## Finite groups with trivial class groups - Project Euclid

Jul 14, 1977 - Since $P\Gamma L(2, p)=PGL(2, p)$ , $G/O(G)$ must be isomorphic to one of ... Then we have $\Sigma=(1+\tau_{1})(1+\tau_{2})\Phi_{p}(\sigma)$ ..... suffices to show that $(1, 1, 1, \overline{a}^{k})\not\in{\rm Im}\nu$ . Define a ...

J. Math. Soc. Japan Vol. 31, No. 1, 1979

Finite groups with trivial class groups By Shizuo ENDO and Yumiko HIRONAKA 14, 1977)

(Received July

be a finite dimensional semisimple Q-algebra and let be a Z-order in $A$ . We mean by the class group of the class group defined by using locally free left be a maximal . Let -modules and denote it by $A$ containing Z-order in . We define to be the kernel of the natural surjection and to be the order of . Let be a finite group and let $ZG$ be the integral group ring of $G$ . Then $ZG$ can be regarded as a Z-order in the semisimple Q-algebra $QG$ . In this paper we will try to determine all finite groups $G$ for which

Let

$A$

$\Lambda$

$\Lambda$

$\Lambda$

$\Omega$

$C(\Lambda)$

$\Lambda$

$D(\Lambda)$

$C(\Lambda)\rightarrow C(\Omega)$

$D(\Lambda)$

$d(\Lambda)$

$G$

$d(ZG)=1$

.

Let $C_{n}(n\geqq 1)$ denote the cyclic group of order and let $D_{n}(n\geqq 2)$ denote the dihedral group of order $2n$ . Let denote the symmetric, alternating group on symbols, respectively. P. Cassou-Nogu\‘es  showed that, for a finite abelian group $G,$ $d(ZG)=1$ ( any prime), or . if and only if Hence we have only to treat the nonabelian case. Our main result is the following: THEOREM. $A$ finite nonabelian group for which $d(ZG)=1$ is isomorphic . to one of the groups: $D_{n}(n\geqq 3),$ $n$

$S_{n},$

$A_{n}$

$n$

$G\cong C_{1},$

$C_{p}$

$C_{4},$

$p$

$C_{6},$

$C_{8},$

$C_{9},$

$C_{10},$

$C_{14}$

$C_{2}\times C_{2}$

$G$

$A_{4},$

$S_{4},$

$A_{5}$

It is well known $(e. g. )$ that $d(ZA_{4})=d(ZS_{4})=d(ZA_{5})=1$ . It is also known that $d(ZD_{n})=1$ in each of the following cases: (i) is an odd prime (); (ii) is a power of an odd regular prime (); or (iii) is a power  (). Cassou-Nogu\‘es Recently 2 of showed that there is an infinite number of pairs $(p, q)$ of distinct odd primes such that $d(ZD_{pq})>1$ . It seems for which $d(ZD_{n})=1$ . difficult to determine all integers $n$

$n$

$n$

$p,$

$q$

$n$

\S 1.

The group

$T(ZG)$

.

be the ideal of $ZG$ generated by be a finite group and let . We define the subgroup $T(ZG)$ of $D(ZG)$ to be the kernel of the natural surjection $D(ZG)\rightarrow D(ZG/(\Sigma))$ and $t(ZG)$ to be the order of $T(ZG)$

Let

$\Sigma=_{\sigma\in G}7^{\urcorner}\sigma$

$G$

$(\Sigma)$

162

S. ENDO and Y. HIRONAKA

() and by We denote by $A(G)$ the Artin exponent of $|G|$ a Sylow -subgroup of for a prime divisor of . Recently S. Ullom  showed some basic and important results on $T(ZG)$ . The following theorem which is a summary of Ullom’s results will play an essential role in the proof of our main result.

(, ).

$G$

$G$

$P$

$p$

THEOREM 1. 1 (). (1) For any subquotient (2)

The exponent

(3)

$t(ZC_{n})=1$

$G^{(p)}$

of

divides

$T(ZG)$

for any

$A(G)$

$H$

of

$G$

$t(ZH)$

divides

$t(ZG)$

.

.

().

$n\geqq 1$

is an odd prime, then divides If is a noncyclic p-group where $t(ZG)$ . If is a noncyclic 2-group which is not dihedral, then 2 divides $t(ZG)$ . (5) If is the metacyclic group defined by (4)

$G$

$p$

$p$

$G$

$G$

$G=\langle\sigma, \tau|\sigma^{p}=\tau^{q}=1, \tau^{-1}\sigma\tau=\sigma^{r}\rangle$

where is an odd prime, is a divisor of of unity modulo , then $t(ZG)=q/(q, 2)$ . $p$

$q$

$p-1$

and

$r$

is a primitive q-th root

$p$

From this we deduce

PROPOSITION 1. 2. Let maximal normal subgroup \langle

$t\geqq 0$

)

$D_{2}t(t\geqq 1),$

$A_{4},$

$S_{4}$

$,$

$G$

be a

of

$O(G)$

or

$A_{5}$

finite group for $G$

of

which $t(ZG)=1$ . Then the odd order is cyclic and $G/O(G)\cong C_{2}c$

.

of odd order is cyclic. Therefore PROOF. By (1. 1) any subgroup of $O(G)$ is cyclic and, for any odd prime $p||G|$ , is also cyclic. Further, by \langle 1. 1) cyclic $G/O(G)$ or dihedral. Hence is is isomorphic to one of the groups: (i) , (ii) , (iii) the subgroups of (, containing $PSL(2, p^{s})$ where , or (iv) is an odd prime and $G$

$G^{(p)}$

$G^{(2)}$

$P\Gamma L(2, p^{s})$

$D_{2^{t}}(t\geqq 1)$

$C_{2^{t}}(t\geqq 0)$

$s\geqq 1$

$p$

$A_{7}$

).

Since contains a subgroup which is a semidirect product of and , by (1. 1) we have $t(ZA_{7})>1$ , and so the case such that acts faithfully on \langle iv) is excluded. Hence we have only to consider the case (iii). It is clear that $PSL(2, p^{s})^{(p)}$ is an elementary abelian -group of order . Therefore from \langle 1. 1) it follows that $s=1$ . The group $PSL(2, p)$ contains a subgroup which is a semidirect product of , then by (1. 1) and $C_{p-1/2}$ such that $C_{p-1/2}$ acts faithfully on . If $P\Gamma L(2, p)=PGL(2, p)$ $t(ZPSL(2, p))>1$ , $G/O(G)$ we have . Since must be isomorphic to one of the groups: $A_{4}(\cong PSL(2,3))$ , $S_{4}(\cong PGL(2,3))$ , $A_{7}$

$C_{7}$

$C_{3}$

$C_{7}$

$C_{3}$

$P$

$C_{p}$

$p^{s}$

$C_{p}$

$p\geqq 7$

$A_{5}(\cong$

or

. But

contains a subgroup which is a semi, and so again by such that acts faithfully on and direct product of (1. 1) we have $t(ZS_{5})>1$ . Thus we conclude that $G/O(G)\cong A_{4},$ or .

$PSL(2,5))$ ,

$S_{5}(\cong PGL(2,5))$ $C_{5}$

$C_{4}$

$S_{5}$

$C_{4}$

$C_{5}$

$S_{4}$

$A_{5}$

Finite groups with trivial class groups

\S 2. The group

$T(Z(C_{2}\times C_{2}\times C_{p}))$

163

.

In this section we give the following:

PROPOSITION 2. 1. Let be an odd prime. Then $t(Z(C_{2}\times C_{2}\times C_{p}))=2$ . We begin with LEMMA 2. 2. Let be an odd prime. Then $t(Z(C_{2}\times C_{2}\times C_{p}))=1$ or 2. (), this follows directly from (1. 1). PROOF. Since be a primitive n-th Let $U(S)$ denote the unit group of a ring . Let be the n-th cyclotomic polynomial. root of unity and let is a LEMMA 2. 3. The natural surjection bijection. $p$

$p$

$A(C_{2}\times C_{2}\times C_{p})=2$

$S$

$\zeta_{n}$

$\Phi_{n}(X)$

$D(Z(C_{2}\times C_{p}))\rightarrow D(ZC_{2}[\zeta_{p}])$

PROOF. For example, see . We refer to  and  for the Mayer-Vietoris sequence which will be used in \S \S 2 and 3. LEMMA 2. 4. The natural surjection is a bijection. , PROOF. Write and hence there is a pullback . Then we have $D(Z(C_{2}\times C_{2}\times C_{p})/(\Sigma))\rightarrow D(Z(C_{2}\times C_{2})[\zeta_{p}])$

$C_{2}\times C_{2}\times C_{p}=\langle\tau_{1},$

$\tau_{2},$

$\sigma|\tau_{1}^{2}=\tau_{2}^{2}=\sigma^{p}=1,$

$\tau_{1}\tau_{2}=\tau_{2}\tau_{1},$

$\tau_{1}\sigma=\sigma\tau_{1}$

$\Sigma=(1+\tau_{1})(1+\tau_{2})\Phi_{p}(\sigma)$

$\tau_{2}\sigma=\sigma\tau_{2}\rangle$

diagram $Z(C_{2}\times C_{2}\times C_{p})/(\Sigma)$

$\rightarrow$

$Z(C_{2}\times C_{p})$

$Z(C_{2}\times C_{2}\times C_{p})/((1+\tau_{2})\Phi_{p}\downarrow(\sigma))\rightarrow Z(C_{2}\times C_{p})/((1+\tau_{2})\Phi_{p}\downarrow(\sigma))$

. From this we get an exact (Mayer-Vietoris) sequence $U(Z(C_{2}\times C_{2}\times C_{p})/((1+\tau_{2})\Phi_{p}(\sigma)))\oplus U(Z(C_{2}\times C_{p}))$

$\rightarrow^{..\alpha}U(Z(C_{2}\times C_{p})/((1+\tau_{2})\Phi_{p}(\sigma)))\rightarrow D(Z(C_{2}\times C_{2}\times C_{p})/(\Sigma))$

$\rightarrow D(Z(C_{2}\times C_{2}\times C_{p})/((1+\tau_{2})\Phi_{p}(\sigma)))\oplus D(Z(C_{2}\times C_{p}))$

$\rightarrow D(Z(C_{2}\times C_{p})/((1+\tau_{2})\Phi_{p}(\sigma)))$

It is clear that

$\mu$

, we have Consequently it follows that

is surjective. Since

$\cong D(Z(C_{2}\times C_{p})/((1+\tau_{2})\Phi_{p}(\sigma)))$

.

.

$t(Z(C_{2}\times C_{p}))=1$

$D(Z(C_{2}\times C_{p}))$

$D(Z(C_{2}\times C_{2}\times C_{p})/(\Sigma))\cong D(Z(C_{2}\times C_{2}\times C_{p})/((1+\tau_{2})\Phi_{p}(\sigma)))$

Further, from the pullback diagram $Z(C_{2}\times C_{2}\times C_{p})/((1+\tau_{2})\Phi_{p}(\sigma))\rightarrow Z(C_{2}\times C_{p})$

$Z(C_{2}\times C_{2})[\zeta_{p}]\downarrow$

$ZC_{2}[\zeta_{p}]\downarrow$

$\rightarrow$

.

164

we

S. ENDO and Y. HIRONAKA

an exact sequence

get

$U(Z(C_{2}\times C_{2})[\zeta_{p}])\oplus U(Z(C_{2}\times C_{p}))\rightarrow^{\mu^{\prime}}U(ZC_{2}[\zeta_{p}])$

$\rightarrow D(Z(C_{2}\times C_{2}\times C_{p})/((1+\tau_{2})\Phi_{p}(\sigma)))$

$\rightarrow D(Z(C_{2}\times C_{2})[\zeta_{p}])\oplus D(Z(C_{2}\times C_{p}))$

$\rightarrow D(ZC_{2}[\zeta_{p}])$

Clearly

.

is surjective and by (2. 3)

$\mu^{\prime}$

$D(Z(C_{2}\times C_{p}))\cong D(ZC_{2}[\zeta_{p}])$

. Therefore

we have $D(Z(C_{2}\times C_{2}\times C_{p})/((1+\tau_{2})\Phi_{p}(\sigma)))\cong D(Z(C_{2}\times C_{2})[\zeta_{p}])$

.

Thus we conclude that $D(Z(C_{2}\times C_{2}\times C_{p})/(\Sigma))\cong D(Z(C_{2}\times C_{2})[\zeta_{p}])$

We denote by (resp. (resp.

$U(ZC_{2}[\zeta_{p}])$

$U^{*}(Z(C_{2}\times C_{p}))$

)

(resp.

) the image of

$U(Z(C_{2}\times C_{p}))$

$U(Z(C_{2}\times C_{p}))\rightarrow U(F_{2}(C_{2}\times C_{p}))$

).

LEMMA 2. 5. If the order of , then

map

under the natural

$U(ZC_{2}[\zeta_{p}])\rightarrow U(F_{2}C_{2}[\zeta_{p}])$

$U^{*}(ZC_{2}[\zeta_{p}])^{(2)}$

$U^{*}(ZC_{2}[\zeta_{p}])$

.

$U^{*}(Z(C_{2}\times C_{p}))^{(2)}$

$t(Z(C_{2}\times C_{2}\times C_{p}))=2$

is equal to the order of

.

PROOF. From the pullback diagrams $Z(C_{2}\times C_{2}\times C_{p})\rightarrow$

$Z(C_{2}\times C_{p})\downarrow$

$Z(C_{2}\times C_{2})[\zeta_{p}]\rightarrow ZC_{2}[\zeta_{p}]$

$Z(C_{2}\times C_{p})$

$\rightarrow F_{2}(C_{2}\times C_{p})\downarrow^{1}$

$\rightarrow F_{2}C_{2}[\zeta_{p}]\downarrow$

$ZC_{2}[\zeta_{p}]\downarrow$

,

we

get

a commutative diagram with exact rows $0\rightarrow U(F_{2}(C_{2}\times C_{p}))/U^{*}(Z(C_{2}\times C_{p}))\rightarrow D(Z(C_{2}\times C_{2}\times C_{p}))$

$U(F_{2}C_{2}[\zeta_{p}]\downarrow)/U^{*}(ZC_{2}[\zeta_{p}])$

$\rightarrow D(Z(C_{2}^{\vee}\times C_{2})[\zeta_{p}])|i^{\prime}$

$0\rightarrow$

$\rightarrow D(Z(C_{2}\times C_{p}))\oplus D(Z(C_{2}\times C_{p}))\rightarrow 0$ $|\lambda$

$D(ZC_{2}[\zeta_{p}])\oplus D(ZC_{2}[\zeta_{p}])$

$\rightarrow$

By (2. 3) $C_{2}\times C_{p}))$

is bijective and by (2. 4) the kernel of . Hence we have $\lambda$

$\lambda^{\prime}$

coincides with

$\rightarrow 0$

.

$T(Z(C_{2}\times$

$t(Z(C_{2}\times C_{2}\times C_{p}))$

$=|U(F_{2}(C_{2}\times C_{p}))||U^{*}(ZC_{2}[\zeta_{p}])|/|U(F_{2}C_{2}[\zeta_{p}])||U^{*}(Z(C_{2}\times C_{p}))|$

It is easy to see that

$|U(F_{2}(C_{2}\times C_{p}))|=2|U(F_{2}C_{2}[\zeta_{p}])|$

$(C_{2}\times C_{p}))^{(2)}|=|U^{*}(ZC_{2}[\zeta_{p}])^{(2)}|$

conclude by

(2. 2) that

, then

2 divides

$t(Z(C_{2}\times C_{2}\times C_{p}))=2$

.

. Therefore, if

$t(Z(C_{2}\times C_{2}\times C_{p}))$

.

. $|U^{*}(Z$

Thus we

Finite

groups with trivial class groups

165

We are now ready to prove (2. 1). PROOF OF (2. 1). Write a commutative diagram

$C_{2}\times C_{p}=\langle\tau, \sigma|\tau^{2}=\sigma^{p}=1, \tau\sigma=\sigma\tau\rangle$

.

Then we have

$\phi_{2}$

$U(Z[\tau, \sigma])\rightarrow U(F_{2}[\tau, \sigma])\cong U(F_{2}[\tau])\oplus U(F_{2}[\tau, \zeta_{p}])$

$U(Z\overline{\lfloor}\tau,\zeta_{p}])\downarrow\psi^{\prime}\rightarrow^{\phi_{1}}U(f_{2}[\tau, \zeta_{p}])\downarrow\psi$

. By (2. 5) it suffices to show that . Let be a primitive root modulo . The map induces a surjection $U(Z[\zeta_{p}+\zeta_{p^{-1}}])\rightarrow U(F_{p})$ . of $U(Z[\zeta_{p}+\zeta_{p^{-1}}])$ such that exist elements $|({\rm Im}\phi_{1})^{(2)}|=|({\rm Im}\phi_{2})^{(2)}|$

defined by Therefore there

$Z[\zeta_{p}]\rightarrow F_{p}$

$p$

$a$

$f(\zeta_{p})\rightarrow f(1)$

$u_{t}(\zeta_{p}),$

$1\leqq i\leqq t$

$U(Z[\zeta_{p}+\zeta_{p^{-1}}])=\langle u_{i}(\zeta_{p})|1\leqq i\leqq t\rangle$

$u_{i}(1)\equiv a(mod p)$

Since the exponent of

$U(F_{2}[\zeta_{p}])$

divides

$u_{i}(\zeta_{p})^{2p- 1-1}=1+2v_{i}(\zeta_{p})$

From the fact that

$u_{i}(1)\equiv a(mod p)$

,

$1\leqq i\leqq t$

$2^{p-1}-1$

,

,

,

.

we can write

$v_{i}(\zeta_{p})\in Z[\zeta_{p}+\zeta_{p}^{-1}]$

it follows that

$\prod_{j=1}^{p- 1/2}u_{i}(\zeta_{p^{j}})=N_{Q(\zeta_{p}+\zeta_{p^{-1}})/Q}(u_{i}(\zeta_{p}))=-1$

Through the inclusion $1-v_{i}(\zeta_{p})(\tau-1)$

with

$U(Z[\zeta_{p}])=\langle\zeta_{p}\rangle\cdot U(Z[\zeta_{p}+\zeta_{p^{-1}}])$

and

.

$2^{p-1}-1$

is odd.

$({\rm Im}\phi_{1})^{(2)}=\langle 1+\overline{v}_{i}(\zeta_{p})(\tau+1)|1\leqq i\leqq t\rangle$

Let $1\leqq i\leqq t\rangle$

$f(\tau, \sigma)$

be an element of

we may identify

. Then it is easy to see that

$\prod_{j=1}^{p-1/2}(1-v_{i}(\zeta_{p^{j}})(\tau-1))=\tau$

(2)

.

$U(Z[\tau, \zeta_{p}])\rightarrow U(Z[\zeta_{p}])\oplus U(Z[\zeta_{p}])$

$(1, u_{i}(\zeta_{p})^{2-1}p- 1)$

(1)

Note that

.

$Z[\tau, \sigma]$

such that

Then we have

.

$f(\tau, \zeta_{p})\in\langle 1-v_{i}(\zeta_{p})(\tau-1)|$

. Then we can write $f(\tau, \sigma)=\prod_{i=1}^{t}(1-v_{i}(\sigma)(\tau-1))^{h_{i}}+(c+d\tau)\Phi_{p}(\sigma)$

are integers. It is clear that if $f(\tau, 1)\in U(Z[\tau]),$ . , if and only if $f(1,1),$

if and only $f(-1,1)\in U(Z)$ . However $f(1,1)=1+p(c+d)$ and $f(-1,1)=\prod_{i}(1+2v_{i}(1))^{h_{i}}+p(c-d)\equiv a^{h}(mod p)$ where where

$h_{i},$

$c,$

$f(\tau, \sigma)\in U(Z[\tau, \sigma])$

$d$

$i$

$e.$

166

S. ENDO and Y. HIRONAKA

$h=(2^{p-1}-1)\sum_{i}h_{i}$

.

Therefore, if

$f(\tau, \sigma)\in U(Z[\tau, \sigma])$

, then $d=-c$ and $p-1/2|h$ .

Conversely, if $d=-c$ and $p-1/2|h$ , then $f(1,1)=1$ , and, for

1 when $f(-1,1)=\{-1$

when

$2h/p-1$

is even

$2h/P-1$

is odd.

some

$c$

,

In this case we have $f(\tau, \sigma)=\prod_{l}(1-v_{i}(\sigma)(\tau-1))^{h_{i}}-c(\tau-1)\Phi_{p}(\sigma)\in U(Z[\tau, \sigma])$

and (li $(1+v_{i}(\zeta_{p})(\tau+1))^{h_{i}}$ ,

1) when $2h/p-1$

$\phi_{2}(f(\tau, \sigma))=\{(\prod_{i}(1+\overline{v}_{i}(\zeta_{p})(\tau+1))^{h_{i}}, \tau)$

If we put we see that

(3)

when

$2h/P-1$

is even is odd.

$V=\{f(\tau, \sigma)\in U(Z[\tau, \sigma])|f(\tau, \zeta_{p})\in\langle 1-v_{i}(\zeta_{p})(\tau-1)|1\leqq i\leqq t\rangle\}$

$({\rm Im}\phi_{2})^{(2)}=\phi_{2}(V)$

.

, then

From this it follows that

$({\rm Im}\phi_{2})^{(2)}=\left\{\begin{array}{l}\langle(1+\overline{v}_{i}(\zeta_{p})(\tau+1), \tau)|1\leqq i\leqq t\rangle when p\equiv 3(mod 4)\\\langle(1, \tau), ((1+\overline{v}_{i}(\zeta_{p})(\tau+1))(1+\overline{v}_{i^{\prime}}(\zeta_{p})(\tau+1)), 1)|1\leqq i1$ for any prime Since $d(ZC_{2p})$ divides $d(Z(C_{p}\times D_{q}))$ , this implies that $d(Z(C_{p}\times D_{q}))>1$ for any . However $d(ZC_{2p})$ is odd. Hence this does not imply that 2 divides $\mu_{2}$

$k$

$P$

$(1, 1, 1, \overline{a}^{k})\not\in{\rm Im}\mu_{2}$

$N_{F_{p}[\zeta q+\zeta_{q^{-1}}]/F_{p}}$

$\overline{\gamma}$

$a$

$\mu_{1}$

$U(F_{p}[\zeta_{q}+\zeta_{q}^{-1}])\rightarrow U(F_{p})$

$U(F_{p}[\zeta_{q}+\zeta_{q}^{-1}])$

$(1, 1, 1, \overline{\gamma}^{k})$

$N_{F_{p}[\zeta_{q}+\zeta_{q^{-1}}]/F_{p}}(\overline{\gamma})=\overline{a}$

$\langle(1,1,1,\overline{\gamma})\rangle$

$\not\in{\rm Im}\mu_{1}$

$\mu_{1}|$

${\rm Im}\mu_{1}$

$p\geqq 11$

$p\geqq 11$

$d(Z(C_{p}\times D_{q}))$

.

PROPOSITION 3. 4. Let $d(Z(D_{p}\times D_{q}))$

$p$

,

$q$

be distinct

odd primes.

Then 2 divides

.

PROOF. Write $D_{p}\times D_{q}=\langle\sigma,$

$\tau_{1},$

$\tau_{2}|\sigma^{pq}=\tau_{1}^{2}=\tau_{2}^{2}=1,$

$\tau_{1}\sigma^{q}=\sigma^{-q}\tau_{1},$

$\tau_{2}\sigma^{q}=\sigma^{q}\tau_{2},$

$\tau_{1}\sigma^{p}=\sigma^{p}\tau_{1}$

$\tau_{2}\sigma^{p}=\sigma^{-p}\tau_{2},$

,

$\tau_{1}\tau_{2}=\tau_{2}\tau_{1}\rangle$

.

170

S. ENDO and Y. HIRONAKA

Then we have a commutative diagram $Z[\tau_{1}, \tau_{2}]\oplus Z[\zeta_{p},\zeta_{q}Z(D_{p\downarrow}\times D_{q})\tau_{1}, \tau_{2}]$

$Z[\zeta_{p}, \tau_{1}, \tau_{2}]\oplus Z[\zeta_{q}, \tau_{1}, \tau_{2}]$

$\downarrow$

.

$\rightarrow F_{p}[\tau_{1}, \tau_{2}]\oplus F_{q}[\zeta_{p}, \tau_{1}, \tau_{2}]\oplus F_{q}[\tau_{1}, \tau_{2}]\oplus F_{p}[\zeta_{q}, \tau_{1}, \tau_{2}]$

From this we get an exact sequence $U(Z[\tau_{1}, \tau_{2}])\oplus U(Z[\zeta_{p}, \zeta_{q}, \tau_{1}, \tau_{2}])\oplus U(Z[\zeta_{p}, \tau_{1}, \tau_{2}])\oplus U(Z[\zeta_{q}, \tau_{1}, \tau_{2}])$ $\nu$

$\rightarrow U(F_{p}[\tau_{1}, \tau_{2}])\oplus U(F_{q}[\zeta_{p}, \tau_{1}, \tau_{2}])\oplus U(F_{q}[\tau_{1}, \tau_{2}])\oplus U(F_{p}[\zeta_{q}, \tau_{1}, \tau_{2}])$

$\rightarrow D(Z(D_{p}\times D_{q}))\rightarrow D(Z[\tau_{1}, \tau_{2}])\oplus D(Z[\zeta_{p} , \zeta_{q}, \tau_{1}, \tau_{2}])$

$\oplus D(Z[\zeta_{p}, \tau_{1}, \tau_{2}])\oplus D(Z[\zeta_{q}, \tau_{1}, \tau_{2}])\rightarrow 0$

.

, Note that and . Since ] is hereditary, it follows that . It is clear that $D(Z[\zeta_{p}, \tau_{1}, \tau_{2}])\cong D(ZD_{2p})$ $D(Z[\tau_{1}, \tau_{2}])=0$ . , Further we see that and ]) $\cong D(ZD_{2q})$ . Therefore we have an exact sequence $\tau_{1}\zeta_{p}=\zeta_{p}^{-1}\tau_{1},$

$\tau_{1},$

$\tau_{1}\zeta_{q}=\zeta_{q}\tau_{1},$

$\tau_{2}\zeta_{p}=\zeta_{p}\tau_{2}$

$Z[\zeta_{p},$

$\tau_{2}\zeta_{q}=\zeta_{q^{-1}}\tau_{2}$

$\zeta_{q}$

$D(Z[\zeta_{p}, \zeta_{q}, \tau_{1}, \tau_{2}])=0$

$\tau_{2}$

$D(Z[\zeta_{q}$

$\tau_{1},$

$\tau_{2}$

$0\rightarrow Coker\nu\rightarrow D(Z(D_{p}\times D_{q}))\rightarrow D(ZD_{2p})\oplus D(ZD_{2q})\rightarrow 0$

Denote by $F_{q},$

$F_{p}$

$S$

one of the rings

and define a map

$N_{s}$

:

$Z[\zeta_{p}, \zeta_{q}],$

$Z[\zeta_{p}],$

$Z[\zeta_{q}],$

$U(S[\tau_{1}, \tau_{2}])\rightarrow U(S)$

$x+y\tau_{1}+z\tau_{2}+w\tau_{1}\tau_{2^{-}}|\tau_{1}\tau_{2}(w)\tau_{1}(y)\tau_{2}(z)x$

$Z,$

$F_{q}[\zeta_{p}],$

. $F_{p}[\zeta_{q}]$

,

by

$\tau_{1}(x)$

$\tau_{1}(w)$

$\tau_{1}\tau_{2}(z)\tau_{2}(w)y$

$\tau_{1}\tau_{2}(y)\tau_{2}(x)z$

$\tau_{1}\tau_{2}(x)\tau_{2}(y)\tau_{1}(z)w|$

. Here it should be noted that $N^{\prime}=(N_{Z}, N_{Z[\zeta_{p},\zeta_{q}]}, N_{ZL\zeta_{p}3}, N_{Z[\zeta_{q}]})$

$N_{z[\zeta_{p},\zeta_{q}]}=NrdQ\zeta\zeta_{p}\zeta_{q},\tau_{1’\cdot 2}- 3/Q\mathfrak{c}\zeta_{p}+\zeta_{p^{-1}’}\zeta_{q}+\zeta_{q^{-13}}$

and

$N=(N_{F_{p}}, N_{p_{q}[\zeta_{p}]}, N_{F_{q}}, N_{p_{p}[\zeta_{q}]})$

.

Let

. Then we

get a commutative diagram $U(Z[\tau_{1}, \tau_{2}])\oplus U(Z[\zeta_{p}, \zeta_{q}, \tau_{1}, \tau_{2}])\oplus U(Z[\zeta_{p}, \tau_{1}, \tau_{2}])\oplus U(Z[\zeta_{q}, \tau_{1}, \tau_{2}])$

$\downarrow N^{\prime}$

$U(Z)\oplus U(Z[\zeta_{p}+\zeta_{p^{-1}}, \zeta_{q}+\zeta_{q}^{-1}])\oplus U(Z[\zeta_{p}+\zeta_{p}^{-1}])\oplus U(Z[\zeta_{q}+\zeta_{q^{-1}}])$ $\nu$

$-U(F_{p}[\tau_{1}, \tau_{2}])\oplus U(F_{q}[\zeta_{p}, \tau_{1}, \tau_{2}])\oplus U(F_{q}[\tau_{1}, \tau_{2}])\oplus U(F_{p}[\zeta_{q}, \tau_{1}, \tau_{2}])$

$\underline{\nu_{1U(F_{p})\oplus U(F_{q}[\zeta_{p}+\zeta_{p^{-1}}])\oplus U(F_{q})\oplus U(F_{p}[\zeta_{q}+\zeta_{q^{-1}}])}}\downarrow N$

Finite

groups with trivial class groups

171

denotes the restriction map of . The map $N$ is surjective and hence $|Coker\nu_{1}|$ divides $d(Z(D_{p}\times D_{q}))$ . Then, in the same way as in the proof of (3. 2), we can show that 2 divides Coker . Thus we conclude that 2 divides

where

$\nu$

$\nu_{1}$

$\nu_{1}|$

$d(Z(D_{p}\times D_{q}))$

.

REMARK 3. 5. In the proof of (3. 4) it was shown that there exists a surjection: $D(Z(D_{p}\times D_{q}))\rightarrow D(ZD_{2p})\oplus D(ZD_{2q})$ . It is seen that, if is one of $d(ZD_{2p})=1$ $p=47,53,59,179$ 19379, , primes: then . the However we do or not. not know whether $d(ZD_{2p})=1$ for any prime , define the group and For any integers as follows: $p$

$p\leqq 31,$

$p$

$n\geqq 3$

$t\geqq 2$

$H_{n.t}$

$H_{n.t}=\langle\sigma, \tau|\sigma^{n}=\tau^{2^{i}}=1, \sigma\tau=\tau\sigma^{-1}\rangle$

PROPOSITION 3. 6. Let

$P$

.

Then 2 divides

be an odd prime.

$d(ZH_{p.2})$

.

PROOF. By Fr\"ohlich  this has already been proved in a more precise form. For completeness we give a simple proof. Now there is an exact sequence $U(Z[\zeta_{p},\overline{\tau}])\oplus U(Z[\overline{\tau}])\rightarrow U(F_{p}[\overline{\tau}])\rightarrow D(ZH_{p.2}/(\vee^{-1))}-2$

$-D(Z[\zeta_{p},\overline{\tau}])\oplus D(Z[\overline{\tau}])\rightarrow 0$

where and . It is clear that Therefore we get an exact sequence $\overline{\tau}^{2}=-1$

$D(Z[\zeta_{p},\overline{\tau}])=D(Z[\overline{\tau}])=0$

$\zeta_{p}\overline{\tau}=\overline{\tau}\zeta_{p}^{-1}$

$U(Z[\zeta_{p},\overline{\tau}])\rightarrow^{\phi}U(F_{p}[\overline{\tau}])\rightarrow D(ZH_{p,2}/(\tau^{2}+1))\rightarrow 0$

.

.

$d\in Z$ be a solution of the be a primitive root modulo and let $(c, d),$ Let congruence $X^{2}+Y^{2}\equiv a(mod p)$ . Then and . Then there exists . Let be an odd integer and suppose that , . Write such that and let . Then $p$

$a$

$c,$

$(\overline{c}+\overline{d}\overline{\tau})(\overline{c}-\overline{d}\overline{\tau})$

$\alpha=\overline{c}+\overline{d}\overline{\sim}\in U(F_{p}[\overline{\tau}])$

$k$

$=\overline{a}$

$\alpha^{k}\in{\rm Im}\phi$

$f(\zeta_{p},\overline{\tau})\in U(Z[\zeta_{p},\overline{\tau}])$

$f_{1}(\zeta_{p}),$

$f_{2}(\zeta_{p})\in Z[\zeta_{p}]$

$\overline{f}(1,\overline{\tau})=\alpha^{k}$

$f(\zeta_{p},\overline{\tau})=f_{1}(\zeta_{p})+f_{2}(\zeta_{p})_{\overline{T}}$

$g(\zeta_{p},\overline{\tau})=f_{1}(\zeta_{p^{-1}})-f_{2}(\zeta_{p}^{-1})\overline{\tau}$

$Nrd_{Q[\zeta_{p}\cdot]/Q\ddagger\zeta_{p}+\zeta_{p^{-11}}}\overline{\tau}(f(\zeta_{p},\overline{\tau}))=f(\zeta_{p},\overline{\tau})g(\zeta_{p},\overline{\tau})=f_{1}(\zeta_{p})f_{1}(\zeta_{p}^{-1})+f_{2}(\zeta_{p})f_{2}(\zeta_{p}^{-1})>0$

Consequently

$N_{Q[\zeta_{p}+\zeta_{p^{-1}}]/Q}(f(\zeta_{p},\overline{\tau})g(\zeta_{p},\overline{\tau}))=1$

and therefore . This implies that Hence cludes that 2 divides $d(ZH_{p,2})$ .

$\overline{(N_{Q\mathfrak{c}\zeta_{p}+\zeta_{p^{-1J/Q}}}(f(\zeta_{p},\overline{\tau})g(\zeta_{p},\overline{\tau}))}$

$\alpha^{k}\not\in{\rm Im}\phi$

\S 4.

$\langle\alpha\rangle^{(2)}$

)

.

However

$f(1,\overline{\tau})\overline{g}(1,\overline{\tau})=\overline{a}^{k}$

$=(d^{k})^{p-1/2}=\overline{-1}$

,

is not contained in

. ,

a contradiction. ${\rm Im}\phi$

, which

con-

The main result.

We are now in a position to prove our main result. THEOREM 4. 1. finite nonabelian group for which . phic to one of the groups: $D_{n}(n\geqq 3),$ $G$

$A$

$A_{4},$

$S_{4},$

$A_{5}$

$d(ZG)=1$ is

isomor-

172

S. ENDO and Y. HIRONAKA

PROOF. By (1. 2) $O(G)$ is cyclic and $G/O(G)\cong C_{2^{t}},$ and let $m=|O(G)|$ . Let $C(G)$ denote the center of Write $A_{4},$

$D_{2^{t}},$

$O(G)=\langle\rho\rangle$

or

$S_{4}$

$G$

$A_{5}$

.

.

and so we may write Assume that $G/O(G)\cong C_{2^{i}}$ . Then by conjugation yields a map . The action of on : nonabelian, [ Aut . Since . If is : Ker ] , we have Ker , then there is a prime $p|m$ such that acts faithfully on . Therefore by (1. 1) $d(ZG/(\tau^{2}s-1, \rho^{m/p}-1))>1$ and so $d(ZG)>1$ . Thus we : Ker ] $=2$ . In this case, suppose that $O(G)\cap C(G)\neq\{1\}$ and must have [ and . Then . Hence let . According to (3. 2) $d(ZG/(\tau^{2}-1))$ and so $>1$ and therefore $d(ZG)>1$ . Consequently we have $O(G)\cap C(G)=\{1\}$ . Then . By virtue of (3. 6), $d(ZG)>1$ if . Hence . we must have $t=1,$ . (i)

$G^{(2)}\cong C_{2^{t}}$

$G^{(2)}=\langle\tau\rangle$

$\langle\tau\rangle$

$\langle\rho\rangle$

$G$

$\langle\rho\rangle$

$\langle\tau\rangle\rightarrow$

$\phi$

$\langle\tau\rangle$

$\phi\subsetneqq\langle\tau\rangle$

$s\geqq 2$

$\langle\tau\rangle/Ker\phi$

$\langle\tau\rangle$

$\langle\rho\rangle^{(p)}$

$\phi$

$m^{\prime\prime}=m/m^{\prime}>1$

$m^{\prime}|m$

$O(G)\cap C(G)=\langle\rho^{m^{\prime}}\rangle,$

$(m^{\prime}, m^{\prime\prime})=1$

$G/\langle\tau^{2}\rangle\cong C_{m^{\prime}}\times D_{m},$

$G=\langle\rho^{m^{1}}\rangle\times\langle\rho^{m^{\prime}}, \tau\rangle$

$t\geqq 2$

$G=\langle\rho, \tau|\rho^{m}=\tau^{2^{t}}=1, \rho\tau=\tau\rho^{-1}\rangle$

$i$

$G\cong D_{m}$

$e.,$

Assume that

(ii)

$G/O(G)\cong D_{2^{t}},$

$t\geqq 2$

. Then

. If write for a a subquotient of which is isomorphic to (1. 1) and (2. 1) it follows that $t(ZG)>1$ . Therefore we have $G$

Let action of

, be all prime divisors of and let by conjugation yields a map : Since Aut is abelian,

$G^{(2)}$

.

$N_{i}=Ker\phi_{i}$

$C_{2}\times C_{2}\times C_{p}$

$1\leqq i\leqq r$

on

and so we may

$G^{(2)}\cong D_{2^{t}}$

$0(G)\cap C(G)\neq\{1\}$

$G^{(2)}=\langle\sigma, \tau|\sigma^{2^{t}}=\tau^{2}=1, \sigma\tau=\tau\sigma^{-1}\rangle$

$p_{i},$

$=2^{s}$

$\phi$

$m$

$O(G)r)C(G)=\{1\}$ .

$[G^{(2)}, G^{(2)}]=\langle\sigma^{2}\rangle\subset N_{i}$

$\langle\rho_{i}\rangle$

. The . Let . Only one of , for some , then

$\langle\rho_{i}\rangle=\langle\rho\rangle^{(p_{i)}}$

$G^{(2)}\rightarrow Aut\langle\rho_{t}\rangle$

$\phi_{i}$

$\langle\rho_{i}\rangle$

, then there exists prime $p|m$ . From

$\sigma$

because $O(G)\cap C(G)=\{1\}$ . If is contained in in , this shows or is . Since contained either which is isomorphic to that there exists a subquotient of . Again from (2. 1) it follows that $t(ZG)>1$ . Thus we must have for . each , which implies that $\tau,$

$\sigma\not\in N_{i}$

$N_{i}$

$\tau\sigma$

$N_{i}$

$\sigma\tau$

$\tau$

$i$

$\langle\sigma^{2}, \tau\rangle\cong\langle\sigma^{2}, \tau\sigma\rangle\cong D_{2^{t-1}}$

$G$

$C_{2}\times C_{2}\times C_{p_{i}}$

$\sigma\in N_{i}$

$G\cong D_{2^{t}m}$

$i$

. Then we have Assume that and . In the same way as in the case (ii) it can be shown that $O(G)\cap C(G)=\{1\}$ . If there exist such that and , then . such that Now suppose that there exist but . Then we can find such that and (iii)

$G/O(G)\cong D_{2}\cong C_{2}\times C_{2}$

$G^{(2)}\cong C_{2}\times C_{2}$

$0(G)\neq\{1\}$

$\tau_{1},$

$\rho\tau_{1}=\tau_{1}\rho^{-1}$

$\rho\tau_{2}=$

$G\cong C_{2}\times D_{m}\cong D_{2m}$

$\tau_{2}\rho$

$\tau_{1},$

$\tau_{2}\rho^{-1},$

$\tau_{2}\in G^{(2)}$

$\rho_{1},$

$\tau_{2}\rho$

$\rho_{2}\tau_{2}=\tau_{2}\rho_{2}$

.

Clearly

$\tau_{2}\in G^{(2)}$

$\rho\tau_{1}=\tau_{1}\rho^{-1}$

$\rho_{2}\in\langle\rho\rangle$

$\rho_{1}\tau_{1}\tau_{2}=\tau_{1}\tau_{2}\rho_{1}$

and

$\rho=\rho_{1}\rho_{2},$

$\rho_{2}\tau_{1}\tau_{2}=\tau_{1}\tau_{2}\rho_{2}^{-1}$

and . Then we have $d(ZG)>1$ . Therefore that, suppose for any $\tau\in G^{(2)}-\{1\},$ Next such that We can Pnd $\times\langle\rho_{2}, \tau_{2}\rangle$

. Let

$m_{1}=|\langle\rho_{1}\rangle|$

$m_{2}=|\langle\rho_{2}\rangle|$

, and

$(m_{1}, m_{2})=1$

$\rho\tau_{2}\neq$

$\rho_{1}\tau_{2}=\tau_{2}\rho_{1}^{-1}$

so

and

$G=\langle\rho_{1}, \tau_{1}\tau_{2}\rangle$

$G\cong D_{m_{1}}\times D_{m_{2}}$

.

$\tau_{1}\in G^{(2)}-\{1\}$

.

by (3. 4)

$\rho\tau\neq\tau\rho^{-1},$

$\rho_{1},$

assumption both Then such that

$\rho_{1}$

$\langle\rho_{1}\rangle$

. Let

and . By and are different from 1. Further let . $O(G)\cap C(G)=\{1\}$ , and hence we can Pnd because . By assumption we have and

$\rho_{2}\in\langle\rho\rangle$

$\rho_{2}\tau_{2}=\tau_{2}\rho_{2}^{-1}$

$\tau\rho$

$\rho_{1}=\rho_{1}^{\prime}\rho_{2}^{\prime},$

$\rho=\rho_{1}\rho_{2},$

$\rho_{1}\tau_{1}=\tau_{1}\rho_{1}^{-1}$

$\rho_{2}\tau_{1}=\tau_{1}\rho_{2}$

$\tau_{2}\in G^{(2)}-\{1, \tau_{1}\}$

$\rho_{2}$

$\rho_{1}^{\prime},$

$\rho_{1}^{\prime}\tau_{2}=\tau_{2}\rho_{1}^{\prime}$

$\rho_{2}^{\prime}\tau_{2}=\tau_{2}\rho_{2^{-1}}^{\prime}$

$\rho_{2}^{\prime}\in$

Finite groups with trivial class groups

$\rho_{1}^{\prime}\neq 1$

. Let

$m_{1}=|\langle\rho_{1}^{\prime}\rangle|$

and

$m_{2}=|\langle\rho_{2}\rangle|$

. Then

173

and therefore,

$G/\langle\rho_{2}^{\prime}\rangle\cong D_{m_{1}}\times D_{m_{2}}$

by (3. 4), $d(ZG)>1$ . (iv) Assume that $G/O(G)\cong A_{4},$ or . If $O(G)=\{1\}$ , then $G\cong A_{4},$ or , as required. Suppose that $O(G)\neq\{1\}$ and let : be the $G$ map defined by the action of by conjugation. Note that on $(G/O(G))^{(2)}$ . Then we see that Ker contains a subgroup which is isomorphic $S_{4}$

$A_{5}$

$S_{4}$

$A_{5}$

$G\rightarrow Aut\langle\rho\rangle$

$\phi$

$G^{(2)}\cong$

$\langle\rho\rangle$

$\phi$

to

. Therefore there exists a subgroup $H$ of . According to (2. 1) we have $t(ZG)>1$ . From $(i)\sim(iv)$ we conclude that $G\cong D_{n}(n\geqq 3),$

$C_{2}\times C_{2}$

$G$

which is isomorphic to

$C_{2}\times C_{2}\times C_{m}$

$A_{4},$

$S_{4}$

or

$A_{5}$

.

for which REMARK 4. 2. It is fairly difficult to determine all integers $d(ZD_{n})=1$ . It should be noted () that there exists an infinite number of pairs $(p, q)$ of distinct odd primes such that $d(ZD_{pq})>1$ . Some further group given $T(ZG)$ in a forthcoming paper. will be results on the $n$

$p,$

$q$

References 

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1972).

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(1976),

116-136.

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31  

(1974),

305-342.

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174

S. ENDO and Y. HIRONAKA

Shizuo ENDO

Yumiko HIRONAKA

Department of Mathematics Tokyo Metropolitan University Fukazawa-cho, Setagaya-ku, Tokyo, 158 Japan

Department of Mathematics University of Tsukuba Sakura.mura, Niihari-gun, Ibaraki, 300-31 Japan