SOOCHOW JOURNAL OF MATHEMATICS
Volume 21, No. 1, pp. 21-27, January 1995
A NOTE ON PRIMITIVE PROPERTIES OF WORDS IN fuv ug
BY
JUNMIN HUANG
Abstract. In this paper, we have provided six results for primitive prop-
erties of words in fuv ug as follows: (1) Let s t 2 N with gcd(s t) = 1:
If there exist proper factors k of s and r of s + t and if k + r > s
then there are two word u v 2 X + with lg(u) = s, lg(v) = t such
that fuv ug \ Q = : (2) Let u v 2 X + j(u)j > 1 2 - lg(u) and
lg(v) = 2i , i 2 N: If lg(u) > 2i then fuv ug \ Q 6= : (3) Let u v 2
X + j(u)j > 1 p - lg(u) and lg(v) = pi i 2 N p 7 is prime. If
lg(u) > 3pi then fuv ug \ Q 6= : (4) Let u v 2 X + j(u)j > 1
p1 p2 pr pr+1 are continuous prime from 2 to pr+1 pi - lg(u),
i = 1 2 r lg(v) = pi11 pi22 pirr : If lg(v) < (pr+1 ; 2)lg(u) then
one of uv u must be primitive. (5) Let u v 2 X + j(u)j > 1 pk
- lg (u)
pk is prime, k = 1 2 n lg(v) = pi11 pi22 pinn i 2 N
= 1 2 n p1 < p2 < < pn lg(u) = q1j1 q2j2 qsjs j 2 N
= 1 2 s q1 < q2 < < qs q1 q2 qs are all primes. Let
lg(uv) = dk11 dk22 dkmm d 2 N = 1 2 m d1 < d2 < < dm
d1 d2 dm are all primes. If lg(u) > (q1 )=(q1 d1 ; q1 ; d1 )]lg(v) =
(q1 )(q1 d1 ; q1 ; d1 )]pi11 pi22 pinn then one of uv u must be primitive.
(6) Let v 2 X n u 2 X + (u) 6= (v) p - lg(u) p 5 is prime. Then
fuv ug \ Q 6= if and only if n = 1 2 3:
1. Basic Concepts and Preliminaries
Let X which contains more than one letter, be an alphabet and let X
the free monoid generated by X: The empty word will be denoted by 1 and
X + = X nf1g: N will be used to represent the set of all positive integers. A
word w 2 X + is primitive if w = f n for some f 2 X + imples that n = 1:
Let Q be the set of all primitive words over X: For each u 2 X + let (u) =
Received April 12, 1993 revised April 10, 1994.
21
22
JUNMIN HUANG
fa 2 X ju = xay for some x y 2 X g and (1) = : For w v 2 X we dene
w p v if and only if v = wx for some x 2 X (2]): For any word x 2 X lg(x)
is the number of letters occurs in the word x: In particular we let lg(1) = 0:
Following 2], for a word u = f n f 2 Q n 1 let (u) = f: Borwein had
given the following: If u 2 X + u 6= an a 2 X then fua ug \ Q 6= see 2].
This fact has been generalized to the following:
Theorem 1. (3]) fuv ug\ Q 6= 8v 2 X n n 1 j(u)j > 1 2 lg(u)
-
( 3 lg(u) 5 lg(u)) if and only if n = 2i 0 i 5 (n = 3i 0 i 2 n =
5i 0 i 1):
-
-
Theorem 2. (3]) fuv ug \ Q 6= 8v 2 X n n 1 u 2 X + (u) 6=
(v) 2 lg(u) (3 lg(u)) if and only if n = 1 2 3 4 6 8 10 14 (n = 1 2 3 5):
-
-
In 3] it didn't discuss for the case lg(v) = pi p a prime greater than seven,
i 2 N: This paper is a continuation of the work of 3] and deals again for the
cases p = 2 3 5 7 with the condition p lg(u) lg(v) = pi i 2 N:
We rst list some known results related to our work:
-
Lemma 1. (3]) Let u v 2 X + be such that lg(u) = i, lg(v) = j gcd(i j ) = d: If um and vn have a common initial segment of length i + j ; d
then u and v are powers of a common word of length not greater than d:
1.
Note that we often assume d = 1 in our arguments when we apply Lemma
Lemma 2. (3]) Let u v 2 X + and gcd(lg(u) lg(v)) = 1: If j(u)j > 1
then (u) 6= (uv):
Lemma 3. (3]) If lg(u) = p is a prime and u 2= Q then u = ap for some
a 2 X i.e., j(u)j = 1:
Lemma 4. (3]) Let u v 2 X + : If (u) 6= (v) then (u) 6= (uv):
Lemma 5. (3]) Let u v 2 X +: If lg(uv) = p is a prime and uv 2= Q
then (u) = (v):
We say w d v if and only if v = wx = yw for some x y 2 X :
A NOTE ON PRIMITIVE PROPERTIES OF WORDS IN fuv
ug
23
Lemma 6. (2]) If v d u then v = (xy)k and u = (xy)k+1 x for some
x y 2 X and k 0:
2. Main Results
Theorem 1 is saying : If v 2 X n u 2 X + j(u)j > 1 2 lg(u) n =
2i i 6 we might have fuv ug \ Q = : But, what condition will ensure us
fuv ug \ Q 6= ? First we indicate a method to nd two words u and v such
that fuv ug \ Q = : under specic conditions.
-
Theorem 3. Let s t 2 N with gcd(s t) = 1: If there exist proper factors
k of s and r of s + t and if k + r > s then there are two words u v 2 X + with
lg(u) = s lg(v) = t such that fuv ug \ Q = :
Proof. It is clear that gcd(s s + t) = 1: Let s = n k and s + t = m r.
Since k + r > s r must be greater than (n ; 1) k:
Suppose r > n k: Then let u = (ak;1 b)n and v = al (ual )m;1 where
a b 2 X and l = r ; s:
Suppose s > r > (n ; 1) k and u = f n uv = gm : Then g = f n;1 f1
where f = f1f2 with f2 d f: From Lemma 6, f2 = (xy)p x p 0 and f1 = xy:
Hence u = (xy)p+1 x]n and v = yx(xy)p+1 x]n;2 xy](xy)p+1 x]n;1 xy]m;2 and
so uv = (xy)p+1 x]n;1 xy]m :
In Theorem 3 if we do not ask k + r > s then we cannot obtain this result
because of Lemma 1.
Theorem 4. Let u v 2 X + j(u)j > 1 2 lg(u) and lg(v) = 2i i 2 N:
-
If lg(u) > 2i then fuv ug \ Q 6= :
Proof. By Theorem 1 it is easy to see that the result is true for the
case 1 i 5: Now let i 6: Since 2 lg(u) and lg(v) = 2i we have
gcd(lg(u) lg(v)) = gcd(lg(u) lg(uv)) = 1: Again assume fuv ug \ Q = u =
f n uv = gm f g 2 Q: From Lemma 2 we see that gcd(n m) = 1 f 6= g:
Here both lg(u) and lg(uv) are reducible and from Lemma 3 we have m n -
24
JUNMIN HUANG
3 lg(u) 2i : Then
lg(f ) + lg(g) =lg(u)=n] + lg(u) + lg(v)]=m 2 lg(u) + lg(v)]=3
=2 lg(u) + 2i ]=3 < lg(u):
Hnece as lg(u) > 2i , we would have f = g this is a contradiction and hence
fuv ug \ Q 6= :
So even it has been pointed out by 3] that as lg(v) = 26 = 64 there are
u and v such that fuv ug \ Q = but we know that as lg(v) = 26 = 64
lg(u) > 26, then fuv ug \ Q 6= from the above theorem.
Example. Let 2 lg(u) and let lg(v) = 26 = 64: If we let lg(u) = 21
-
then the following expressions holds:
lg(u) = 21 = 3 7:
lg(uv) = 21 + 64 = 85 = 5 7 21 = lg(u) < lg(v) = 26 :
So lg(f )+ lg(g) 7+17 <
6 21 = lg(u) or lg(f )+ lg(g) lg(u)=n +lg(u)+
lg(v)]=m 2 lg(u) + lg(v)]=3 = 2 lg(u) + 64]= 3 6 lg(u):
In this time we may have fuv ug \ Q = :
Let u = f n n = 3 lg(f ) = 7 uv = gm m = 5 lg(g) = 17 lg(f ) < lg(g)
u =a1 a2 a3 a4 a5a6a7 a1 a2a3a4a5 a6a7 a1 a2 a3 a4a5 a6 a7
uv =a1 a2 a3 a4 a5a6a7 a1 a2a3a4a5 a6a7 a1 a2 a3 a4a5 a6 a7 b5 b6 b17 (b1 b2 b17)3
=b1 b2 b17 b1 b2 b3 b4 b5 b6 b7 b17 (b1 b2 b17 )3 :
So that we have
a = a1 = b1 = a4 = b4 = a7 = b7 = b8 = b15 = b11 = b14
b = a2 = b2 = a5 = b5 = b9 = b16 = b12
c = a3 = b3 = a6 = b6 = b10 = b17 = b13:
Then
u =abcabcaabcabcaabcabca
v =bcaabcabcaabc(abcabcaabcabcaabc)3
uv =(abcabcaabcabcaabc)5
f =abcabca g = abcabcaabcabcaabc a b c 2 X:
Latter on we shall see that this theorem may be generalized to general prime
p:
A NOTE ON PRIMITIVE PROPERTIES OF WORDS IN fuv
ug
25
Theorem 5. Let u v 2 X + j(u)j > 1 p lg(u) and lg(v) = pi i 2 N
p 7 is prime. If lg(u) > 3pi then fuv ug \ Q 6= :
-
Proof. For these two words u and v with the given conditions, since
p lg(u) and lg(v) = pi we have gcd(lg(u) lg(v)) = gcd(lg(u) lg(uv)) = 1:
Thus, from Lemma 2, we have f 6= g:
Assume fuv ug \ Q = and let u = f n uv = gm m n 2: Since
lg(u) = n lg(f ) lg(uv) = m lg(g) and (m n) = 1 we have m or n must be
odd. Then by Lemma 3, both lg(u) and lg(uv) must be reducibles.
(a) Let m n 3 lg(u) > 3pi (> 2pi > pi ):
Then lg(f ) + lg(g) = lg(u)=n + lg(u) + pi ]=m 2 lg(u) + pi ]=3 < lg(u):
(b) Let n = 2 m 3 lg(u) > 3pi (> 2pi > pi ):
Then lg(f ) + lg(g) lg(u)=2 + lg(u) + pi ]=3 = 5 lg(u) + 2pi ]=6 < lg(u):
(c) Let m = 2 n 3 lg(u) > 3pi (> 2pi > pi ):
Then lg(f ) + lg(g) lg(u)=3 + lg(u) + pi ]=2 = 5 lg(u) + 3 pi ]=6 < lg(u):
Therefore, by Lemma 1 f = g and so fuv ug \ Q 6= : Here the prime p 7
may generalized to p 3:
-
Theorem 6. Let u v 2 X + j(u)j > 1 p1 p2 , pl+1 are continuous
primes from 2 to pl+1 p1 < p2 < < pl < pl+1 pi pi+1 pi lg(u) i =
1 2 l lg(v) = pi11 pi22 pill : If lg(v) < (pi+1 ; 2)lg(u) then fuv ug\ Q 6= :
-
-
Proof. Let u = f n uv = gm f g 2 Q: Because of pi lg(u), i = 1 2 l
-
and lg(v) = pi11 pi22 pill we have gcd(lg(u) lg(v)) = gcd(lg(u) lg(uv)) =
1: Then f 6= g follows from Lemma 2. Assume fuv ug \ Q = so that
gcd(lg(f ) lg(g)) = gcd(m n) = 1 and we have limit to lg(u) and lg(v) and
p1 p2 pl+1 are continuous primes from 2 to pl+1 so that it must be m n pl+1 and hence lg(f ) + lg(g) = lg(u)=n + lg(u) + lg(v)]=m 2 lg(u) +
lg(v)]=pl+1 < lg(u) then f = g because of Lemma 1 and thereby fuv ug\ Q 6=
:
Theorem 7. Let u v 2 X + j(u)j > 1 the prime pk lg(u) k =
1 2 l lg(v) = pi1 pi2 pill i 2 N = 1 2 l p1 < p2 < < pl
-
1
2
26
JUNMIN HUANG
lg(u) = q1j1 q2j2 qsjs j 2 N = 1 2 s q1 < q2 < < qs q1 q2 qs
are all primes, lg(uv) = r1k1 r2r2 rtkt k 2 N = 1 2 t r1 < r2 <
< rt r1 r2 rt are all primes. Then fuv ug \ Q 6= as lg(u) >
q1 (lg(v) ; r1)=q1 r1 ; q1 ; r1 ]):
Proof. First assume that u = f n, uv = gm , gcd(lg(u), lg(v)) = gcd(lg(u),
lg(uv)) = 1, because of the given conditions. Then we have gcd(m n) = 1
f 6= g follows from Lemma 2.
Assume fuv ug\ = m n 2: Then both lg(u) and lg(uv) are reducibles, because of Lemma 3.
Let lg(u) = q1j1 q2j2 qsjs j 2 N = 1 2 s q1 < q2 < < qs
q1 q2 qs are all primes,
lg(uv) = r1k1 r2k2 rtkt kd 2 N d = 1 2 t r1 < r2 < < rt
r1 r2 rt are all primes.
So these q r and p are all inequality and that
n lg(f ) = lg(u) so that n = q1J1 q2J2 qsJs q1 J j = 1 2 s
m lg(g) = lg(uv) so that m = r1K1 r2K2 rsKt r1 K k = 1 2 t
and hence 1=n 1=q1 1=m 1=r1
lg(f ) + lg(g) ; 1
= lg(u)=n + lg(u) + lg(v)]=m ; 1
lg(u)=q1 + lg(u) + lg(v)]=r1 ; 1
= (q1 + r1 )lg(u) + q1 (lg(v) ; 1)]=(q1 r1 )
< (q1 + r1)lg(u) + (q1 r1 ; q1 ; r1 )lg(u)=q1 r1
= lg(u):
We shall have f = g follows from Lemma 1 and hence fuv ug \Q 6= :
The signicance of Theorem 6 is that we have a sucient condition of
fuv ug \ Q 6= for gcd(lg(u) lg(v)) = 1:
A NOTE ON PRIMITIVE PROPERTIES OF WORDS IN fuv ug
27
Propositions 3.10 and 3.11 of 3] are generalized to general situation. Latter we generalize Theorem 3 to general situation of prime p( 5):
Theorem 8. Let v 2 X n u 2 X + (u) 6= (v) p lg(u): Then fuv ug \
-
Q 6= if and only if n = 1 2 3:
Proof. (() As n = 1 2 3 fuv ug \ Q 6= because of 3].
()) Let n be a positive integer such that n 6= 1 2 3:
(1) Assume that 2jn n 4: Set u = a2 v = bm a2 bm m = (n ; 2)=2: Then
we have u = a2 uv = (a2 bm )2 and fuv ug\ = :
(2) Assume that 2 n n 5: Set u = a3 v = bm a3 bm m = (n ; 3)=2:
Then we have u = a3 uv = (a3 bm )2 and fuv ug \ Q = : Hence as n 6= 1 2 3
there exist u v 2 X + such that fuv ug \ Q = :
Here we had generalized Borwein's result to general situation. As j(u)j >
1 we criticize primitive properties of fuv ug shall be very easy by Theorem 6
-
for general situation.
References
1] R. C. Lyndon and M. P. Schutzenberger, TheEquationaM = bN cP inaFreeGroup, Michigan
Math. J., 9 (1962), 289-295.
2] H. J. Shyr, Free Monoids and Languages, Institute of Applied Mathematics, Taichung,
Taiwan, 1991.
3] Y. S. Tsai and H. L. Wu, PrimitivePropertiesofWordsin fuv ug Soochow Journal of
Mathematics, 20:1 (1994), 83-100.
Department of Mathematics, Nanjing University, Nanjing, China.