Associative words in the symmetric group of degree three
Let G be a group. An element w(x, y) of the absolutely free group on free generators x, y is called an associative word in G if the equality w(w(g₁, g₂), g₃)=w(g₁, w(g₂, g₃)) holds for all g₁, g₂ ∈ G. In this paper we determine all associative words in the symmetric group on three letters....
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Plonka, E. 2019-06-09T13:54:07Z 2019-06-09T13:54:07Z 2013 Associative words in the symmetric group of degree three / E. Plonka // Algebra and Discrete Mathematics. — 2013. — Vol. 15, № 1. — С. 83–95. — Бібліогр.: 9 назв. — англ. 1726-3255 2010 MSC:20B30, 08A40,20F12. https://nasplib.isofts.kiev.ua/handle/123456789/152265 Let G be a group. An element w(x, y) of the absolutely free group on free generators x, y is called an associative word in G if the equality w(w(g₁, g₂), g₃)=w(g₁, w(g₂, g₃)) holds for all g₁, g₂ ∈ G. In this paper we determine all associative words in the symmetric group on three letters. en Інститут прикладної математики і механіки НАН України Algebra and Discrete Mathematics Associative words in the symmetric group of degree three Article published earlier |
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Associative words in the symmetric group of degree three |
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Associative words in the symmetric group of degree three Plonka, E. |
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Let G be a group. An element w(x, y) of the absolutely free group on free generators x, y is called an associative word in G if the equality w(w(g₁, g₂), g₃)=w(g₁, w(g₂, g₃)) holds for all g₁, g₂ ∈ G. In this paper we determine all associative words in the symmetric group on three letters.
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Associative words in the symmetric group of degree three / E. Plonka // Algebra and Discrete Mathematics. — 2013. — Vol. 15, № 1. — С. 83–95. — Бібліогр.: 9 назв. — англ. |
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AT plonkae associativewordsinthesymmetricgroupofdegreethree |
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2025-11-24T23:55:19Z |
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1850501178492715008 |
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Algebra and Discrete Mathematics RESEARCH ARTICLE
Volume 15 (2013). Number 1. pp. 83 – 95
c© Journal “Algebra and Discrete Mathematics”
Associative words in the symmetric group
of degree three
Ernest Płonka
Communicated by V. I. Sushchansky
Abstract. Let G be a group. An element w(x, y) of the
absolutely free group on free generators x, y is called an associative
word in G if the equality w(w(g1, g2), g3) = w(g1, w(g2, g3)) holds
for all g1, g2 ∈ G. In this paper we determine all associative words
in the symmetric group on three letters.
1. Introduction
Let G be a group and let F = F (x, y) be the absolutely free group on
free generators x, y. Let V = V (G) be the subgroup of F consisting of all
words w such that w(g1, g2) = 1 for all g1, g2 ∈ G. An element w ∈ F is
said to be associative in G if the equality
w(w(g1, g2), g3) = w(g1, w(g2, g3)) (1.1)
holds for all elements g1, g2, g3 ∈ G. The words 1, x, y, xy and yx are,
of course, associative (trivial words) for any group. It is known that in
the absolutely free group ([6,7]) and in the class of all abelian groups
([4]) there are no other associative words. In other groups (G; ·) such
nontrivial word w can exist, however. Moreover, in some free nilpotent
groups there are nontrivial associative words w(x, y) = x ◦ y such that
(G; ◦) is a group and the group operation x · y can be expressed as a
2010 MSC: 20B30, 08A40,20F12.
Key words and phrases: associative words, symmetric group S3.
84 Associative words in the symmetric group
word in the group (G; ◦): see [1,2,5,9]. In this paper we are looking for
the associative words in the symmetric group of degree three which is
metabelian but not nilpotent. We show that each associative word in S3
is equivalent modulo V (S3) to one of the five words mentioned above or
to one of x3, y3, x4, y4, x4y4, [x, y]x+y, [y, x]x+y.
2. Preliminaries
We use standard notations:
x−1yx = yx, [y, x] = y−1x−1yx, [y, x]−1 = [x, y] xβy = (xy)β ,
xα+βy+z+0 = xα(xβ)yxz
for arbitrary group elements x, y, z and all integers α, β.
Let us recall the following simple facts about the identities in S3.
(i) The relations
[xy, z] = [x, z]y[y, z], [x, yz] = [x, z][x, y]z
are identities in any group.
(ii) The commutator subgroup S′
3 of S3 consists of all even permuta-
tions and the square of each element from S3 is in S′
3.
This yields
(iii) For all products C of commutators the equality
C(1−x)(1+x) = 1
is an identity in S3.
(iv) The equalities
x6 = [y, x]3 = 1, [[y, x], [u, v]] = 1, [x2, [y, z]] = 1,
[y2, x] = [y, x]y+1, [y3, x] = [y, x]y−1, [y4, x] = [y, x]−y−1
are identities in the group S3.
From (ii) and (iv) one can derived
(v) The equality
[y, x]xy = [y, x]−1−x−y
is an identity in S3.
E. Płonka 85
The following consequence of Corollary 2 in [8] plays very important
role in our considerations.
(vi) If for some A, B, C ∈ Z3 the equality
[y, x]A+Bx+Cy = 1,
holds for all x, y ∈ S3, then A = B = C = 0.
Proposition 2.1. Any 2−word in S3 is equivalent (mod V ) to some
word of the form
w(x, y) = xαyβ[y, x]A+Bx+Cy (2.1)
where α, β ∈ Z6 and A, B, C ∈ Z3.
Proof. It is enough to apply Hall’s classical collection process from [3].
Proposition 2.2. The word
w(x, y) = xαyβ[y, x]A+Bx+Cy
is asociative in S3, then α, β ∈ {0, 1, 3, 4}.
Proof. By putting y = z = 1 and x = y = 1 into (2.2) we get
xα2
= xα, and xβ2
= xβ (2.2)
and therefore α(α − 1) ≡ β(β − 1) ≡ 0 (mod 6).
Proposition 2.3. If w(x, y) is associative in a group G, then the word
u(x, y) = w(y, x) is also associative in G.
Proof. We have
u(u(x, y), z) = w(z, w(y, x)) = w(w(z, y), x) = u(x, u(y, z)).
3. Associative words
First of all we show that for some pairs (α, β) no word of the form
(2.1) is associative in S3. It what follows we shall always assume that
A, B, C ∈ Z3 and sometimes we write γ(s, t) instead of A + Bs + Ct.
Theorem 3.1. There are no associative words in the group S3 which are
of the form
xαyβ [y, x]A+Bx+Cy, (3.1)
where (α, β) is one of the following pairs
(1, 3), (3, 1), (1, 4), (4, 1), (3, 4), (4, 3), (3, 3)
86 Associative words in the symmetric group
Proof. Case α = 1, β = 3.
Let us begin with an auxiliary result
w(x, y)3 = xy3[y, x]γxy3[y, x]γxy3[y, x]γ
= xy3x2[y, x]γy3[y3, x][y, x]xγy3[y, x]γ
= xy3x2[y, x]γ [y3, x]y[y, x]xyγ [y, x]γ
= xy3x3[y, x]x+1[y, x]γ [y3, x](y−1)y[y, x]xyγ [y, x]γ
= xy3x3[y, x]x+1[y, x]γ [y3, x](y−1)y[y, x]xyγ [y, x]γ .
We have thus established
(xy3[y, x]γ)3 = x3y3[y, x]−1+x−y+(1−x−y)γ . (3.2)
Further we have
L = w(w(1, y), z) = w(y3, z) = y3z3[z, y3]γ(y,z) = y3z3[z, y](y−1)γ(y,z),
R = w(1, w(y, z)) = w(y, z)3 = (yz3[z, y]γ(y,z))3
= yz3(z3y[y, z3])yz3[z, y](yz−1)γ(y,z) = y3z3[z, y]−1−z+y+(yz−1)γ(y,z).
Thus L = R is equivalent to the equality
[z, y](−1−A−B+C)+(1+A−C)y+(−1−A+B+C))z = 1.
By (vi) we have −1−A−B+C = 0, 1+A−C = 0 and −1−A+B+C =
0, which has the solution B = 0, C = A + 1. Therefore the associative
word (3.1) has to be of the form w(x, y) = xy3[y, x]C−1+Cy. Let us put
z = x into the associative low (1.1). We get
L = w(w(x, y), x) = xy3[y, x]γ(x,y)w(x, y)x3[x, xy3[y, x]γ(x,y)]γ(xy,x)
x4y3[y, x]x+y[y, x]xγ(x,y)[y, x](1−x+y)γ(xy,x)[y, x](1−x)γ(x,y)γ(xy,x),
which in the case B = 0, A = C − 1 gives
L = x4y3[y, x]x+y+x(C−1+Cy)+(1−x+y)(C−1+Cx)+(1−x)((C+Cx)−1)(C−1+Cy)
x4y3[y, x]x+y+x(C−1+Cy)+(1−x+y)(C−1+Cx)+(x−1)(C−1+Cy).
After some calculations we get
L = x4y3[y, x]C+C(x+1)y.
E. Płonka 87
Similarly we have
R = w(x, w(y, x)) = xw(y, x)3[yx3[x, y]γ(y,x), x]γ(x,xy)
= xw(y, x)3[yx3, x]γ(x,xy)[[x, y]γ(y,x), x]γ(x,xy)
= xw(y, x)3[y, x]xγ(x,xy)[y, x](1−x)γ(y,x)γ(x,xy),
which in the case B = 0 and A = C − 1 implies
R = xw(y, x)3[y, x]x(C−1+Cxy[y, x](1−x)(C+Cx−1)(C+1+Cxy)
= xw(y, x)3[y, x]−C+Cx+(1+C)y.
Now from the equality L = R we obtain
w(x, y)3 = x3y3[y, x]−x.
We get a contradiction, because formula (3.2) for γ = C − 1 + Cy
gives
w(x, y)3 = x3y3[y, x]C+1−x+(C+1)y.
Case α = 1, β = 4.
We have
L = w(w(1, y), z) = w(y4, z) = y4z4[z, y4]γ(1,z) = y4z4[z, y]−(y+1)γ(1,z).
Similarly
R = w(1, w(y, z)) = (yz4[z, y]yγ(y,z)yz4[z, y]γ(y,z))2
= (y2z4[z4, y][z, y]yγ(y,z)yz4[z, y]γ(y,z))2 = y4z4[z, y](z+1)−(y+1)γ(y,z).
Hence L = R yields [z, y]1+z = 1 which, by (vi), is not an identity in S3 .
Case α = 3, β = 4.
We have
L = w(w(1, y), z) = w(y4, z) = z4[z, y4]γ(1,z) = z4[z, y]−(y+1)γ(1,z).
Since y2 commutes both z4 and [z, y]γ(y,z), we can use of the previous
case. We obtain
R = w(1, w(y, z)) = w(y, z)4 = (y2yz4[z, y]γ(y,z)y2yz4[z, y]γ(y,z))2 =
= y2(yz4[z, y]γ(y,z)yz4[z, y]γ(y,z))2 = z4[z, y]−(z+1)−(y+1)γ(y,z).
88 Associative words in the symmetric group
Thus the condition L = R yields the equality
[z, y]z+1 = 1,
which is not an identity in S3.
Case α = 3, β = 3. We have
L = w(w(x, 1), z) = x3z3[z, x3]γ(x,z) = x3z3[z, x](x−1)(A+Bx+Cz),
R = w(x, w(1, z)) = x3z3[z3, x]γ(x,z) = x3z3[z, x](z−1)(A+Bx+Cz).
Thus the equality L = R implies, in view of (vi),
C − B ≡ A − C + B ≡ B − A − C ≡ 0 (mod 3)
which yields A = 0 and B − C = 0. So every word of the form w(x, y) =
x3y3[x, y]B(x+y) satisfies the equation w(w(x, 1), z) = w(x, w(1, z)) but
none of them is associative. Indeed,for such words we have
L = w(w(1, y), z) = w(y3, z) = y3z3[z3, y3]B(y+z)
= y3z3[z, y]B(y−1)(z−1)(y+z) = y3z3,
R = w(1, w(y, z)) = w(y, z)3
= y3z3[z, y]B(y+z)y3z3[z, y]B(y+z)y3z3[z, y]B(y+z)
= y3z3[z, y]B(y+z)y3z3y3z3[z, y]B(yz+1)(y+z)
= y3z3[z, y]B(y+z)[z, y](z−1)(y−1)[z, y]−B(y+z)
= y3z3[z, y](B+1)(y+z).
Thus L = R implies the equation [z, y]y+z = 1, which is not an identity
in S3.
Now by Proposition 2.3 we know that if the word w(x, y) of the form
(2.1) is associative in S3, then w(y, x) is also associative in S3. Since
w(y, x) = yixj [x, y]A+Bx+Cz = xjyi[y, x]A
′+B′x+C′y
for some A′, B′, C ′ ∈ Z3 the proof of Theorem 3.1 is complete.
In the following lemmas we consider the cases of pairs (α, β) for which
there exist associative words in S3.
Lemma 3.2. The word
w(x, y) = x[y, x]A+Bx+Cy = x[y, x]γ(x,y) (3.3)
is associative in S3 if and only if A = B = C = 0.
E. Płonka 89
Proof. Using the identities (ii), (iv) and (v) we have
w(x, w(y, y)) = x[y, x]γ(x,y),
w(w(x, y), y) = x[y, x]γ(x,y)[y, x[y, x]γ(x,y)]γ(x,y)
= x[y, x]γ(x,y)[y, x]γ(x,y)[y, x](1−y)γ(x,y)γ(x,y).
Taking into account (iii) we see that if w is associative, then
[y, x]A+Bx+Cy+(1−y)(A−C+Bx)2
= 1,
which, by (vi) ensures the following system of congruences
A + (A − C)2 + B2 + 2(A − C)B ≡ 0 (mod 3),
B + 2(A − C)B + 2(A − C)B ≡ 0 (mod 3),
C − (A − C)2 − B2 + 2(A − C)B ≡ 0 (mod 3).
The solution of the system are four triples (A, B, C) of the form (0, 0, 0),
(2, 2, 0), (2, 0, 1) and (0, 1, 1). In order to exclude the last three cases we
put y = x into (3.3). Then we get
L = w(w(x, x), z) = x[z, x]γ(x,z)
R = w(x, w(x, z)) = x[x[z, x]γ(x,z), x]γ(x,x)
= x[z, x](x−1)γ(x,x)γ(x,z).
Thus the condition L = R together with (iii) gives the equality
[z, x](x−1)(A−B−C)(A+Bx+Cz) = [z, x]A+Bx+Cz.
The equality is, by (vi), an identity in S3 if and only if the triples (A, B, C)
satisfies the following system of congruences
(A − B − C)(B − A − C) ≡ A (mod 3),
(A − B − C)(A − B − C) ≡ B (mod 3),
(A − B − C)(B − A − C) ≡ C (mod 3).
The proof of the lemma is complete, because none of the triples (2, 2, 0),
(2, 0, 1) and (0, 1, 1) do satisfy the system.
By Proposition 2.3 we have also
90 Associative words in the symmetric group
Corollary 3.3. The word
y[y, x]A+Bx+Cy
satisfies the associativity low if and only if A = B = C = 0.
Lemma 3.4. The word
w(x, y) = xy[y, x]A+Bx+Cy
is associative in S3 if and only if B = C = A = 0 or A − 1 = B = C = 0.
Proof. We have
w(w(x, y), z) = xy[y, x]γ(x,y)z[z, xy[y, x]γ(x,y)]γ(xy,z)
= xyz[y, x]zγ(x,y)+(z−1)γ(x,y)γ(xy,z)[z, x]yγ(xy,z)[z, y]γ(xy,z)
and
w(x, w(y, z)) = xyz[z, y]γ(y,z)[yz[z, y]γ(y,z), x]γ(x,yz)
= xyz[y, x]zγ(x,yz)[z, x]γ(x,yz)[z, y]γ(y,z)+(x−1)γ(y,z)γ(xy,z).
Hence we get
(w(x, w(y, z)))−1w(w(x, y), z) (3.4)
= [y, x](1−z){−Cy+γ(x,y)γ(xy,z)}[z, x](1−y)(−A)[z, y](1−x){−By+γ(y,z)γ(xy,z)}
By putting z = y into (3.4) we obtain
(w(x, w(y, y)))−1w(w(x, y), y) =
[y, x](1−y){−A−Cy+(A+Bx+Cy)(A+Bxy+Cy)},
which in view of (iii) and (v) can be rewritten as
[y, x](1−y){(A−C)2−(A−C)−B2)}.
Now we put y = x into (3.4). This gives
w(x, w(x, z))−1w(w(x, x), z) =
[z, x](1−x){(A−B)2−(A−B)−C2}}
E. Płonka 91
In view of (vi) if the word w(x, y) is associative in S3, then the following
system of congruences
{
(A − C)2 − (A − C) − B2 ≡ 0 (mod 3),
(A − B)2 − (A − B) − C2 ≡ 0 (mod 3).
has to satisfy. The solution of the system is B = C = 0 and A = 0 or
A = 1. Since the words xy and yx are associative, Lemma 3.4 follows.
Lemma 3.5. The 2−word
w(x, y) = x3[y, x]A+Bx+Cy (3.5)
is associative in S3 if and only if A = B = C = 0.
Proof. Clearly, the word x3 is associative in the group S3. We have
R = w(x, w(1, z)) = x3,
L = w(w(x, 1)z) = w(x3, z) = x3[z, x]γ(x,z) = x3[z, x](x−1)(A+Bx+Cz)
[z, x](−A+B−C)+(A−B−C)x+Cz.
So the equality R = L is equivalent to the conditions C = 0 and A = B.
Further we have
w(w(x, x), z) = x3[z, x3]A+Bx+Cz = x3[z, x](x−1)(A+Bx+Cz),
w(x, w(x, z)) = x3[x3[z, x]A+Bx+Cz, x]A+Bx+Cx
= x3[z, x](A−B−C)(x−1)(A−B+Cz)
Hence the equality w(w(x, x), z) = w(x, w(x, z)) after using (v) and (vi),
yields the system of equalities
(A − B − C)(B − A − C) ≡ 2A + B − C (mod 3),
(A − B − C)2 ≡ A − B − C (mod 3),
C(A − B − C) ≡ C (mod 3).
The system has four solutions for (A, B, C):(0,0,0),(1,0,0),(1,1,0) and
(2,2,0). We check that the last three triple do not produce associative
words of the form w(x, y) = x3[y, x]γ(x, y). To do this let us calculate
w(x, y)3 = x3[y, x]γ(x,y)(x3[y, x]γ(x,y)x3)[y, x]γ(x,y)
= x3[y, x]γ(x,y)[y, x]xγ(x,y)[y, x]γ(x,y) = x3[y, x](x−1)γ(x,y)
92 Associative words in the symmetric group
Taking this into account we get
L(A, B, C) = w(w(x, y), y) = w(x, y)3[y, w(x, y)γ(x,y) =
x3[y, x](x−1)γ(x,y)+(1−y)γ(x,y)γ(x,y),
R(A, B, C) = w((x, w(y, y)) = x3[y3, x]γ(x,y) = x3[y, x](y−1)γ(x,y)
Now it easy to check the following equalities
L(1, 0, 0) = x3[y, x]x−y, R(1, 0, 0) = x3[y, x]y−1
L(1, 1, 0) = x3[y, x]x+y, R(1, 1, 0) = x3[y, x]x+1
L(2, 2, 0) = x3[y, x]−x−y, R(2, 2, 0) = x3[y, x]−x−1.
The proof is thus complete.
Lemma 3.6. The 2−word
w(x, y) = x4[y, x]A+Bx+Cy
is associative in S3 if and only if A = B = C = 0.
Proof. We put z = 1 into the associativity law and we make use of the
formulas (i), (ii), (iii) and (iv). We have
L = w(w(x, y), 1) = w(x, y)4 = (x4[y, x]A+Bx+Cy
R = w(x, w(y, 1)) = w(x, y4) = x4[y4, x]γ(x,y) = x4[y, x]−(y+1)γ(x,y)
= x4[y, x](B−A−C)+y(B−A−C).
Therefore the equality L = R ensures B = 0 and A = C. Taking this into
account we get
w(w(x, x), z) = w(x4, z) = x4[z, x4]γ(1,z) = x4[z, x]−(x+1)(A+Az) = x4
w(x, w(x, z)) = x4[x4[z, x]γ(x,z), x]γ(1,z) = x4[z, x]A(1−x),
which shows that A = B = C = 0 and Lemma 3.6 follows.
Lemma 3.7. The word
w(x, y) = x4y4[y, x]A+Bx+Cy
is associative in S3 if and only if A = B = C = 0.
E. Płonka 93
Proof. We have
L = w(w(1, y), z) = w(y4, z) = y4z4[z, y4]γ(1,z)
= y4z4[z, y]−(y+1)(A+B+Cz) = y4z4[z, y](C−A−B)+(C−A−B)y
and
R = w(1, w(y, z)) = w(y, z)4 = y4z4[z, y]A+By+Cz.
Hence C = 0 and A = B. Taking this into account we check
L = w(w(x, y), x) = x2y4[y, x]γ(x,y)[x, x4y4[y, x]γ(x,y)]γ(1,x)
= x2y4[y, x]γ(x,y)[y, x](y+1)A(x+1))[y, x](1−x)A(x+1)γ(x,y)
= x2y4[y, x]A(1+x)
and also
R = w(x, w(y, x)) = x4y4x4[x, y]γ(y,x)[y4x4[x, y]γ(y,x), x]γ(x,1)
x4y4x4[x, y]γ(y,x)[y, x]−(y−1)γ(x,1)[x, y](x−1)γ(x,1)γ(x,y)
= x2y4[y, x]A(1+y).
By (vi) L = R if and only if A = 0. Clearly, x4y4 is associative word in
S3. The proof is thus completed.
Lemma 3.8. If the word
w(x, y) = [y, x]A+Bx+Cy = [y, x]γ(x,y)
is associative, then A = B − C = 0. Conversely, the word
w(x, y) = [y, x]B(x+y) (3.6)
satisfies the associativity law for all B ∈ Z3.
Proof. Using the identities (i), (ii),(ii) and (iv) we have
L = w(w(x, y), z)) = [z, [y, x]γ(x,y)]γ(1,z) = [y, x](1−z)(A+Bx+Cy)(A+B+Cz)
and similarly
R = w(x, w(y, z)) = [w(y, z), x]γ(x,1)
= [z, y](x−1)(A+By+Cz)(A+Bx+C).
94 Associative words in the symmetric group
Thus if w is an associative word in S3, then in the case y = x, we get
[z, x](x−1)(A−B+C)(A−B+Cz) = 1, (3.7)
because of (iii) and (v). Similarly, in the case z = y we obtain the equation
[y, x](1−y)(A+B−C)(A−C+Bx) = 1. (3.8)
Now (3.7), (3.8) and (vi) imply the system of congruences
(A + C − B)2 ≡ 0 (mod 3),
(A + C − B)(A − B − C) ≡ 0 (mod 3),
(A + C − B)C ≡ 0 (mod 3),
(A + B − C)2 ≡ 0 (mod 3),
(A + B − C)B ≡ 0 (mod 3),
(A + B − C)(B + C − A) ≡ 0 (mod 3),
which have the solution A = B − C = 0.
Conversely, we check that the word w(x, y) = [y, x]Bx+By is associative.
Indeed, by (ii) and (iii) we have
w(w(x, y), z) = [z, [y, x]B(x+y)]B(1+z) = [y, x]B
2(1−z)(1+z)(x+y) = 1
and
w(x, w(y, z)) = [[z, y]B(z+y), x]B(1+x) = [z, y]B
2(y+z)(x−1)(x+1) = 1,
as required.
We have thus established our main result
Theorem 3.9. There are precisely (modulo V (S3)) twelve associative
words in the group S3. Namely 1, x, x3, x4, y, y3, y4, xy, yx, x4y4, [y, x]x+y
and [x, y]x+y.
References
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Berlin 1974, pp. 221–225.
E. Płonka 95
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Contact information
Ernest Płonka Institute of Mathematics, Silesian University of
Technology, ul. Kaszubska 23, 44-100 Gliwice,
Poland
E-Mail: eplonka@polsl.pl
Received by the editors: 19.10.2011
and in final form 26.06.2012.
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