Hence δ = 1 and ε = 2. Therefore we have

ec = 2b + 2e_{1} + e.

On the other hand 45 = (ec, ec) = (e^{2}, c^{2}) = (5.1 + c + ⋯, 5.1 + 2c + 2d). So we must have e^{2} = 5.1 + c + d + · · · . This shows that e^{2} = d^{2}, by the equality (bc)^{2} = b^{2}c^{2}. Furthermore, (bc)e = 3be + 2e^{2} and (ec)b = 2b^{2} + be + 2be_{1} which imply that e^{2} = 5.1 + c + d + v + x, by the equality (bc)e = (ec)b. This implies that v = x. Hence

(d^{2}, e^{2}) = (5.1 + c + d + 2x, 5.1 + c + d + 2x) = 55.

But (de, de) ≤ 45. Since (d^{2}, e^{2}) = (de, de) we have a contradiction.

Now we assume that λ = γ = 3. Hence be = dc = 2c + 3d. We have 15 = (dc, d) = (d^{2}, c), this implies that c appears thrice in the decomposition of d^{2}. Also 15 = (be, d) = (bd, e) implies that e appears thrice in the decomposition of bd. Hence we have d^{2} = 5.1+3c+g, bd = 3e + b + h, for some h, g ∈ B − {1, b, c}. Therefore (bd, bd) = 55 and

(b^{2}, d^{2}) = (5.1 + 3c + d, 5.1 + 3c + g) = \{ \begin{matrix}75 & if g = d, \\ 70 & if g ≠ d. \end{matrix}

Since (b^{2}, d^{2}) = (bd, bd) we have a contradiction.

If the case (2) holds, then bc = 3b+e_{1} +e_{2}, e_{1} ≠ e_{2}. Let c^{2} = 5.1+λc+γd+⋯, for some non-negative integers λ, γ. We have (b^{2}, c^{2}) = (5.1+3c+d, 5.1+λc+γd+⋯) = 25+5(3λ+γ), and (bc, bc) = 55. Since (b^{2}, c^{2}) = (bc, bc), we have 6 = 3λ + γ, such that λ + γ ≤ 4 by f(c^{2}) = 25. Hence, λ = 1, γ = 3 or λ = 2, γ = 0.

If λ = 1, γ = 3 then c^{2} = 5.1+3d+c. This shows that c appears thrice in the decomposition of dc. Since b^{2}c = b(bc), we must have be_{1} + be_{2} + c = 6d + dc. Hence be_{1} = α′d + ⋯ and be_{2} = β′d + ⋯, where α′ + β′ = 6. Since 5α′ = (be_{1}, d) = (bd, e_{1}) and 5β′ = (be_{2}, d) = (bd, e_{2}), we have that e_{1}, e_{2} appear α′ and β′ times in the decomposition of bd respectively. But by the equalities f(bd) = 25 and f(α′e_{1} + β′e_{2}) = 5(α′ + β′) = 30 that is a contradiction.

If λ = 2, γ = 0, then bc = 3b+e_{1} +e_{2} and c^{2} = 5.1+2c+x+y for some x, y ∈ B −{1, c, d} By the equality b(bc) = b^{2}c we have

be_{1} + be_{2} + 3d = dc + 2c + 3x + 3y.

By the above equality we can see that d appears δ times in the decomposition of dc such that 3 ≤ δ ≤ 5.

If δ = 3 then dc = 3d + · · · . Hence d^{2} = 5.1 + 3c + x. Therefore

(bd, bd) = (b^{2}, d^{2}) = \{ \begin{matrix}75 & if x = d, \\ 70 & if x ≠ d. \end{matrix}

On the other hand (bd, b) = (b^{2}, d) = 5. So bd = b + · · · . This implies that (bd, bd) cannot be equal to 70 or 75, a contradiction.

If δ = 4 then dc = 4d + x, for some x ∈ B − {1, d}. Hence 20 = (dc, d) = (d^{2}, c), and so d^{2} = 5.1+4c. Therefore (d^{2}, c^{2}) = (5.1+4c, 5.1+c+3d) = 65, and (dc, dc) = (4d+x, 4d+x) = 85. Since (d^{2}, c^{2}) = (dc, dc) we have a contradiction.

If δ = 5 then dc = 5d. Hence 25 = (dc, d) = (d^{2}, c), i.e., c appears 5 times in the decomposition of d^{2}. Since f(d^{2}) = 25 we have a contradiction. This completes the proof of the proposition. □

Proposition 2.5. There is no homogeneous standard integral table algebra (A, B) of degree 5 such that B contains a faithful element b where

b\overline{b} = 5.1 + 3c + d, c = \overline{c}, d = \overline{d}.

Proof. Let b ∈ B such that b\overline{b} = 5.1 + 3c + d, for some c, d ∈ B − \{1, b\} and c = \overline{c}, d = \overline{d}. Hence (b\overline{b}, b\overline{b}) = 75. Since (b\overline{b}, b\overline{b}) = (b^{2}, b^{2}), we have (b^{2}, b^{2}) = 75. But this is a contradiction, because b^{2} has one of the following cases:

b^{2} = \{ \begin{matrix}5c, & then (b^{2}, b^{2}) = 125. \\ 4c + d, & then (b^{2}, b^{2}) = 85. \\ 3c + 2d, & then (b^{2}, b^{2}) = 65. \\ 3c + d + e, & then (b^{2}, b^{2}) = 55. \\ 2c + 2d + e, & then (b^{2}, b^{2}) = 45. \\ 2c + d + e + g, & then (b^{2}, b^{2}) = 35. \\ c + d + e + g + h, & then (b^{2}, b^{2}) = 25. \end{matrix}

This completes the proof of the proposition. □

Acknowledgment. Both authors would like to thank the referee for his suggestions. The first author would like to thank the Institute for Studies in Physics and Mathematics for its financial support. The second author would like to thank the Institute for Advanced Studies in Basic Sciences, Gava Zang, Zanjan, for its financial support.

References

[1] Arad, Z.: Homogeneous Integral Table Algebra of Degree Two, Three, and Four with a Faithful Element. Preprint

[2] Arad, Z.; Blau, H. I.: On table algebras and applications to finite group theory. J. Algebra 138 (1991), 137–185.

[3] Arad, Z.; Fisman, E.; Husam, A.; Muzychuk, M.: Homogeneous integral table algebras of degree four. In preparation.

[4] Blau, H. I.: Quotient Structures in C-algebras. J. Algebra 177 (1995), 297–337.

[5] Blau, H. I.: Homogeneous integral table algebras of degree two. Algebra Colloquium, to appear.

[6] Blau H. I.; Xu, B.: On homogeneous integral table algebras. J. Algebra, to appear.

[7] Blau, H. I.; Xu, Bangteng: Homogeneous integral table algebras of degree three with a faithful real element. Part 1, submitted.

Received March 9, 1998