- Open Access
Solutions of the third order Cauchy difference equation on groups
© Guo and Li; licensee Springer. 2014
- Received: 6 February 2014
- Accepted: 30 June 2014
- Published: 25 July 2014
Let be a function, where is a group and is an abelian group. In this paper, the following third order Cauchy difference of = − − − − + + + + + + − − − − (), is studied. We first give some special solutions of on free groups. Then sufficient and necessary conditions on finite cyclic groups and symmetric groups are also obtained.
- Cauchy difference
- free group
- symmetric group
- cyclic group
The first order Cauchy difference will be abbreviated as Cf. In , by using the reduction formulas and relations, as given in [2, 3], the general solution of the second order Cauchy difference equation was provided on free groups. Particularly, the authors also gave the expression of general solutions on symmetric group and finite cyclic group.
Remark 1 (1) is an abelian group under the pointwise addition of functions; (2) .
for all and .
Hence, the above relations imply that is a homomorphism. Similarly, the fact is also true for both and . This proves (2.4).
from (2.4) and the above claim for . This confirms (2.5) for . □
The function ;
is a homomorphism;
is a homomorphism;
is a homomorphism.
Before presenting Proposition 1, we first introduce the following useful lemma, which was given in .
Lemma 2 (Lemma 2.4 in )
for and all , such that , .
which is (2.7). This completes the proof. □
Remark 3 In particular, if , then Proposition 1 holds.
In this section, we study the solutions on a free group. We first solve (1.5) for the free group G on a single letter a.
Proof Necessity. It can be obtained from (2.5) in Lemma 1.
which leads to the result that is a homomorphism with respect to each variable. □
At the end of this section, for the free group on an alphabet with , we discuss some special solutions of (1.5).
belongs to ;
belongs to ;
belongs to .
Proof Claim (i) follows from the fact that is a morphism from to ℤ by (3.6).
This concludes assertion (ii). Claim (iii) follows from (3.7) directly. □
The symmetric group on a finite set X is the group whose elements are all bijective functions from X to X and whose group operation is that of function composition. If , then it is called the symmetric group of degree n and denoted .
In this section, we consider (1.5) for .
for all , , and all rearrangements π.
This proves (4.1). By a similar procedure, we can also verify (4.2)-(4.3). □
which implies that since . This completes the proof. □
In particular, (4.8) follows from (4.9)-(4.11). □
for every .
which gives (4.12).
by (4.12). In particular, taking in (4.13), we obtain (4.14). This completes the proof. □
According to Lemma 6, we give the following main result in this section.
This proves that f must have the form (4.15) with .
Sufficiency. Let be defined by (4.15), where is a constant with . In order to prove the identity of (1.5), by the symmetry of , , , it suffices to verify the following four cases: case (i) is odd, and , , are even; case (ii) , are odd, and , are even; case (iii) , , are odd, and is even; case (iv) , , and are odd.
In fact, for case (i) it is easy to see that , , , , , , , are odd, and , , , , , , are even, which leads to the equality of (1.5). The proofs of the other cases are similar. □
Let be a cyclic group of order n with generator a. In this section, we study the general solution on the finite cyclic group .
This proves (5.2)-(5.3).
where the last identity is obtained because , (5.2), and n is odd.
which, after a long and tedious computation, gives 0. Consequently, . This completes the proof. □
The authors are grateful to the editor and the referees for their valuable comments and suggestions, especially for the improvement of Theorem 2. This paper is supported by the National Science Foundation of China (11301226, 61202109), Zhejiang Provincial Natural Science Foundation of China under Grant No. LQ13A010017.
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