Integer group determinants for ${\rm C}_{2}^{4}$

Yuka Yamaguchi and Naoya Yamaguchi

Abstract.

We determine all possible values of the integer group determinant of ${\rm C}_{2}^{4}$ , where ${\rm C}_{2}$ is the cyclic group of order $2$ .

Key words and phrases:

Group determinant, Integer group determinant, Cyclic group

1991 Mathematics Subject Classification:

11C20, 11E76, 20C15

1. Introduction

For a finite group $G$ , assigning a variable $x_{g}$ for each $g\in G$ , the group determinant of $G$ is defined as $\det{\left(x_{gh^{-1}}\right)}_{g,h\in G}$ . When the variables $x_{g}$ are all integers, the group determinant is called an integer group determinant of $G$ . For a broad context of group determinants, see their use in the identification of a group [3], [7], [11]; the related abelian question of circulants [6], [9]; the Lind-Lehmer problem [1], [2], [9]; and the representation theoretic background [5].

Let $S(G)$ denote the set of all possible values of the integer group determinant of $G$ :

S(G):=\left\{\det{\left(x_{gh^{-1}}\right)}_{g,h\in G}\mid x_{g}\in\mathbb{Z}\right\}.

Let ${\rm C}_{2}$ be the cyclic group of order $2$ . The complete descriptions of $S\left({\rm C}_{2}^{2}\right)$ and $S\left({\rm C}_{2}^{3}\right)$ are obtained in [1, Theorem 5.3] and [10, Theorem 3.1], respectively:

•

$S\left({\rm C}_{2}^{2}\right)=\left\{4m+1,\>2^{4}(2m+1),\>2^{6}m\mid m\in\mathbb{Z}\right\}$ ;
•

$S\left({\rm C}_{2}^{3}\right)=\left\{8m+1,\>2^{8}(4m+1),\>2^{12}m\mid m\in\mathbb{Z}\right\}$ .

In this paper, we determine $S\left({\rm C}_{2}^{4}\right)$ .

Theorem 1.1.

Let $A:=\left\{(8k-3)(8l+3)\mid k,l\in\mathbb{Z}\right\}\subsetneq\left\{8m-1\mid m\in\mathbb{Z}\right\}$ . Then we have

\displaystyle S\left({\rm C}_{2}^{4}\right)

\displaystyle=\left\{16m+1,\>2^{16}(4m+1),\>2^{24}(4m+1),\>2^{24}(8m+3),\>2^{24}m^{\prime},\>2^{26}m\mid m\in\mathbb{Z},\>m^{\prime}\in A\right\}.

Let $\mathbb{Z}_{\rm odd}$ be the set of all odd numbers. We remark that all of the odd values in $S\left({\rm C}_{2}^{4}\right)$ are already known: $S\left({\rm C}_{2}^{n}\right)\cap\mathbb{Z}_{\rm odd}=\left\{2^{n}m+1\mid m\in\mathbb{Z}\right\}$ holds for any $n$ . For example, see [2, Lemmas 2.1 and 2.2]. Although $S\left({\rm C}_{2}^{4}\right)\cap\mathbb{Z}_{\rm odd}=\left\{16m+1\mid m\in\mathbb{Z}\right\}$ is not new result, we give a proof in our method.

For every group $G$ of order at most $15$ , $S(G)$ was determined (see [8], [10]). For the groups of order $16$ , the complete descriptions of $S(G)$ were obtained for the dihedral group [1, Theorem 5.3] and the cyclic group [13]. There are fourteen groups of order $16$ up to isomorphism [4], [14]. Theorem 1.1 determines $S(G)$ for one of the unsolved twelve groups.

2. Relations with group determinants of subgroups

For any $g=(\overline{\varepsilon_{0}},\overline{\varepsilon_{1}},\ldots,\overline{\varepsilon_{n-1}})\in{\rm C}_{2}^{n}$ with $\varepsilon_{i}\in\{0,1\}$ , we denote the variable $x_{g}$ by $x_{j}$ , where $j:=\varepsilon_{n-1}\cdot 2^{n-1}+\varepsilon_{n-2}\cdot 2^{n-2}+\cdots+\varepsilon_{0}\cdot 2^{0}$ , and let

D_{n}(x_{0},x_{1},\ldots,x_{2^{n}-1}):=\det{\left(x_{gh^{-1}}\right)}_{g,h\in{\rm C}_{2}^{n}}.

From the $H={\rm C}_{2}^{n-1}$ and $K={\rm C}_{2}$ case of [12, Theorem 1.1], we have the following corollary.

Corollary 2.1.

For any positive integer $n$ ,

D_{n+1}(x_{0},\ldots,x_{2^{n+1}-1})=D_{n}(x_{0}+x_{2^{n}},\ldots,x_{2^{n}-1}+x_{2^{n+1}-1})D_{n}(x_{0}-x_{2^{n}},\ldots,x_{2^{n}-1}-x_{2^{n+1}-1}).

Throughout this paper, we assume that $a_{0},a_{1},\ldots,a_{15}\in\mathbb{Z}$ and, for any $0\leq i\leq 3$ , put

	$\displaystyle b_{i}$	$\displaystyle:=(a_{i}+a_{i+8})+(a_{i+4}+a_{i+12}),$	$\displaystyle c_{i}$	$\displaystyle:=(a_{i}+a_{i+8})-(a_{i+4}+a_{i+12}),$
	$\displaystyle d_{i}$	$\displaystyle:=(a_{i}-a_{i+8})+(a_{i+4}-a_{i+12}),$	$\displaystyle e_{i}$	$\displaystyle:=(a_{i}-a_{i+8})-(a_{i+4}-a_{i+12}).$

The following relations will be frequently used in this paper:

	$\displaystyle D_{4}(a_{0},a_{1},\ldots,a_{15})$	$\displaystyle=D_{3}(a_{0}+a_{8},\ldots,a_{7}+a_{15})D_{3}(a_{0}-a_{8},\ldots,a_{7}-a_{15}),$
		$\displaystyle=D_{2}(b_{0},b_{1},b_{2},b_{3})D_{2}(c_{0},c_{1},c_{2},c_{3})D_{2}(d_{0},d_{1},d_{2},d_{3})D_{2}(e_{0},e_{1},e_{2},e_{3}).$

Remark 2.2.

The following hold:

$(1)$

$b_{i}\equiv c_{i}\equiv d_{i}\equiv e_{i}\pmod{2}\>\>\text{for}\>\>0\leq i\leq 3$ ;
$(2)$

$b_{i}+c_{i}+d_{i}+e_{i}\equiv 0\pmod{4}\>\>\text{for}\>\>0\leq i\leq 3$ .

Noting that $\det{(\alpha_{st})_{s,t}}\equiv\det{(\beta_{st})_{s,t}}\pmod{2}$ holds if $\alpha_{st}\equiv\beta_{st}\pmod{2}$ for any $s$ and $t$ , we have the following lemma.

Lemma 2.3.

The following hold:

$(1)$

$D_{4}(a_{0},a_{1},\ldots,a_{15})\equiv D_{3}(a_{0}+a_{8},\ldots,a_{7}+a_{15})\equiv D_{3}(a_{0}-a_{8},\ldots,a_{7}-a_{15})\pmod{2}$ ;
$(2)$

$D_{4}(a_{0},a_{1},\ldots,a_{15})\equiv D_{2}(b_{0},b_{1},b_{2},b_{3})\equiv D_{2}(c_{0},c_{1},c_{2},c_{3})$
$\equiv D_{2}(d_{0},d_{1},d_{2},d_{3})\equiv D_{2}(e_{0},e_{1},e_{2},e_{3})\pmod{2}$ .

Lemma 2.4.

We have $D_{4}(a_{0},a_{1},\ldots,a_{15})\in 2\mathbb{Z}\iff b_{0}+b_{2}\equiv b_{1}+b_{3}\pmod{2}$ .

Proof.

From Lemma 2.3 (2) and

	$\displaystyle D_{2}(b_{0},b_{1},b_{2},b_{3})$	$\displaystyle=D_{1}(b_{0}+b_{2},b_{1}+b_{3})D_{1}(b_{0}-b_{2},b_{1}-b_{3})$
		$\displaystyle=\left\{(b_{0}+b_{2})^{2}-(b_{1}+b_{3})^{2}\right\}\left\{(b_{0}-b_{2})^{2}-(b_{1}-b_{3})^{2}\right\}$
		$\displaystyle\in\begin{cases}2\mathbb{Z},&b_{0}+b_{2}\equiv b_{1}+b_{3}\pmod{2},\\ \mathbb{Z}_{\rm odd},&b_{0}+b_{2}\not\equiv b_{1}+b_{3}\pmod{2},\end{cases}$

the lemma is proved. ∎

3. Group determinant of ${\rm C}_{2}^{2}$

In this section, we prove five lemmas which give properties of the group determinant of ${\rm C}_{2}^{2}$ . These lemmas are used in the next section.

By direct calculation, we have the following lemma.

Lemma 3.1.

The identity $D_{2}(x_{0},x_{1},x_{2},x_{3})=\sum_{i=0}^{3}x_{i}^{4}-2\sum_{0\leq i<j\leq 3}x_{i}^{2}x_{j}^{2}+8x_{0}x_{1}x_{2}x_{3}$ holds. Therefore, $D_{2}(x_{0},x_{1},x_{2},x_{3})$ is a symmetric polynomial in $x_{0},x_{1},x_{2},x_{3}$ .

Lemma 3.2.

For any $k,l,m,n\in\mathbb{Z}$ , the following hold:

$(1)$

$D_{2}(2k,2l,2m,2n+1)\equiv 8(k+l+m)+1\pmod{16}$ ;
$(2)$

$D_{2}(2k,2l+1,2m+1,2n+1)\equiv 8k-3\pmod{16}$ .

Proof.

We have

	$\displaystyle D_{2}(2k,2l,2m,2n+1)$
	$\displaystyle\hskip 14.22636pt=D_{1}(2k+2m,2l+2n+1)D_{1}(2k-2m,2l-2n-1)$
	$\displaystyle\hskip 14.22636pt=\left\{4(k+m)^{2}-4(l+n)^{2}-4(l+n)-1\right\}\left\{4(k-m)^{2}-4(l-n)^{2}+4(l-n)-1\right\}$
	$\displaystyle\hskip 14.22636pt\equiv-4(k+m)^{2}+4(l+n)^{2}+4(l+n)-4(k-m)^{2}+4(l-n)^{2}-4(l-n)+1$
	$\displaystyle\hskip 14.22636pt\equiv-8k^{2}-8m^{2}+8l^{2}+8n(n+1)+1$
	$\displaystyle\hskip 14.22636pt\equiv 8(k+l+m)+1\pmod{16},$
	$\displaystyle D_{2}(2k,2l+1,2m+1,2n+1)$
	$\displaystyle\hskip 14.22636pt=D_{1}(2k+2m+1,2l+2n+2)D_{1}(2k-2m-1,2l-2n)$
	$\displaystyle\hskip 14.22636pt=\left\{4(k+m)^{2}+4(k+m)+1-4(l+n+1)^{2}\right\}\left\{4(k-m)^{2}-4(k-m)+1-4(l-n)^{2}\right\}$
	$\displaystyle\hskip 14.22636pt\equiv 4(k+m)^{2}+4(k+m)-4(l+n+1)^{2}+4(k-m)^{2}-4(k-m)-4(l-n)^{2}+1$
	$\displaystyle\hskip 14.22636pt\equiv 8k^{2}+8m(m+1)-8l(l+1)-8n(n+1)-3$
	$\displaystyle\hskip 14.22636pt\equiv 8k-3\pmod{16}.$

∎

Lemma 3.3.

For any $k,l,m,n\in\mathbb{Z}$ , let $\alpha:=D_{2}(2k,2l,2m,2n)$ . The following hold:

$(1)$

If $k+m\not\equiv l+n,\>km\equiv ln\pmod{2}$ , then $\alpha\in\left\{2^{4}(8a+1)\mid a\in\mathbb{Z}\right\}$ ;
$(2)$

If $k+m\not\equiv l+n,\>km\not\equiv ln\pmod{2}$ , then $\alpha\in\left\{2^{4}(8a-3)\mid a\in\mathbb{Z}\right\}$ ;
$(3)$

If $k,l,m,n$ are even and $k+m\not\equiv l+n\pmod{4}$ , then $\alpha\in\left\{2^{8}(4a+1)\mid a\in\mathbb{Z}\right\}$ ;
$(4)$

If $k,l,m,n$ are odd and $k+m\not\equiv l+n\pmod{4}$ , then $\alpha\in\left\{2^{8}(4a-1)\mid a\in\mathbb{Z}\right\}$ ;
$(5)$

If $k\equiv l\equiv m\equiv n\pmod{2}$ and $k+m\equiv l+n\pmod{4}$ , then $\alpha\in 2^{12}\mathbb{Z}$ ;
$(6)$

If exactly two of $k,l,m,n$ are even numbers, then $\alpha\in 2^{10}\mathbb{Z}$ .

Proof.

First, we prove (1) and (2). We have $(k+m)^{2}-(l+n)^{2}=(k-m)^{2}-(l-n)^{2}+4(km-ln)$ . Therefore, if $k+m\not\equiv l+n\pmod{2}$ , then

	$\displaystyle 2^{-4}\alpha$	$\displaystyle=\left\{(k+m)^{2}-(l+n)^{2}\right\}\left\{(k-m)^{2}-(l-n)^{2}\right\}$
		$\displaystyle\equiv\begin{cases}1\pmod{8},&km\equiv ln\pmod{2},\\ -3\pmod{8},&km\not\equiv ln\pmod{2}.\end{cases}$

Second, we prove (3). If $k,l,m,n$ are even and $k+m\not\equiv l+n\pmod{4}$ , then there exist $k^{\prime},l^{\prime},m^{\prime},n^{\prime}\in\mathbb{Z}$ satisfying $k=2k^{\prime}$ , $l=2l^{\prime}$ , $m=2m^{\prime}$ , $n=2n^{\prime}$ and $k^{\prime}+m^{\prime}\not\equiv l^{\prime}+n^{\prime}\pmod{2}$ . Therefore,

2^{-8}\alpha=\left\{(k^{\prime}+m^{\prime})^{2}-(l^{\prime}+n^{\prime})^{2}\right\}\left\{(k^{\prime}-m^{\prime})^{2}-(l^{\prime}-n^{\prime})^{2}\right\}\equiv 1\pmod{4}.

Third, we prove (4). If $k,l,m,n$ are odd and $k+m\not\equiv l+n\pmod{4}$ , then there exist $k^{\prime},l^{\prime},m^{\prime},n^{\prime}\in\mathbb{Z}$ satisfying $k=2k^{\prime}+1$ , $l=2l^{\prime}+1$ , $m=2m^{\prime}+1$ , $n=2n^{\prime}+1$ and $k^{\prime}+m^{\prime}\not\equiv l^{\prime}+n^{\prime}\pmod{2}$ . Therefore,

\displaystyle 2^{-8}\alpha=\left\{(k^{\prime}+m^{\prime}+1)^{2}-(l^{\prime}+n^{\prime}+1)^{2}\right\}\left\{(k^{\prime}-m^{\prime})^{2}-(l^{\prime}-n^{\prime})^{2}\right\}\equiv-1\pmod{4}.

Fourth, we prove (5). If $k\equiv l\equiv m\equiv n\pmod{2}$ and $k+m\equiv l+n\pmod{4}$ , then $(k+m)^{2}-(l+n)^{2}\equiv(k-m)^{2}-(l-n)^{2}\equiv 0\pmod{16}$ . Therefore,

\displaystyle\alpha=2^{4}\left\{(k+m)^{2}-(l+n)^{2}\right\}\left\{(k-m)^{2}-(l-n)^{2}\right\}\in 2^{12}\mathbb{Z}.

Finally, we prove (6). If $(k,l,m,n)\equiv(0,0,1,1)\pmod{2}$ , then $(k+m)^{2}-(l+n)^{2}\equiv(k-m)^{2}-(l-n)^{2}\equiv 0\pmod{8}$ . Therefore,

\displaystyle\alpha=2^{4}\left\{(k+m)^{2}-(l+n)^{2}\right\}\left\{(k-m)^{2}-(l-n)^{2}\right\}\in 2^{10}\mathbb{Z}.

Lemma 3.1 completes the proof of (6). ∎

Lemma 3.4.

For any $k,l,m,n\in\mathbb{Z}$ , let $\alpha:=D_{2}(2k,2l+1,2m,2n+1)$ . The following hold:

$(1)$

If $k\equiv m\equiv 0\pmod{2}$ and (I) hold, then $\alpha\in\left\{2^{6}(8a-1)(4b-1)\mid a,b\in\mathbb{Z}\right\}$ ;
$(2)$

If $k\equiv m\equiv 0\pmod{2}$ and (II) hold, then $\alpha\in\left\{2^{6}(8a+3)(4b+1)\mid a,b\in\mathbb{Z}\right\}$ ;
$(3)$

If $k\equiv m\equiv 1\pmod{2}$ and (I) hold, then $\alpha\in\left\{2^{6}(8a+3)(4b-1)\mid a,b\in\mathbb{Z}\right\}$ ;
$(4)$

If $k\equiv m\equiv 1\pmod{2}$ and (II) hold, then $\alpha\in\left\{2^{6}(8a-1)(4b+1)\mid a,b\in\mathbb{Z}\right\}$ ,

where

(I)

$k+m\equiv 1-l-n\equiv 0\pmod{4}$ or $k-m\equiv 2-l+n\equiv 0\pmod{4}$ ;
(II)

$k+m\equiv 1-l-n\equiv 2\pmod{4}$ or $k-m\equiv 2-l+n\equiv 2\pmod{4}$ .

Proof.

Note that

	$\displaystyle\alpha$	$\displaystyle=D_{1}(2k+2m,2l+2n+2)D_{1}(2k-2m,2l-2n)$
		$\displaystyle=2^{4}\left\{(k+m)^{2}-(l+n+1)^{2}\right\}\left\{(k-m)^{2}-(l-n)^{2}\right\}.$

If $k\equiv m\equiv 0\pmod{2}$ , then

	$\displaystyle(k+m)^{2}-(l+n+1)^{2}\equiv\begin{cases}-4\pmod{16},&\text{if}\>\>k+m\equiv 1-l-n\equiv 0\pmod{4},\\ 4\pmod{16},&\text{if}\>\>k+m\equiv 1-l-n\equiv 2\pmod{4},\\ -1\pmod{8},&\text{if}\>\>k-m\equiv 2-l+n\equiv 0\pmod{4},\\ 3\pmod{8},&\text{if}\>\>k-m\equiv 2-l+n\equiv 2\pmod{4},\end{cases}$
	$\displaystyle(k-m)^{2}-(l-n)^{2}\equiv\begin{cases}-1\pmod{8},&\text{if}\>\>k+m\equiv 1-l-n\equiv 0\pmod{4},\\ 3\pmod{8},&\text{if}\>\>k+m\equiv 1-l-n\equiv 2\pmod{4},\\ -4\pmod{16},&\text{if}\>\>k-m\equiv 2-l+n\equiv 0\pmod{4},\\ 4\pmod{16},&\text{if}\>\>k-m\equiv 2-l+n\equiv 2\pmod{4}.\end{cases}$

If $k\equiv m\equiv 1\pmod{2}$ , then

	$\displaystyle(k+m)^{2}-(l+n+1)^{2}\equiv\begin{cases}-4\pmod{16},&\text{if}\>\>k+m\equiv 1-l-n\equiv 0\pmod{4},\\ 4\pmod{16},&\text{if}\>\>k+m\equiv 1-l-n\equiv 2\pmod{4},\\ 3\pmod{8},&\text{if}\>\>k-m\equiv 2-l+n\equiv 0\pmod{4},\\ -1\pmod{8},&\text{if}\>\>k-m\equiv 2-l+n\equiv 2\pmod{4},\end{cases}$
	$\displaystyle(k-m)^{2}-(l-n)^{2}\equiv\begin{cases}3\pmod{8},&\text{if}\>\>k+m\equiv 1-l-n\equiv 0\pmod{4},\\ -1\pmod{8},&\text{if}\>\>k+m\equiv 1-l-n\equiv 2\pmod{4},\\ -4\pmod{16},&\text{if}\>\>k-m\equiv 2-l+n\equiv 0\pmod{4},\\ 4\pmod{16},&\text{if}\>\>k-m\equiv 2-l+n\equiv 2\pmod{4}.\end{cases}$

From the above, the lemma is proved. ∎

Note that for any integers $a,b$ and $c$ , where $b$ is odd, it holds that

ab\equiv c\pmod{8}\iff a\equiv bc\pmod{8}

since $b^{2}\equiv 1\pmod{8}$ . From this, we have the following.

Remark 3.5.

For any $k,l,m,n\in\mathbb{Z}$ , the following hold:

$(1)$

$(2k+2l+1)(2m+2n+1)\equiv 1\pmod{8}\iff k-m\equiv-l+n\pmod{4}$ ;
$(2)$

$(2k+2l+1)(2m+2n+1)\equiv-1\pmod{8}\iff k+m\equiv-l-n-1\pmod{4}$ ;
$(3)$

$(2k+2l+1)(2m+2n+1)\equiv 3\pmod{8}\iff k+m\equiv 1-l-n\pmod{4}$ ;
$(4)$

$(2k+2l+1)(2m+2n+1)\equiv-3\pmod{8}\iff k-m\equiv 2-l+n\pmod{4}$ .

Lemma 3.6.

For any $k,l,m,n\in\mathbb{Z}$ , the following hold:

$(1)$

$D_{2}(2k,2l,2m,2n)\in\begin{cases}2^{4}\mathbb{Z}_{\rm odd},&k+m\not\equiv l+n\pmod{2},\\ 2^{8}\mathbb{Z}_{\rm odd}\cup 2^{10}\mathbb{Z},&k+m\equiv l+n\pmod{2};\end{cases}$
$(2)$

$D_{2}(2k+1,2l+1,2m+1,2n+1)\in\begin{cases}2^{4}\mathbb{Z}_{\rm odd},&k+m\not\equiv l+n\pmod{2},\\ 2^{9}\mathbb{Z},&k+m\equiv l+n\pmod{2};\end{cases}$
$(3)$

$D_{2}(2k,2l+1,2m,2n+1)$
$\in\begin{cases}2^{7}\mathbb{Z},&k\not\equiv m\pmod{2},\\ 2^{8}\mathbb{Z},&k\equiv m\pmod{2}\>\>\text{and}\>\>(2k+2l+1)(2m+2n+1)\equiv\pm 1\pmod{8},\\ 2^{6}\mathbb{Z}_{\rm odd},&k\equiv m\pmod{2}\>\>\text{and}\>\>(2k+2l+1)(2m+2n+1)\equiv\pm 3\pmod{8}.\end{cases}$

Proof.

From Lemma 3.3, it follows that (1) holds. Also, we obtain (2) from

	$\displaystyle D_{2}(2k+1,2l+1,2m+1,2n+1)$	$\displaystyle=D_{1}(2k+2m+2,2l+2n+2)D_{1}(2k-2m,2l-2n)$
		$\displaystyle=2^{4}\left\{(k+m+1)^{2}-(l+n+1)^{2}\right\}\left\{(k-m)^{2}-(l-n)^{2}\right\}.$

Below, we prove (3). Let $\alpha:=D_{2}(2k,2l+1,2m,2n+1)$ . Note that

	$\displaystyle\alpha$	$\displaystyle=D_{1}(2k+2m,2l+2n+2)D_{1}(2k-2m,2l-2n)$
		$\displaystyle=2^{4}\left\{(k+m)^{2}-(l+n+1)^{2}\right\}\left\{(k-m)^{2}-(l-n)^{2}\right\}.$

If $k\not\equiv m\pmod{2}$ , then $\alpha\in 2^{7}\mathbb{Z}$ . If $k\equiv m\pmod{2}$ and $(2k+2l+1)(2m+2n+1)\equiv 1\pmod{8}$ , then $(k-m)^{2}-(l-n)^{2}\in 2^{4}\mathbb{Z}$ from Remark 3.5 $(1)$ . Thus, $\alpha\in 2^{8}\mathbb{Z}$ . If $k\equiv m\pmod{2}$ and $(2k+2l+1)(2m+2n+1)\equiv-1\pmod{8}$ , then $(k+m)^{2}-(l+n+1)^{2}\in 2^{4}\mathbb{Z}$ from Remark 3.5 $(2)$ . Thus, $\alpha\in 2^{8}\mathbb{Z}$ . The cases of $k\equiv m\pmod{2}$ and $(2k+2l+1)(2m+2n+1)\equiv\pm 3\pmod{8}$ are proved from Lemma 3.4 and Remark 3.5 $(3)$ and $(4)$ . ∎

4. Impossible values

In this section, we consider impossible values.

Lemma 4.1.

We have $S\left({\rm C}_{2}^{4}\right)\cap\mathbb{Z}_{\rm odd}\subset\left\{16m+1\mid m\in\mathbb{Z}\right\}$ .

Proof.

Let $D_{4}(a_{0},\ldots,a_{15})=D_{2}(b_{0},b_{1},b_{2},b_{3})D_{2}(c_{0},c_{1},c_{2},c_{3})D_{2}(d_{0},d_{1},d_{2},d_{3})D_{2}(e_{0},e_{1},e_{2},e_{3})$ be an odd number. Then, $b_{0}+b_{2}\not\equiv b_{1}+b_{3}\pmod{2}$ holds from Lemma 2.4. We divide the proof into the following cases:

(i)

Exactly three of $b_{0},b_{1},b_{2},b_{3}$ are even;
(ii)

Exactly one of $b_{0},b_{1},b_{2},b_{3}$ is even.

First, we consider the case of (i). If $(b_{0},b_{1},b_{2},b_{3})\equiv(0,0,0,1)\pmod{2}$ , then there exist $k_{i},l_{i},m_{i}\in\mathbb{Z}$ satisfying $(b_{0},b_{1},b_{2})=(2k_{0},2l_{0},2m_{0})$ , $(c_{0},c_{1},c_{2})=(2k_{1},2l_{1},2m_{1})$ , $(d_{0},d_{1},d_{2})=(2k_{2},2l_{2},2m_{2})$ , $(e_{0},e_{1},e_{2})=(2k_{3},2l_{3},2m_{3})$ and $\sum_{i=0}^{3}k_{i}\equiv\sum_{i=0}^{3}l_{i}\equiv\sum_{i=0}^{3}m_{i}\equiv 0\pmod{2}$ from Remark 2.2. Therefore, from Lemma 3.2 $(1)$ , we have

\displaystyle D_{4}(a_{0},a_{1},\ldots,a_{15})

\displaystyle\equiv\prod_{i=0}^{3}\left\{8(k_{i}+l_{i}+m_{i})+1\right\}\equiv 8\sum_{i=0}^{3}(k_{i}+l_{i}+m_{i})+1\equiv 1\pmod{16}.

Lemma 3.1 completes the proof for the case (i). Next, we consider the case of (ii). If $(b_{0},b_{1},b_{2},b_{3})\equiv(0,1,1,1)\pmod{2}$ , then there exist $k^{\prime}_{i}\in\mathbb{Z}$ satisfying $b_{0}=2k^{\prime}_{0}$ , $c_{0}=2k^{\prime}_{1}$ , $d_{0}=2k^{\prime}_{2}$ , $e_{0}=2k^{\prime}_{3}$ and $\sum_{i=0}^{3}k^{\prime}_{i}\equiv 0\pmod{2}$ from Remark 2.2. Therefore, from Lemma 3.2 $(2)$ , we have

\displaystyle D_{4}(a_{0},a_{1},\ldots,a_{15})\equiv\prod_{i=0}^{3}(8k^{\prime}_{i}-3)\equiv 8\sum_{i=0}^{3}k^{\prime}_{i}+1\equiv 1\pmod{16}.

Lemma 3.1 completes the proof for the case (ii). ∎

Lemma 4.2.

We have $S\left({\rm C}_{2}^{4}\right)\cap 2\mathbb{Z}\subset 2^{16}\mathbb{Z}_{\rm odd}\cup 2^{24}\mathbb{Z}_{\rm odd}\cup 2^{26}\mathbb{Z}$ .

Proof.

Let $D_{4}(a_{0},\ldots,a_{15})=D_{2}(b_{0},b_{1},b_{2},b_{3})D_{2}(c_{0},c_{1},c_{2},c_{3})D_{2}(d_{0},d_{1},d_{2},d_{3})D_{2}(e_{0},e_{1},e_{2},e_{3})$ be an even number. Then, $b_{0}+b_{2}\equiv b_{1}+b_{3}\pmod{2}$ holds from Lemma 2.4. We prove the following:

(i)

If $b_{0},b_{1},b_{2},b_{3}$ are even, then $D_{4}(a_{0},\ldots,a_{15})\in 2^{16}\mathbb{Z}_{\rm odd}\cup 2^{24}\mathbb{Z}_{\rm odd}\cup 2^{26}\mathbb{Z}$ ;
(ii)

If $b_{0},b_{1},b_{2},b_{3}$ are odd, then $D_{4}(a_{0},\ldots,a_{15})\in 2^{16}\mathbb{Z}_{\rm odd}\cup 2^{26}\mathbb{Z}$ ;
(iii)

If exactly two of $b_{0},b_{1},b_{2},b_{3}$ are even, then $D_{4}(a_{0},\ldots,a_{15})\in 2^{24}\mathbb{Z}_{\rm odd}\cup 2^{26}\mathbb{Z}$ .

First, we prove (i). If $b_{0},b_{1},b_{2},b_{3}$ are even, then there exist $k_{i},l_{i},m_{i},n_{i}\in\mathbb{Z}$ satisfying

	$\displaystyle(b_{0},b_{1},b_{2},b_{3})$	$\displaystyle=(2k_{0},2l_{0},2m_{0},2n_{0}),$	$\displaystyle(c_{0},c_{1},c_{2},c_{3})=(2k_{1},2l_{1},2m_{1},2n_{1}),$
	$\displaystyle(d_{0},d_{1},d_{2},d_{3})$	$\displaystyle=(2k_{2},2l_{2},2m_{2},2n_{2}),$	$\displaystyle(e_{0},e_{1},e_{2},e_{3})=(2k_{3},2l_{3},2m_{3},2n_{3})$

and $\sum_{i=0}^{3}k_{i}\equiv\sum_{i=0}^{3}l_{i}\equiv\sum_{i=0}^{3}m_{i}\equiv\sum_{i=0}^{3}n_{i}\equiv 0\pmod{2}$ from Remark 2.2. Let $N$ denote the cardinal number of the set $\left\{i\mid 0\leq i\leq 3,\>\>k_{i}+m_{i}\equiv l_{i}+n_{i}\pmod{2}\right\}$ . Then, $N\in\{0,2,4\}$ holds from $\sum_{i=0}^{3}(k_{i}+m_{i}-l_{i}-n_{i})\equiv 0\pmod{2}$ . Therefore, from Lemma 3.6 $(1)$ , we obtain (i). Also, in the same way, we can prove (ii) by using Lemma 3.6 (2). Finally, we prove (iii). If $(b_{0},b_{1},b_{2},b_{3})\equiv(0,1,0,1)\pmod{2}$ , then there exist $k^{\prime}_{i},m^{\prime}_{i}\in\mathbb{Z}$ satisfying $(b_{0},b_{2})=(2k^{\prime}_{0},2m^{\prime}_{0})$ , $(c_{0},c_{2})=(2k^{\prime}_{1},2m^{\prime}_{1})$ , $(d_{0},d_{2})=(2k^{\prime}_{2},2m^{\prime}_{2})$ , $(e_{0},e_{2})=(2k^{\prime}_{3},2m^{\prime}_{3})$ and $\sum_{i=0}^{3}k^{\prime}_{i}\equiv\sum_{i=0}^{3}m^{\prime}_{i}\equiv 0\pmod{2}$ from Remark 2.2. Let $N^{\prime}$ denote the cardinal number of the set $\left\{i\mid 0\leq i\leq 3,\>\>k^{\prime}_{i}\equiv m^{\prime}_{i}\pmod{2}\right\}$ . Then, $N^{\prime}\in\{0,2,4\}$ holds from $\sum_{i=0}^{3}(k^{\prime}_{i}-m^{\prime}_{i})\equiv 0\pmod{2}$ . Therefore, from Lemma 3.6 $(3)$ , we have $D_{4}(a_{0},\ldots,a_{15})\in 2^{24}\mathbb{Z}_{\rm odd}\cup 2^{26}\mathbb{Z}$ . Lemma 3.1 completes the proof of (iii). ∎

Lemma 4.3.

We have $S\left({\rm C}_{2}^{4}\right)\cap 2^{16}\mathbb{Z}_{\rm odd}\subset\left\{2^{16}(4m+1)\mid m\in\mathbb{Z}\right\}$ .

Proof.

Note that we have $S\left({\rm C}_{2}^{3}\right)\cap 2\mathbb{Z}=\left\{2^{8}(4m+1),2^{12}m\mid m\in\mathbb{Z}\right\}$ from the description in the introduction. Let $D_{4}(a_{0},\ldots,a_{15})=D_{3}(a_{0}+a_{8},\ldots,a_{7}+a_{15})D_{3}(a_{0}-a_{8},\ldots,a_{7}-a_{15})\in 2^{16}\mathbb{Z}_{\rm odd}$ . Then, from Lemma 2.3 $(1)$ , there exist $k,l\in\mathbb{Z}$ satisfying $D_{3}(a_{0}+a_{8},\ldots,a_{7}+a_{15})=2^{8}(4k+1)$ and $D_{3}(a_{0}-a_{8},\ldots,a_{7}-a_{15})=2^{8}(4l+1)$ . Therefore, we have $D_{4}(a_{0},\ldots,a_{15})\in\left\{2^{16}(4m+1)\mid m\in\mathbb{Z}\right\}$ . ∎

Let $A:=\left\{(8k-3)(8l+3)\mid k,l\in\mathbb{Z}\right\}$ .

Lemma 4.4.

If $m\equiv-1\pmod{8}$ and $m\not\in A$ , then $2^{24}m\not\in S\left({\rm C}_{2}^{4}\right)$ .

To prove Lemma 4.4, we use the following two lemmas.

Lemma 4.5.

If $m\equiv-1\pmod{8}$ and $m\not\in A$ , then $D_{4}(a_{0},a_{1},\ldots,a_{15})\neq 2^{24}m$ for any $a_{0},a_{1},\ldots,a_{15}\in\mathbb{Z}$ with $b_{0}\equiv b_{1}\equiv b_{2}\equiv b_{3}\equiv 0\pmod{2}$ .

Proof.

We prove by contradiction. Assume that there exist $a_{0},a_{1},\ldots,a_{15}\in\mathbb{Z}$ with $b_{0}\equiv b_{1}\equiv b_{2}\equiv b_{3}\equiv 0\pmod{2}$ satisfying $D_{4}(a_{0},a_{1},\ldots,a_{15})=2^{24}m$ . Then there exist $k_{i},l_{i},m_{i},n_{i}\in\mathbb{Z}$ satisfying

	$\displaystyle D_{2}^{(0)}$	$\displaystyle:=D_{2}(2k_{0},2l_{0},2m_{0},2n_{0})=D_{2}(b_{0},b_{1},b_{2},b_{3}),$
	$\displaystyle D_{2}^{(1)}$	$\displaystyle:=D_{2}(2k_{1},2l_{1},2m_{1},2n_{1})=D_{2}(c_{0},c_{1},c_{2},c_{3}),$
	$\displaystyle D_{2}^{(2)}$	$\displaystyle:=D_{2}(2k_{2},2l_{2},2m_{2},2n_{2})=D_{2}(d_{0},d_{1},d_{2},d_{3}),$
	$\displaystyle D_{2}^{(3)}$	$\displaystyle:=D_{2}(2k_{3},2l_{3},2m_{3},2n_{3})=D_{2}(e_{0},e_{1},e_{2},e_{3})$

and $\sum_{i=0}^{3}k_{i}\equiv\sum_{i=0}^{3}l_{i}\equiv\sum_{i=0}^{3}m_{i}\equiv\sum_{i=0}^{3}n_{i}\equiv 0\pmod{2}$ from Remark 2.2. Note that $\sum_{i=0}^{3}(k_{i}+m_{i}-l_{i}-n_{i})\equiv 0\pmod{2}$ . Since $D_{2}^{(0)}D_{2}^{(1)}D_{2}^{(2)}D_{2}^{(3)}\in 2^{24}\mathbb{Z}_{\rm odd}$ , one of the following holds: (i) Three of $D_{2}^{(i)}$ are of type (1) or (2) in Lemma 3.3 and the other one is of type (5) in Lemma 3.3; (ii) Two of $D_{2}^{(i)}$ are of type (1) or (2) in Lemma 3.3 and the others are of type (3) or (4) in Lemma 3.3. In the case of (i), we have $\sum_{i=0}^{3}(k_{i}+m_{i}-l_{i}-n_{i})\not\equiv 0\pmod{2}$ . This is a contradiction. In the case of (ii), since $m\equiv-1\pmod{8}$ and $m\not\in A$ , we have

\displaystyle D_{2}^{(\alpha)},D_{2}^{(\beta)}\in\left\{2^{4}(8a+1)\mid a\in\mathbb{Z}\right\},D_{2}^{(\gamma)}\in\left\{2^{8}(4a+1)\mid a\in\mathbb{Z}\right\},D_{2}^{(\delta)}\in\left\{2^{8}(4a-1)\mid a\in\mathbb{Z}\right\},

where $\left\{\alpha,\beta,\gamma,\delta\right\}=\left\{0,1,2,3\right\}$ . This implies that

	$\displaystyle(k_{\alpha},l_{\alpha},m_{\alpha},n_{\alpha})$	$\displaystyle\equiv(0,0,0,1),\>(0,0,1,0),\>(0,1,0,0)\>\text{or}\>(1,0,0,0)\pmod{2},$
	$\displaystyle(k_{\beta},l_{\beta},m_{\beta},n_{\beta})$	$\displaystyle\equiv(0,0,0,1),\>(0,0,1,0),\>(0,1,0,0)\>\text{or}\>(1,0,0,0)\pmod{2},$
	$\displaystyle(k_{\gamma},l_{\gamma},m_{\gamma},n_{\gamma})$	$\displaystyle\equiv(0,0,0,0)\pmod{2},$
	$\displaystyle(k_{\delta},l_{\delta},m_{\delta},n_{\delta})$	$\displaystyle\equiv(1,1,1,1)\pmod{2}.$

Therefore, at least two of $\sum_{i=0}^{3}k_{i}$ , $\sum_{i=0}^{3}l_{i}$ , $\sum_{i=0}^{3}m_{i}$ , $\sum_{i=0}^{3}n_{i}$ are odd. This is a contradiction. ∎

Lemma 4.6.

If $m\equiv-1\pmod{8}$ and $m\not\in A$ , then $D_{4}(a_{0},a_{1},\ldots,a_{15})\neq 2^{24}m$ for any $a_{0},a_{1},\ldots,a_{15}\in\mathbb{Z}$ , where exactly two of $b_{0},b_{1},b_{2},b_{3}$ are even.

Proof.

We prove by contradiction. Assume that there exist $a_{0},a_{1},\ldots,a_{15}\in\mathbb{Z}$ , where exactly two of $b_{0},b_{1},b_{2},b_{3}$ are even, satisfying $D_{4}(a_{0},a_{1},\ldots,a_{15})=2^{24}m$ . If $(b_{0},b_{1},b_{2},b_{3})\equiv(0,1,0,1)\pmod{2}$ , then there exist $k_{i},l_{i},m_{i},n_{i}\in\mathbb{Z}$ satisfying

	$\displaystyle D_{2}^{(0)}$	$\displaystyle:=D_{2}(2k_{0},2l_{0}+1,2m_{0},2n_{0}+1)=D_{2}(b_{0},b_{1},b_{2},b_{3}),$
	$\displaystyle D_{2}^{(1)}$	$\displaystyle:=D_{2}(2k_{1},2l_{1}+1,2m_{1},2n_{1}+1)=D_{2}(c_{0},c_{1},c_{2},c_{3}),$
	$\displaystyle D_{2}^{(2)}$	$\displaystyle:=D_{2}(2k_{2},2l_{2}+1,2m_{2},2n_{2}+1)=D_{2}(d_{0},d_{1},d_{2},d_{3}),$
	$\displaystyle D_{2}^{(3)}$	$\displaystyle:=D_{2}(2k_{3},2l_{3}+1,2m_{3},2n_{3}+1)=D_{2}(e_{0},e_{1},e_{2},e_{3})$

and $\sum_{i=0}^{3}k_{i}\equiv\sum_{i=0}^{3}l_{i}\equiv\sum_{i=0}^{3}m_{i}\equiv\sum_{i=0}^{3}n_{i}\equiv 0\pmod{2}$ from Remark 2.2. Since $D_{2}^{(0)}D_{2}^{(1)}D_{2}^{(2)}D_{2}^{(3)}\in 2^{24}\mathbb{Z}_{\rm odd}$ , it follows from Lemma 3.6 $(3)$ that $k_{i}\equiv m_{i}\pmod{2}$ and $(2k_{i}+2l_{i}+1)(2m_{i}+2n_{i}+1)\equiv\pm 3\pmod{8}$ for any $0\leq i\leq 3$ . Then, from Remark 3.5 $(3)$ and $(4)$ , we find that every $D_{2}^{(i)}$ is of type $(1)$ or $(4)$ in Lemma 3.4 since $m\not\in A$ . In addition, since $m\equiv-1\pmod{8}$ , exactly one or three of $D_{2}^{(i)}$ are of type $(1)$ in Lemma 3.4. Then we have $\sum_{i=0}^{3}k_{i}\equiv 1\pmod{2}$ . This is a contradiction. Lemma 3.1 completes the proof. ∎

Proof of Lemma 4.4.

Let $m\equiv-1\pmod{8}$ and $m\not\in A$ . From Lemma 2.4, it is sufficient to prove that $D_{4}(a_{0},\ldots,a_{15})\neq 2^{24}m$ for any $a_{0},\ldots,a_{15}\in\mathbb{Z}$ with $b_{0}+b_{2}\equiv b_{1}+b_{3}\pmod{2}$ . If $b_{0},b_{1},b_{2},b_{3}$ are odd, then as mentioned in the proof of Lemma 4.2, it holds that $D_{4}(a_{0},\ldots,a_{15})\in 2^{16}\mathbb{Z}_{\rm odd}\cup 2^{26}\mathbb{Z}$ . Thus, $D_{4}(a_{0},\ldots,a_{15})\neq 2^{24}m$ . If all or exactly two of $b_{0},b_{1},b_{2},b_{3}$ are even, then we have $D_{4}(a_{0},\ldots,a_{15})\neq 2^{24}m$ from Lemmas 4.5 and 4.6. ∎

5. Possible values

In this section, we determine all possible values. Lemmas 4.1–4.4 imply that $S\left({\rm C}_{2}^{4}\right)$ does not include every integer that is not mentioned in the following lemma.

Lemma 5.1.

For any $m,n\in\mathbb{Z}$ , the following hold:

$(1)$

$16m+1\in S\left({\rm C}_{2}^{4}\right)$ ;
$(2)$

$2^{16}(4m+1)\in S\left({\rm C}_{2}^{4}\right)$ ;
$(3)$

$2^{24}(4m+1)\in S\left({\rm C}_{2}^{4}\right)$ ;
$(4)$

$2^{24}(4m+1)(8n+3)\in S\left({\rm C}_{2}^{4}\right)$ ;
$(5)$

$2^{26}m\in S\left({\rm C}_{2}^{4}\right)$ .

Proof.

We obtain (1) from

	$\displaystyle D_{4}(m+1,m,m,\ldots,m)$	$\displaystyle=D_{3}(2m+1,2m,2m,\ldots,2m)D_{3}(1,0,0,\ldots,0)$
		$\displaystyle=D_{2}(4m+1,4m,4m,4m)D_{2}(1,0,0,0)^{3}$
		$\displaystyle=D_{1}(8m+1,8m)D_{1}(1,0)^{7}$
		$\displaystyle=(8m+1)^{2}-(8m)^{2}$
		$\displaystyle=16m+1.$

We obtain (2) from

	$\displaystyle D_{4}(k+2,k,k,\ldots,k)$	$\displaystyle=D_{3}(2k+2,2k,2k,\ldots,2k)D_{3}(2,0,0,\ldots,0)$
		$\displaystyle=D_{2}(4k+2,4k,4k,4k)D_{2}(2,0,0,0)^{3}$
		$\displaystyle=D_{1}(8k+2,8k)D_{1}(2,0)^{7}$
		$\displaystyle=\left\{(8k+2)^{2}-(8k)^{2}\right\}\cdot 4^{7}$
		$\displaystyle=2^{16}(8k+1)$

and

	$\displaystyle D_{4}(1-k,k,1-k,k,-k,k,1-k,k-1,1-k,k,-k,k-1,-k,k,-k,k)$
	$\displaystyle\quad=D_{3}(2-2k,2k,1-2k,2k-1,-2k,2k,1-2k,2k-1)D_{3}(0,0,1,1,0,0,1,-1)$
	$\displaystyle\quad=D_{2}(2-4k,4k,2-4k,4k-2)D_{2}(2,0,0,0)D_{2}(0,0,2,0)D_{2}(0,0,0,2)$
	$\displaystyle\quad=D_{1}(4-8k,8k-2)D_{1}(0,2)^{2}D_{1}(2,0)^{3}D_{1}(-2,0)D_{1}(0,-2)$
	$\displaystyle\quad=\left\{(4-8k)^{2}-(8k-2)^{2}\right\}(-4)^{3}\cdot 4^{4}$
	$\displaystyle\quad=2^{16}(8k-3).$

We obtain (3) from

	$\displaystyle D_{4}(m+3,m+1,m,m,\ldots,m)$	$\displaystyle=D_{3}(2m+3,2m+1,2m,2m,\ldots,2m)D_{3}(3,1,0,0,\ldots,0)$
		$\displaystyle=D_{2}(4m+3,4m+1,4m,4m)D_{2}(3,1,0,0)^{3}$
		$\displaystyle=D_{1}(8m+3,8m+1)D_{1}(3,1)^{7}$
		$\displaystyle=\left\{(8m+3)^{2}-(8m+1)^{2}\right\}\cdot 8^{7}$
		$\displaystyle=2^{24}(4m+1).$

We obtain (4) by substituting

	$\displaystyle a_{0}=m-n+2,\>\>a_{1}=-(m+n+1),\>\>a_{2}=m+n+1,\>\>a_{3}=n-m,$
	$\displaystyle a_{4}=m-n-1,\>\>a_{5}=-(m+n),\>\>a_{6}=m+n+1,\>\>a_{7}=n-m,$
	$\displaystyle a_{8}=m-n-1,\>\>a_{9}=-(m+n),\>\>a_{10}=m+n+1,\>\>a_{11}=n-m,$
	$\displaystyle a_{12}=m-n-1,\>\>a_{13}=-(m+n),\>\>a_{14}=m+n+1,\>\>a_{15}=n-m.$

In fact,

	$\displaystyle D_{4}(a_{0},\ldots,a_{15})$	$\displaystyle=D_{2}(b_{0},b_{1},b_{2},b_{3})D_{2}(c_{0},c_{1},c_{2},c_{3})D_{2}(d_{0},d_{1},d_{2},d_{3})D_{2}(e_{0},e_{1},e_{2},e_{3})$
		$\displaystyle=D_{2}(4m-4n-1,-4m-4n-1,4m+4n+4,4n-4m)D_{2}(3,-1,0,0)^{3}$
		$\displaystyle=D_{1}(8m+3,-8m-1)D_{1}(-8n-5,-8n-1)D_{1}(3,-1)^{6}$
		$\displaystyle=\left\{(8m+3)^{2}-(8m+1)^{2}\right\}\left\{(8n+5)^{2}-(8n+1)^{2}\right\}\cdot 8^{6}$
		$\displaystyle=2^{24}(4m+1)(8n+3).$

We obtain (5) from

	$\displaystyle D_{4}(k+1,1-k,k+1,-k,k-1,-k,k+1,-k,k,-k,k+1,-k,k-2,1-k,k+1,-k)$
	$\displaystyle\quad=D_{3}(2k+1,1-2k,2k+2,-2k,2k-3,1-2k,2k+2,-2k)D_{3}(1,1,0,0,1,-1,0,0)$
	$\displaystyle\quad=D_{2}(4k-2,2-4k,4k+4,-4k)D_{2}(4,0,0,0)D_{2}(2,0,0,0)D_{2}(0,2,0,0)$
	$\displaystyle\quad=D_{1}(8k+2,2-8k)D_{1}(-6,2)D_{1}(4,0)^{2}D_{1}(2,0)^{2}D_{1}(0,2)^{2}$
	$\displaystyle\quad=\left\{(8k+2)^{2}-(2-8k)^{2}\right\}\cdot 32\cdot 16^{2}\cdot 4^{2}\cdot(-4)^{2}$
	$\displaystyle\quad=2^{26}(2k)$

and

	$\displaystyle D_{4}(k-1,k+1,k+1,k+2,k+1,k+2,k+2,k,k,k,k,k,k,k,k,k)$
	$\displaystyle\quad=D_{3}(2k-1,2k+1,2k+1,2k+2,2k+1,2k+2,2k+2,2k)D_{3}(-1,1,1,2,1,2,2,0)$
	$\displaystyle\quad=D_{2}(4k,4k+3,4k+3,4k+2)D_{2}(-2,-1,-1,2)^{2}D_{2}(0,3,3,2)$
	$\displaystyle\quad=D_{1}(8k+3,8k+5)D_{1}(-3,1)^{4}D_{1}(-1,-3)^{2}D_{1}(3,5)$
	$\displaystyle\quad=\left\{(8k+3)^{2}-(8k+5)^{2}\right\}\cdot 8^{4}\cdot(-8)^{2}\cdot(-16)$
	$\displaystyle\quad=2^{26}(2k+1).$

∎

Remark that Lemma 5.1 $(4)$ implies that $2^{24}(8m+3),2^{24}m^{\prime}\in S\left({\rm C}_{2}^{4}\right)$ for any $m\in\mathbb{Z}$ and $m^{\prime}\in A$ . From Lemmas 4.1–4.4 and 5.1, Theorem 1.1 is proved.

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Yuka Yamaguchi

Faculty of Education

University of Miyazaki

1-1 Gakuen Kibanadai-nishi

Miyazaki 889-2192

JAPAN

[email protected]

Naoya Yamaguchi

Faculty of Education

University of Miyazaki

1-1 Gakuen Kibanadai-nishi

Miyazaki 889-2192

JAPAN

[email protected]

Integer group determinants for C24{\rm C}_{2}^{4}

Abstract.

Key words and phrases:

1991 Mathematics Subject Classification:

1. Introduction

Theorem 1.1.

2. Relations with group determinants of subgroups

Corollary 2.1.

Remark 2.2.

Lemma 2.3.

Lemma 2.4.

Proof.

3. Group determinant of C22{\rm C}_{2}^{2}

Lemma 3.1.

Lemma 3.2.

Proof.

Lemma 3.3.

Proof.

Lemma 3.4.

Proof.

Remark 3.5.

Lemma 3.6.

Proof.

4. Impossible values

Lemma 4.1.

Proof.

Lemma 4.2.

Proof.

Lemma 4.3.

Proof.

Lemma 4.4.

Lemma 4.5.

Proof.

Lemma 4.6.

Proof.

Proof of Lemma 4.4.

5. Possible values

Lemma 5.1.

Proof.

References

Integer group determinants for ${\rm C}_{2}^{4}$

3. Group determinant of ${\rm C}_{2}^{2}$