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 [1] 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 $(e. g. [14])$ 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 ([9]); (ii) is a power of an odd regular prime ([7]); or (iii) is a power [2] ([4]). 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
([8]) and by We denote by $A(G)$ the Artin exponent of $|G|$ a Sylow -subgroup of for a prime divisor of . Recently S. Ullom [14] 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.
([13], [14]).
$G$
$G$
$P$
$p$
THEOREM 1. 1 ([14]). (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
$t(ZG)$
.
.
([13]).
$n\geqq 1$
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 ([12], containing $PSL(2, p^{s})$ where , or (iv) is an odd prime and $G$
$G^{(p)}$
$G^{(2)}$
$P\Gamma L(2, p^{s})$
$D_{2^{t}}(t\geqq 1)$
$C_{2^{t}}(t\geqq 0)$
$s\geqq 1$
$p$
$A_{7}$
[5]).
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. ([8]), 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 [11]. We refer to [10] and [11] 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 [3] 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 ([2]) 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 [1]
P. Cassou-Nogu\‘es, Classes d’id\’eaux de l’alg\‘ebre d’un groupe ab\’elien, C. R. Acad. Sci. Paris, 276 (1973), A973-A975. (Doctorat de Sp\’ecialit\’e, Bordeaux,
[2]
P. Cassou-Nogu\‘es,
1972).
J. Algebra, 41
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(1976),
116-136.
A Fr\"ohlich, Module invariants and root numbers for quaternion fields of degree , Proc. Cambridge Philos. Soc., 76 (1974), 393-399. [4] A. Fr\"ohlich, M. E. Keating and S. M. J. Wilson, The class group of quaternion and dihedral 2-groups, Mathematika, 21 (1974), 64-71. [5] D. Gorenstein, Finite groups, Harper & Row Publ., New York, 1968. [6] H. Hasse, \"Uber die Klassenzahl abelscher Zahlk\"orper, Akademie Verlag, Berlin,
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$4l^{r}$
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M. E. Keating, Class groups of metacyclic groups of order ,$ p $a regular prime, Mathematika, 21 (1974), 90-95. T. Y. Lam, Artin exponent of finite groups, J. Algebra, 9 (1968), 94-119. M. P. Lee, Integral representations of dihedral groups of order $2p$ , Trans. Amer. Math. Soc., 110 (1964) , 213-231. J. Milnor, Introduction to algebraic $K$-theory, Ann. of Math. Studies, Princeton Univ. Press, Princeton, 1971. I. Reiner and S. Ullom, A Mayer-Vietoris sequence for class groups, J. Algebra, $p^{r}q$
31 [12] [13]
(1974),
305-342.
M. Suzuki, On finite groups with cyclic Sylow subgroups for all odd primes, Amer. J. Math., 77 (1955), 657-691. R. G. Swan, Periodic resolutions for finite groups, Ann. of Math., 72 (1960),
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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