Solutions of the nth-order Cauchy difference equation on groups

Abstract

Let $(G, \cdot)$ be a group, $(H, +)$ be an Abelian group, and $f:G\rightarrow H$ be a function. In this paper, for a positive integer n, we first give a representation of nth-order Cauchy difference of f via the function as

$$ C^{(n)}f(x_{1},x_{2},\ldots,x_{n},x_{n+1})= \sum_{1\leq m \leq n+1}(-1)^{n+1-m}\sum _{1\leq i_{1}< i_{2}< \cdots< i_{m}\leq n+1}f(x_{i_{1}}x_{i_{2}}\cdots x_{i_{m}}), $$

where $x_{1}, x_{2},\ldots, x_{n+1}\in G$. Then, based on the representation, we get some special solutions of $C^{(n)}f=0$ on free groups. Moreover, sufficient and necessary conditions on symmetric groups and finite cyclic groups are also obtained.

1 Introduction

It is well known that the solutions to Jensen’s functional equation

$$ f(x+y)+f(x-y)=2f(x) $$

(1.1)

are just all linear functions $f(x)=cx+d$ if we assume that f are continuous, and they are the set of all homomorphisms (when $d=0$) on real line $\mathbb {R}$ as an additive group. Let $(G, \cdot)$ be a group, and $(H, +)$ be an Abelian group. Denote by $e\in G$ and $0\in H$ the identity elements, respectively. In [1], it was pointed out that the set of solutions is not equivalent to all homomorphisms on a general group. Therefore, finding out the solutions to equation (1.1) on groups becomes an interesting problem (see [1, 2] and references therein). Note that these solutions are related to their Cauchy differences [3–6]. For a function $f:G\rightarrow H$, its Cauchy differences $C^{(m)}f$ are defined by

$$\begin{aligned}& C^{(0)}f=f, \end{aligned}$$

(1.2)

$$\begin{aligned}& C^{(1)}f(x_{1}, x_{2})=f(x_{1}x_{2})-f(x_{1})-f(x_{2}) , \\& C^{(m+1)}f(x_{1}, x_{2}, \ldots, x_{m+2}) \\& \quad =C^{(m)}f(x_{1}x_{2},x_{3}, \ldots, x_{m+2})-C^{(m)}f(x_{1},x_{3}, \ldots, x_{m+2})-C^{(m)}f(x_{2},x_{3}, \ldots, x_{m+2}). \end{aligned}$$

(1.3)

The first-order Cauchy difference $C^{(1)}f$ will be abbreviated as Cf. In general, $C^{(m+1)}f=0$ if and only if $C^{(m)}f$ is m-additive, and $C^{(m)}f=0$ implies $C^{(n)}f=0$ for $n\geq m$ [7]. Fischer and Heuvers [7–9] mentioned that any generalized polynomial f with $C^{(m)}f=0$ has a unique representation of the form $f(x)=f_{1}(x)+\cdots+f_{m}(x)$, where each $f_{j}(x)$, $j=1,\ldots,m$, is a j-form. Besides the properties of solutions satisfying (1.3), we are also concerned with its general solutions. In [10, 11], by using the reduction formulas and relations given in [1, 2], Ng provided the general solution of the second-and third-order Cauchy difference equations on free groups. Some previous results on second order were extended to high order in [11]. Moreover, for the general solution of the third-order Cauchy difference equation on symmetric groups and finite cyclic groups, see reference [12].

Plentiful results were also devoted to the generalized Cauchy difference functional equations of the form

$$ f(x)+f(y)-f(x+y)=g\bigl(H(x,y)\bigr), $$

(1.4)

where H is given, and f, g are unknown. Under regularity assumptions on H and a particular solution $f_{0}$, $g_{0}$, Ebanks [5, 6] investigated the general solution $(f,g)$ of equation (1.4). For particular forms of $g(H(x,y))$, the existence and stability of solutions to equation (1.4) are studied extensively in, e.g., [13–23].

Note that equations (1.3) and (1.4) are concerning functions of one variable. Also, for a map $F: G^{n}\to H$, we define the mth partial Cauchy difference with respect to the ith variable ($1\leq i\leq n$), which is connected to the representation of the generalized polynomial F (see [7–9, 24]).

It is natural to consider the general expression of nth-order Cauchy difference $C^{(n)}f$ and determine the solutions to equation

$$ C^{(n)}f=0, $$

(1.5)

which becomes the motivation of this paper. Remark that equation (1.5) for free group with just one generator has been solved in [11]. In our paper, the purpose is to determine all solutions to equation (1.5) on some given groups. For simplicity, the general solution to equation (1.5) will be denoted by

$$ \operatorname{Ker} C^{(n)}(G, H)=\bigl\{ f:G\rightarrow H \mid C^{(n)}f=0 \bigr\} . $$

(1.6)

Obviously, $\operatorname{Ker} C^{(n)}(G, H)$ is an Abelian group under the pointwise addition of functions, and $\operatorname{Hom}(G, H)\subseteq \operatorname{Ker} C^{(n)}(G, H)$.

2 Some properties on solutions

In this section, we study properties of solutions for nth-order Cauchy differences.

Proposition 1

For a positive integer n, the nth-order Cauchy difference $C^{(n)}f$ can be expressed in terms of f as

$$ C^{(n)}f(x_{1},x_{2}, \ldots,x_{n},x_{n+1})=\sum_{m=1}^{n+1}(-1)^{n+1-m} \sum_{1\leq i_{1}< i_{2}< \cdots< i_{m}\leq n+1}f(x_{i_{1}}x_{i_{2}}\cdots x_{i_{m}}). $$

(2.1)

Proof

Claim (2.1) is true for $n=1, 2$ from (1.2)-(1.3). Suppose that (2.1) holds for $n-1$. Note that

$$\begin{aligned}& C^{(n)}f(x_{1},x_{2},\ldots,x_{n},x_{n+1}) \\& \quad =C^{(n-1)}f(x_{1}x_{2},x_{3},\ldots, x_{n},x_{n+1}) \\& \qquad {}-C^{(n-1)}f(x_{1},x_{3}, \ldots,x_{n},x_{n+1})-C^{(n-1)}f(x_{2},x_{3}, \ldots, x_{n},x_{n+1}) \end{aligned}$$

by (1.3). Then, we have

$$\begin{aligned}& C^{(n-1)}f(x_{1}x_{2},x_{3},\ldots, x_{n},x_{n+1}) \\ & \quad =(-1)^{n-1}f(x_{1}x_{2})+(-1)^{n-1} \sum_{3\leq i_{1} \leq n+1}f(x_{i_{1}}) \\ & \qquad {} + (-1)^{n-2}\sum_{3\leq i_{1}\leq n+1}f(x_{1}x_{2}x_{i_{1}})+ (-1)^{n-2}\sum_{3\leq i_{1}< i_{2}\leq n+1}f(x_{i_{1}}x_{i_{2}}) \\ & \qquad {}+\cdots+(-1)^{2}\sum_{3\leq i_{1}< i_{2}< \cdots< i_{n-3}\leq n+1}f(x_{1}x_{2}x_{i_{1}}x_{i_{2}} \cdots x_{i_{n-3}}) \\ & \qquad {}+(-1)^{2}\sum_{3\leq i_{1}< i_{2}< \cdots< i_{n-2}\leq n+1}f(x_{i_{1}}x_{i_{2}} \cdots x_{i_{n-2}}) \\ & \qquad {}+(-1)^{1}\sum_{3\leq i_{1}< i_{2}< \cdots< i_{n-2}\leq n+1}f(x_{1}x_{2}x_{i_{1}}x_{i_{2}} \cdots x_{i_{n-2}})+(-1)^{1}f(x_{3}x_{4}\cdots x_{n+1}) \\ & \qquad {}+f(x_{1}x_{2}x_{3}\cdots x_{n+1}), \\ & C^{(n-1)}f(x_{1},x_{3},\ldots,x_{n},x_{n+1}) \\ & \quad =(-1)^{n-1}f(x_{1})+(-1)^{n-1}\sum _{3\leq i_{1} \leq n+1}f(x_{i_{1}}) \\ & \qquad {}+ (-1)^{n-2}\sum_{3\leq i_{1}\leq n+1}f(x_{1}x_{i_{1}})+(-1)^{n-2} \sum_{3\leq i_{1}< i_{2}\leq n+1}f(x_{i_{1}}x_{i_{2}}) \\ & \qquad {}+\cdots+ (-1)^{2}\sum_{3\leq i_{1}< i_{2}< \cdots< i_{n-3}\leq n+1}f(x_{1}x_{i_{1}}x_{i_{2}} \cdots x_{i_{n-3}}) \\ & \qquad {}+(-1)^{2}\sum_{3\leq i_{1}< i_{2}< \cdots< i_{n-2}\leq n+1}f(x_{i_{1}}x_{i_{2}} \cdots x_{i_{n-2}}) \\ & \qquad {}+(-1)^{1}\sum_{3\leq i_{1}< i_{2}< \cdots< i_{n-2}\leq n+1}f(x_{1}x_{i_{1}}x_{i_{2}} \cdots x_{i_{n-2}})+(-1)^{1}f(x_{3}x_{4}\cdots x_{n+1}) \\ & \qquad {}+f(x_{1}x_{3}\cdots x_{n+1}), \end{aligned}$$

and

$$\begin{aligned}& C^{(n-1)}f(x_{2},x_{3},\ldots, x_{n},x_{n+1}) \\& \quad =(-1)^{n-1}f(x_{2})+(-1)^{n-1}\sum _{3\leq i_{1} \leq n+1}f(x_{i_{1}}) \\& \qquad {}+ (-1)^{n-2}\sum_{3\leq i_{1}\leq n+1}f(x_{2}x_{i_{1}})+(-1)^{n-2} \sum_{3\leq i_{1}< i_{2}\leq n+1}f(x_{i_{1}}x_{i_{2}}) \\& \qquad {}+ \cdots+ (-1)^{2}\sum_{3\leq i_{1}< i_{2}< \cdots< i_{n-3}\leq n+1}f(x_{2}x_{i_{1}}x_{i_{2}} \cdots x_{i_{n-3}}) \\& \qquad {}+(-1)^{2}\sum_{3\leq i_{1}< i_{2}< \cdots< i_{n-2}\leq n+1}f(x_{i_{1}}x_{i_{2}} \cdots x_{i_{n-2}}) \\& \qquad {}+(-1)^{1}\sum_{3\leq i_{1}< i_{2}< \cdots< i_{n-2}\leq n+1}f(x_{2}x_{i_{1}}x_{i_{2}} \cdots x_{i_{n-2}})+(-1)^{1}f(x_{3}x_{4}\cdots x_{n+1}) \\& \qquad {}+f(x_{2}x_{3}\cdots x_{n+1}). \end{aligned}$$

Summing over these three equalities, we get

$$C^{(n)}f(x_{1},x_{2},\ldots,x_{n},x_{n+1})= \sum_{m=1}^{n+1}(-1)^{n+1-m}\sum _{1\leq i_{1}< i_{2}< \cdots< i_{m}\leq n+1}f(x_{i_{1}}x_{i_{2}}\cdots x_{i_{m}}), $$

and this gives (2.1). □

Proposition 2

If $f\in\operatorname{Ker} C^{(n)}(G, H)$, then the following properties are valid.

• (i) For $i=1,2,\ldots,n-1$ and $j=1,2,\ldots, i+1$, we have

  $$ C^{(i)}f(x_{1}, x_{2}, \ldots,x_{j-1}, e, x_{j+1}, \ldots, x_{i+1})=0. $$

  (2.2)

  In particular,

  $$ f(e)=0. $$

  (2.3)

• (ii) $C^{(n-1)}f$ is a homomorphism with respect to each variable.

Proof

We first check (2.3). For $f\in\operatorname{Ker} C^{(n)}(G, H)$, we take $x_{1}=e$ in (2.1). Then, it follows from (1.3) that

$$\begin{aligned} 0&= C^{(n)}f(e, x_{2}, \ldots, x_{n+1}) \\ &=C^{(n-1)}f(e,x_{2},x_{3},\ldots, x_{n+1})-C^{(n-1)}f(x_{2},x_{3}, \ldots, x_{n+1})-C^{(n-1)}f(e,x_{3}, \ldots, x_{n+1}) \\ &=(-1)C^{(n-1)}f(e,x_{3}, \ldots, x_{n+1})=(-1)^{2}C^{(n-2)}f(e,x_{4}, \ldots, x_{n+1})=\cdots \\ &=(-1)^{n-1}Cf(e, x_{n+1})=(-1)^{n}f(e), \end{aligned}$$

which gives (2.3).

Obviously, (2.2) is true for $i=1, 2$ by (1.3) and (2.3). Assume that (2.2) holds for all numbers smaller than $i\geq3$. By induction, for $j=1$, we have

$$\begin{aligned}& C^{(i)}f(e, x_{2},x_{3},\ldots, x_{i+1}) \\& \quad =C^{(i-1)}f(x_{2}, x_{3},\ldots, x_{i+1})-C^{(i-1)}f(x_{2},x_{3}, \ldots, x_{i+1})-C^{(i-1)}f(e,x_{3}, \ldots, x_{i+1}) \\& \quad = -C^{(i-1)}f(e,x_{3}, \ldots, x_{i+1})=0; \end{aligned}$$

for $j=2$,

$$\begin{aligned} \begin{aligned} &C^{(i)}f(x_{1},e,x_{3},\ldots, x_{i+1}) \\ &\quad =C^{(i-1)}f(x_{1}, x_{3},\ldots, x_{i+1})-C^{(i-1)}f(e,x_{3}, \ldots, x_{i+1})-C^{(i-1)}f(x_{1},x_{3}, \ldots, x_{i+1}) \\ &\quad = -C^{(i-1)}f(e,x_{3}, \ldots, x_{i+1})=0; \end{aligned} \end{aligned}$$

and

$$\begin{aligned}& C^{(i)}f(x_{1}, x_{2}, \ldots,x_{j-1}, e, x_{j+1}, \ldots, x_{i+1}) \\& \quad = C^{(i-1)}f(x_{1}x_{2}, x_{3}, \ldots,x_{j-1}, e, x_{j+1}, \ldots, x_{i+1}) \\& \qquad {} - C^{(i-1)}f(x_{1}, x_{3}, \ldots,x_{j-1}, e, x_{j+1}, \ldots, x_{i+1}) \\& \qquad {} - C^{(i-1)}f(x_{2}, x_{3}, \ldots,x_{j-1}, e, x_{j+1}, \ldots, x_{i+1}) \\& \quad =0 \end{aligned}$$

in the case $j\geq3$. This confirms (2.2).

We infer from (1.3) and (1.6) that $C^{(n-1)}f$ is a homomorphism with respect to the first variable. Then by the symmetry among the variables the Cauchy difference of $C^{(n-1)}f$ in its first variable is equivalent to the other variable, and therefore, (ii) is proved. □

Remark 1

For any function $f:G\rightarrow H$, the following statements are equivalent:

• (i) The function $f\in \operatorname{Ker}C^{(n)}(G, H)$.

• (ii) $C^{(n-1)}f$ is a homomorphism with respect to the j-variable, $j\in\{1, 2, \ldots, n+1\}$.

Next, we give two useful lemmas.

Lemma 1

(Lemma 2.4 in [10])

The following identity is valid for any function $f: G \rightarrow H$ and $\ell\in\mathbb{N}$:

$$ f(x_{1}x_{2}\cdots x_{\ell})=\sum _{m\leq\ell}\sum_{1\leq i_{1}< i_{2}< \cdots< i_{m}\leq\ell}C^{(m-1)}f(x_{i_{1}}, x_{i_{2}},\ldots, x_{i_{m}}). $$

(2.4)

Lemma 2

(Proposition 2.4 in [11])

Let n be a positive integer. If $f\in \operatorname{Ker} C^{(n)}(G, H)$, then for all $x\in G$ and $p\in\mathbb{Z}$, we have

$$ f\bigl(x^{p}\bigr)= \textstyle\begin{cases} \sum_{j=0}^{n-1}\frac{p(p-1)\cdots (p-j)}{(j+1)!}C^{(j)}f(\underbrace{x,x,\ldots,x}_{j+1}) , & p\leq0 \textit{ or } p\geq n, \\ \sum_{j=0}^{p-1}\frac{p(p-1)\cdots (p-j)}{(j+1)!}C^{(j)}f(\underbrace{x,x,\ldots,x}_{j+1}) , & 0< p< n. \end{cases} $$

(2.5)

The following statement is a vision of Lemma 1 under the restriction $f\in\operatorname{Ker} C^{(n)}(G, H)$.

Theorem 1

Suppose that $f\in\operatorname{Ker} C^{(n)}(G, H)$. Then the following identities are valid.

• (i) If $l\geq n$, then, for $m_{k}\in\mathbb{Z}$ and $x_{k}\in G$, $i=1, 2, \ldots, l$, such that $x_{k}\neq x_{k+1}$, we have

  $$\begin{aligned} \begin{aligned}[b] &f\bigl(x_{1}^{m_{1}}x_{2}^{m_{2}} \cdots x_{l}^{m_{l}}\bigr) \\ &\quad = \sum_{k=1}^{l} \Biggl( \sum _{j=0}^{n-1}\frac{m_{k}(m_{k}-1)\cdots (m_{k}-j)}{(j+1)!}C^{(j)}f( \underbrace{x_{k},x_{k},\ldots,x_{k}}_{j+1}) \Biggr) \\ &\qquad {} +\sum_{i=2}^{n-1}\sum _{1\leq k_{1}< k_{2}< \cdots < k_{i}\leq l}C^{(i-1)}f\bigl(x_{k_{1}}^{m_{k_{1}}}, x_{k_{2}}^{m_{k_{2}}}, \ldots, x_{k_{i}}^{m_{k_{i}}}\bigr) \\ &\qquad {} + m_{k_{1}}m_{k_{2}}\cdots m_{k_{n}}\sum _{1\leq k_{1}< k_{2}< \cdots < k_{n}\leq l}C^{(n-1)}f(x_{k_{1}}, x_{k_{2}}, \ldots, x_{k_{n}}). \end{aligned} \end{aligned}$$

  (2.6)

• (ii) If $l< n$, then, for $m_{k}\in\mathbb{Z}$ and all $x_{k}\in G$, $k=1, 2, \ldots, l$, such that $x_{k}\neq x_{k+1}$, we have

  $$\begin{aligned}& f\bigl(x_{1}^{m_{1}}x_{2}^{m_{2}} \cdots x_{l}^{m_{l}}\bigr) \\& \quad = \sum_{k=1}^{l} \Biggl( \sum _{j=0}^{n-1}\frac{m_{k}(m_{k}-1)\cdots (m_{k}-j)}{(j+1)!}C^{(j)}f( \underbrace{x_{k},x_{k},\ldots,x_{k}}_{j+1}) \Biggr) \\& \qquad {} +\sum_{i=2}^{l}\sum _{1\leq k_{1}< k_{2}< \cdots< k_{i}\leq l}C^{(i-1)}f\bigl(x_{k_{1}}^{m_{k_{1}}}, x_{k_{2}}^{m_{k_{2}}}, \ldots, x_{k_{i}}^{m_{k_{i}}}\bigr). \end{aligned}$$

  (2.7)

Proof

Replacing $x_{k}$ in (2.4) by $x_{k}^{m_{k}}$, we have

$$ f\bigl(x_{1}^{m_{1}}x_{2}^{m_{2}} \cdots x_{l}^{m_{l}}\bigr)=\sum_{i=1}^{l} \sum_{1\leq k_{1}< k_{2}< \cdots< k_{i}\leq l }C^{(i-1)}f\bigl(x_{k_{1}}^{m_{k_{1}}}, x_{k_{2}}^{m_{k_{2}}},\ldots, x_{k_{i}}^{m_{k_{i}}}\bigr). $$

(2.8)

We first consider the case $l\geq n$. Vanishing of $C^{(i-1)}f$ for $i>n$ yields

$$\begin{aligned} f\bigl(x_{1}^{m_{1}}x_{2}^{m_{2}}\cdots x_{l}^{m_{l}}\bigr) =&\sum_{k=1}^{l}f \bigl(x_{k}^{m_{k}}\bigr)+ \sum_{i=2}^{n-2} \sum_{1\leq k_{1}< k_{2}< \cdots< k_{i}\leq l}C^{(i-1)}f(x_{k_{1}}, x_{k_{2}}, \ldots, x_{k_{i}}) \\ &{}+\sum_{1\leq k_{1}< k_{2}< \cdots< k_{n}\leq l}C^{(n-1)}f \bigl(x_{k_{1}}^{m_{k_{1}}}, x_{k_{2}}^{m_{k_{2}}}, \ldots, x_{k_{n}}^{m_{k_{n}}}\bigr). \end{aligned}$$

Therefore, by (2.5) in Lemma 2 and (ii) of Proposition 2 we have

$$\begin{aligned}& f\bigl(x_{k}^{m_{k}}\bigr)=\sum_{j=0}^{n-1} \frac{m_{k}(m_{k}-1) \cdots(m_{k}-j)}{(j+1)!}C^{(j)} f(\underbrace{x_{k},x_{k}, \ldots,x_{k}}_{j+1}), \end{aligned}$$

(2.9)

$$\begin{aligned}& C^{(n-1)}f\bigl(x_{k_{1}}^{m_{k_{1}}},x_{k_{2}}^{m_{k_{2}}}, \ldots, x_{k_{n}}^{m_{k_{n}}}\bigr)=m_{k_{1}}m_{k_{2}} \cdots m_{k_{n}}C^{(n-1)}f(x_{k_{1}}, x_{k_{2}}, \ldots, x_{k_{n}}), \end{aligned}$$

(2.10)

which is formula (2.6).

In the case $l< n$, (2.7) is obtained by (2.8)-(2.9) directly. This completes the whole proof. □

3 Solutions on free groups

In this section, we discuss some special solutions of $C^{(n)}f=0$ for the free group on an alphabet $\langle\mathscr{A}\rangle$ with $|\mathscr{A}|\geq2$.

An element $x\in\mathscr {A}$ can be written in the form

$$ x=a_{1}^{n_{1}}a_{2}^{n_{2}}\cdots a_{l}^{n_{l}}, \quad \mbox{where } a_{i}\in \mathscr {A}, n_{i}\in\mathbb{Z}. $$

(3.1)

For each fixed $a\in\mathscr {A}$ and fixed pair of distinct $a, b\in\mathscr {A}$, define the functions W, $W_{2}$, $W_{3}$ by

$$\begin{aligned}& W(x; a) = \sum_{a_{i}=a}n_{i}, \end{aligned}$$

(3.2)

$$\begin{aligned}& W_{2}(x; a, b) = \sum_{i< j, a_{i}=a, a_{j}=b}n_{i}n_{j}, \end{aligned}$$

(3.3)

$$\begin{aligned}& W_{3}(x; a, b) = \sum_{i>j, a_{i}=a, a_{j}=b}n_{i}n_{j} \end{aligned}$$

(3.4)

along with (3.1). Referred from [1, 2], these functions are well defined. Furthermore, they satisfy the following relations:

$$\begin{aligned}& W(xy; a) = W(x; a)+W(y; a), \end{aligned}$$

(3.5)

$$\begin{aligned}& W_{2}(x; a, b) = W_{3}(x; b, a). \end{aligned}$$

(3.6)

Proposition 3

For any fixed $a\in\mathscr {A}$ and fixed pair of distinct a, b in $\mathscr {A}$, the following assertions hold:

• (i) $W(\cdot; a)$ belongs to $\operatorname{Ker} C^{(n)}(\langle \mathscr {A}\rangle, \mathbb{Z})$;

• (ii) $W_{2}(\cdot; a, b)$ belongs to $\operatorname{Ker} C^{(n)}(\langle \mathscr {A}\rangle, \mathbb{Z})$;

• (iii) $W_{3}(\cdot; a, b)$ belongs to $\operatorname{Ker} C^{(n)}(\langle \mathscr {A}\rangle, \mathbb{Z})$.

Proof

Statement (i) follows from the fact that $x\mapsto W(x; a)$ is a morphism from $\langle\mathscr {A}\rangle$ to $\mathbb{Z}$ by (3.5).

Now we consider statement (ii). Let $x_{1}, x_{2},\ldots, x_{n+1}$ in the free group $\langle\mathscr {A}\rangle$ be written as

$$\begin{aligned}& x_{1}=a_{11}^{t_{11}}a_{12}^{t_{12}} \cdots a_{1r_{1}}^{t_{1r_{1}}},\qquad x_{2}=a_{21}^{t_{21}}a_{22}^{t_{22}} \cdots a_{2r_{2}}^{t_{2r_{2}}},\qquad \ldots, \\& x_{n+1}=a_{n+1,1}^{t_{n+1,1}}a_{n+1,2}^{t_{n+1,2}} \cdots a_{n+1,r_{n+1}}^{t_{n+1,r_{n+1}}}. \end{aligned}$$

Then by (1.6) we have

$$\begin{aligned}& C^{(n)}W_{2}(x_{1}, x_{2}, \ldots, x_{n+1}; a, b) \\& \quad =\sum_{m=1}^{ n+1}(-1)^{n+1-m} \sum_{1\leq l_{1}< l_{2}< \cdots< l_{m}\leq n+1}W_{2}(x_{l_{1}} x_{l_{2}}\cdots x_{l_{m}}; a, b) \\& \quad =(-1)^{n}\sum_{l=1}^{n+1}W_{2}(x_{l}; a, b)+(-1)^{n-1}\sum_{1\leq l_{1}< l_{2}\leq n+1}W_{2}(x_{l_{1}}x_{l_{2}}; a, b) \\& \qquad {} + (-1)^{n-2}\sum_{1\leq l_{1}< l_{2}< l_{3}\leq n+1}W_{2}(x_{l_{1}}x_{l_{2}}x_{l_{3}}; a, b)+ \cdots \\& \qquad {} + (-1)^{1}\sum_{1\leq l_{1}< l_{2}< \cdots< l_{n} \leq n+1}W_{2}(x_{l_{1}}x_{l_{2}} \cdots x_{l_{n}}; a, b) \\& \qquad {} + (-1)^{0} W_{2}(x_{1}x_{2} \cdots x_{n+1}; a, b) \\& \quad =(-1)^{n}\sum_{k=1}^{n+1} \sum_{i< j, a_{ki}=a, a_{kj}=b} t_{ki}t_{kj} \\& \qquad {} +(-1)^{n-1}\sum_{1\leq l_{1}< l_{2}\leq n+1} \biggl( \sum_{1\leq k\leq2} \sum_{i< j, a_{l_{k}i}=a, a_{l_{k} j}=b} t_{l_{k} i}t_{l_{k} j} +\sum_{a_{l_{1} i}=a, a_{l_{2} j}=b}t_{l_{1} i}t_{l_{2} j} \biggr) \\& \qquad {} +(-1)^{n-2}\sum_{1\leq l_{1}< l_{2}< l_{3}\leq n+1} \biggl( \sum_{1\leq k\leq3}\sum_{i< j, a_{l_{k}i}=a, a_{l_{k}j}=b}t_{l_{k}i}t_{l_{k}j} \\& \qquad {} +\sum_{1\leq p< q\leq3}\sum _{a_{l_{p}i}=a, a_{l_{q}j}=b}t_{l_{p}i}t_{l_{q}j} \biggr)+\cdots \\ & \qquad {} +(-1)^{1} \sum_{1\leq l_{1}< l_{2}\cdots< l_{n}\leq n+1} \biggl( \sum_{1\leq k\leq n}\sum_{i< j, a_{l_{k}i}=a, a_{l_{k}j}=b}t_{l_{k}i}t_{l_{k}j} \\ & \qquad {} +\sum_{1\leq p< q\leq n}\sum _{a_{l_{p}i}=a, a_{l_{q}j}=b}t_{l_{p}i}t_{l_{q}j} \biggr) \\ & \qquad {} +(-1)^{0} \biggl(\sum_{1\leq k \leq n+1}\sum _{i< j, a_{ki}=a, a_{kj}=b}t_{ki}t_{kj}+\sum _{1\leq p< q\leq n+1}\sum_{a_{pi}=a, a_{qj}=b}t_{pi}t_{qj} \biggr) \\ & \quad = \biggl((-1)^{n} \sum_{1\leq k \leq n+1}\sum _{i< j, a_{ki}=a, a_{kj}=b}t_{ki}t_{kj} \\ & \qquad {} +(-1)^{n-1}\sum_{1\leq l_{1}< l_{2}\leq n+1}\sum _{1\leq k\leq2}\sum_{i< j, a_{l_{k}i}=a, a_{l_{k}j}=b}t_{l_{k}i}t_{l_{k}j} \\ & \qquad {} +(-1)^{n-2}\sum_{1\leq l_{1}< l_{2}< l_{3}\leq n+1}\sum _{1\leq k\leq3}\sum_{i< j, a_{l_{k}i}=a, a_{l_{k}j}=b}t_{l_{k}i}t_{l_{k}j}+ \cdots \\ & \qquad {} + (-1)^{1}\sum_{1\leq l_{1}< l_{2}< \cdots< l_{n}\leq n+1}\sum _{1\leq k\leq n}\sum_{i< j, a_{l_{k}i}=a, a_{l_{k}j}=b}t_{l_{k}i}t_{l_{k}j} \\ & \qquad {} +(-1)^{0} \sum_{1\leq k\leq n+1}\sum _{i< j, a_{ki}=a, a_{kj}=b}t_{ki}t_{kj} \biggr) \\ & \qquad {} + \biggl((-1)^{n-1}\sum_{1\leq l_{1}< l_{2}\leq n+1} \sum_{a_{l_{1}i}=a,a_{l_{2}j}=b}t_{l_{1}i}t_{l_{2}j} \\ & \qquad {} +(-1)^{n-2}\sum_{1\leq l_{1}< l_{2}< l_{3}\leq n+1}\sum _{1\leq p< q\leq 3}\sum_{a_{l_{p}i}=a,a_{l_{q}j}=b}t_{l_{p}i}t_{l_{q}j} \\ & \qquad {} +\cdots+(-1)^{1} \sum_{1\leq l_{1}< l_{2}< \cdots< l_{n}\leq n+1}\sum _{1\leq p< q\leq n}\sum_{a_{l_{p}i}=a,a_{l_{q}j}=b}t_{l_{p}i}t_{l_{q}j} \\ & \qquad {} +(-1)^{0}\sum_{1\leq p< q\leq n+1}t_{pi}t_{qj} \biggr) \\ & \quad \triangleq I_{1}+I_{2}, \end{aligned}$$

where

$$\begin{aligned} \begin{aligned}[b] I_{1}={}&(-1)^{n} \sum _{1\leq k \leq n+1}\sum_{i< j, a_{ki}=a, a_{kj}=b}t_{ki}t_{kj} \\ &{} +(-1)^{n-1}\sum_{1\leq l_{1}< l_{2}\leq n+1}\sum _{1\leq k\leq2}\sum_{i< j, a_{l_{k}i}=a, a_{l_{k}j}=b}t_{l_{k}i}t_{l_{k}j} \\ &{} +(-1)^{n-2}\sum_{1\leq l_{1}< l_{2}< l_{3}\leq n+1}\sum _{1\leq k\leq3}\sum_{i< j, a_{l_{k}i}=a, a_{l_{k}j}=b}t_{l_{k}i}t_{l_{k}j}+ \cdots \\ &{} + (-1)^{1}\sum_{1\leq l_{1}< l_{2}< \cdots< l_{n}\leq n+1}\sum _{1\leq k\leq n}\sum_{i< j, a_{l_{k}i}=a, a_{l_{k}j}=b}t_{l_{k}i}t_{l_{k}j} \\ &{} +(-1)^{0} \sum_{1\leq k\leq n+1}\sum _{i< j, a_{ki}=a, a_{kj}=b}t_{ki}t_{kj}. \end{aligned} \end{aligned}$$

(3.7)

By the symmetry we can see that for any $1\leq k \leq n+1$ and $i< j$, the coefficient of the item $t_{ki}t_{kj}$ in (3.7) is identical and equals

$$\begin{aligned}& (-1)^{n} \binom{n}{0}+(-1)^{n-1}\binom{n}{1}+(-1)^{n-2} \binom{n}{2} \\& \qquad {}+\cdots +(-1)^{1}\binom{n}{n-1}+(-1)^{0}\binom{n}{n} \\& \quad =(-1+1)^{n}=0, \end{aligned}$$

which gives $I_{1}=0$. Now compute

$$\begin{aligned} I_{2} =& (-1)^{n-1}\sum _{1\leq l_{1}< l_{2}\leq n+1}\sum_{a_{l_{1}i}=a,a_{l_{2}j}=b}t_{l_{1}i}t_{l_{2}j} \\ &{} +(-1)^{n-2}\sum_{1\leq l_{1}< l_{2}< l_{3}\leq n+1}\sum _{1\leq p< q\leq 3}\sum_{a_{l_{p}i}=a,a_{l_{q}j}=b}t_{l_{p}i}t_{l_{q}j} \\ &{} +\cdots+(-1)^{1} \sum_{1\leq l_{1}< l_{2}< \cdots< l_{n}\leq n+1}\sum _{1\leq p< q\leq n}\sum_{a_{l_{p}i}=a,a_{l_{q}j}=b}t_{l_{p}i}t_{l_{q}j} \\ &{} +(-1)^{0}\sum_{1\leq p< q\leq n+1}t_{pi}t_{qj}. \end{aligned}$$

(3.8)

Obviously, for any $1\leq p< q \leq n+1$, the coefficient of the item $t_{pi}t_{qj}$ in (3.8) is identical and equals

$$\begin{aligned}& (-1)^{n-1} \binom{n-1}{0}+(-1)^{n-2}\binom{n-1}{1}+(-1)^{n-3} \binom {n-1}{2} \\& \qquad {}+\cdots +(-1)^{1}\binom{n-1}{n-2}+(-1)^{0} \binom{n-1}{n-1} \\& \quad =(-1+1)^{n-1}=0, \end{aligned}$$

which gives $I_{2}=0$. This concludes assertion (ii).

Statement (iii) follows from (3.6) directly. □

4 Solutions on symmetric groups

The symmetric group on a finite set X is a group whose elements are all bijective maps from X to itself and whose group operation is that of the map composition. If $X=\{1, 2, \ldots, m\}$, then it is called a symmetric group of degree m, denoted by $S_{m}$. In this section, we consider $C^{(n)}f=0$ for $G=S_{m}$.

Proposition 4

If $f\in\operatorname{Ker} C^{(n)}(S_{m}, H)$, then for any $i=1,2,\ldots, n$, we have

$$\begin{aligned}& C^{(n-1)}f(x_{1}, x_{2}, \ldots, x_{i-1}, y_{1}y_{2}\cdots y_{p}, x_{i+1}, \ldots, x_{n}) \\& \quad = C^{(n-1)}f(x_{1}, x_{2}, \ldots, x_{i-1}, y_{\pi(1)}y_{\pi(2)}\cdots y_{\pi(p)}, x_{i+1},\ldots, x_{n}) \end{aligned}$$

(4.1)

for all $x_{i}, y_{j}\in S_{n}$, $i=1,2,\ldots,n$, $j=1, 2, \ldots, p$, and all rearrangements π.

Proof

Note that $C^{(n-1)}f$ is a homomorphism with respect to each variable and H is an Abelian group, which yields

$$\begin{aligned}& C^{(n-1)}f(x_{1}, x_{2}, \ldots, x_{i-1}, y_{1}y_{2}\cdots y_{p}, x_{i+1}, \ldots, x_{n}) \\ & \quad =C^{(n-1)}f(x_{1}, x_{2}, \ldots, x_{i-1}, y_{1}, x_{i+1},\ldots, x_{n}) \\ & \qquad {} +C^{(n-1)}f(x_{1}, x_{2}, \ldots, x_{i-1}, y_{2}, x_{i+1},\ldots, x_{n}) \\ & \qquad {} +\cdots+C^{(n-1)}f(x_{1}, x_{2}, \ldots, x_{i-1}, y_{p}, x_{i+1},\ldots, x_{n}) \\ & \quad = C^{(n-1)}f(x_{1}, x_{2}, \ldots, x_{i-1}, y_{\pi(1)}y_{\pi(2)}\cdots y_{\pi(p)}, x_{i+1},\ldots, x_{n}). \end{aligned}$$

This proves (4.1). □

Proposition 5

Let τ be an arbitrary 2-cycle in $S_{m}$, and $f\in\operatorname{Ker} C^{(n)}(S_{m}, H)$. Then we have

$$\begin{aligned}& f\bigl(\tau ^{2}\bigr)=0, \end{aligned}$$

(4.2)

$$\begin{aligned}& C^{(n)}f(\tau , \tau , \ldots, \tau )=(-2)^{n} f(\tau )= \textstyle\begin{cases} 2^{n}f(\tau )& \textit{if } n \textit{ is even}, \\ -2^{n}f(\tau ) & \textit{if } n \textit{ is odd}, \end{cases}\displaystyle \end{aligned}$$

(4.3)

$$\begin{aligned}& 2^{n} f(\tau )=0 . \end{aligned}$$

(4.4)

Proof

We only need to prove (4.3). To this end, we need the following facts.

• (i) If n is even, then

  $$\begin{aligned}& \binom{n+1}{1}+\binom{n+1}{3}+\binom{n+1}{5}+\cdots+ \binom{n+1}{ n+1} \\& \quad =\binom{n+1}{0}+\binom{n+1}{2}+\binom{n+1}{4}+ \cdots+\binom{n+1}{ n}. \end{aligned}$$

  (4.5)

• (ii) If n is odd, then

  $$\begin{aligned}& \binom{n+1}{1}+\binom{n+1}{3}+\binom{n+1}{5}+\cdots+ \binom{n+1}{ n} \\& \quad =\binom{n+1}{0}+\binom{n+1}{2}+\binom{n+1}{4}+ \cdots+\binom {n+1}{n+1}. \end{aligned}$$

  (4.6)

Note that by (2.1)

$$\begin{aligned}& C^{(n)}f(\tau , \tau , \ldots, \tau ) \\& \quad = (-1)^{n}\binom{n+1}{1}f(\tau )+(-1)^{n-1} \binom{n+1}{2}f\bigl(\tau ^{2}\bigr)+(-1)^{n-2} \binom{n+1}{3}f\bigl(\tau ^{3}\bigr) \\& \qquad {} +\cdots+(-1)^{1}\binom{n+1}{n}f\bigl(\tau ^{n} \bigr)+\binom{n+1}{n+1} f\bigl(\tau ^{n+1}\bigr). \end{aligned}$$

(4.7)

We first prove the even case of (4.3). When n is even, by (4.5), (4.7) becomes

$$\begin{aligned}& C^{(n)}f(\tau , \tau , \ldots, \tau ) \\& \quad =\binom{n+1}{1}f(\tau )+\binom{n+1}{3}f(\tau )+\binom{n+1}{5} f(\tau )+\cdots +\binom{n+1}{n+1}f(\tau ) \\& \quad = \biggl(\binom{n+1}{1}+\binom{n+1}{3}+\binom{n+1}{5}+\cdots+ \binom {n+1}{n+1} \biggr)f(\tau ) \\& \quad =\frac{1}{2}\sum_{j=0}^{n+1} \binom{n+1}{j} f(\tau )=\frac {1}{2}(1+1)^{n+1}f(\tau ) \\& \quad =2^{n}f(\tau ), \end{aligned}$$

which confirms the even case. When n is odd, by (4.6), (4.7) becomes

$$\begin{aligned}& C^{(n)}f(\tau , \tau , \ldots, \tau ) \\& \quad = -\binom{n+1}{1}f(\tau )-\binom{n+1}{3}f(\tau )- \binom{n+1}{5}f(\tau )-\cdots -\binom{n+1}{n}f(\tau ) \\& \quad = \biggl(-\binom{n+1}{1}-\binom{n+1}{3}-\binom{n+1}{5}-\cdots- \binom {n+1}{n} \biggr)f(\tau ) \\& \quad =-\frac{1}{2}\sum_{j=0}^{n+1} \binom{n+1}{j} f(\tau )=-\frac {1}{2}(1+1)^{n+1}f(\tau ) \\& \quad =-2^{n}f(\tau ). \end{aligned}$$

This completes the proof. □

Proposition 6

For any 2-cycle $\sigma _{1}, \sigma _{2}, \ldots, \sigma _{n}$ and $f\in\operatorname{Ker} C^{(n)}(S_{m}, H)$, we have

$$ C^{(n-1)}f(\sigma _{1}, \sigma _{2}, \ldots, \sigma _{n})=C^{(n-1)}f\bigl((12),(12), \ldots , (12) \bigr). $$

(4.8)

Proof

For any 2-cycle $\sigma _{1}$, there exists $z\in S_{m}$ such that $\sigma _{1}=z(12)z^{-1}$. Hence, for any $x_{2}, x_{3}, \ldots, x_{n}\in S_{m}$, by (4.1) we have

$$\begin{aligned}& C^{(n-1)}f(\sigma _{1},x_{2}, x_{3}, \ldots, x_{n}) \\& \quad =C^{(n-1)}f\bigl(z(12)z^{-1}, x_{2}, x_{3}, \ldots, x_{n}\bigr) \\& \quad =C^{(n-1)}f\bigl((12)zz^{-1}, x_{2}, x_{3}, \ldots, x_{n}\bigr) \\& \quad =C^{(n-1)}f\bigl((12), x_{2}, x_{3}, \ldots, x_{n}\bigr). \end{aligned}$$

(4.9)

Similarly, for any $2\leq i\leq n$,

$$\begin{aligned}& C^{(n-1)}f(x_{1},x_{2}, \ldots,x_{i-1},\sigma _{i},x_{i+1},\ldots ,x_{n}) \\& \quad =C^{(n-1)}f\bigl(x_{1},x_{2}, \ldots,x_{i-1},z(12)z^{-1},x_{i+1}, \ldots,x_{n}\bigr), \\& \quad =C^{(n-1)}f\bigl(x_{1},x_{2}, \ldots,x_{i-1},(12)zz^{-1},x_{i+1}, \ldots,x_{n}\bigr), \\& \quad =C^{(n-1)}f\bigl(x_{1},x_{2}, \ldots,x_{i-1},(12),x_{i+1},\ldots,x_{n}\bigr). \end{aligned}$$

(4.10)

In particular, (4.8) follows from (4.9)-(4.10). □

Lemma 3

Assume that

$$ C^{(j)}f(\sigma _{1}, \sigma _{2}, \ldots, \sigma _{j+1})=C^{(j)}f\bigl((12), (12), \ldots, (12) \bigr) $$

(4.11)

for every 2-cycle $\sigma _{1}, \sigma _{2}, \ldots, \sigma _{j+1}\in S_{m}$, $j=1,2,\ldots, n-2$. Then for any $x, y, \beta , \tau _{i}\in S_{m}$ and rearrangements π, where β, $\tau _{i}$ are 2-cycles, we have

$$\begin{aligned}& f(\tau _{1} \tau _{2}\cdots \tau _{p}) =f(\tau _{\pi(1)}\tau _{\pi(2)}\cdots\tau _{\pi (p)}), \end{aligned}$$

(4.12)

$$\begin{aligned}& f(x\beta y) =f\bigl(x(12)y\bigr), \end{aligned}$$

(4.13)

$$\begin{aligned}& f(\beta ) = f\bigl((12)\bigr) \end{aligned}$$

(4.14)

for every $f\in\operatorname{Ker} C^{(n)}(S_{m}, H)$.

Proof

First, for any 2-cycle $\tau _{i}\in S_{m}$, $i=1, 2, \ldots ,p$, and rearrangement π, it follows from (4.11), (2.4), and (4.8) that when $p\geq n$,

$$\begin{aligned}& f(\tau _{1}\tau _{2}\cdots\tau _{p}) \\& \quad = \sum_{i=1}^{p} f(\tau _{i})+\sum_{1\leq i_{1}< i_{2}\leq p}Cf(\tau _{i_{1}}, \tau _{i_{2}})+\sum_{1\leq i_{1}< i_{2}< i_{3}\leq p}C^{(2)}f( \tau _{i_{1}}, \tau _{i_{2}}, \tau _{i_{3}}) \\& \qquad {} +\cdots+\sum_{1\leq i_{1}< i_{2}< \cdots< i_{n}\leq p}C^{(n-1)}f(\tau _{i_{1}}, \tau _{i_{2}},\ldots, \tau _{i_{n}}) \\& \quad = \sum_{i=1}^{p} f(\tau _{\pi(i)})+\frac{p(p-1)}{2}Cf\bigl((12), (12)\bigr)+ \frac{p(p-1)(p-2)}{6}C^{(2)}f\bigl((12), (12), (12)\bigr) \\& \qquad {} +\cdots+\frac{p(p-1)(p-2)\cdots (p-n+1)}{(n)!}C^{(n-1)}f\bigl((12),(12),\ldots,(12) \bigr) \\& \quad = f(\tau _{\pi(1)}\tau _{\pi(2)}\cdots\tau _{\pi(p)}), \end{aligned}$$

which gives the case of $p\geq n$ in (4.12). The case of $p< n$ is similar to verify. This confirms the proof of (4.12).

On the other hand, for all $x, y, \beta \in S_{m}$, there exist 2-cycles $\sigma _{i}, \tau _{j}, z\in S_{m}$, $i=1,2,\ldots,p$, $j=1,2,\ldots,q$, such that $x=\sigma _{1}\sigma _{2}\cdots\sigma _{p}$, $y=\tau _{1}\tau _{2}\cdots \tau _{q}$, and $\beta =z(12)z^{-1}$. Noting that $z=\delta_{1}\delta_{2}\cdots\delta_{r}$ for some 2-cycles $\delta_{1},\delta_{2},\ldots, \delta_{r}\in S_{m}$, we obtain

$$\begin{aligned} f(x\beta y)&= f\bigl(\sigma _{1}\sigma _{2}\cdots\sigma _{p}\delta_{1}\delta_{2}\cdots\delta _{r}(12)\delta _{r}^{-1}\delta_{r-1}^{-1} \cdots\delta_{1}^{-1}\tau _{1}\tau _{2} \cdots \tau _{q}\bigr) \\ &= f\bigl(\sigma _{1}\sigma _{2}\cdots\sigma _{p}(12)\delta_{1}\delta_{2}\cdots \delta_{r}\delta _{r}^{-1}\delta_{r-1}^{-1} \tau _{1}\tau _{2}\cdots\tau _{q}\bigr) \\ &= f\bigl(x(12)y\bigr) \end{aligned}$$

by (4.12). In particular, taking $x=y=e$ in (4.13), we get formula (4.14). This completes the proof. □

According to Lemma 3, we give the following main result in this section.

Theorem 2

Assume that (4.11) holds. Then f is a solution to the equation $C^{(n)}(S_{m}, H)=0$ if and only if there is $h_{0}\in H$ such that $2^{n} h_{0}=0$ and

$$ f(x)= \textstyle\begin{cases} 0 & \textit{if } x \textit{ is even}, \\ h_{0} & \textit{if } x \textit{ is odd}. \end{cases} $$

(4.15)

Proof

We first deal with the necessity. Let $f\in \operatorname{Ker} C^{(n)}(S_{m}, H)$. Then, for any $x\in S_{m}$, there exist 2-cycles $\alpha _{i}\in S_{m}$, $i=1, 2, \ldots, p$, such that $x=\alpha _{1}\alpha _{2}\cdots\alpha _{p}$. In view of (4.1), (4.3), and (4.14), we get that for $p\geq n$,

$$\begin{aligned} f(x) =&f(\alpha _{1}\alpha _{2}\cdots\alpha _{p}) \\ =&\sum_{i=1}^{p} f(\alpha _{i})+\sum_{1\leq i_{1}< i_{2}\leq p}Cf(\alpha _{i_{1}}, \alpha _{i_{2}})+\sum_{1\leq i_{1}< i_{2}< i_{3}\leq p}C^{(2)}f( \alpha _{i_{1}}, \alpha _{i_{2}}, \alpha _{i_{3}}) \\ &{} +\cdots+\sum_{1\leq i_{1}< i_{2}< \cdots< i_{n}\leq p}C^{(n-1)}f(\alpha _{i_{1}}, \alpha _{i_{2}},\ldots, \alpha _{i_{n}}) \\ =& p f\bigl((12)\bigr)+\frac{p(p-1)}{2}Cf\bigl((12), (12)\bigr)+ \frac{p(p-1)(p-2)}{6}C^{(2)}f\bigl((12), (12), (12)\bigr) \\ &{} +\cdots+\frac{p(p-1)(p-2)\cdots (p-n+1)}{n!}C^{(n-1)}f\bigl((12),(12),\ldots,(12) \bigr) \\ =&\binom{p}{1}f\bigl((12)\bigr)+\binom{p}{2} \bigl((-2)^{1}f\bigl((12) \bigr) \bigr)+\binom{p}{3} \bigl((-2)^{2}f\bigl((12)\bigr) \bigr) \\ &{} +\cdots+\binom{p}{n} \bigl((-2)^{n-1}f\bigl((12)\bigr) \bigr) \\ =& \bigl(\binom{p}{1}+(-2)^{1}\binom{p}{2}+ (-2)^{2}\binom{p}{3}+ \cdots+(-2)^{n-1}\binom{p}{n} \bigr)f\bigl((12)\bigr). \end{aligned}$$

(4.16)

Let

$$g(p)=\binom{p}{1}+(-2)^{1}\binom{p}{2}+ (-2)^{2}\binom{p}{3}+ \cdots+(-2)^{n-1}\binom{p}{n}. $$

We claim that

$$ g(p)\in \textstyle\begin{cases} 2^{n} \mathbb{Z} & \mbox{if } p \mbox{ is even}, \\ 2^{n}\mathbb{Z}+1 & \mbox{if } p \mbox{ is odd}. \end{cases} $$

(4.17)

We first prove the even case. Obviously, (4.17) is true for $p=2$ since $g(2)=0$. For an inductive proof, suppose that (4.17) holds for $p=2t$, $t\in{\mathbb{N}}$. Then we have

$$\begin{aligned}& g(2t+2)-g(2t) \\& \quad = \biggl(\binom{2t+2}{1}-\binom{2t}{1} \biggr)+(-2)^{1} \biggl(\binom{2t+2}{2} -\binom{2t}{2} \biggr) \\& \qquad {}+(-2)^{2} \biggl( \binom{2t+2}{3}-\binom{2t}{3} \biggr)+\cdots +(-2)^{n-1} \biggl(\binom{2t+2}{n}-\binom{2t}{n} \biggr) \\& \quad =2+(-2)^{1} \biggl(2\binom{2t}{1}+\binom{2t}{0} \biggr)+ (-2)^{2} \biggl(2\binom {2t}{2}+\binom{2t}{1} \biggr)+ \cdots \\& \qquad {} +(-2)^{n-1} \biggl(2\binom{2t}{n-1}+\binom{2t}{n-2} \biggr) \\& \quad = \biggl(2+(-2)^{1}\binom{2t}{0}+ (-2)^{2} \binom{2t}{1}+\cdots+(-2)^{n-1}\binom{2t}{n-2} \biggr) \\& \qquad{} + \biggl((-2)^{1}\cdot2\binom{2t}{1}+ (-2)^{2} \cdot2\binom{2t}{2} +\cdots +(-2)^{n-1}\cdot2\binom{2t}{n-1} \biggr) \\& \quad \triangleq J_{1}+J_{2}, \end{aligned}$$

where

$$\begin{aligned} J_{1}&=2+(-2)^{1}\binom{2t}{0}+ (-2)^{2} \binom{2t}{1}+\cdots+(-2)^{n-1}\binom{2t}{n-2} \\ &=(-2)^{2} \biggl(\binom{2t}{1}+(-2)^{1} \binom{2t}{2}+\cdots+(-2)^{n-3}\binom {2t}{n-2} \biggr) \\ &=(-2)^{2} \biggl(g(2t)-(-2)^{n-2}\binom{2t}{n-1}-(-2)^{n-1} \binom{2t}{ n} \biggr) \\ &=4g(2t)-(-2)^{n}\binom{2t}{n-1}-(-2)^{n+1} \binom{2t}{n} \end{aligned}$$

and

$$\begin{aligned} J_{2}&=(-2)^{1}\cdot2\binom{2t}{1}+ (-2)^{2} \cdot2\binom{2t}{2}+\cdots +(-2)^{n-1}\cdot2\binom{2t}{n-1} \\ &=-2^{2} \biggl(\binom{2t}{1}+(-2)^{1} \binom{2t}{2}+\cdots+ (-2)^{n-2}\cdot \binom{2t}{n-1} \biggr) \\ &=-2^{2} \biggl(g(2t)-(-2)^{n-1}\binom{2t}{n} \biggr) \\ &=-4g(2t)+(-2)^{n+1}\binom{2t}{n}, \end{aligned}$$

which yields $g(2t+2)-g(2t)=-(-2)^{n}\binom{2t}{n-1}$ and $g(2t+2)\in2^{n} \mathbb{Z}$. This confirms the even case of (4.17). When p is odd, (4.17) is true for $p=1$ because of $g(1)=1$. Suppose that (4.17) holds for $p=2t-1$, and then we get

$$\begin{aligned}& g(2t+1)-g(2t-1) \\& \quad = \biggl(\binom{2t+1}{1}-\binom{2t-1}{1} \biggr)+(-2)^{1} \biggl(\binom{2t+1}{2} -\binom{2t-1}{2} \biggr) \\& \qquad {} +(-2)^{2} \biggl(\binom{2t+1}{3}-\binom{2t-1}{3} \biggr)+\cdots +(-2)^{n-1} \biggl(\binom{2t+1}{n}-\binom{2t-1}{n} \biggr) \\& \quad =2+(-2)^{1} \biggl(2\binom{2t-1}{1}+\binom{2t-1}{0} \biggr)+ (-2)^{2} \biggl(2\binom{2t-1}{2}+\binom{2t-1}{1} \biggr)+ \cdots{} \\& \qquad {} +(-2)^{n-1} \biggl(2\binom{2t-1}{n-1}+\binom{2t-1}{n-2} \biggr) \\& \quad = \biggl(2+(-2)^{1}\binom{2t-1}{0}+ (-2)^{2} \binom{2t-1}{1}+\cdots+(-2)^{n-1}\binom{2t-1}{n-2} \biggr) \\& \qquad {} + \biggl((-2)^{1}\cdot2\binom{2t-1}{1}+ (-2)^{2}\cdot2\binom{2t-1}{2} +\cdots +(-2)^{n-1}\cdot2 \binom{2t-1}{n-1} \biggr) \\& \quad \triangleq J_{3}+J_{4}, \end{aligned}$$

where

$$\begin{aligned} J_{3}&=2+(-2)^{1}\binom{2t-1}{0}+ (-2)^{2} \binom{2t-1}{1}+\cdots+(-2)^{n-1}\binom{2t-1}{n-2} \\ &=(-2)^{2} \biggl(\binom{2t-1}{1}+(-2)^{1} \binom{2t-1}{2}+\cdots +(-2)^{n-3}\binom{2t-1}{n-2} \biggr) \\ &=(-2)^{2} \biggl(g(2t-1)-(-2)^{n-2}\binom{2t-1}{n-1}-(-2)^{n-1} \binom {2t-1}{n} \biggr) \\ &=4g(2t-1)-(-2)^{n}\binom{2t-1}{n-1}-(-2)^{n+1} \binom{2t-1}{n} \end{aligned}$$

and

$$\begin{aligned} J_{4} =&(-2)^{1}\cdot2\binom{2t-1}{1}+ (-2)^{2}\cdot2\binom{2t-1}{2} +\cdots +(-2)^{n-1}\cdot2 \binom{2t-1}{n-1} \\ =&-2^{2} \biggl(\binom{2t-1}{1}+(-2)^{1} \binom{2t-1}{2}+\cdots+ (-2)^{n-2}\cdot \binom{2t-1}{n-1} \biggr) \\ =&-2^{2} \biggl(g(2t-1)-(-2)^{n-1}\binom{2t-1}{n} \biggr) \\ =&-4g(2t-1)+(-2)^{n+1}\binom{2t-1}{n}, \end{aligned}$$

which yields $g(2t+1)-g(2t-1)=-(-2)^{n}\binom{2t-1}{n-1}$ and $g(2t-1)\in2^{n} \mathbb{Z}+1$. This completes the proof of (4.17). According to (4.17) and (4.4), equation (4.16) becomes

$$ f(x)= \textstyle\begin{cases} 0 & \mbox{if } x \mbox{ is even}, \\ f((12)) & \mbox{if } x \mbox{ is odd}. \end{cases} $$

This proves that when $p\geq n$, f must be of the form (4.15) with $h_{0}=f((12))$.

When $p< n$,

$$\begin{aligned} f(x) =&f(\alpha _{1}\alpha _{2}\cdots\alpha _{p}) \\ =&\sum_{i=1}^{p} f(\alpha _{i})+\sum_{1\leq i_{1}< i_{2}\leq p}Cf(\alpha _{i_{1}}, \alpha _{i_{2}})+\sum_{1\leq i_{1}< i_{2}< i_{3}\leq p}C^{(2)}f( \alpha _{i_{1}}, \alpha _{i_{2}}, \alpha _{i_{3}}) \\ &{} +\cdots+\sum_{1\leq i_{1}< i_{2}< \cdots< i_{p-1}\leq p}C^{(p-2)}f(\alpha _{i_{1}}, \alpha _{i_{2}},\ldots, \alpha _{i_{p-1}}) \\ &{} +C^{(p-1)}f(\alpha _{1}, \alpha _{2},\ldots, \alpha _{p}) \\ =& p f\bigl((12)\bigr)+\frac{p(p-1)}{2}Cf\bigl((12), (12)\bigr)+ \frac{p(p-1)(p-2)}{6}C^{(2)}f\bigl((12), (12), (12)\bigr) \\ &{} +\cdots +\frac{p(p-1)(p-2)\cdots2}{(p-1)!}C^{(p-2)}f\bigl((12),(12),\ldots,(12) \bigr) \\ &{} +C^{(p-1)}f\bigl((12),(12),\ldots,(12)\bigr) \\ =&\binom{p}{1}f\bigl((12)\bigr)+\binom{p}{2} \bigl((-2)^{1}f\bigl((12) \bigr) \bigr)+\binom{p}{3} \bigl((-2)^{2}f\bigl((12)\bigr) \bigr)+\cdots \\ &{} +\binom{p}{p-1} \bigl((-2)^{p-2}f\bigl((12)\bigr) \bigr)+ (-2)^{p-1}f\bigl((12)\bigr) \\ =& \biggl(\binom{p}{1}+(-2)^{1}\binom{p}{2}+ (-2)^{2}\binom{p}{3}+ \cdots+(-2)^{p-2}\binom{p}{p-1}+(-2)^{p-1} \biggr)f\bigl((12) \bigr) \\ =& \biggl(-\frac{1}{2} \biggl((-2)^{0}\binom{p}{0}+(-2)^{1}\binom{p}{1}+(-2)^{2}\binom{p}{2}+ (-2)^{3} \binom{p}{3} \\ &{} +\cdots+(-2)^{p-1}\binom{p}{p-1}+(-2)^{p} \biggr)+ \frac{1}{2} \biggr) f\bigl((12)\bigr) \\ =& \biggl(-\frac{1}{2}(-2+1)^{p}+\frac{1}{2} \biggr)f \bigl((12)\bigr) \\ =& \textstyle\begin{cases} 0 & \mbox{if } p \mbox{ is even}, \\ f((12)) & \mbox{if } p \mbox{ is odd}. \end{cases}\displaystyle \end{aligned}$$

This implies that when $p< n$, f must be of the form (4.15) with $h_{0}=f((12))$.

Now we turn to the sufficiency. Let $f: S_{m}\rightarrow H$ be defined by (4.15), where $h_{0}$ is a constant with $2^{n} h_{0}=0$. In order to prove $C^{(n)}f=0$, by the symmetry of $x_{1}, x_{2}, \ldots, x_{n+1}$, it suffices to verify that it holds when $x_{1}, x_{2}, \ldots, x_{k}$ are odd and $x_{k+1}, x_{k+2}, \ldots, x_{n+1}$ are even for any $k=1, 2, \ldots, n+1$. To this end, we only need to verify the following two cases: (i) k is odd; (ii) k is even. In fact, for case (i), by (2.1) we have

$$\begin{aligned}& C^{(n)}f(x_{1},x_{2},\ldots,x_{n+1}) \\& \quad =\sum_{1\leq m \leq n+1}(-1)^{n+1-m}\sum _{1\leq i_{1}< i_{2}< \cdots< i_{m}\leq n+1}f(x_{i_{1}}x_{i_{2}}\cdots x_{i_{m}}) \\& \quad = \biggl((-1)^{n} \binom{k}{1} +(-1)^{n-1}\binom{k}{1} \binom{n+1-k}{1}+(-1)^{n-2}\binom{k}{1}\binom {n+1-k}{2} \\& \qquad {} +\cdots+(-1)^{k-1}\binom{k}{1}\binom{n+1-k}{n+1-k} \biggr)h_{0} \\& \qquad {} + \biggl((-1)^{n-2} \binom{k}{3} +(-1)^{n-3}\binom{k}{3} \binom{n+1-k}{1}+(-1)^{n-4}\binom{k}{3} \binom {n+1-k}{2} \\& \qquad {} +\cdots+(-1)^{k-3}\binom{k}{3} \binom{n+1-k}{n+1-k} \biggr)h_{0} \\& \qquad {} + \biggl((-1)^{n-4} \binom{k}{5} +(-1)^{n-5}\binom{k}{5} \binom{n+1-k}{1}+(-1)^{n-6}\binom{k}{5} \binom {n+1-k}{2} \\& \qquad {} +\cdots+(-1)^{k-5}\binom{k}{5} \binom{n+1-k}{n+1-k} \biggr)h_{0} \\ & \qquad {} +\cdots \\ & \qquad {} + \biggl((-1)^{n-k+1} \binom{k}{k} +(-1)^{n-k}\binom{k}{k} \binom{n+1-k}{1}+(-1)^{n-k-1}\binom{k}{k} \binom {n+1-k}{2} \\& \qquad {} +\cdots+(-1)^{0}\binom{k}{k} \binom{n+1-k}{n+1-k} \biggr)h_{0} \\& \quad = \biggl( \binom{k}{1}(-1)^{k-1}(-1+1)^{n+1-k} \biggr)h_{0}+ \biggl( \binom{k}{3}(-1)^{k-3}(-1+1)^{n+1-k} \biggr)h_{0} \\& \qquad {} + \biggl( \binom{k}{5}(-1)^{k-5}(-1+1)^{n+1-k} \biggr)h_{0}+\cdots \\& \qquad {} + \biggl( \binom{k}{k}(-1)^{k-k}(-1+1)^{n+1-k} \biggr)h_{0} \\& \quad = 0, \end{aligned}$$

which confirms the odd case. With a similar discussion, the proof of the even case is also obtained. □

5 Solutions on finite cyclic groups

Let $C_{m}=\langle a\mid a^{m}=e \rangle$ be a cyclic group of order m with generator a. In this section, we study a general solution on the finite cyclic group $C_{m}$.

Theorem 3

Assume that $mC^{(k)}f(\underbrace{a, a, \ldots, a}_{k+1})=0$, $k=1,2,\ldots, n-2$, and $m=n$, $mf(a)=0$. Then f is a solution to the equation $C^{(n)}(C_{m}, H)=0$ if and only if it is given by

$$ f\bigl(a^{p}\bigr)= \textstyle\begin{cases} \sum_{j=0}^{n-1}\frac{p(p-1) \cdots(p-j)}{(j+1)!}C^{(j)}f(\underbrace{a,a,\ldots,a}_{j+1}) &\textit{if } p\leq0 \textit{ or } p\geq n, \\ \sum_{j=0}^{p-1}\frac{p(p-1) \cdots(p-j)}{(j+1)!}C^{(j)}f(a,a,\ldots,a) &\textit{if } 0< p< n, \end{cases} $$

(5.1)

where $p\in\mathbb{Z}$ and $f(a)$, $C^{(n-1)}f(a, a,\ldots, a)$ satisfy

$$\begin{aligned}& m C^{(n-1)}f(a, a,\ldots, a)=0, \end{aligned}$$

(5.2)

$$\begin{aligned}& mf(a)=0 \quad \textit{if } m< n \textit{ or } m>n, \end{aligned}$$

(5.3)

$$\begin{aligned}& C^{(m-1)}f(a, a, \ldots, a)=0 \quad \textit{if } m=n. \end{aligned}$$

(5.4)

Proof

Let $f:C_{m}\rightarrow H$ be a function satisfying $C^{(n)}f=0$. Then, by (2.5) we see that f also satisfies (5.1). Now using (2.2), (ii) of Proposition 2, and the fact $a^{m}=e$, we get

$$ m C^{(n-1)}f(a, a,\ldots, a)=C^{(n-1)}f\bigl(a^{m}, a, \ldots, a\bigr)=C^{(n-1)}f(e, a,\ldots, a)=0, $$

which gives (5.2). Furthermore, let $p=m$ in (5.1), according to the assumptions of $mC^{(k)}f(a, a, \ldots, a)=0$, $k=1, 2, \ldots, n-2$, $mf(a)=0$, and $m=n$, (5.3)-(5.4) are obtained. This proves (5.2)-(5.4) and implies the necessity.

To give out the sufficiency, we claim that (5.1)-(5.4) give a well-defined function on $C_{m}$. Indeed, when $m>n$, it suffices to verify the following four cases: (i) $p+m\geq n$ and $n-m\leq p\leq0$ or $p\geq n$; (ii) $p+m\geq n$ and $0< p< n$; (iii) $p+m\leq0$ and $p\leq-m<0$; (iv) $0< p+m< n$ and $-m< p< n-m<0$.

For case (i), by (5.1) we have

$$\begin{aligned}& f\bigl(a^{p+m}\bigr)-f\bigl(a^{p}\bigr) \\& \quad =\sum_{j=0}^{n-1}\frac{(p+m)(p+m-1)\cdots (p+m-j)}{(j+1)!}C^{(j)}f(a,a, \ldots,a) \\& \qquad {} -\sum_{j=0}^{n-1} \frac{p(p-1)\cdots (p-j)}{(j+1)!}C^{(j)}f(a,a,\ldots,a) \\& \quad =mf(a)+\sum_{j=1}^{n-1} \biggl( \frac{(p+m)(p+m-1) \cdots(p+m-j)}{(j+1)!} -\frac{p(p-1) \cdots(p-j)}{(j+1)!} \biggr) \\& \qquad {} \times C^{(j)}f(a,a,\ldots,a). \end{aligned}$$

(5.5)

It is easy to see that the coefficient $C^{(j)}f(a,a,\ldots,a)$ is an integer multiple of m for $j=1,2,\ldots, n-1$, and therefore, by (5.2)-(5.3) and the assumption of $mC^{(k)}f(a, a, \ldots, a)=0$, $k=1,2,\ldots, n-2$, (5.5) equals 0.

For case (ii), we compute that

$$\begin{aligned}& f\bigl(a^{p+m}\bigr)-f\bigl(a^{p}\bigr) \\& \quad =\sum_{j=0}^{n-1}\frac{(p+m)(p+m-1) \cdots (p+m-j)}{(j+1)!}C^{(j)}f( \underbrace{a,a,\ldots,a}_{j+1}) \\& \qquad {} -\sum_{j=0}^{p-1} \frac{p(p-1) \cdots (p-j)}{(j+1)!}C^{(j)}f(\underbrace{a,a,\ldots,a}_{j+1}) \\& \quad = mf(a)+\sum_{j=1}^{p-1} \biggl( \frac{(p+m)(p+m-1) \cdots (p+m-j)}{(j+1)!} \\& \qquad {} -\frac{p(p-1) \cdots(p-j)}{(j+1)!} \biggr)C^{(j)}f(\underbrace {a,a, \ldots,a}_{j+1}) \\& \qquad {} +\sum_{j=p}^{n-1} \frac{(p+m)(p+m-1) \cdots(p+m-j)}{(j+1)!}C^{(j)}f(\underbrace{a,a,\ldots,a}_{j+1}). \end{aligned}$$

(5.6)

With the same discussion as in case (i), (5.6) equals 0. The proofs of the other two cases are similar to case (i).

When $m< n$, it suffices to verify the following five cases: (i) $p+m\geq n$ and $n-m< p< n$; (ii) $p+m\geq n$ and $p\geq n$; (iii) $p+m\leq0$ and $p\leq-m$; (iv) $0< p+m< n$ and $-m< p<0$; (v) $0< p+m< n$ and $0< p< n-m$. Since the proof is similar to the case of $m>n$, we omit the details.

When $m=n$, it suffices to verify the following four cases: (i) $p+m\geq n$ and $0< p< n$; (ii) $p+m\geq n$ and $p\geq n$; (iii) $p+m\leq0$ and $p\leq-m$; (iv) $0< p+m< n$ and $-m< p<0$. Similarly to the cases of $m>n$ and $m< n$, the proof of $m=n$ is omitted. □