PRIMITIVE PROPERTIES OF WORDS IN fuv ug
BY
Y. S. TSAI AND H. L. WU
Abstract. Here in this paper, we have proved two interesting and fascinating results for primitive properties of words in f
g
uv u
(1)
as follows:
for all v2X u2X + with j(u)j>1 and 2 ^ lg(u) (in resp. 3 ^ lg(u) and
5 ^ lg(u)) if and only if n=2 0i5 (in resp. n=3 0i2 and n=5 0i1).
(2) fuv ug\Q6= for all v2X u2X + with (u)6=(v) and 2 ^ lg(u) (in resp. 3 ^ lg(u))
if and only if n=1 2 3 4 6 8 10 or 14 (in resp. n=1 2 3 or 5).
As is known, Borwein proved that if u2X + u6=a a2X then one of ua u must
be primitive. If one gives a close observation to the above results (1) and
(2), one immmediately realizes that Borwein's result is a special case which is
derived from two dierent directions.
fuv ug\Q6=
n
i
i
i
n
n
1. Introduction
Let X which contains more than one element, be an alphabet and 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 positive integers. A
word w 2 X + is primitive if w = f n for some f 2 X + implies n = 1. Let
Q be the set of all primitive words over X (as in 2]). For each u 2 X + let
(u) = fa 2 X ju = xay for some x y 2 X g and (1) = (as in 1],3]).
For w v 2 X we dene w p v if and only if v = wx for some x 2 X : For
any word x in X lg(x) is the number of letters occur in x: In particular,
Received December 15, 1992 revised April 13, 1993.
83
let lg(1) = 0: The terminologies which are not dened here could be found in
Shyr 2].
The main purpose of this paper is to generalize Borwein's result which
states that if u 2 X + u 6= an a 2 X then one of ua u must be primitive.
We want to discuss that if the surx word a of ua is not only a letter, then
what happens. But in the general cases, the above result of Borwein has
two dierent meanings. One is that if u v 2 X + lg(v) < j(u)j and lg(u)
or lg(uv) is odd, then one of uv u must be primitive. The other is that if
u v 2 X + (u) 6= (v) lg(v) j(u)j and lg(u) or lg(uv) is odd, then one of
uv u must be primitive. In addition, we obtain that fuv ug \ Q 6= for all
v 2 X n u 2 X + with j(u)j > 1 and 2 ^ lg(u) if and only if n = 2i 0 i 5
in section 3 and that fuv ug \ Q 6= for all v 2 X n u 2 X + with (u) 6= (v)
and 2 ^ lg(u) if and only if n = 1 2 3 4 6 8 10 or 14 in section 4.
First we present some well-known results which are needed later.
Proposition 1.1. (4]) Let f g 2 Q: If f n = gmx n m 2 and x p g
then f = g:
Proposition 1.2. (1],4]) 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:
Note that we often assume d = 1 in our arguments when we apply Proposition 1.2.
2. Generalization of a Borwein's result
In this section, we shall discuss the primitive words in fuv ug from two
dierent directions (u) 6= (v) and j(u)j > 1 respectively. If one require that
j(u)j > lg(v) hold in each case simultaneously, then the condition j(u)j > 1
just a special situation of the condition (u) 6= (v):
We shall prove four lemmas below and use these lemmas to prove main
results of this section in Proposition 2.5.
Lemma 2.1. Let u v 2 X + with lg(v) lg(u): If uv is not primitive,
then (uv) = (u):
Proof. (u) (uv) is obvious. On the other hand, let uv = f i for some
f 2 Q and i 2: Since lg(v) lg(u) we have f p u: Hence (uv) = (f ) (u): Therefore, (uv) = (u):
For any u 2 X + there exist uniquely f 2 Q and n 2 N such that u = f n
(see 2]). We dene (u) = f:
Lemma 2.2. Let u v 2 X + and uv 62 Q: If lg(v) j(u)j then
lg(v) minflg((u)) lg((uv))g: Moreover, If lg(v) < j(u)j then lg(v) <
minflg((u)) lg((uv))g:
Proof. Let u = f i uv = gj for some f g 2 Q i 2 N and j 2: Since
lg(v) j(u)j lg(u) from Lemma 2.1 we obtain that (uv) = (u): Hence
(f ) = (u) = (uv) = (g): Therefore, lg(v) j(u)j = j(f )j = j(g)j:
This implies lg(v) minflg(f ) lg(g)g = minflg((u)) lg((uv))g: Using the
same argument, we obtain lg(v) < minflg((u)) lg((uv))g if lg(v) < j(u)j:
Lemma 2.3. Let u v 2 X + (u) 6= (v) and lg(v) j(u)j: If neither
u nor uv is primitive, then uv = g2 for some g 2 Q:
Proof. Let u = f i uv = gj f g 2 Q and i j 2: Then lg(v) lg(f ) 1 lg(u)
2
from Lemma 2.2. If j 3 then u = gm x where x p g and m 2:
Therefore u = f i = gm x: By proposition 1.1, we obtain f = g. This implies
that v = f j;i : This contradicts (u) 6= (v): Hence uv = g2 :
The following lemma can be proved directly from Lemma 2.3.
Lemma 2.4. Let u v 2 X + and lg(v) < j(u)j: If neither u nor uv is
primitive, then uv = g2 for some g 2 Q:
Proposition 2.5. Let u v 2 X + (u) 6= (v) and lg(v) = k j(u)j: If
lg(u) or lg(uv) is odd, then one of uv u must be primitive.
Proof. If lg(uv) is odd, then the result follows from Lemma 2.3.
Assume lg(u) is odd and u = f i uv = gj for some f g 2 Q i j 2: Then
f i and gj have u as a left common factor. Evidently, i 3 and j = 2: Now
lg(f ) + l(g) = lg(f ) + ilg(f2) + k = (i + 2)lg2(f ) + k (i + 3)2 lg(f ) lg(u):
The rst inequality holds because of Lemma 2.2. This implies f = g by
Proposition 1.2 and hence v = f j;i : This contradicts (u) 6= (v): Hence
either u or uv is primitive.
Corollary 2.6. Let u v 2 X + and lg(v) = k < j(u)j: If lg(u) or lg(uv)
is odd, then one of uv u must be primitive.
Now, we give another proof of Borwein's result.
Corollary 2.7. (2]) Let u 2 X + u 6= an a 2 X: Then one of ua u must
be primitive.
Proof. It is clear that (u) 6= (a) lg(a) = 1 j(u)j and either lg(u)
or lg(ua) is odd. Hence either u or ua is a primitive by Proposition 2.5.
3. Conditions for primitive words in fuv ug with j(u)j > 1
In section 2, we found that one could investigate the primitive properties in
fuv ug from two dierent directions. In this section, we consider the primitive
words in fuv ug with a condition j(u)j > 1: We begin with the following two
lemmas.
Lemma 3.1. Let u v 2 X + and gcd(lg(u) lg(v)) = 1: If j(u)j > 1 then
(u) 6= (uv):
Proof. If (u) = (uv) i.e., u = gm uv = gn for some g 2 Q then
d = lg(g) j(g)j = j(u)j > 1: Now, djlg(u) and djlg(uv) imply djlg(v) a
contradiction. Hence (u) 6= (uv):
The next lemma is obvious.
Lemma 3.2. If lg(u) = p is prime and u 62 Q then u = ap for some
a 2 X i:e: j(u)j = 1:
We need other conditions for our discussion on primitive words in fuv ug:
In Proposition 3.3, 3.4 and 3.5, we assume 2 6 j lg(u) 3 6 j lg(u) and 5 6 j lg(u)
respectively.
Proposition 3.3. Let u v 2 X + and j(u)j > 1: If lg(u) is odd and
lg(v) = 2i 0 i 5 then one of uv u must be primitive.
Proof. If lg(v) = 1 the result follows directly from Corollary 2.7. We
assume that lg(v) = 2i i 1: Suppose that neither u nor uv is primitive and
let u = f n uv = gm for some f g 2 Q m n 3: Then f n and gm have u as a
left common factor. It is clear that f 6= g from Lemma 3.1 and that lg(u) and
lg(uv) are not prime from Lemma 3.2. From these facts and the hypothesis
that lg(u) is odd, we have:
(1) lg(v) = 2: By counting, we have lg(u) 25: Hence
lg(f ) + lg(g) = lgn(u) + lg(um) + 2 2lg(u3) + 2 < lg(u):
(2) lg(v) = 4: By counting, we have lg(u) 21: Hence
lg(f ) + lg(g) = lgn(u) + lg(um) + 4 2lg(u3) + 4 < lg(u):
(3) lg(v) = 8: By counting, we have lg(u) 25: Hence
lg(f ) + lg(g) = lgn(u) + lg(um) + 8 2lg(u3) + 8 < lg(u):
(4) lg(v) = 16: By counting, we have two cases:
Case 1. lg(u) = 9: Clearly, u = f 3 and uv = g5 : Hence
lg(f ) + lg(g) = 3 + 5 < lg(u):
Case 2. lg(u) 33:
) + 16 2lg(u) + 16 < lg(u):
lg(f ) + lg(g) = lgn(u) + lg(um
3
(5) lg(v) = 32: By counting, we have two cases:
Case 1. lg(u) = 25: Clearly, u = f 5 and uv = gm where m = 3 or 19:
Hence
lg(f ) + lg(g) 5 + 19 < lg(u):
Case 2. lg(u) 33.
) + 32 2lg(u) + 32 < lg(u):
lg(f ) + lg(g) = lgn(u) + lg(um
3
In all of the above cases, we have f = g by Proposition 1.2, which is a
contradiction. Therefore, one of uv u must be primitive.
Remark 1. The proposition is false if lg(v) = 64: For example, let X =
fa b dg u = (abbabba)3 and v = (bbaabbabbaabb)(abbabbaabbabbaabb)3 :
Then uv = (abbabbaabbabbaabb)5 :
Proposition 3.4 Let u v 2 X + and j(u)j > 1: If 3 6 j lg(u) and lg(v) =
3i 0 i 2 then one of uv u must be primitive.
Proof. If lg(v) = 1 the result follows directly from Corollary 2.7. We
assume that lg(v) = 3i i 1: Suppose that neither u nor uv is primitive and
let u = f n uv = gm for some f g 2 Q m n 2: Then f n and gm have u as
a left common factor. It is clear that f 6= g from Lemma 3.1, that both lg(u)
and lg(uv) are not prime from Lemma 3.2 and that 3 6 j n 3 6 j m and n or m
is odd. From these facts and the hypothesis that 3 6 j lg(u), we have:
(1) lg(v) = 3: By counting, we obtain lg(u) 22:
Case 1. n m > 3:
lg(f ) + lg(g) = lgn(u) + lg(um) + 3 < 2lg(u3) + 3 < lg(u):
Case 2. m = 2 or n = 2: If m = 2 then n 5: Hence
lg(f ) + lg(g) = lgn(u) + lg(u2) + 3 < lg(5u) + lg(u2) + 3 = 7lg(u10) + 15 < lg(u):
Similarly, we have the same conclusion if n = 2:
(2) lg(v) = 9: By counting, we obtain lg(u) 16:
Case 1. n m > 3:
lg(f ) + lg(g) = lgn(u) + lg(um) + 9 < 2lg(u3) + 9 < lg(u):
Case 2. m = 2 or n = 2: If m = 2 then n 5: We have
lg(f ) + lg(g) = lgn(u) + lg(u2) + 9 < lg(5u) + lg(u2) + 9
+ 3lg(u) = lg(u):
= 7lg(u10) + 45 < 7lg(u) 10
Similarly, we have the same conclusion if n = 2:
In all of the above cases, we have f = g by Proposition 1.2, which is a
contradiction. Therefore, one of uv u must be primitive.
Remark 2. The Proposition 3.4 is false if lg(v) = 27: For example,
let X = fa b dg u = (abba)2 and v = (bbaabb)(abbaabb)3 : Then uv =
(abbaabb)5 :
Proposition 3.5. Let u v 2 X + and j(u)j > 1: If 5 6 j lg(u) and
lg(v) = 5i 0 i 1 then one of uv u must be primitive.
Proof. If lg(v) = 1 the result follows directly from Corollary 2.7. We
assume that lg(v) = 5: Suppose that neither u nor uv is primitive, let u = f n
and uv = gm for some f g 2 Q m n 2: Then f n and gm have u as a left
common factor. It is clear that f 6= g from Lemma 3.1, that lg(u) and lg(uv)
are not prime from Lemma 3.2 and that n or m is odd. Thus we have four
cases:
Case 1. lg(u) = 4: We have
lg(f ) + lg(g) ; 1 = 2 + 3 ; 1 = lg(u):
Case 2. lg(u) = 9: We have
lg(f ) + lg(g) ; 1 3 + 7 ; 1 = lg(u):
Case 3. lg(u) 16 and m n 3: We obtain
lg(f ) + lg(g) = lgn(u) + lg(um) + 5 < 2lg(u3) + 5 < lg(u):
Case 4. lg(u) 16 and n = 2 or m = 2: If n = 2 then m 3: We have
lg(f ) + lg(g) = lgm(u) + lg(u2) + 5 lg(3u) + lg(u2) + 5
= 5lg(u) + 15 < lg(u):
6
Similarly, we have the same conclusion if m = 2:
In all of the above cases, we have f = g by Proposition 1.2, which is a
contradiction. Therefore, one of uv u must be primitive.
Proposition 3.6. Let u v 2 X + j(u)j > 1 and lg(u) 6= 9: If 7 6 j lg(u)
and lg(v) = 7 then one of uv u must be primitive.
Proof. Suppose that neither u nor uv is primitive and let u = f n uv = gm
for some f g 2 Q m n 2: Then f n and gm have u as a left common factor.
It is clear that f 6= g from Lemma 3.1, that lg(u) and lg(uv) are not prime
from Lemma 3.2 and that n or m is a odd. Thus we have six cases:
Case 1. lg(u) = 8: We have
lg(f ) + lg(g) ; 1 4 + 5 ; 1 = lg(u):
Case 2. lg(u) = 15: We have
lg(f ) + lg(g) ; 1 5 + 11 ; 1 = lg(u):
Case 3. lg(u) = 18: It is clear that m = 5: Thus lg(f ) + lg(g) 9 + 5 <
lg(u):
Case 4. lg(u) = 20: We have lg(f ) + lg(g) 10 + 9 < lg(u):
Case 5. lg(u) 25 and m = 2(or n = 2). If m = 2 then n 3: Thus
lg(f ) + lg(g) lg(3u) + lg(u2) + 7 = 5lg(u6) + 21 < lg(u):
Similarly, we have the same conclusion if n = 2:
Case 6. lg(u) 25 and m n 3: We obtain
lg(f ) + lg(g) = lgn(u) + 2lg(um) + 7 2lg(u3) + 7 < lg(u):
In all of the above cases, we have f = g by Proposition 1.2, which is a
contradiction. Therefore, one of uv u must be primitive.
Remark 3. The result does not hold if lg(u) = 9: For example, let
X = fa bg and let u = (aba)3 v = baabaab: Then uv = (abaabaab)2 :
The converses of Proposition 3.3, Proposition 3.4 and Proposition 3.5 are
as follow:
Proposition 3.7. If m 6= 2i 0 i 5 then there exist u v 2 X + with
2 6 j lg(u) j(u)j > 1 and lg(v) = m such that neither u nor uv is primitive.
Proof. (1) Suppose m = 2j k where j 2 N0 k 6= 1 and 2 6 j k: Put
f 2 (X k \ Q) and let u = f 3 v = f (2 ) : Then neither u nor uv is primitive.
(2) Suppose that m = 2j j 6: By Remark 1, we only consider j > 6:
Case 1. The last digit of m is 8 e:g: m = 128: Put f 2 (X 9 \ Q) and
3
3 4
g 2 X n for which n = m+27
5 ; 27: Let u = f and v = g(f g) : Then neither
u nor uv is primitive.
Case 2. The last digit of m is 6 e:g: m = 256: Put f 2 (X 3 \ Q) and
g 2 X n for which n = m5+9 ; 9: Let u = f 3 and v = g(f 3 g)4 : Then neither u
nor uv is primitive.
Case 3. The last digit of m is 2 e:g: m = 512: Put f 2 (X 11 \ Q) and
3
3 4
g 2 X n for which n = m+33
5 ; 33: Let u = f and v = g(f g) : Then neither
u nor uv is primitive.
Case 4. The last digit of m is 4 e:g: m = 1024: Put f 2 (X 7 \ Q) and
3
3 4
g 2 X n for which n = m+21
5 ; 21: Let u = f and v = g(f g) : Then neither
u nor uv is primitive.
j
Proposition 3.8. If m 6= 3i 0 i 2 then there exist u v 2 X + with
3 6 j lg(u) j(u)j > 1 and lg(v) = m such that neither u nor uv is primitive.
Proof. (1) Suppose m = 3j k where j 2 N0 k 6= 1 and 3 6 j k: Put
f 2 (X k \ Q) and let u = f 2 v = f (3 ): Then neither u nor uv is primitive.
(2) Suppose that m = 3j j 3: By Remark 2, we only consider j > 3:
j
Case 1. The last digit of m is 1 e:g: m = 81: Put f 2 (X 2 \ Q) and
g 2 X n for which n = m5+4 ; 4: Let u = f 2 and v = g(f 2 g)4 : Then neither u
nor uv is primitive.
Case 2. The last digit of m is 3 e:g: m = 243: Put f 2 (X 11 \ Q) and
2
2 4
g 2 X n for which n = m+22
5 ; 22: Let u = f and v = g(f g) : Then neither
u nor uv is primitive.
Case 3. The last digit of m is 9 e:g: m = 729: Put f 2 (X 8 \ Q) and
2
2 4
g 2 X n for which n = m+16
5 ; 16: Let u = f and v = g(f g) : Then neither
u nor uv is primitive.
Case 4. The last digit of m is 7 e:g: m = 2187: Put f 2 (X 4 \ Q) and
g 2 X n for which n = m5+8 ; 8: Let u = f 2 and v = g(f 2 g)4 : Then neither u
nor uv is primitive.
Proposition 3.9. If m 6= 5i 0 i 1 then there exist u v 2 X + with
5 6 j lg(u) j(u)j > 1 and lg(v) = m such that neither u nor uv is primitive.
Proof. (1) Suppose that m = 5j k where j 2 N0 k 6= 1 and 5 6 j k: Put
f 2 (X k \ Q) and let u = f 2 v = f (5 ) : Then neither u nor uv is primitive.
(2) Suppose that m = 5j j 2: Put f 2 (X 3 \ Q) and g 2 X n for which
n = m2+9 ; 9: Let u = f 3 and v = g(f 3 g): Then neither u nor uv is primitive.
j
The following results are obtained by using the same arguments.
Proposition 3.10. Let u v 2 X + j(u)j > 1 and 2 and 3 are not
factors of lg(u): If lg(v) = 2i 3j i j 2 N such that lg(v) 72 then one of
uv u must be primitive.
Proof. Suppose that neither u nor uv is primitive, let u = f n and uv = gm
for some f g 2 Q m n 5: Then f n and gm have u as a left common factor.
It is clear that f 6= g from Lemma 3.1 and that lg(u) and lg(uv) are not prime
from Lemma 3.2. By counting, we have lg(u) 25: Hence
+ lg(v) 2lg(u) + 72
lg(f ) + lg(g) lgn(u) + lg(u) m
5
2
lg
(
u
)
+
3
lg
(
u
)
<
= lg(u):
5
Hence f = g by Proposition 1.2, a contradiction. Therefore, one of uv u must
be primitive.
Proposition 3.11. Let u v 2 X + j(u)j > 1 and 2 and 3 are not
factors of lg(u): If lg(v) = 96 108 144 or 162 then one of uv u must be
primitive.
Proof. suppose that both u and uv are non-primitive, let u = f n and
uv = gm for some f g 2 Q m n 5: Then f n and gm have u as a left common
factor. It is clear that f 6= g from Lemma 3.1 and that both lg(u) and lg(uv)
are not prime from Lemma 3.2.
(1) Suppose lg(v) = 96: By counting, we obtain two cases.
Case 1. lg(u) = 25: We have
lg(f ) + lg(g) = 5 + 11 < lg(u):
Case 2. lg(u) 49: We have
lg(f ) + lg(g) 2lg(u5) + 96 < 4lg5(u) < lg(u):
(2) Suppose lg(v) = 108: By counting, we obtain three cases.
Case 1. lg(u) = 25: We have
lg(f ) + lg(g) 5 + 19 lg(u):
Case 2. lg(u) = 35: We have
lg(f ) + lg(g) 7 + 13 lg(u):
Case 3. lg(u) 77: We have
lg(f ) + lg(g) 2lg(u)5+ 108 < 4lg5(u) < lg(u):
(3) Suppose lg(v) = 144: By counting, we obtain two cases.
Case 1. lg(u) = 25: We have
lg(f ) + lg(g) = 5 + 13 < lg(u):
Case 2. lg(u) 55: We have
lg(f ) + lg(g) 2lg(u)5+ 144 < 5lg5(u) = lg(u):
(4) Suppose lg(v) = 162: By counting, we obtain two cases.
Case 1. lg(u) = 25: We have
lg(f ) + lg(g) 5 + 17 < lg(u):
Case 2. lg(u) 55: We have
lg(f ) + lg(g) 2lg(u)5+ 162 < 5lg5(u) = lg(u):
In all of the above cases, we have f = g by Proposition 1.2, which is a
contradiction. Therefore, one of uv u must be primitive.
From Propositions 3.3, 3.7, 3.4, 3.8, 3.5 and 3.9, we have the following
theorems.
Theorem 3.12. fuv ug \ Q 6= for all v 2 X n u 2 X + with j(u)j > 1
and 2 6 j lg(u) if and only if n = 2i 0 i 5:
Theorem 3.13. fuv ug \ Q 6= for all v 2 X n u 2 X + with j(u)j > 1
and 3 6 j lg(u) if and only if n = 3i 0 i 2:
Theorem 3.14. fuv ug \ Q 6= for all v 2 X n u 2 X + with j(u)j > 1
and 5 6 j lg(u) if and only if n = 5i 0 i 1:
4. Conditions for primitive words in fuv ug with (u) 6= (v)
In this section, we consider the primitive words in fuv ug with (u) 6=
(v): For the similar reason that the hypotheses are not enough, we assume 2
6 j lg(u) and 3 6 j lg(u) in Proposition 4.4 and 4.6 respectively.
The following two lemmas are easy to see.
Lemma 4.1. Let u v 2 X +: If (u) 6= (v) then (u) 6= (uv):
Proof. If u = f i and uv = f j for some f 2 Q then v = f j;i : This implies
(v) = (f ) = (u) a contradiction.
Lemma 4.2. Let u v 2 X + : If lg(uv) = p is a prime and uv 62 Q then
(u) = (v):
fag:
Proof. It is obvious that uv = ap for some a 2 X: Hence (u) = (v) =
The following result is also a generalization of Browein's.
Proposition 4.3. Let u v 2 X + and (u) 6= (v): If lg(v) = 1 2 or 3
then one of uv u must be primitive.
Proof. If lg(v) = 1 the result follows directly from Corollary 2.7. We
assume that lg(v) > 1: Suppose that neither u nor uv is primitive and let
u = f m uv = gn for some f g 2 Q and m n 2: Then f m and gn have u as
a left common factor. It is clear that f 6= g from Lemma 4.1, and that lg(uv)
is not prime.
(1) lg(v) = 2: We have
lg(f ) + lg(g) ; 1 lg(2u) + lg(u2) + 2 ; 1 = lg(u):
(2) lg(v) = 3: It is obvious that one of m n is odd and that lg(u) 3:
Without loss of generality, assume that m 3: Now,
lg(f ) + lg(g) ; 1 lg(3u) + lg(u2) + 3 ; 1 = 5lg(u6) + 3 lg(u):
In all of the above cases, we have f = g by Proposition 1.2, which is a
contradiction. Therefore, one of uv u must be primitive.
Proposition 4.4. Let u v 2 X + and (u) 6= (v): If lg(u) is odd and
lg(v) = 1 2 3 4 6 8 10 or 14 then one of uv u must be primitive.
Proof. If lg(v) = 1 2 or 3 the result follows directly from Proposition
4.3. We assume that lg(v) > 3: Suppose that neither u nor uv is primitive,
let u = f m uv = gn for some f g 2 Q m n 3: Then f m and gn have u as a
left common factor. It is clear that f 6= g from Lemma 4.1 and that lg(uv) is
not prime from Lemma 4.2. If lg(v) = 4 or 8 we can assume f 2 X because
of Proposition 3.3.
(1) lg(v) = 4: By counting, we have lg(u) 5: Hence
lg(f ) + lg(g) 1 + lg(u3) + 4 = lg(u3) + 7 < lg(u):
(2) lg(v) = 6: By counting, we have two cases.
Case 1. lg(u) = 3: We have
lg(f ) + lg(g) ; 1 1 + 3 ; 1 = lg(u):
Case 2. lg(u) 9: We have
lg(f ) + lg(g) = lgm(u) + lg(un) + 6 2lg(u3) + 6 < lg(u):
(3) lg(v) = 8: By counting, we have lg(u) 7: Hence
lg(f ) + lg(g) = 1 + lg(u3) + 8 = lg(u)3+ 11 < lg(u):
(4) lg(v) = 10: By counting, we have two cases:
Case 1. lg(u) = 5 We have
lg(f ) + lg(g) ; 1 1 + 5 ; 1 = lg(u):
Case 2. lg(u) 11: We have
lg(f ) + lg(g) 2lg(u3) + 10 < lg(u):
(5) lg(v) = 14: By counting, we have two cases.
Case 1. lg(u) = 7: We have
lg(f ) + lg(g) ; 1 1 + 7 ; 1 = lg(u):
Case 2. lg(u) 11: We have
lg(f ) + lg(g) ; 1 2lg(u3) + 14 ; 1 = 2lg(u3) + 11 lg(u):
In all of the above cases, we have f = g by Proposition 1.2, which is a
contradiction. Therefore, one of uv u must be primitive.
The following proposition is the converse of Proposition 4.4.
Proposition 4.5. Suppose m 6= 1 2 3 4 6 8 10 or 14: Then there exist
u v 2 X + with 2 6 j lg(u) lg(v) = m and (u) 6= (v) such that both uv and
u are non-primitive.
Proof. Let X = fa b dg:
(1) Suppose that 2 6 j m and m 5: Choose k 2 N with 2 6 j k and
1 < k < m: Put u = ak v = bn ak bn where n = m2+k ; k: Then neither uv nor
u is primitive.
(2) The last digit of m is 0 and m 20:
Case 1. m = 20: Put u = a7 v = b2 (a7 b2 )2 : Then neither uv nor u is
primitive.
Case 2. m 30: Put u = a5 v = bn (a5 bn )4 where n = m5+5 ; 5: Then
neither uv nor u is primitive.
(3) The last digit of m is 2 and m 12:
Case 1. m = 12: Put u = a3 v = b2 (a3 b2 )2 : Then neither uv nor u is
primitive.
Case 2. m 22: Put u = a3 v = bn (a3 bn )4 where n = m5+3 ; 3: Then
neither uv nor u is primitive.
(4) The last digit of m is 4 and m 24:
Case 1. m = 24: Put u = a3 v = b6 (a3 b6 )2 : Then neither uv nor u
primitive.
Case 2. m = 34: Put u = a5 v = b8 (a5 b8 )2 : Then neither uv nor u is
primitive.
Case 3. m = 44: Put u = a5 v = b2 (a5 b2 )6 : Then neither uv nor u is
primitive.
Case 4. m 54: Put u = a11 v = bn (a11 bn )4 where n = m+11
5 ; 11:
Then neither uv nor u is primitive.
(5) The last digit of m is 6 and m 16:
Case 1. m = 16: Put u = a5 v = b2 (a5 b2 )2 : Then neither uv nor u is
primitive.
Case 2. m = 26: Put u = a7 v = b4 (a7 b4 )2 : Then neither uv nor u is
primitive.
Case 3. m = 36: Put u = a3 v = b10 (a3 b10 )2 : Then neither uv nor u is
primitive.
Case 4. m 46: Put u = a9 v = bn (a9 bn )4 where n = m5+9 ; 9: Then
neither uv nor u is primitive.
(6) The last digit of m is 8 and m 18:
Case 1. m = 18: Put u = a3 v = b4 (a3 b4 )2 : Then neither uv nor u is
primitive.
Case 2. m = 28: Put u = a5 v = b6 (a5 b6 )2 : Then neither uv nor u is
primitive.
Case 3. m 38: Put u = a7 v = bn (a7 bn )4 where n = m5+7 ; 7: Then
neither uv nor u is primitive.
Proposition 4.6. Let u v 2 X + and (u) 6= (v): If 3 6 j lg(u) and
lg(v) = 1 2 3 or 5 then one of uv u must be primitive.
Proof. If lg(v) = 1 2 or 3 the result follows directly from Proposition
4.3. We assume that lg(v) = 5: Suppose that neither u nor uv is primitive, let
u = f m uv = gn for some f g 2 Q and m n 2: It is clear that f 6= g from
Lemma 4.1 and that lg(uv) is not prime from Lemma 4.2. By counting, we
have four cases:
Case 1. lg(u) = 4: It is obvious that m = 2 and n = 3: Hence
lg(f ) + lg(g) ; 1 = 2 + 3 ; 1 = lg(u):
Case 2. lg(u) = 5: It is obvious that m = 5: Hence
lg(f ) + lg(g) ; 1 1 + 5 ; 1 = lg(u):
Case 3. lg(u) = 7: It is obvious that m = 7: Hence
lg(f ) + lg(g) 1 + 6 = lg(u):
Case 4. lg(u) 10 and m = 3 or n = 3: Assume m = 3: Then
lg(f ) + lg(g) ; 1 = lg(3u) + lg(u2) + 5 ; 1 = 5lg(u6) + 9 < lg(u):
Similarly, we have the same conclusion if n = 2:
In all of the above cases, we have f = g by Proposition 1.2, which is a
contradiction. Therefore, one of uv u must be primitive.
Similarly, the following proposition is the converse of Proposition 4.6.
Proposition 4.7. Suppose m 6= 1 2 3 or 5: Then there exist u v 2 X +
with 3 6 j lg(u) lg(v) = m and (u) 6= (v) such that both uv and u are
non-primitive.
Proof. Let X = fa b dg:
(1) Suppose 2jm and m 4: Put u = a2 v = bn a2 bn where n = m2;2 :
Then neither uv nor u is primitive.
(2) Suppose m is odd and m 7: Put u = a5 v = bn a5 bn where n = m2;5 :
Then neither uv nor u is primitive.
From 4.4, 4.5, 4.6 and 4.7, we have the following theorems.
Theorem 4.8. fuv ug \ Q 6= for all v 2 X n u 2 X + with (u) 6= (v)
and 2 6 j lg(u) if and only if n = 1 2 3 4 6 8 10 or 14:
Theorem 4.9. fuv ug \ Q 6= for all v 2 X n u 2 X + with (u) 6= (v)
and 3 6 j lg(u) if and only if n = 1 2 3 or 5:
5. Conclusion
One might ask whether we can investigate primitive words in fuv ug
by dierent conditions to obtain similar results. If j(u)j = 1 (assuming
j(u)j > 1 in section 3) and (u) = (v) (assuming (u) 6= (v) in section
4), then u v 2 a for some a 2 X: And then nothing need to say. Hence
our conclusion is that any signicant hypotheses that we need for primitive
properties in fuv ug must meet the conditions j(u)j > 1 or (u) 6= (v):
References
1] Lothaire, M, Combinatorics on Words, Addison-Wesley Publishing Company, 1983.
2] Shyr, H. J, Free Monoids and Languages, Institute of Applied Mathematics National
Chung-Hsing University, Taichung, Taiwan, 1991.
3] Tsai, Y. S. and Wu, H. L, Countings of Several Kinds of Primitive Words and d-primitive
Words in Some Specic Sets, (To appear).
4] Tsai, Y. S. and Wu, H. L, Properties on the Product of Primitive Words, (To appear).
Department of Mathematics, Chung-Yuan Christian University, Chung Li, Taiwan.