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Generalized Dedekind’s theorem and its application to integer group determinants

Naoya Yamaguchi and Yuka Yamaguchi
Abstract

In this paper, we give a refinement of a generalized Dedekind’s theorem. In addition, we show that all possible values of integer group determinants of any group are also possible values of integer group determinants of its any abelian subgroup. By applying the refinement of a generalized Dedekind’s theorem, we determine all possible values of integer group determinants of the direct product group of the cyclic group of order 88 and the cyclic group of order 22.

1 Introduction

For a finite group GG, let xgx_{g} be an indeterminate for each gGg\in G and let [xg]\mathbb{Z}[x_{g}] be the multivariate polynomial ring in xgx_{g} over \mathbb{Z}. The group matrix MG(xg)M_{G}(x_{g}) and the group determinant ΘG(xg)\Theta_{G}(x_{g}) of GG were defined by Dedekind as follows:

MG(xg):=(xgh1)g,hG,ΘG(xg):=detMG(xg)[xg].M_{G}(x_{g}):=\left(x_{gh^{-1}}\right)_{g,h\in G},\quad\Theta_{G}(x_{g}):=\det{M_{G}(x_{g})}\in\mathbb{Z}[x_{g}].

When the elements of GG are reordered arbitrarily, the group matrix MM formed according to this reordering is of the form M=P1MG(xg)PM=P^{-1}M_{G}(x_{g})P, where PP is an appropriate permutation matrix. Thus, ΘG(xg)\Theta_{G}(x_{g}) is invariant under any reordering of the elements of GG. For a finite group GG, let G^\widehat{G} be a complete set of representatives of the equivalence classes of irreducible representations of GG over \mathbb{C}. Around 1880, for the case that GG is abelian, Dedekind gave the irreducible factorization of ΘG(xg)\Theta_{G}(x_{g}) over \mathbb{C}: Let GG be a finite abelian group. Then

ΘG(xg)=χG^gGχ(g)xg.\Theta_{G}(x_{g})=\prod_{\chi\in\widehat{G}}\sum_{g\in G}\chi(g)x_{g}.

This is called Dedekind’s theorem. In 1896, Frobenius [3] gave the irreducible factorization of ΘG(xg)\Theta_{G}(x_{g}) over \mathbb{C} for any finite group: Let GG be a finite group. Then

ΘG(xg)=φG^det(gGφ(g)xg)degφ.\Theta_{G}(x_{g})=\prod_{\varphi\in\widehat{G}}\det{\left(\sum_{g\in G}\varphi(g)x_{g}\right)^{\deg{\varphi}}}.

This is the most well known generalization of Dedekind’s theorem. This generalization is obtained from the decomposition of the regular representation LL of GG as a direct sum of irreducible representations and the expression MG(xg)=gGxgL(g)M_{G}(x_{g})=\sum_{g\in G}x_{g}L(g). Frobenius created the character theory of finite groups in the process of obtaining the irreducible factorization. For the history on the theory, see, e.g., [2, 4, 5, 6, 20]. On the other hand, another generalization of Dedekind’s theorem was given in [21]: Let GG be a finite abelian group and let HH be a subgroup of GG. For every hHh\in H, there exists a homogeneous polynomial Ah[xg]A_{h}\in\mathbb{C}[x_{g}] satisfying degAh=|G/H|\deg{A_{h}}=\left|G/H\right| and

ΘG(xg)=χH^hHχ(h)Ah=ΘH(Ah).\displaystyle\Theta_{G}(x_{g})=\prod_{\chi\in\widehat{H}}\sum_{h\in H}\chi(h)A_{h}=\Theta_{H}(A_{h}). (1)

If H=GH=G, then we can take Ah=xhA_{h}=x_{h} for each hHh\in H. This generalization shows that the group determinant of an abelian group can be written by the group determinant of any subgroup. Let Cn={0¯,1¯,,n1¯}{\rm C}_{n}=\left\{\overline{0},\overline{1},\ldots,\overline{n-1}\right\} be the cyclic group of order nn. The matrix MCn(xg)M_{{\rm C}_{n}}(x_{g}) is similar to the circulant matrix of order nn. That is, the circulant determinant is a special case of the group determinant. For the circulant determinant, Laquer [12, Theorem 2] gave the following factorization in 1980: Let n=rsn=rs, where rr and ss are relatively prime, and let xj:=xj1¯x_{j}:=x_{\overline{j-1}} for any 1jn1\leq j\leq n. Then

ΘCn(xj)=l=0s1ΘCr(yjl),yjl:=k=0s1ζsl(kr+j1)xkr+j,\displaystyle\Theta_{{\rm C}_{n}}{\left(x_{j}\right)}=\prod_{l=0}^{s-1}\Theta_{{\rm C}_{r}}{\left(y_{j}^{l}\right)},\quad y_{j}^{l}:=\sum_{k=0}^{s-1}\zeta_{s}^{l(kr+j-1)}x_{kr+j},

where ζs\zeta_{s} is a primitive ss-th root of unity. We call this theorem Laquer’s theorem. In recently, Laquer’s theorem was generalized as follows [24, Theorem 1.1]: Let G=H×KG=H\times K be a direct product of finite abelian groups. Then we have

ΘG(xg)=χK^ΘH(yhχ),yhχ=kKχ(k)xhk.\displaystyle\Theta_{G}(x_{g})=\prod_{\chi\in\widehat{K}}\Theta_{H}(y_{h}^{\chi}),\quad y_{h}^{\chi}=\sum_{k\in K}\chi(k)x_{hk}. (2)

In this paper, we give a refinement of (1), which is a generalization of (2).

Theorem 1.1.

Let GG be a finite abelian group, let HH be a subgroup of GG and let

G^H:={χG^χ(h)=1,hH},G=tTtH,G^=χXχG^H.\displaystyle\widehat{G}_{H}:=\left\{\chi\in\widehat{G}\mid\chi(h)=1,\>h\in H\right\},\quad G=\displaystyle\bigsqcup_{t\in T}tH,\quad\widehat{G}=\displaystyle\bigsqcup_{\chi\in X}\chi\widehat{G}_{H}.

Then we have

ΘG(xg)=χXΘG/H(ytHχ)=ΘH(zh),\displaystyle\Theta_{G}(x_{g})=\prod_{\chi\in X}\Theta_{G/H}{\left(y_{tH}^{\chi}\right)}=\Theta_{H}(z_{h}),

where

ytHχ:=hHχ(th)xth,zh:=1|H|χXχ(h1)ΘG/H(ytHχ).y_{tH}^{\chi}:=\sum_{h\in H}\chi(th)x_{th},\quad z_{h}:=\frac{1}{|H|}\sum_{\chi\in X}\chi(h^{-1})\Theta_{G/H}{\left(y_{tH}^{\chi}\right)}.

For any f(xg)[xg]f(x_{g})\in\mathbb{Z}[x_{g}], we denote by f(xg)hf(x_{g})_{h} the sum of all monomials cxg1xg2xgkcx_{g_{1}}x_{g_{2}}\cdots x_{g_{k}} in f(xg)f(x_{g}) satisfying g1g2gk=hg_{1}g_{2}\cdots g_{k}=h. The following theorem gives another expression for zhz_{h} in Theorem 1.1. When calculating zhz_{h}, the expression for zhz_{h} in the following theorem might be more useful than one in Theorem 1.1 (see Example 2.3).

Theorem 1.2.

Let ytH:=hHxthy_{tH}:=\sum_{h\in H}x_{th}. Then we have

zh=ΘG/H(ytH)h[xg].z_{h}=\Theta_{G/H}{\left(y_{tH}\right)_{h}}\in\mathbb{Z}[x_{g}].

Theorem 1.1 is a refinement of (1) since {χ|HχX}=H^\left\{\chi|_{H}\mid\chi\in X\right\}=\widehat{H} holds and we can take Ah=zhA_{h}=z_{h} in (1). Note that for a finite abelian group GG and any subgroup KK, there exists a subgroup HH of GG satisfying KG/HK\cong G/H. That is, Theorem 1.1 implies that ΘG(xg)\Theta_{G}(x_{g}) can also be expressed as a product of the group determinants of any subgroup. Thus, Theorem 1.1 derives (2). We apply Theorems 1.1 and 1.2 to the study of the integer group determinant.

A group determinant called an integer group determinant when its variables are integers. For a finite group GG, let

S(G):={ΘG(xg)|xg}.\displaystyle S(G):=\left\{\Theta_{G}(x_{g})\>|\>x_{g}\in\mathbb{Z}\right\}.

It immediately follows from MG(xg)=gGxgL(g)M_{G}(x_{g})=\sum_{g\in G}x_{g}L(g) that S(G)S(G) is a monoid. Determining S(G)S(G) is an open problem. For the G=CnG={\rm C}_{n} cases, determining S(G)S(G) is called Olga Taussky-Todd’s circulant problem since Olga Taussky-Todd suggested it at the meeting of the American Mathematical Society in Hayward, California [14]. Even Olga Taussky-Todd’s circulant problem remains as an open problem.

For S(Cn)S({\rm C}_{n}), the following relation is known [10, Lemma 3.6]: Let n,q1n,q\geq 1. If qnq\mid n, then

S(Cn)S(Cq).\displaystyle S({\rm C}_{n})\subset S({\rm C}_{q}). (3)

From Theorems 1.1 and 1.2, we obtain a generalization of (3).

Corollary 1.3.

Let GG be a finite abelian group and let HH be a subgroup of GG. Then

S(G)S(H).S(G)\subset S(H).

Corollary 1.3 is generalized as follows.

Theorem 1.4.

Let GG be a finite group and let HH be an abelian subgroup of GG. Then

{α[G:H]αS(H)}S(G)S(H),\left\{\alpha^{[G:H]}\mid\alpha\in S(H)\right\}\subset S(G)\subset S(H),

where [G:H][G:H] is the index of HH in GG.

For some types of groups, the problem was solved in [1, 10, 12, 13, 14, 15, 18, 19, 25, 28]. As a result, for every group GG of order at most 1515, S(G)S(G) is determined (see [15, 19]). For the groups of order 1616, the complete descriptions of S(G)S(G) were obtained for D16{\rm D}_{16} [1, Theorem 5.3], C16{\rm C}_{16} [25] and C24{\rm C}_{2}^{4} [28], where Dn{\rm D}_{n} denotes the dihedral group of order nn.

Laquer [12] determined S(C2p)S({\rm C}_{2p}), where pp is an odd prime, by using Laquer’s theorem which provides an expression for the integer circulant determinant of C2p{\rm C}_{2p} as a product of two integer circulant determinants of Cp{\rm C}_{p}. In [28], S(C24)S({\rm C}_{2}^{4}) is determined by using (2) which provides an expression for the integer group determinant of C2n{\rm C}_{2}^{n} as a product of two integer group determinants of C2n1{\rm C}_{2}^{n-1}. We can generalize these approaches by using Theorem 1.1 to determine S(G)S(G) for any abelian groups. There are fourteen groups of order 1616 up to isomorphism [7, 30], and five of them are abelian. The unsolved abelian groups of order 1616 are C8×C2{\rm C}_{8}\times{\rm C}_{2}, C42{\rm C}_{4}^{2} and C4×C22{\rm C}_{4}\times{\rm C}_{2}^{2}. By applying Theorem 1.1, we determine S(C8×C2)S({\rm C}_{8}\times{\rm C}_{2}).

Theorem 1.5.

Let A:={(8k3)(8l3)k,l,kl(mod2)}{16m7m}A:=\left\{(8k-3)(8l-3)\mid k,l\in\mathbb{Z},\>k\equiv l\pmod{2}\right\}\subsetneq\left\{16m-7\mid m\in\mathbb{Z}\right\}. Then we have

S(C8×C2)\displaystyle S\left({\rm C}_{8}\times{\rm C}_{2}\right) ={16m+1,m, 210(2m+1), 212mm,mA}\displaystyle=\left\{16m+1,\>m^{\prime},\>2^{10}(2m+1),\>2^{12}m\mid m\in\mathbb{Z},\>m^{\prime}\in A\right\}
{211p(2m+1)p=a2+b21,a+b±3(mod 8),m}\displaystyle\quad\cup\left\{2^{11}p(2m+1)\mid p=a^{2}+b^{2}\equiv 1,\>a+b\equiv\pm 3\>\>({\rm mod}\>{8}),\>m\in\mathbb{Z}\right\}
{211p(2m+1)p3(mod 8),m}\displaystyle\quad\cup\left\{2^{11}p(2m+1)\mid p\equiv-3\>\>({\rm mod}\>{8}),\>m\in\mathbb{Z}\right\}
{211p2(2m+1)p3(mod 8),m},\displaystyle\quad\cup\left\{2^{11}p^{2}(2m+1)\mid p\equiv 3\>\>({\rm mod}\>{8}),\>m\in\mathbb{Z}\right\},

where pp denotes a prime.

The remaining two abelian groups could also be solved by using Theorem 1.1. (While this paper under review, it have been solved in [26, 29]. Also, as for non-abelian groups of order 16, D8×C2{\rm D}_{8}\times{\rm C}_{2}, Q8×C2{\rm Q}_{8}\times{\rm C}_{2} [16, Theorems 3.1 and 4.1], Q16{\rm Q}_{16} [17] and C22C4{\rm C}_{2}^{2}\rtimes{\rm C}_{4} [27] have been solved, where Qn{\rm Q}_{n} denotes the generalized quaternion group of order nn.)

Pinner and Smyth [19, p.427] noted the following inclusion relations for all groups of order 88:

S(C23)S(C4×C2)S(Q8)S(D8)S(C8).S({\rm C}_{2}^{3})\subsetneq S({\rm C}_{4}\times{\rm C}_{2})\subsetneq S({\rm Q}_{8})\subsetneq S({\rm D}_{8})\subsetneq S({\rm C}_{8}).

From preceding results and Theorem 1.5, we have

S(C24)S(C8×C2)S(D16)S(C16).S({\rm C}_{2}^{4})\subsetneq S({\rm C}_{8}\times{\rm C}_{2})\subsetneq S({\rm D}_{16})\subsetneq S({\rm C}_{16}).

Determining the integer group determinants aims to investigate the structure of a group by means of the group determinant. It is expected that individual new results will help us understand more about groups.

This paper is organized as follows. In Section 2, we prove Theorems 1.1 and 1.2. In Sections 3 and 4, we prove Theorems 1.4 and 1.5, respectively.

2 Proofs of Theorems 1.1 and 1.2

For a finite group GG, let xgx_{g} be an indeterminate for each gGg\in G, let [xg]\mathbb{C}[x_{g}] be the multivariate polynomial ring in xgx_{g} over \mathbb{C}, let G\mathbb{C}G the group algebra of GG over \mathbb{C}, and let [xg]G:=[xg]G={gGAggAg[xg]}\mathbb{C}[x_{g}]G:=\mathbb{C}[x_{g}]\otimes\mathbb{C}G=\left\{\sum_{g\in G}A_{g}g\mid A_{g}\in\mathbb{C}[x_{g}]\right\} be the group algebra of GG over [xg]\mathbb{C}[x_{g}]. Also, for a finite abelian group GG and a subgroup HH of GG, let

G^H:={χG^|χ(h)=1,hH}.\widehat{G}_{H}:=\left\{\chi\in\widehat{G}\>|\>\chi(h)=1,\>h\in H\right\}.

It is easily verified that G^H={φπ|φG/H^}\widehat{G}_{H}=\left\{\varphi\circ\pi\>|\>\varphi\in\widehat{G/H}\right\}, where π:GG/H\pi:G\rightarrow G/H is the canonical homomorphism. To prove Theorem 1.1, we use the following lemma.

Lemma 2.1 ([21, Lemma 3.6]).

Let GG be a finite abelian group and HH be a subgroup of GG. For every hHh\in H, there exists a homogeneous polynomial Ah[xg]A_{h}\in\mathbb{C}[x_{g}] satisfying degAh=|G/H|\deg{A_{h}}=\left|G/H\right| and

χG^HgGχ(g)xgg=hHAhh[xg]H.\prod_{\chi\in\widehat{G}_{H}}\sum_{g\in G}\chi(g)x_{g}g=\sum_{h\in H}A_{h}h\in\mathbb{C}[x_{g}]H.

If H=GH=G, then we can take Ah=xhA_{h}=x_{h} for each hHh\in H.

Proof of Theorem 1.1.

From Dedekind’s theorem, we have

ΘG(xg)\displaystyle\Theta_{G}(x_{g}) =χG^gGχ(g)xg\displaystyle=\prod_{\chi\in\widehat{G}}\sum_{g\in G}\chi(g)x_{g}
=χXχG^HtThH(χχ)(th)xth\displaystyle=\prod_{\chi\in X}\prod_{\chi^{\prime}\in\widehat{G}_{H}}\sum_{t\in T}\sum_{h\in H}\left(\chi\chi^{\prime}\right)(th)x_{th}
=χXχG^HtTχ(t)hHχ(th)xth\displaystyle=\prod_{\chi\in X}\prod_{\chi^{\prime}\in\widehat{G}_{H}}\sum_{t\in T}\chi^{\prime}(t)\sum_{h\in H}\chi(th)x_{th}
=χXχG/H^tHG/Hχ(tH)ytHχ\displaystyle=\prod_{\chi\in X}\prod_{\chi^{\prime}\in\widehat{G/H}}\sum_{tH\in G/H}\chi^{\prime}(tH)y_{tH}^{\chi}
=χXΘG/H(ytHχ).\displaystyle=\prod_{\chi\in X}\Theta_{G/H}{\left(y_{tH}^{\chi}\right)}.

Next, we show that for any χX\chi\in X, there exists Ah[xg]A_{h}\in\mathbb{C}[x_{g}] satisfying

ΘG/H(ytHχ)=hHχ(h)Ah.\Theta_{G/H}{\left(y_{tH}^{\chi}\right)}=\sum_{h\in H}\chi(h)A_{h}.

For any χX\chi\in X, let Fχ:[xg]G[xg]GF_{\chi}\colon\mathbb{C}[x_{g}]G\to\mathbb{C}[x_{g}]G be the [xg]\mathbb{C}[x_{g}]-algebra homomorphism defined by Fχ(g)=χ(g)gF_{\chi}(g)=\chi(g)g. Then, from Lemma 2.1, there exists Ah[xg]A_{h}\in\mathbb{C}[x_{g}] satisfying

hHχ(h)Ahh\displaystyle\sum_{h\in H}\chi(h)A_{h}h =Fχ(hHAhh)\displaystyle=F_{\chi}\left(\sum_{h\in H}A_{h}h\right)
=Fχ(χG^HgGχ(g)xgg)\displaystyle=F_{\chi}\left(\prod_{\chi^{\prime}\in\widehat{G}_{H}}\sum_{g\in G}\chi^{\prime}(g)x_{g}g\right)
=Fχ(χG^HtThHχ(t)xthth)\displaystyle=F_{\chi}\left(\prod_{\chi^{\prime}\in\widehat{G}_{H}}\sum_{t\in T}\sum_{h\in H}\chi^{\prime}(t)x_{th}th\right)
=χG^HtThHχ(t)χ(th)xthth.\displaystyle=\prod_{\chi^{\prime}\in\widehat{G}_{H}}\sum_{t\in T}\sum_{h\in H}\chi^{\prime}(t)\chi(th)x_{th}th.

Let F:[xg]G[xg]F\colon\mathbb{C}[x_{g}]G\to\mathbb{C}[x_{g}] be the [xg]\mathbb{C}[x_{g}]-algebra homomorphism defined by F(g)=1F(g)=1. Applying FF to the both sides of the above, we have

hHχ(h)Ah=χG^HtThHχ(t)χ(th)xth=χG/H^tHG/Hχ(tH)ytHχ=ΘG/H(ytHχ).\sum_{h\in H}\chi(h)A_{h}=\prod_{\chi^{\prime}\in\widehat{G}_{H}}\sum_{t\in T}\sum_{h\in H}\chi^{\prime}(t)\chi(th)x_{th}=\prod_{\chi^{\prime}\in\widehat{G/H}}\sum_{tH\in G/H}\chi^{\prime}(tH)y_{tH}^{\chi}=\Theta_{G/H}{\left(y_{tH}^{\chi}\right)}.

From the above, it follows that there exists Ah[xg]A_{h}\in\mathbb{C}[x_{g}] satisfying

ΘG(xg)=χXΘG/H(ytHχ)=χXhHχ(h)Ah=χH^hHχ(h)Ah=ΘH(Ah).\displaystyle\Theta_{G}(x_{g})=\prod_{\chi\in X}\Theta_{G/H}{\left(y_{tH}^{\chi}\right)}=\prod_{\chi\in X}\sum_{h\in H}\chi(h)A_{h}=\prod_{\chi\in\widehat{H}}\sum_{h\in H}\chi(h)A_{h}=\Theta_{H}(A_{h}). (4)

Finally, we show that AhA_{h} in (4) is expressed as

Ah=1|H|χXχ(h1)ΘG/H(ytHχ)A_{h}=\frac{1}{|H|}\sum_{\chi\in X}\chi(h^{-1})\Theta_{G/H}{\left(y_{tH}^{\chi}\right)}

for any hHh\in H. From orthogonality relations for characters, for any hHh\in H, we have

χXχ(h1)ΘG/H(ytHχ)\displaystyle\sum_{\chi\in X}\chi(h^{-1})\Theta_{G/H}{\left(y_{tH}^{\chi}\right)} =χXχ(h1)hHχ(h)Ah\displaystyle=\sum_{\chi\in X}\chi(h^{-1})\sum_{h^{\prime}\in H}\chi(h^{\prime})A_{h^{\prime}}
=χXhHχ(h1h)Ah\displaystyle=\sum_{\chi\in X}\sum_{h^{\prime}\in H}\chi(h^{-1}h^{\prime})A_{h^{\prime}}
=hHχH^χ(h1h)Ah\displaystyle=\sum_{h^{\prime}\in H}\sum_{\chi\in\widehat{H}}\chi(h^{-1}h^{\prime})A_{h^{\prime}}
=|H|Ah.\displaystyle=|H|A_{h}.

Remark 2.2.

From the proof of Theorem 1.1, AhA_{h} in Lemma 2.1 equals to zhz_{h} in Theorem 1.1.

Proof of Theorem 1.2.

From Lemma 2.1 and Remark 2.2, we have

hHzhh=χG^HgGχ(g)xgg=χG/H^gGχ(gH)xgg=χG/H^tHG/HhHχ(tH)xthth.\displaystyle\sum_{h\in H}z_{h}h=\prod_{\chi\in\widehat{G}_{H}}\sum_{g\in G}\chi(g)x_{g}g=\prod_{\chi\in\widehat{G/H}}\sum_{g\in G}\chi(gH)x_{g}g=\prod_{\chi\in\widehat{G/H}}\sum_{tH\in G/H}\sum_{h^{\prime}\in H}\chi(tH)x_{th^{\prime}}th^{\prime}.

Therefore, we have

zh=(χG/H^tHG/Hχ(tH)hHxth)h=ΘG/H(ytH)h[xg].\displaystyle z_{h}=\left(\prod_{\chi\in\widehat{G/H}}\sum_{tH\in G/H}\chi(tH)\sum_{h^{\prime}\in H}x_{th^{\prime}}\right)_{h}=\Theta_{G/H}{\left(y_{tH}\right)}_{h}\in\mathbb{Z}[x_{g}].

From Theorems 1.1 and 1.2, we can take zh[xg]z_{h}\in\mathbb{Z}[x_{g}] satisfying ΘG(xg)=ΘH(zh)\Theta_{G}(x_{g})=\Theta_{H}(z_{h}). Thus, Corollary 1.3 is obtained.

Example 2.3.

Using Theorems 1.1 and 1.2, we calculate ΘC4(xg)\Theta_{{\rm C}_{4}}(x_{g}). Let G=C4G={\rm C}_{4} and H={0¯,2¯}H=\{\overline{0},\overline{2}\}. Then, G/H={0¯H,1¯H}G/H=\left\{\overline{0}H,\overline{1}H\right\}. We write xi¯x_{\overline{i}} as xix_{i} for any 0i30\leq i\leq 3. From Theorem 1.2, we have

z0¯=ΘG/H(ytH)0¯=x02+x222x1x3,z2¯=ΘG/H(ytH)2¯=2x0x2x12x32\displaystyle z_{\overline{0}}=\Theta_{G/H}(y_{tH})_{\overline{0}}=x_{0}^{2}+x_{2}^{2}-2x_{1}x_{3},\quad z_{\overline{2}}=\Theta_{G/H}(y_{tH})_{\overline{2}}=2x_{0}x_{2}-x_{1}^{2}-x_{3}^{2}

since y0¯H=x0+x2y_{\overline{0}H}=x_{0}+x_{2}, y1¯H=x1+x3y_{\overline{1}H}=x_{1}+x_{3} and

ΘG/H(ytH)=y0¯H2y1¯H2=(x02+2x0x2+x22)(x12+2x1x3+x32).\displaystyle\Theta_{G/H}(y_{tH})=y_{\overline{0}H}^{2}-y_{\overline{1}H}^{2}=\left(x_{0}^{2}+2x_{0}x_{2}+x_{2}^{2}\right)-\left(x_{1}^{2}+2x_{1}x_{3}+x_{3}^{2}\right).

Therefore, from Theorem 1.1, we have

ΘG(xg)=ΘH(zh)=z0¯2z2¯2=(x02+x222x1x3)2(2x0x2x12x32)2.\displaystyle\Theta_{G}(x_{g})=\Theta_{H}(z_{h})=z_{\overline{0}}^{2}-z_{\overline{2}}^{2}=\left(x_{0}^{2}+x_{2}^{2}-2x_{1}x_{3}\right)^{2}-\left(2x_{0}x_{2}-x_{1}^{2}-x_{3}^{2}\right)^{2}.

3 Proof of Theorem 1.4

The lower bound in Theorem 1.4 is derived from [23, Lemma 3.2]. Also, the upper bound immediately follows from the following lemma essentially provided in [22, Theorem 1.4].

Lemma 3.1.

Let GG be a finite group and let HH be an abelian subgroup of GG. Then, there exists a homogeneous polynomial Ah[xg]A_{h}\in\mathbb{Z}[x_{g}] satisfying degAh=[G:H]\deg{A_{h}}=\left[G:H\right] and

ΘG(xg)=ΘH(Ah),\Theta_{G}(x_{g})=\Theta_{H}(A_{h}),

where [G:H][G:H] is the index of HH in GG.

The proof in [22] is not concise. We give a brief proof of Lemma 3.1. For the purpose, we use the following identity [9, p. 82, Theorem 2.6]; see also [8, 11]: Given the block matrix MM of the form

(M11M12M1nM21M22M2nMn1Mn2Mnn),\begin{pmatrix}M_{11}&M_{12}&\cdots&M_{1n}\\ M_{21}&M_{22}&\cdots&M_{2n}\\ \vdots&\vdots&\ddots&\vdots\\ M_{n1}&M_{n2}&\cdots&M_{nn}\end{pmatrix},

where the matrices MijM_{ij} are pairwise commuting of size m×mm\times m then

detM=det(σSnsgn(σ)M1σ(1)M2σ(2)Mnσ(n)).\det{M}=\det{\left(\sum_{\sigma\in S_{n}}\operatorname{sgn}(\sigma)M_{1\sigma(1)}M_{2\sigma(2)}\cdots M_{n\sigma(n)}\right)}.
Proof of Lemma 3.1.

Let G={g1,g2,,gmn}G=\{g_{1},g_{2},\ldots,g_{mn}\}, let H={h1,h2,,hm}H=\{h_{1},h_{2},\ldots,h_{m}\} and let G=t1Ht2HtnHG=t_{1}H\sqcup t_{2}H\sqcup\cdots\sqcup t_{n}H, where gi=tkhlGg_{i}=t_{k}h_{l}\in G with i=(k1)m+li=(k-1)m+l for 1kn1\leq k\leq n and 1lm1\leq l\leq m. Then, the group matrix (xgigj1)1i,jmn\left(x_{g_{i}g_{j}^{-1}}\right)_{1\leq i,j\leq mn} of GG can be expressed as the block matrix:

(xgigj1)1i,jmn=(Mkl)1k,ln,\left(x_{g_{i}g_{j}^{-1}}\right)_{1\leq i,j\leq mn}=\left(M_{kl}\right)_{1\leq k,l\leq n},

where MklM_{kl} is the matrix obtained by replacing each xhihj1x_{h_{i}h_{j}^{-1}} in the group matrix (xhihj1)1i,jm\left(x_{h_{i}h_{j}^{-1}}\right)_{1\leq i,j\leq m} of HH to x(tkhi)(tlhj)1x_{(t_{k}h_{i})(t_{l}h_{j})^{-1}}. That is, Mkl=(x(tkhi)(tlhj)1)1i,jmM_{kl}=\left(x_{(t_{k}h_{i})(t_{l}h_{j})^{-1}}\right)_{1\leq i,j\leq m}. Since HH is abelian, MklM_{kl} are pairwise commuting. Therefore, there exists Ah[xg]A_{h}\in\mathbb{Z}[x_{g}] satisfying

ΘG(xg)=det(xgigj1)1i,jmn=det(σSnsgn(σ)M1σ(1)M2σ(2)Mnσ(n))=ΘH(Ah)\Theta_{G}(x_{g})=\det{\left(x_{g_{i}g_{j}^{-1}}\right)_{1\leq i,j\leq mn}}=\det{\left(\sum_{\sigma\in S_{n}}\operatorname{sgn}(\sigma)M_{1\sigma(1)}M_{2\sigma(2)}\cdots M_{n\sigma(n)}\right)}=\Theta_{H}(A_{h})

since σSnsgn(σ)M1σ(1)M2σ(2)Mnσ(n)\sum_{\sigma\in S_{n}}\operatorname{sgn}(\sigma)M_{1\sigma(1)}M_{2\sigma(2)}\cdots M_{n\sigma(n)} is also of the form of a group matrix of HH. ∎

4 Proof of Theorem 1.5

In this section, by applying Theorem 1.1, we determine S(C8×C2)S({\rm C}_{8}\times{\rm C}_{2}).

4.1 Relations with group determinants of subgroups

We denote the variable xi¯x_{\overline{i}} by xix_{i} for any i¯Cn\overline{i}\in{\rm C}_{n} and let Dn(x0,x1,,xn1):=ΘCn(xg)D_{n}(x_{0},x_{1},\ldots,x_{n-1}):=\Theta_{{\rm C}_{n}}(x_{g}). Also, for any g=(r¯,s¯)C8×C2g=(\overline{r},\overline{s})\in{\rm C}_{8}\times{\rm C}_{2} with r{0,1,,7}r\in\{0,1,\ldots,7\} and s{0,1}s\in\{0,1\}, we denote the variable ygy_{g} by yjy_{j}, where j:=r+8sj:=r+8s, and let D8×2(y0,y1,,y15):=ΘC8×C2(yg)D_{8\times 2}(y_{0},y_{1},\ldots,y_{15}):=\Theta_{{\rm C}_{8}\times{\rm C}_{2}}(y_{g}). From the G=C8×C2G={\rm C}_{8}\times{\rm C}_{2} and H={(0¯,0¯),(0¯,1¯)}H=\left\{(\overline{0},\overline{0}),(\overline{0},\overline{1})\right\} case of Theorem 1.1, we have

D8×2(y0,,y15)=D8(y0+y8,,y7+y15)D8(y0y8,,y7y15).\displaystyle D_{8\times 2}(y_{0},\ldots,y_{15})=D_{8}(y_{0}+y_{8},\ldots,y_{7}+y_{15})D_{8}(y_{0}-y_{8},\ldots,y_{7}-y_{15}).

Let ζn\zeta_{n} be a primitive nn-th root of unity. From the G=C8G={\rm C}_{8} and H={0¯,4¯}H=\left\{\overline{0},\overline{4}\right\} case of Theorem 1.1, we have

D8(x0,x1,,x7)\displaystyle D_{8}(x_{0},x_{1},\ldots,x_{7}) =D4(x0+x4,x1+x5,x2+x6,x3+x7)\displaystyle=D_{4}(x_{0}+x_{4},x_{1}+x_{5},x_{2}+x_{6},x_{3}+x_{7})
×D4(x0x4,ζ8(x1x5),ζ82(x2x6),ζ83(x3x7)).\displaystyle\quad\times D_{4}(x_{0}-x_{4},\zeta_{8}(x_{1}-x_{5}),\zeta_{8}^{2}(x_{2}-x_{6}),\zeta_{8}^{3}(x_{3}-x_{7})).

From the G=C4G={\rm C}_{4} and H={0¯,2¯}H=\left\{\overline{0},\overline{2}\right\} case of Theorem 1.1, we have

D4(x0,x1,x2,x3)\displaystyle D_{4}(x_{0},x_{1},x_{2},x_{3}) =D2(x0+x2,x1+x3)D2(x0x2,ζ4(x1x3))\displaystyle=D_{2}(x_{0}+x_{2},x_{1}+x_{3})D_{2}(x_{0}-x_{2},\zeta_{4}(x_{1}-x_{3}))
=D2(x02+x222x1x3,x12x32+2x0x2).\displaystyle=D_{2}(x_{0}^{2}+x_{2}^{2}-2x_{1}x_{3},-x_{1}^{2}-x_{3}^{2}+2x_{0}x_{2}).

Let D~4(x0,x1,x2,x3):=D4(x0,ζ8x1,ζ82x2,ζ83x3)\widetilde{D}_{4}(x_{0},x_{1},x_{2},x_{3}):=D_{4}(x_{0},\zeta_{8}x_{1},\zeta_{8}^{2}x_{2},\zeta_{8}^{3}x_{3}). Then we have the following lemma.

Lemma 4.1.

The following hold:

  1. (1)(1)

    D4(x0,x1,x2,x3)={(x0+x2)2(x1+x3)2}{(x0x2)2+(x1x3)2}D_{4}(x_{0},x_{1},x_{2},x_{3})=\left\{(x_{0}+x_{2})^{2}-(x_{1}+x_{3})^{2}\right\}\left\{(x_{0}-x_{2})^{2}+(x_{1}-x_{3})^{2}\right\};

  2. (2)(2)

    D~4(x0,x1,x2,x3)=(x02x22+2x1x3)2+(x12x322x0x2)2\widetilde{D}_{4}(x_{0},x_{1},x_{2},x_{3})=(x_{0}^{2}-x_{2}^{2}+2x_{1}x_{3})^{2}+(x_{1}^{2}-x_{3}^{2}-2x_{0}x_{2})^{2}.

Lemma 4.1 (2) shows that D~4(x0,x1,x2,x3)\widetilde{D}_{4}(x_{0},x_{1},x_{2},x_{3})\in\mathbb{Z} holds for any x0,x1,x2,x3x_{0},x_{1},x_{2},x_{3}\in\mathbb{Z}. Throughout this paper, we assume that a0,a1,,a15a_{0},a_{1},\ldots,a_{15}\in\mathbb{Z}, and for any 0i30\leq i\leq 3, put

bi\displaystyle b_{i} :=(ai+ai+8)+(ai+4+ai+12),\displaystyle:=(a_{i}+a_{i+8})+(a_{i+4}+a_{i+12}), ci\displaystyle c_{i} :=(ai+ai+8)(ai+4+ai+12),\displaystyle:=(a_{i}+a_{i+8})-(a_{i+4}+a_{i+12}),
di\displaystyle d_{i} :=(aiai+8)+(ai+4ai+12),\displaystyle:=(a_{i}-a_{i+8})+(a_{i+4}-a_{i+12}), ei\displaystyle e_{i} :=(aiai+8)(ai+4ai+12).\displaystyle:=(a_{i}-a_{i+8})-(a_{i+4}-a_{i+12}).

Also, let 𝒂:=(a0,a1,,a15)\bm{a}:=(a_{0},a_{1},\ldots,a_{15}) and let

𝒃:=(b0,b1,b2,b3),𝒄:=(c0,c1,c2,c3),𝒅:=(d0,d1,d2,d3),𝒆:=(e0,e1,e2,e3).\displaystyle\bm{b}:=(b_{0},b_{1},b_{2},b_{3}),\quad\bm{c}:=(c_{0},c_{1},c_{2},c_{3}),\quad\bm{d}:=(d_{0},d_{1},d_{2},d_{3}),\quad\bm{e}:=(e_{0},e_{1},e_{2},e_{3}).

The following relations will be frequently used in this paper:

D8×2(𝒂)\displaystyle D_{8\times 2}(\bm{a}) =D8(a0+a8,a1+a9,,a7+a15)D8(a0a8,a1a9,,a7a15)\displaystyle=D_{8}(a_{0}+a_{8},a_{1}+a_{9},\ldots,a_{7}+a_{15})D_{8}(a_{0}-a_{8},a_{1}-a_{9},\ldots,a_{7}-a_{15})
=D4(𝒃)D~4(𝒄)D4(𝒅)D~4(𝒆).\displaystyle=D_{4}(\bm{b})\widetilde{D}_{4}(\bm{c})D_{4}(\bm{d})\widetilde{D}_{4}(\bm{e}).
Remark 4.2.

For any 0i30\leq i\leq 3, the following hold:

  1. (1)(1)

    bicidiei(mod2)b_{i}\equiv c_{i}\equiv d_{i}\equiv e_{i}\pmod{2};

  2. (2)(2)

    bi+ci+di+ei0(mod4)b_{i}+c_{i}+d_{i}+e_{i}\equiv 0\pmod{4}.

Lemma 4.3.

We have D8×2(𝐚)D4(𝐛)D~4(𝐜)D4(𝐝)D~4(𝐞)(mod2)D_{8\times 2}(\bm{a})\equiv D_{4}(\bm{b})\equiv\widetilde{D}_{4}(\bm{c})\equiv D_{4}(\bm{d})\equiv\widetilde{D}_{4}(\bm{e})\pmod{2}.

Proof.

From Lemma 4.1, we have D4(x0,x1,x2,x3)x0+x1+x2+x3D~4(x0,x1,x2,x3)(mod2)D_{4}(x_{0},x_{1},x_{2},x_{3})\equiv x_{0}+x_{1}+x_{2}+x_{3}\equiv\widetilde{D}_{4}(x_{0},x_{1},x_{2},x_{3})\pmod{2}. Therefore, from Remark 4.2 (1), the lemma is proved. ∎

4.2 Impossible odd numbers

Let odd\mathbb{Z}_{\rm odd} be the set of all odd numbers and A:={(8k3)(8l3)k,l,kl(mod2)}A:=\left\{(8k-3)(8l-3)\mid k,l\in\mathbb{Z},\>k\equiv l\pmod{2}\right\}.

Lemma 4.4.

We have S(C8×C2)odd{16m+1m}AS({\rm C}_{8}\times{\rm C}_{2})\cap\mathbb{Z}_{\rm odd}\subset\left\{16m+1\mid m\in\mathbb{Z}\right\}\cup A.

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

Lemma 4.5.

We have

D4(𝒃)D~4(𝒄)D4(𝒅)D~4(𝒆)\displaystyle D_{4}(\bm{b})\widetilde{D}_{4}(\bm{c})D_{4}(\bm{d})\widetilde{D}_{4}(\bm{e})
=D4(b1,b2,b3,b0)D~4(c1,c2,c3,c0)D4(d1,d2,d3,d0)D~4(e1,e2,e3,e0)\displaystyle\quad=D_{4}(b_{1},b_{2},b_{3},b_{0})\widetilde{D}_{4}(c_{1},c_{2},c_{3},-c_{0})D_{4}(d_{1},d_{2},d_{3},d_{0})\widetilde{D}_{4}(e_{1},e_{2},e_{3},-e_{0})
=D4(b2,b3,b0,b1)D~4(c2,c3,c0,c1)D4(d2,d3,d0,d1)D~4(e2,e3,e0,e1)\displaystyle\quad=D_{4}(b_{2},b_{3},b_{0},b_{1})\widetilde{D}_{4}(c_{2},c_{3},-c_{0},-c_{1})D_{4}(d_{2},d_{3},d_{0},d_{1})\widetilde{D}_{4}(e_{2},e_{3},-e_{0},-e_{1})
=D4(b3,b0,b1,b2)D~4(c3,c0,c1,c2)D4(d3,d0,d1,d2)D~4(e3,e0,e1,e2).\displaystyle\quad=D_{4}(b_{3},b_{0},b_{1},b_{2})\widetilde{D}_{4}(c_{3},-c_{0},-c_{1},-c_{2})D_{4}(d_{3},d_{0},d_{1},d_{2})\widetilde{D}_{4}(e_{3},-e_{0},-e_{1},-e_{2}).
Proof.

From Lemma 4.1, we have

D4(x0,x1,x2,x3)=D4(x1,x2,x3,x0),D~4(x0,x1,x2,x3)\displaystyle D_{4}(x_{0},x_{1},x_{2},x_{3})=-D_{4}(x_{1},x_{2},x_{3},x_{0}),\quad\widetilde{D}_{4}(x_{0},x_{1},x_{2},x_{3}) =D~4(x1,x2,x3,x0).\displaystyle=\widetilde{D}_{4}(x_{1},x_{2},x_{3},-x_{0}).

Therefore, the lemma is proved. ∎

Lemma 4.6.

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

  1. (1)(1)

    D4(2k+1,2l,2m,2n)8m+1(mod16)D_{4}(2k+1,2l,2m,2n)\equiv 8m+1\pmod{16};

  2. (2)(2)

    D~4(2k+1,2l,2m,2n)8m+1(mod16)\widetilde{D}_{4}(2k+1,2l,2m,2n)\equiv 8m+1\pmod{16}.

Proof.

Let D:=D4(2k+1,2l,2m,2n)D:=D_{4}(2k+1,2l,2m,2n) and D~:=D~4(2k+1,2l,2m,2n)\widetilde{D}:=\widetilde{D}_{4}(2k+1,2l,2m,2n). Then we have

D\displaystyle D ={4(k+m)2+4(k+m)+14(l+n)2}{4(km)2+4(km)+1+4(ln)2}\displaystyle=\left\{4(k+m)^{2}+4(k+m)+1-4(l+n)^{2}\right\}\left\{4(k-m)^{2}+4(k-m)+1+4(l-n)^{2}\right\}
8m+1(mod16),\displaystyle\equiv 8m+1\pmod{16},
D~\displaystyle\widetilde{D} ={4k(k+1)+14m2+8ln}2+{4l24n28km4m}2\displaystyle=\left\{4k(k+1)+1-4m^{2}+8ln\right\}^{2}+\left\{4l^{2}-4n^{2}-8km-4m\right\}^{2}
8m+1(mod16).\displaystyle\equiv 8m+1\pmod{16}.

Lemma 4.7.

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

  1. (1)(1)

    D4(2k,2l+1,2m+1,2n+1)8(k+l+n)3(mod16)D_{4}(2k,2l+1,2m+1,2n+1)\equiv 8(k+l+n)-3\pmod{16};

  2. (2)(2)

    D~4(2k,2l+1,2m+1,2n+1)8(k+l+n)+1(mod16)\widetilde{D}_{4}(2k,2l+1,2m+1,2n+1)\equiv 8(k+l+n)+1\pmod{16}.

Proof.

Let D:=D4(2k,2l+1,2m+1,2n+1)D:=D_{4}(2k,2l+1,2m+1,2n+1) and D~:=D~4(2k,2l+1,2m+1,2n+1)\widetilde{D}:=\widetilde{D}_{4}(2k,2l+1,2m+1,2n+1). Then,

D\displaystyle D ={4(k+m)2+4(k+m)+14(l+n+1)2}{4(km)24(km)+1+4(ln)2}\displaystyle=\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\}
8(k+l+n)3(mod16),\displaystyle\equiv 8(k+l+n)-3\pmod{16},
D~\displaystyle\widetilde{D} ={4k24m(m+1)+8ln+4l+4n+1}2+{4l(l+1)4n(n+1)8km4k}2\displaystyle=\left\{4k^{2}-4m(m+1)+8ln+4l+4n+1\right\}^{2}+\left\{4l(l+1)-4n(n+1)-8km-4k\right\}^{2}
8(k+l+n)+1(mod16).\displaystyle\equiv 8(k+l+n)+1\pmod{16}.

Proof of Lemma 4.4.

Let D8×2(𝒂)=D4(𝒃)D~4(𝒄)D4(𝒅)D~4(𝒆)oddD_{8\times 2}(\bm{a})=D_{4}(\bm{b})\widetilde{D}_{4}(\bm{c})D_{4}(\bm{d})\widetilde{D}_{4}(\bm{e})\in\mathbb{Z}_{\rm odd}. Then, b0+b2b1+b3(mod2)b_{0}+b_{2}\not\equiv b_{1}+b_{3}\pmod{2} holds since D4(𝒃)D_{4}(\bm{b}) is odd. We prove the following:

  1. (i)

    If exactly three of b0,b1,b2,b3b_{0},b_{1},b_{2},b_{3} are even, then D8×2(𝒂){16m+1m}D_{8\times 2}(\bm{a})\in\left\{16m+1\mid m\in\mathbb{Z}\right\};

  2. (ii)

    If exactly one of b0,b1,b2,b3b_{0},b_{1},b_{2},b_{3} is even, then D8×2(𝒂)AD_{8\times 2}(\bm{a})\in A.

First, we prove (i). If 𝒃(1,0,0,0)(mod2)\bm{b}\equiv(1,0,0,0)\pmod{2}, then there exist mim_{i}\in\mathbb{Z} satisfying b2=2m0b_{2}=2m_{0}, c2=2m1c_{2}=2m_{1}, d2=2m2d_{2}=2m_{2}, e2=2m3e_{2}=2m_{3} and i=03mi0(mod2)\sum_{i=0}^{3}m_{i}\equiv 0\pmod{2} from Remark 4.2. Therefore, from Lemma 4.6, D8×2(𝒂)i=03(8mi+1)8i=03mi+11(mod16)D_{8\times 2}(\bm{a})\equiv\prod_{i=0}^{3}(8m_{i}+1)\equiv 8\sum_{i=0}^{3}m_{i}+1\equiv 1\pmod{16}. From this and Lemma 4.5, the remaining three cases are also proved. Next, we prove (ii). If 𝒃(0,1,1,1)(mod2)\bm{b}\equiv(0,1,1,1)\pmod{2}, then there exist ki,li,nik_{i},l_{i},n_{i}\in\mathbb{Z} satisfying (b0,b1,b3)=(2k0,2l0+1,2n0+1)(b_{0},b_{1},b_{3})=(2k_{0},2l_{0}+1,2n_{0}+1), (c0,c1,c3)=(2k1,2l1+1,2n1+1)(c_{0},c_{1},c_{3})=(2k_{1},2l_{1}+1,2n_{1}+1), (d0,d1,d3)=(2k2,2l2+1,2n2+1)(d_{0},d_{1},d_{3})=(2k_{2},2l_{2}+1,2n_{2}+1), (e0,e1,e3)=(2k3,2l3+1,2n3+1)(e_{0},e_{1},e_{3})=(2k_{3},2l_{3}+1,2n_{3}+1) and i=03kii=03lii=03ni0(mod2)\sum_{i=0}^{3}k_{i}\equiv\sum_{i=0}^{3}l_{i}\equiv\sum_{i=0}^{3}n_{i}\equiv 0\pmod{2} from Remark 4.2. Therefore, from Lemma 4.7, we have D4(𝒃)D~4(𝒄)(8r03)(8r1+1)8r0+8r13(mod16)D_{4}(\bm{b})\widetilde{D}_{4}(\bm{c})\equiv(8r_{0}-3)(8r_{1}+1)\equiv 8r_{0}+8r_{1}-3\pmod{16} and D4(𝒅)D~4(𝒆)(8r23)(8r3+1)8r2+8r33(mod16)D_{4}(\bm{d})\widetilde{D}_{4}(\bm{e})\equiv(8r_{2}-3)(8r_{3}+1)\equiv 8r_{2}+8r_{3}-3\pmod{16}, where ri:=ki+li+nir_{i}:=k_{i}+l_{i}+n_{i}. Thus, there exist s0,s1s_{0},s_{1}\in\mathbb{Z} satisfying D4(𝒃)D~4(𝒄)=16s0+8r0+8r13D_{4}(\bm{b})\widetilde{D}_{4}(\bm{c})=16s_{0}+8r_{0}+8r_{1}-3, D4(𝒅)D~4(𝒆)=16s1+8r2+8r33D_{4}(\bm{d})\widetilde{D}_{4}(\bm{e})=16s_{1}+8r_{2}+8r_{3}-3. Let k:=2s0+r0+r1k:=2s_{0}+r_{0}+r_{1} and l:=2s1+r2+r3l:=2s_{1}+r_{2}+r_{3}. Then D8×2(𝒂)=(8k3)(8l3)AD_{8\times 2}(\bm{a})=(8k-3)(8l-3)\in A since kl(mod2)k\equiv l\pmod{2} holds from i=03ri0(mod2)\sum_{i=0}^{3}r_{i}\equiv 0\pmod{2}. From this and Lemma 4.5, the remaining three cases are also proved. ∎

4.3 Impossible even numbers

We will use Kaiblinger’s [10, Theorem 1.1] results S(C4)=odd24S({\rm C}_{4})=\mathbb{Z}_{\rm odd}\cup 2^{4}\mathbb{Z} and S(C8)=odd25S({\rm C}_{8})=\mathbb{Z}_{\rm odd}\cup 2^{5}\mathbb{Z}.

Lemma 4.8.

We have S(C8×C2)2210S\left({\rm C}_{8}\times{\rm C}_{2}\right)\cap 2\mathbb{Z}\subset 2^{10}\mathbb{Z}.

Proof.

Let D8×2(𝒂)=D8(a0+a8,,a7+a15)D8(a0a8,,a7a15)2D_{8\times 2}(\bm{a})=D_{8}(a_{0}+a_{8},\ldots,a_{7}+a_{15})D_{8}(a_{0}-a_{8},\ldots,a_{7}-a_{15})\in 2\mathbb{Z}. Since D8(a0+a8,,a7+a15)D8(a0a8,,a7a15)(mod2)D_{8}(a_{0}+a_{8},\ldots,a_{7}+a_{15})\equiv D_{8}(a_{0}-a_{8},\ldots,a_{7}-a_{15})\pmod{2} holds from ai+ai+8aiai+8(mod2)a_{i}+a_{i+8}\equiv a_{i}-a_{i+8}\pmod{2}, we have D8(a0+a8,,a7+a15),D8(a0a8,,a7a15)S(C8)2=25D_{8}(a_{0}+a_{8},\ldots,a_{7}+a_{15}),D_{8}(a_{0}-a_{8},\ldots,a_{7}-a_{15})\in S({\rm C}_{8})\cap 2\mathbb{Z}=2^{5}\mathbb{Z}. Therefore, D8×2(𝒂)210D_{8\times 2}(\bm{a})\in 2^{10}\mathbb{Z}. ∎

Lemma 4.9.

Let pi=ai2+bi21(mod8)p_{i}=a_{i}^{2}+b_{i}^{2}\equiv 1\pmod{8} be a prime with ai±bi{8m±1m}a_{i}\pm b_{i}\in\left\{8m\pm 1\mid m\in\mathbb{Z}\right\} for each 1ir1\leq i\leq r, let pr+1,,pr+s1(mod8)p_{r+1},\ldots,p_{r+s}\equiv-1\pmod{8} be primes, let q1,,qt3(mod8)q_{1},\ldots,q_{t}\equiv 3\pmod{8} be distinct primes, and let k1,,kr+sk_{1},\ldots,k_{r+s} be non-negative integers. Then

211p1k1prkrpr+1kr+1pr+skr+sQS(C8×C2)2^{11}p_{1}^{k_{1}}\cdots p_{r}^{k_{r}}p_{r+1}^{k_{r+1}}\cdots p_{r+s}^{k_{r+s}}Q\not\in S({\rm C}_{8}\times{\rm C}_{2})

for any Q{±1,±q1qt}Q\in\left\{\pm 1,\>\pm q_{1}\cdots q_{t}\right\}.

Let

α0\displaystyle\alpha_{0} :=(b0+b2)2(b1+b3)2,\displaystyle:=(b_{0}+b_{2})^{2}-(b_{1}+b_{3})^{2}, α1\displaystyle\alpha_{1} :=(b0b2)2+(b1b3)2,\displaystyle:=(b_{0}-b_{2})^{2}+(b_{1}-b_{3})^{2},
α2\displaystyle\alpha_{2} :=(d0+d2)2(d1+d3)2,\displaystyle:=(d_{0}+d_{2})^{2}-(d_{1}+d_{3})^{2}, α3\displaystyle\alpha_{3} :=(d0d2)2+(d1d3)2,\displaystyle:=(d_{0}-d_{2})^{2}+(d_{1}-d_{3})^{2},
β\displaystyle\beta :=(c02c22+2c1c3)ζ4(c12c322c0c2),\displaystyle:=(c_{0}^{2}-c_{2}^{2}+2c_{1}c_{3})-\zeta_{4}(c_{1}^{2}-c_{3}^{2}-2c_{0}c_{2}), γ\displaystyle\gamma :=(e02e22+2e1e3)ζ4(e12e322e0e2).\displaystyle:=(e_{0}^{2}-e_{2}^{2}+2e_{1}e_{3})-\zeta_{4}(e_{1}^{2}-e_{3}^{2}-2e_{0}e_{2}).

Then we have α0α1=D4(𝒃)\alpha_{0}\alpha_{1}=D_{4}(\bm{b}), α2α3=D4(𝒅)\alpha_{2}\alpha_{3}=D_{4}(\bm{d}), ββ¯=D~4(𝒄)\beta\overline{\beta}=\widetilde{D}_{4}(\bm{c}), γγ¯=D~4(𝒆)\gamma\overline{\gamma}=\widetilde{D}_{4}(\bm{e}), where x¯\overline{x} denotes the complex conjugate of xx\in\mathbb{C}. To prove Lemma 4.9, we use the following remark and two lemmas.

Remark 4.10.

From Remark 4.2 (1)(1) and

α12odd\displaystyle\alpha_{1}\in 2\mathbb{Z}_{\rm odd} b0+b2b1+b31(mod2),\displaystyle\iff b_{0}+b_{2}\equiv b_{1}+b_{3}\equiv 1\pmod{2},
α32odd\displaystyle\alpha_{3}\in 2\mathbb{Z}_{\rm odd} d0+d2d1+d31(mod2),\displaystyle\iff d_{0}+d_{2}\equiv d_{1}+d_{3}\equiv 1\pmod{2},
ββ¯2odd\displaystyle\beta\overline{\beta}\in 2\mathbb{Z}_{\rm odd} c0+c2c1+c31(mod2),\displaystyle\iff c_{0}+c_{2}\equiv c_{1}+c_{3}\equiv 1\pmod{2},
γγ¯2odd\displaystyle\gamma\overline{\gamma}\in 2\mathbb{Z}_{\rm odd} e0+e2e1+e31(mod2),\displaystyle\iff e_{0}+e_{2}\equiv e_{1}+e_{3}\equiv 1\pmod{2},

we have α12oddα32oddββ¯2oddγγ¯2odd\alpha_{1}\in 2\mathbb{Z}_{\rm odd}\iff\alpha_{3}\in 2\mathbb{Z}_{\rm odd}\iff\beta\overline{\beta}\in 2\mathbb{Z}_{\rm odd}\iff\gamma\overline{\gamma}\in 2\mathbb{Z}_{\rm odd}.

Lemma 4.11.

If D8×2(𝐚)211oddD_{8\times 2}(\bm{a})\in 2^{11}\mathbb{Z}_{\rm odd}, then we have (αi,αj)23odd×24odd(\alpha_{i},\alpha_{j})\in 2^{3}\mathbb{Z}_{\rm odd}\times 2^{4}\mathbb{Z}_{\rm odd} and α1\alpha_{1}, α3\alpha_{3}, ββ¯\beta\overline{\beta}, γγ¯2odd\gamma\overline{\gamma}\in 2\mathbb{Z}_{\rm odd}, where {i,j}={0,2}\{i,j\}=\{0,2\}.

Proof.

Let D8×2(𝒂)=α0α1α2α3ββ¯γγ¯211oddD_{8\times 2}(\bm{a})=\alpha_{0}\alpha_{1}\alpha_{2}\alpha_{3}\beta\overline{\beta}\gamma\overline{\gamma}\in 2^{11}\mathbb{Z}_{\rm odd}. Then, from Lemma 4.3, α0α1α2α3ββ¯γγ¯0(mod2)\alpha_{0}\alpha_{1}\equiv\alpha_{2}\alpha_{3}\equiv\beta\overline{\beta}\equiv\gamma\overline{\gamma}\equiv 0\pmod{2} holds. In particular, α0α1,α2α3S(C4)2=24\alpha_{0}\alpha_{1},\alpha_{2}\alpha_{3}\in S({\rm C}_{4})\cap 2\mathbb{Z}=2^{4}\mathbb{Z}. From this and Remark 4.10, we have ββ¯,γγ¯2odd\beta\overline{\beta},\gamma\overline{\gamma}\in 2\mathbb{Z}_{\rm odd}. Therefore, α1,α32odd\alpha_{1},\alpha_{3}\in 2\mathbb{Z}_{\rm odd}. ∎

Lemma 4.12.

Let b0+b2b1+b31(mod2)b_{0}+b_{2}\equiv b_{1}+b_{3}\equiv 1\pmod{2}. Then the following hold:

  1. (1)(1)

    α0α1+4(b0b2+b1b3)2(mod16)\alpha_{0}\equiv\alpha_{1}+4(b_{0}b_{2}+b_{1}b_{3})-2\pmod{16};

  2. (2)(2)

    α2α3+4(d0d2+d1d3)2(mod16)\alpha_{2}\equiv\alpha_{3}+4(d_{0}d_{2}+d_{1}d_{3})-2\pmod{16};

  3. (3)(3)

    Re(β)(1)c2+2(c0c2+c1c3)(mod8)\operatorname{Re}(\beta)\equiv(-1)^{c_{2}}+2(c_{0}c_{2}+c_{1}c_{3})\pmod{8};

  4. (4)(4)

    Re(γ)(1)e2+2(e0e2+e1e3)(mod8)\operatorname{Re}(\gamma)\equiv(-1)^{e_{2}}+2(e_{0}e_{2}+e_{1}e_{3})\pmod{8};

  5. (5)(5)

    (b0b2+b1b3)+(c0c2+c1c3)+(d0d2+d1d3)+(e0e2+e1e3)0(mod4)(b_{0}b_{2}+b_{1}b_{3})+(c_{0}c_{2}+c_{1}c_{3})+(d_{0}d_{2}+d_{1}d_{3})+(e_{0}e_{2}+e_{1}e_{3})\equiv 0\pmod{4}.

Proof.

We obtain (1) from α0=α1+4(b0b2+b1b3)2(b1+b3)2\alpha_{0}=\alpha_{1}+4(b_{0}b_{2}+b_{1}b_{3})-2(b_{1}+b_{3})^{2}. In the same way, we can obtain (2). We obtain (3) from Re(β)=(c0c2)22c22+2(c0c2+c1c3)\operatorname{Re}(\beta)=(c_{0}-c_{2})^{2}-2c_{2}^{2}+2(c_{0}c_{2}+c_{1}c_{3}). In the same way, we can obtain (4). We prove (5). There are four cases:

𝒃(0,0,1,1),(0,1,1,0),(1,1,0,0)or(1,0,0,1)(mod2).\bm{b}\equiv(0,0,1,1),\>(0,1,1,0),\>(1,1,0,0)\>\>\text{or}\>\>(1,0,0,1)\pmod{2}.

If 𝒃(0,0,1,1)(mod2)\bm{b}\equiv(0,0,1,1)\pmod{2}, then

(b0b2+b1b3)+(c0c2+c1c3)+(d0d2+d1d3)+(e0e2+e1e3)\displaystyle(b_{0}b_{2}+b_{1}b_{3})+(c_{0}c_{2}+c_{1}c_{3})+(d_{0}d_{2}+d_{1}d_{3})+(e_{0}e_{2}+e_{1}e_{3})
(b0+b1)+(c0+c1)+(d0+d1)+(e0+e1)\displaystyle\qquad\equiv(b_{0}+b_{1})+(c_{0}+c_{1})+(d_{0}+d_{1})+(e_{0}+e_{1})
0(mod4)\displaystyle\qquad\equiv 0\pmod{4}

from Remark 4.2. In the same way, the remaining three cases can also be proved. ∎

Proof of Lemma 4.9.

We prove by contradiction. Assume that there exist a0,a1,,a15a_{0},a_{1},\ldots,a_{15}\in\mathbb{Z} satisfying D8×2(𝒂)=α0α1α2α3ββ¯γγ¯=211p1k1prkrpr+1kr+1pr+skr+sQD_{8\times 2}(\bm{a})=\alpha_{0}\alpha_{1}\alpha_{2}\alpha_{3}\beta\overline{\beta}\gamma\overline{\gamma}=2^{11}p_{1}^{k_{1}}\cdots p_{r}^{k_{r}}p_{r+1}^{k_{r+1}}\cdots p_{r+s}^{k_{r+s}}Q, where QQ is ±1\pm 1 or ±q1qt\pm q_{1}\cdots q_{t}. Since α1\alpha_{1}, α3\alpha_{3}, ββ¯\beta\overline{\beta} and γγ¯\gamma\overline{\gamma} are integers expressible in the form x2+y2x^{2}+y^{2}, in the prime factorization of them, every prime of the form 4k+34k+3 occurs an even number of times. From this fact and Lemma 4.11, there exist lf,mf,nf,uf,vf0l_{f},m_{f},n_{f},u_{f},v_{f}\geq 0 satisfying

αi\displaystyle\alpha_{i} =23p1k1l1w1prkrlrwrpr+1kr+1lr+12wr+1pr+skr+slr+s2wr+sQ1,\displaystyle=2^{3}p_{1}^{k_{1}-l_{1}-w_{1}}\cdots p_{r}^{k_{r}-l_{r}-w_{r}}p_{r+1}^{k_{r+1}-l_{r+1}-2w_{r+1}}\cdots p_{r+s}^{k_{r+s}-l_{r+s}-2w_{r+s}}Q_{1},
αj\displaystyle\alpha_{j} =24p1l1prlrpr+1lr+1pr+slr+sQ2,\displaystyle=2^{4}p_{1}^{l_{1}}\cdots p_{r}^{l_{r}}p_{r+1}^{l_{r+1}}\cdots p_{r+s}^{l_{r+s}}Q_{2},
α1\displaystyle\alpha_{1} =2p1m1prmrpr+12mr+1pr+s2mr+s,\displaystyle=2p_{1}^{m_{1}}\cdots p_{r}^{m_{r}}p_{r+1}^{2m_{r+1}}\cdots p_{r+s}^{2m_{r+s}},
α3\displaystyle\alpha_{3} =2p1n1prnrpr+12nr+1pr+s2nr+s,\displaystyle=2p_{1}^{n_{1}}\cdots p_{r}^{n_{r}}p_{r+1}^{2n_{r+1}}\cdots p_{r+s}^{2n_{r+s}},
ββ¯\displaystyle\beta\overline{\beta} =2p1u1prurpr+12ur+1pr+s2ur+s,\displaystyle=2p_{1}^{u_{1}}\cdots p_{r}^{u_{r}}p_{r+1}^{2u_{r+1}}\cdots p_{r+s}^{2u_{r+s}},
γγ¯\displaystyle\gamma\overline{\gamma} =2p1v1prvrpr+12vr+1pr+s2vr+s,\displaystyle=2p_{1}^{v_{1}}\cdots p_{r}^{v_{r}}p_{r+1}^{2v_{r}+1}\cdots p_{r+s}^{2v_{r+s}},

where {i,j}={0,2}\{i,j\}=\{0,2\}, wf:=mf+nf+uf+vfw_{f}:=m_{f}+n_{f}+u_{f}+v_{f} and Q1Q2=QQ_{1}Q_{2}=Q. From the above, we have αi8,αj0,α1α32(mod16)\alpha_{i}\equiv 8,\alpha_{j}\equiv 0,\alpha_{1}\equiv\alpha_{3}\equiv 2\pmod{16}. Therefore, from Lemma 4.12 (1) and (2),

(b0b2+b1b3,d0d2+d1d3)(2,0)or(0,2)(mod4).(b_{0}b_{2}+b_{1}b_{3},d_{0}d_{2}+d_{1}d_{3})\equiv(2,0)\>\>\text{or}\>\>(0,2)\pmod{4}.

Note that c0c2+c1c3e0e2+e1e30(mod2)c_{0}c_{2}+c_{1}c_{3}\equiv e_{0}e_{2}+e_{1}e_{3}\equiv 0\pmod{2} since b0+b2b1+b31(mod2)b_{0}+b_{2}\equiv b_{1}+b_{3}\equiv 1\pmod{2}. From [25, Lemma 4.8], we have Re(β),Re(γ){8m±1m}\operatorname{Re}(\beta),\operatorname{Re}(\gamma)\in\{8m\pm 1\mid m\in\mathbb{Z}\}. From this and Lemma 4.12 (3) and (4), it follows that c0c2+c1c3e0e2+e1e30(mod4)c_{0}c_{2}+c_{1}c_{3}\equiv e_{0}e_{2}+e_{1}e_{3}\equiv 0\pmod{4}. Therefore, we have

(b0b2+b1b3)+(c0c2+c1c3)+(d0d2+d1d3)+(e0e2+e1e3)2(mod4).(b_{0}b_{2}+b_{1}b_{3})+(c_{0}c_{2}+c_{1}c_{3})+(d_{0}d_{2}+d_{1}d_{3})+(e_{0}e_{2}+e_{1}e_{3})\equiv 2\pmod{4}.

This contradicts Lemma 4.12 (5)(5). ∎

4.4 Possible numbers

Lemmas 4.4, 4.8 and 4.9 imply that S(C8×C2)S\left({\rm C}_{8}\times{\rm C}_{2}\right) does not include every integer that is not mentioned in Lemmas 4.13 and 4.14.

Lemma 4.13.

For any m,nm,n\in\mathbb{Z}, the following are elements of S(C8×C2)S\left({\rm C}_{8}\times{\rm C}_{2}\right):

  1. (1)(1)

    16m+116m+1;

  2. (2)(2)

    (16m3)(16n3)(16m-3)(16n-3);

  3. (3)(3)

    (16m+5)(16n+5)(16m+5)(16n+5);

  4. (4)(4)

    210(2m+1)2^{10}(2m+1);

  5. (5)(5)

    212(2m+1)2^{12}(2m+1);

  6. (6)(6)

    212(2m)2^{12}(2m).

Lemma 4.14.

The following hold:

  1. (1)(1)

    Suppose that pp is a prime with p3(mod8)p\equiv-3\pmod{8}, then 211p(2m+1)S(C8×C2)2^{11}p(2m+1)\in S({\rm C}_{8}\times{\rm C}_{2});

  2. (2)(2)

    Suppose that pp is a prime with p3(mod8)p\equiv 3\pmod{8}, then 211p2(2m+1)S(C8×C2)2^{11}p^{2}(2m+1)\in S({\rm C}_{8}\times{\rm C}_{2});

  3. (3)(3)

    Suppose that pp is a prime with p=a2+b21(mod8)p=a^{2}+b^{2}\equiv 1\pmod{8} and a+b±3(mod8)a+b\equiv\pm 3\pmod{8}, then 211p(2m+1)S(C8×C2)2^{11}p(2m+1)\in S({\rm C}_{8}\times{\rm C}_{2}).

Proof of Lemma 4.13.

We obtain (1) from D8×2(m+1,m,,m)=16m+1D_{8\times 2}(m+1,m,\ldots,m)=16m+1. From

D8×2(m+n,,m+n5,m+n1,m+n1,m+n1,mn,,mn)\displaystyle D_{8\times 2}(\overbrace{m+n,\ldots,m+n}^{5},m+n-1,m+n-1,m+n-1,m-n,\ldots,m-n)
=(16m3)(16n3),\displaystyle\quad=(16m-3)(16n-3),
D8×2(m+n+1,,m+n+15,m+n,m+n,m+n,mn,,mn)\displaystyle D_{8\times 2}(\overbrace{m+n+1,\ldots,m+n+1}^{5},m+n,m+n,m+n,m-n,\ldots,m-n)
=(16m+5)(16n+5),\displaystyle\quad=(16m+5)(16n+5),

we obtain (2) and (3), respectively. We obtain (4) from

D8×2(m+1,m+1,m+1,m,m,m,m+1,m,,m)=210(4m+1),\displaystyle D_{8\times 2}(m+1,m+1,m+1,m,m,m,m+1,m,\ldots,m)=2^{10}(4m+1),
D8×2(m,,m6,m+1,m1,m,m,m1,m,m1,m1,m,m1)=210(4m1).\displaystyle D_{8\times 2}(\overbrace{m,\ldots,m}^{6},m+1,m-1,m,m,m-1,m,m-1,m-1,m,m-1)=2^{10}(4m-1).

We obtain (5) from D8×2(m+2,m,m+1,,m+16,m,,m)=212(2m+1)D_{8\times 2}(m+2,m,\overbrace{m+1,\ldots,m+1}^{6},m,\ldots,m)=2^{12}(2m+1). From

D8×2(m+1,m,m,m+1,m+1,m,m+1,m,m1,m1,m,m1,m,m,m1,m)\displaystyle D_{8\times 2}(m+1,m,m,m+1,m+1,m,m+1,m,m-1,m-1,m,m-1,m,m,m-1,m)
=212(2m),\displaystyle\quad=2^{12}(2m),

we obtain (6). ∎

To prove Lemma 4.14, we use the following lemma.

Lemma 4.15 ([25, Proof of Theorem 5.1]).

The following hold:

  1. (1)(1)

    Suppose that p3(mod8)p\equiv-3\pmod{8} is a prime, then there exist k,lk,l\in\mathbb{Z} satisfying

    2p=(8k+3)2+(8l+1)2;2p=(8k+3)^{2}+(8l+1)^{2};
  2. (2)(2)

    Suppose that p3(mod8)p\equiv 3\pmod{8} is a prime, then there exist k,lk,l\in\mathbb{Z} satisfying

    p=(4k1)2+2(4l1)2;p=(4k-1)^{2}+2(4l-1)^{2};
  3. (3)(3)

    Suppose that p=a2+b21(mod8)p=a^{2}+b^{2}\equiv 1\pmod{8} is a prime with a+b±3(mod8)a+b\equiv\pm 3\pmod{8}, then there exist k,l,m,nk,l,m,n\in\mathbb{Z} satisfying

    2p={(4k1)2(4m2)2+2(2l1)(4n)}2\displaystyle 2p=\left\{(4k-1)^{2}-(4m-2)^{2}+2(2l-1)(4n)\right\}^{2}
    +{(2l1)2(4n)22(4k1)(4m2)}2.\displaystyle\qquad+\left\{(2l-1)^{2}-(4n)^{2}-2(4k-1)(4m-2)\right\}^{2}.
Proof of Lemma 4.14.

First, we prove (1). Let

a0\displaystyle a_{0} =k+m+2,\displaystyle=k+m+2, a1\displaystyle a_{1} =l+m+1,\displaystyle=l+m+1, a2\displaystyle a_{2} =k+m,\displaystyle=-k+m, a3\displaystyle a_{3} =l+m+1,\displaystyle=-l+m+1,
a4\displaystyle a_{4} =k+m+1,\displaystyle=k+m+1, a5\displaystyle a_{5} =l+m,\displaystyle=l+m, a6\displaystyle a_{6} =k+m+1,\displaystyle=-k+m+1, a7\displaystyle a_{7} =l+m,\displaystyle=-l+m,
a8\displaystyle a_{8} =km,\displaystyle=k-m, a9\displaystyle a_{9} =lm,\displaystyle=l-m, a10\displaystyle a_{10} =km,\displaystyle=-k-m, a11\displaystyle a_{11} =lm1,\displaystyle=-l-m-1,
a12\displaystyle a_{12} =km,\displaystyle=k-m, a13\displaystyle a_{13} =lm,\displaystyle=l-m, a14\displaystyle a_{14} =km1,\displaystyle=-k-m-1, a15\displaystyle a_{15} =lm.\displaystyle=-l-m.

Then D8×2(𝒂)=210{(8k+3)2+(8l+1)2}(2m+1)D_{8\times 2}(\bm{a})=2^{10}\left\{(8k+3)^{2}+(8l+1)^{2}\right\}(2m+1). Therefore, from Lemma 4.15 (1), we obtain (1). We prove (2). Let

a0\displaystyle a_{0} =k+l+m,\displaystyle=k+l+m, a1\displaystyle a_{1} =kl+m,\displaystyle=k-l+m, a2\displaystyle a_{2} =l+m+1,\displaystyle=-l+m+1, a3\displaystyle a_{3} =l+m+1,\displaystyle=-l+m+1,
a4\displaystyle a_{4} =kl+m+2,\displaystyle=-k-l+m+2, a5\displaystyle a_{5} =k+l+m+1,\displaystyle=-k+l+m+1, a6\displaystyle a_{6} =l+m+1,\displaystyle=l+m+1, a7\displaystyle a_{7} =l+m,\displaystyle=l+m,
a8\displaystyle a_{8} =k+lm,\displaystyle=k+l-m, a9\displaystyle a_{9} =klm,\displaystyle=k-l-m, a10\displaystyle a_{10} =lm,\displaystyle=-l-m, a11\displaystyle a_{11} =lm,\displaystyle=-l-m,
a12\displaystyle a_{12} =klm,\displaystyle=-k-l-m, a13\displaystyle a_{13} =k+lm1,\displaystyle=-k+l-m-1, a14\displaystyle a_{14} =lm1,\displaystyle=l-m-1, a15\displaystyle a_{15} =lm\displaystyle=l-m

and s:=4k1s:=4k-1, t:=4l1t:=4l-1. Then D8×2(𝒂)=211(s2+2t2)2(2m+1)D_{8\times 2}(\bm{a})=2^{11}(s^{2}+2t^{2})^{2}(2m+1). Therefore, (2) is proved from Lemma 4.15 (2). We prove (3). Let l:=l2l^{\prime}:=\frac{l}{2} if ll is even; l12\frac{l-1}{2} if ll is odd and

a0\displaystyle a_{0} =k+r+1,\displaystyle=k+r+1, a1\displaystyle a_{1} =l+r,\displaystyle=l^{\prime}+r, a2\displaystyle a_{2} =m+r,\displaystyle=m+r, a3\displaystyle a_{3} =n+r,\displaystyle=n+r,
a4\displaystyle a_{4} =k+r+1,\displaystyle=-k+r+1, a5\displaystyle a_{5} =l+r+(1)l+12,\displaystyle=-l^{\prime}+r+\frac{(-1)^{l}+1}{2}, a6\displaystyle a_{6} =m+r+2,\displaystyle=-m+r+2, a7\displaystyle a_{7} =n+r+1,\displaystyle=-n+r+1,
a8\displaystyle a_{8} =kr1,\displaystyle=k-r-1, a9\displaystyle a_{9} =lr,\displaystyle=l^{\prime}-r, a10\displaystyle a_{10} =mr,\displaystyle=m-r, a11\displaystyle a_{11} =nr,\displaystyle=n-r,
a12\displaystyle a_{12} =kr,\displaystyle=-k-r, a13\displaystyle a_{13} =lr+(1)l12,\displaystyle=-l^{\prime}-r+\frac{(-1)^{l}-1}{2}, a14\displaystyle a_{14} =mr,\displaystyle=-m-r, a15\displaystyle a_{15} =nr1.\displaystyle=-n-r-1.

Then D8×2(𝒂)=210D~4(4k1,2l1,4m2,4n)(2r+1)D_{8\times 2}(\bm{a})=2^{10}\widetilde{D}_{4}(4k-1,2l-1,4m-2,4n)(2r+1). Therefore, (3) is proved from Lemma 4.15 (3). ∎

From Lemmas 4.4, 4.8, 4.9, 4.13 and 4.14, Theorem 1.5 is proved.

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Faculty of Education, University of Miyazaki, 1-1 Gakuen Kibanadai-nishi, Miyazaki 889-2192, Japan

Email address, Naoya Yamaguchi: [email protected]

Email address, Yuka Yamaguchi: [email protected]