Sharp characters with irrational values - Project Euclid

J. Math. Soc. Japan Vol. 48, No. 3, 1996

Sharp characters with irrational values By Dean ALVIS and

$S6hei$

NOZAWA

(Received Apr. 1, 1993) (Revised Apr. 11, 1995)

\S 1. Introduction. of degree . Let be a finite group, and let be a character of Let on the nonidentity elements of . We call the be the set of values of is a . It is known that type of the pair . Put $|G|$ [3], [4]). (see pair is The integer of is a divisor and rational said to be sharp if $f_{L}(n)=|G|$ . The notion of sharpness was first introduced for permutation characters by Ito and Kiyota in [5] as a generalization of sharply -transitive permutation representations. Indeed, if is a sharply -transitive permutation group and is sharp of type $\{0,1,$ , is the associated permutation character, then $t-1\}$ . In [4], Cameron and Kiyota extended the definition of sharp pair to of that given above and posed the problem of classifying, for a given set algebraic integers, the sharp pairs of type . The pair is said to be normalized if does not contain the principal character of as a constituent. If does contain the principal character as a constituent with multiplicity $m$ , and if we put $x’=x-m\cdot 1_{G},$ $L’=\{\alpha-m|\alpha\in L\}$ , is sharp of type then if and only if $(G, x’)$ is sharp of type $L’$ . Thus when classiit is no loss of generality to consider only normalized pairs fying sharp pairs. In this paper we give a complete classification of sharp pairs of type contains an irrational number. Several special cases have appeared in when the literature. For example, we have the following results. (See also [1], [9], $G$

$G$

$\chi$

$L$

$G$

$\chi$

$f_{L}(n)$

$f_{L}(x)=\Pi_{\alpha\in L}(x-\alpha)$

$(G, \chi)$

$L$

$n$

$(G, \chi)$

$f_{L}(n)$

$t$

$G$

$\chi$

$t$

$(G, \chi)$

$\cdots$

$L$

$L$

$(G, \chi)$

$(G, \chi)$

$\chi$

$G$

$\chi$

$1_{G}$

$L$

$(G, \chi)$

$(G, \chi)$

$L$

$L$

[6], [7], [8].)

is sharp and normalized THEOREM 1.1 (Cameron-Kiyota, [4]). SuPPose , algebraic conjugacy where class of irrational single tyPe a consists of of $G$ values. Then is either linear or the sum is cyclic of odd prime order, and of two complex conjugate linear characters of . $(G, \chi)$

$L$

$L$

$\chi$

$G$

is sharp THEOREM 1.2 (Alvis-Kiyota-Lenstra-Nozawa, [2]). SupPose . Then is a cyclic group of odd and normalized of type , where order, and is either a linear character or the sum of two complex cmjugate $(G, \chi)$

$L$

$\chi$

$L\cap Z=\emptyset$

$G$

568

D. ALVIS and S. NOZAWA

of .

linear characters

$G$

Our main result is the following. is sharp and normalized and THEOREM 1.3. Suppose assumes an irrational value. Then me of the following holds. (i) and is linear, or else is cyclic of order $m$ , and either and is the sum of two complex conjugate linear characters of . (ii) is dihedral of order $2m$ , where is odd, and is irreducible of degree 2. (iii) is dihedral or generalized quaternion of order $2m$ , where is $x=\psi$ even, and , where is irreducible of degree 2 and or is linear with cyclic kernel of order $m$ . (iv) is isomorphic to the binary octahedral group and is irredudble of $(G, \chi)$

$\chi$

$m\geqq 3$

$G$

$m\geqq 5$

$\chi$

$G$

$\chi$

$G$

$m\geqq 5$

$\chi$

$G$

$m\geqq 8$

$x=\psi+\epsilon$

$\psi$

$\epsilon$

$G$

$\chi$

degree 2. is isomorphic to is isomorphic to

(v)

$G$

(vi)

$G$

and is irreducible of degree 2. and is irreducible of degree 3.

$SL(2,5)$ $A_{6}$

$\chi$

$\chi$

\S 2. The proof of Theorem 1.3. We require the following two results, the second of which is proved in the next section. THEOREM 2.1 (Blichfeldt [3], Cameron-Kiyota [4]). Let be the set of values of and let be a character of G. Let . is a rational integer, and $|G|$ dizndes

$G$

$L$

$\chi$

$\chi$

finite group, on . Then

be a

$G\backslash \{1\}$

$f_{L}(n)$

$f_{L}(n)$

THEOREM 2.2. Suppose $(G, \chi)$ is sharp and normalized, $g\in G$ , and is $G$ $L(g)$ be a representation of affording and let irrational. Let be the set following the Then on . hold. of values of (i) $C_{G}(g)$ is cyclic, and assumes irrational values on the generators of $C_{G}(g)$ . (ii) Any nonidentity eigenvalue of occurs with multiplicity 1. Moreover, , then . if and are distinct nonidentity eigenvalues of (iii) If or $o(g)$ is odd, then $f_{L(g)}(n)=o(g)$ . If and $o(g)$ is even, then $f_{L(\emptyset)}(n)=2o(g)$ . $\chi(g)$

$\chi$

$\rho$

$\chi$

$\langle g\rangle\backslash \{1\}$

$\chi$

$\rho(g)$

$\omega’$

$\omega$

$\rho(g)$

$\chi\neq\overline{\chi}$

$\omega’=\overline{\omega}$

$x=\overline{x}$

PROOF OF THEOREM 1.3. Throughout this proof we suppose $(G, \chi)$ is sharp contains an irrational number. Also, we and normalized of type , where By Theorem 2.2, there is some affording a representation of denote by $g\in G$ such that is irrational and . We fix such an element for $H=C_{G}(g),$ $N=N_{G}(H)$ proof. this . Put the remainder of $L$

$L$

$G$

$\rho$

$\chi(g)$

STEP 1.

If

$x\neq\overline{x}$

$\chi$

.

$C_{G}(g)=\langle g\rangle$

, then

$G$

is cyclic and

$\chi$

is linear.

$g$

569

Sharp characters with irrational values

has a single nonidentity eigenvalue . Then PROOF. Suppose $N=C_{G}(g)=H$. Also, $C_{G}(h)=H$ for Thus by multiplicity 2.2. 1 Theorem with is a Frobenius group . Thus , so and $xH\cap H=\{1\}$ for $K$ $H$ is the be the Frobenius kernel, so with Frobenius complement . Let $K$ is normalas ranges over . Since disjoint union of the sets ized, we have $\rho(g)$

$\chi\neq\overline{\gamma_{\vee}}$

$G$

$x\in G\backslash H$

$h\in H\backslash \{1\}$

$G\backslash K$

$xH\backslash \{1\}$

$0=|G$

so

:

$(G, \chi)$

$x$

$K|\cdot(\chi 1)_{G}=(\chi 1)_{K}+|H|\cdot(\chi 1)_{H}-n$

,

and hence $n=1$ since $|H|\geqq 3$ . Therefore is faithful. Thus $G=C_{G}(g)=\langle g\rangle$ as is abelian since

$(\chi 1)_{K}=1-(|H|-1)(n-1)\geqq 0$ ,

is linear, and so required.

$G$

$\chi$

$\chi$

$\square$

be a nonidentity . Let We suppose for the rest of this proof that $H$ such that . , and let be the linear character of eigenvalue of of a is of such that constituent unique character irreducible Let be the has both and the induced character . Put $m=|H|$ and $d=\psi(1)$ . Since only two nonidentity eigenvalues by Theorem 2.2, $|N:H|$ S2. Also, $C_{G}(h)=H$ . for any $h\in H$ such that $x=\overline{x}$

$\omega$

$\lambda(g)=\omega$

$\lambda$

$\rho(g)$

$G$

$\psi$

$\psi$

$\lambda^{G}$

$\chi$

$\rho(g)$

$h^{2}\neq 1$

. STEP 2. Suppose conjugate linear characters.

Then

$\psi\neq\phi$

$G$

is cyclic and

$\chi$

is the sum of two complex

, so $N=H$. Let $K=$ is not conjugate to In this case $H$ . Then $K$ contains , and hence $|K|>|G$ : $H|\cdot(|H|-2)>|G|/2$ since $(G, \chi)$ is normalized. It since $|H|$ Z5. Therefore $K=G$ and is faithful. follows that $\psi_{H}=(d-1)\cdot 1_{H}+\lambda$ . , we have Since Note

PROOF.

$g^{-1}$

$g$

$ker(\chi-\psi-\overline{\psi})$

$x=\psi+\overline{\psi}$

$\psi$

$(\psi, \psi)_{G}=1$

$|G:H| \sum_{x\in H.x^{2}\neq 1}|d-1+\lambda(x)|^{2}|G|/3$ since $>|G|/3$ since . In either case we have $|G:K|\leqq 2$ . Observe that $x=(n-d)\epsilon+\psi$, where with kernel K. (If $K=G$ , then $n=d$ and $x=\psi.$ ) is the linear character of is the Sylow -subgroup of $H$, If $m$ is divisible by an odd prime and then $C_{G}(P)=H$, so $N_{G}(P)=N$ , and thus $P\in Sy1_{p}(G)$ since $|N:H|\leqq 2$ . excbanges the non, then Suppose $m$ is a power of 2. If , and thus , so has a pair of eigenvalues identity eigenspaces of $G$

$d\geqq 2$

$\psi=\overline{\psi}$

$H\subseteqq K$

$m\geqq 8$

$m\geqq 5$

$G$

$\epsilon$

$p$

$P$

$x\in N\backslash H$

$\rho(g)$

$\rho(x)$

$P$

$\rho(x)$

$\zeta,$

$-\zeta$

570


is rational by Theorem 2.2. In particular, if $X$ is the conjugacy class of in , then . Therefore $N\in Sy1_{2}(G)$ since $N_{G}(N)=N$ .

$\chi(x)$

$g$

$G$

$N\cap X\subseteqq\{g, g^{-1}\}$

is dihedral of STEP 3. Supp $oseH$ is normal in K. If $m$ is odd, then $2m$ is urther and $x=\psi$ is irreducrble of degree 2. If $m$ is even, then order $2m$ $x=\psi$ , and where or dihedral or generalized quaternion of order is irreductble of degree 2. $G$

$G$

$x=\psi+\epsilon$

$\psi$

PROOF. Suppose that $m$ is divisible by an odd prime . Let $P$ be the Sylow -subgroup of $H$. Then $P$ is the unique Sylow -subgroup of since $H$ is normal in $K$, and thus $H=C_{G}(P)$ is normal in . Now suppose $m$ is a power of 2, so $N\in Sy1_{2}(G)$ . Then $H$ is normal in since $G=N\cdot K$. in either case, so $G=N$ and $|G$ : $H|=2$ Thus Therefore $H$ is normal in is dihedral or generalized quaternion if $m$ is even, is dihedral if $m$ is odd, , then $\chi(x)=-(n-2)$ . Let $L(g)$ and $d=2$ in either case. Note that if be as in Theorem 2.2. If $m$ is odd, then $f_{L(Q)}(n)=m=|G|/2$ , and hence $-(n-2)=n-2$ since $(c, x)$ is sharp, so $n=2$ and $x=\psi$ . If $m$ is even, then $p$

$G$

$P$

$P$

$G$

$G$

.

$G$

$G$

$G$

$x\in G\backslash H$

$f_{L(g)}(n)=2m=|G|$ $(G, \chi)$

,

is sharp, so

so

$-(n-2)\in x(H)\cap Z\subseteqq\{n-4, n-3, n-2, n-1, n\}$

or

$n=2$

$n=3$ ,

and therefore

$x=\psi$

or

$x=\psi+\epsilon$

.

since

$\square$

For the remainder of this proof we suppose $H$ is not normal in $K$. Let $M=$ $M_{0}=M\cap Z,$ $M^{*}=M\backslash Z$. is faithful on $K$ since is faithful, Note $-1$ . Since as an eigenvalue if is irrational by cannot have so 2.2, only assumes values rational . Theorem on $\psi(K\backslash \{1\}),$

$\chi$

$\psi$

$d\not\in M_{0}$

$\chi(x)$

$\rho(x)$

$G\backslash K$

$\chi$

STEP 4. PROOF.

$M^{*}\subseteqq\psi(H)$

.

Suppose not, so there is some . By Theorem 2.2 we have

such that is irrational , where and is irrational $N’=N_{G}(H’),$ $m’=|H’|$ . Let and . Put be a nonidentity $H’$ eigenvalue of , and let be the linear character of . defined by $z\in G$ Also, , , Note if . and is contained in a conjugate of $H$ and a conjugate of $H’$ , then both $H$ and $H’$ are conjugate to , which is impossible since . Therefore we have $\psi(x)\not\in\psi(H)$

$C_{G}(g’)=\langle g’\rangle$

$x\in K$

$\psi(x)$

$C_{G}(x)=\langle g’\rangle$

$\chi(g’)$

$H’=\langle g’\rangle,$

$\rho(g’)$

$\omega’$

$\lambda’$

$\lambda’(g’)=\omega’$

$\psi_{H’}=(d-2)\cdot 1_{H’}+\lambda’+\overline{\lambda}’$

$z^{2}\neq 1$

$C_{G}(z)$

$z$

$\psi(x)\not\in\psi(H)$

$\frac{1}{|N|}\sum_{x\in Hx^{2}\neq 1}|d-2+\lambda(x)+\overline{\lambda(x)}|^{2}+\frac{1}{|N’|}\sum_{y\in H’,y^{2}\neq 1}|d-2+\lambda’(y)+\overline{\lambda’(y)}|^{2}3$

. Then

, $\nu(\xi\delta)=0$

576


$

$\sum_{\xi\in x+}(\nu(\xi)+\nu(\xi\omega))$

$n \leqq\sum_{\xi\in x+}(\nu(\xi)+\nu(\xi\omega))/\varphi(\xi)$

and hence $n=\nu(1)+\nu(\omega)$ and $\nu(\xi)=0$ for claim holds. NOW suppose $1>1$ . Then $n-p^{s}=\nu_{0}(1)-\nu_{0}(\delta)$

for any

.

$\delta\in\langle\delta_{p}\rangle\backslash \{1\}$

$\xi\not\in\{1, \omega\}$

,

.

,

Thus $o(g)=o(y)=p$ , and the

$p^{s}=\nu_{0}(\omega)-\nu_{0}(\delta\omega)$

Therefore

$n \leqq\nu^{+}(1)+\nu^{+}(\omega)+\nu^{-}(\delta)+\nu^{-}(\delta\omega)\leqq\sum_{\xi\in x+}(\nu(\xi)+\nu(\xi\omega))+\sum_{\xi\in x-}(\nu(\xi\delta)+\nu(\xi\delta\omega))$

and summing over


,

gives

$(P-1) \sum_{\delta}\sum_{\xi\in x-}(\nu(\xi\delta)+\nu(\xi\delta\omega))\leqq(p-1)(n-\sum_{\xi\in X^{+}}(\nu(\xi)+\nu(\xi\omega)))$

$\leqq\sum_{\delta}\sum_{\xi\in x-}(\nu(\xi\delta)+\nu(\xi\delta\omega))$

If

$p>2$ ,

then

$\nu(\xi\delta)=\nu(\xi\delta\omega)=0$

for

$\xi\in X^{-}$

and


,

$n \leqq\nu^{+}(1)+\nu^{+}(\omega)\leqq\sum_{\xi\in X+}(\nu(\xi)+\nu(\xi\omega))/\varphi(\xi)$

On the other hand, if

$p=2$

.

so

.

(3.3)

then

$n=\nu_{0}(1)-\nu_{0}(-1)+\nu_{0}(\omega)-\nu_{0}(-\omega)$

$\leqq\nu^{+}(1)+\nu^{+}(\omega)+\nu^{-}(-1)+\nu^{-}(-\omega)$

$\leqq\sum_{\xi\in x+}(\nu(\xi)+\nu(\xi\omega))+\sum_{\xi\in x-}(\nu(-\xi)+\nu(-\xi\omega))/2$

since

for , and hence

$\xi\in X^{-}$

$\varphi(\xi)\geqq 2$

. Therefore

$\nu(-\xi)=\nu(-\xi\omega)=0$

holds if $p=2$ . , so $o(g)=o(y)=p^{\iota}$ as claimed. for is real. Note Case 2. Suppose $\Sigma_{\xi\in\langle\zeta\rangle}\nu(\xi)$

(3.3) also

Thus

for

$\xi\in X^{-}$

$n=\nu(1)+\nu(\omega)$

since

and

$n=$

$\nu(\xi)=0$

$\xi\not\in\{1, \omega\}$

$\chi(g)$

$Tr(\chi(g))=\varphi(\zeta_{p’})(n-2p^{s})+\varphi(\zeta_{p’})p^{s}\omega+\varphi(\zeta_{p’})p^{s}\overline{\omega}$

We claim $o(g)=o(y)=p^{\iota}$ or else and (3.4) we have

$p^{\iota}=5$

$n-2p^{s}=\nu_{0}(1)-\nu_{0}(\delta)$

for any

$\delta\in\langle\delta_{p}\rangle\backslash$

{

$1,$

$\omega$

,

,

. To see this, first suppose $1=1$ . By

$p^{s}=\nu_{0}(\omega)-\nu_{0}(\delta)=\nu_{0}(\overline{\omega})-\nu_{0}(\delta)$

to}. Therefore $n=\nu_{0}(1)+\nu_{0}(\omega)+\nu_{0}(\overline{\omega})-3\nu_{0}(\delta)$

51

.

$\nu^{+}(1)+\nu^{+}(\omega)+\nu^{+}(\overline{\omega})+3\nu^{-}(\delta)$

$\leqq\sum_{\xi\in x+}(\nu(\xi)+\nu(\xi\omega)+\nu(\xi\overline{\omega}))+3\sum_{\xi\in x-}\nu(\xi\delta)$

.

(3.4) (3.1)

577


Summing over , we obtain $\delta$

$(p-3) \sum_{\delta}\sum_{\xi\in x-}\nu(\xi\delta)\leqq(p-3)(n-\sum_{\xi\in X^{+}}(\nu(\xi)+\nu(\xi\omega)+\nu(\xi\overline{\omega})))$

$\leqq 3\sum_{\delta}\sum_{\xi\in x-}\nu(\xi\delta)$

If

$p>5$ ,

then

$\nu(\xi\delta)=0$

for

$\delta\in\langle\delta_{p}\rangle\backslash \{1, \omega,\overline{\omega}\}$

.

and

$\xi\in\langle\zeta_{p’}\rangle$

so

,

,

$\sum_{\xi\in X+}(\nu(\xi)+\nu(\xi\omega)+\nu(\xi\overline{\omega}))\leqq n\leqq\sum_{\xi\in x+}(\nu(\xi)+\nu(\xi\omega)+\nu(\xi\overline{\omega}))/\varphi(\xi)$

and hence and $o(y)=p$ if $p>5$ , as claimed. NOW suppose $1>1$ . Then $n=\nu(1)+\nu(\omega)+\nu(\overline{\omega})$

$n-2p^{s}=\nu_{0}(1)-\nu_{0}(\delta)$

for any


,

$\nu(\xi)=0$

for

$\xi\not\in\{1, \omega,\overline{\omega}\}$

. We

conclude

$o(g)=$

$p^{s}=\nu_{0}(\omega)-\nu_{0}(\delta\omega)=\nu_{0}(\overline{\omega})-\nu_{0}(\delta\overline{\omega})$

. Thus

$n=\nu_{0}(1)+\nu_{0}(\omega)+\nu_{0}(\overline{\omega})-\nu_{0}(\delta)-\nu_{0}(\delta\omega)-\nu_{0}(\delta\overline{\omega})$

;il

$\nu^{+}(1)+\nu^{+}(\omega)+\nu^{+}(\overline{\omega})+\nu^{-}(\delta)+\nu^{-}(\delta\omega)+\nu^{-}(\delta\overline{\omega})$

$\leqq\sum_{\xi\in X+}(\nu(\xi)+\nu(\xi\omega)+\nu(\xi\overline{\omega}))+\sum_{\xi\in x-}(\nu(\xi\delta)+\nu(\xi\delta\omega)+\nu(\xi\delta\overline{\omega}))$

Summing over

$\delta$

.

gives $(P-1) \sum_{\delta}\sum_{\xi\in x-}(\nu(\xi\delta)+\nu(\xi\delta\omega)+\nu(\xi\delta\overline{\omega}))$

$

$(p-1)(n- \sum_{\xi\in X^{+}}(\nu(\xi)+\nu(\xi\omega)+\nu(\ovalbox{\tt\small REJECT}^{-})))$

$\leqq\sum_{\delta}\sum_{\xi\in X^{-}}(\nu(\xi\delta)+\nu(\xi\delta\omega)+\nu(\xi\delta\overline{\omega}))$

$If_{r}^{*}p>2$

, then

$\nu(\xi\delta)=\nu(\xi\delta\omega)=\nu(\xi\delta\overline{\omega})=0$

for

.


and

$\xi\in X^{-}$

$n \leqq\nu^{+}(1)+\nu^{+}(\omega)+\nu^{+}(\overline{\omega})=\sum_{\xi\in X+}(\nu(\xi)+\nu(\xi\omega)+\nu(\xi\overline{\omega}))/\varphi(\xi)$

Suppose

$p=2$

,

so

.

(3.5)

. Then $n=\nu_{0}(1)+\nu_{0}(\omega)+\nu_{0}(\overline{\omega})-\nu_{0}(-1)-\nu_{0}(-\omega)-\nu_{0}(-\overline{\omega})$

$\leqq\nu^{+}(1)+\nu^{+}(\omega)+\nu^{+}(\overline{\omega})+\nu^{-}(-1)+\nu^{-}(-\omega)+\nu^{-}(-\overline{\omega})$

$\leqq\nu^{+}(1)+\nu^{+}(\omega)+\nu^{+}(\overline{\omega})+\sum_{\xi\in X-}(\nu(-\xi)+\nu(-\xi\omega)+\nu(-\ovalbox{\tt\small REJECT}^{-}))/2$

. for so (3.5) also holds if

since

It follows that $p=2$ . Therefore , so $o(g)=o(y)=p^{l}$ , as required.

$\varphi(\xi)\geqq 2$

$\xi\not\in\{1, \omega,\overline{\omega}\}$

lemma.

$\square$

$\xi\in X^{-}$

$\nu(-\xi)=\nu(-\xi\omega)=\nu(-\ovalbox{\tt\small REJECT}^{-})=0$

for

$\xi\in X^{-}$

,

and $\nu(\xi)=0$ for This completes the proof of the

$n=v(1)+\nu(\omega)+\nu(\overline{\omega})$

578


LEMMA 3.6. Suppose is a -srngular element of and whenever is -srngular and . If $h\in C_{G}(g)$ is a -element for some prime and $gh$ , then is not -srngular. $G$

$\chi$

$g$

$x$

$\chi$

$\langle g\rangle\subseteqq\langle x\rangle$

$\langle g\rangle=\langle x\rangle$

$P$

$P$

$\chi$

$h\not\in\langle g\rangle$

be the -part of . Let PROOF. Suppose not, so $gh$ is -singular. Let , and let on be the eigenvalue of be a nonidentity eigenvalue of the -eigenspace of . $\chi(g)=n-1+\omega$ Suppose . Since all nonidentity eigenvalues of $\rho(gh)$ have the same order, the only nonidentity eigenvalue of $\rho(gh)$ is , and the only non, a contradiction. identity eigenvalue of is . If $o(h)\leqq o(g_{p})$ , then $o(h)>o(g_{p})$ , so , another con, then On the other hand, if tradiction. Suppose on the be the eigenvalue of . Let -eigenspace of . Then are the only nonidentity eigenvalues of and $\rho(gh)$ , and therefore , so if $o(h)\leqq o(g_{p})$ , As above, we have and if $o(h)>o(g_{p})$ , both of which are contradictions. This completes the proof of the lemma. $\chi$

$P$

$g_{p}$

$\rho(g)$

$\omega$

$g$

$\rho(h)$

$\lambda$

$\rho(g)$

$\omega$

$\lambda\omega$

$\lambda$

$\rho(h)$

$h\in\langle g_{p}\rangle$

$\langle g\rangle\subset\langle gh\rangle$

$g_{p}\in\langle h\rangle$

$\chi(g)=n-2+\omega+\overline{\omega}$

$\lambda\omega$

$\rho(g)$

$\overline{\omega}$

$\lambda’$

$\lambda’=\overline{\lambda}$

$\overline{\lambda\omega}=\lambda’\overline{\omega}$

$\rho(h)$

$\lambda’\overline{\omega}$

.

$h\in\langle g_{p}\rangle$


$\square$

Suppose is -singular. If $h\in C_{G}(g)$ is a $\chi(gh)$ is not equal to for any -element

LEMMA 3.7. prime , then $p$

$\chi$

$g$

$\chi(y)$

$P$

-element

for some

of . for some -element

$P$

$G$

$y$

PROOF. Suppose to the contrary that $\chi(gh)=x(y)$ . Let be a primitive $|G|- th$ root of unity, and let be the Then $n-\chi(g)=(\chi(1)-\chi(y))+(\chi(gh)-X(g))\in(1-\omega_{0})Z[\omega]$ , and hence product of the algebraic conjugates of $n-\chi(g)$ , a contradiction.

$P$

$G$

$\omega$

$P$

$\omega_{0}$

$P$

$y$

-part of

of $\omega$

.

divides the

$\square$

Suppose now that are complex roots of unity whose orders are and , relatively prime. Assume and and are $Q$ -bases for , respectively, and $1\in S\cap T$ . Then $ST=\{\sigma\tau|\sigma\in S, \tau\in T\}$ is a $Q$ -basis for is and it follows that no nonzero linear combination of elements of $Q$ $Q(\alpha)+Q(\beta)$ of . This observation is used an element of the -subspace in the proofs of the following two results. $\alpha$

$\beta$

$S\subseteqq\langle\alpha\rangle$

$Q(\alpha)$

$T\subseteqq\langle\beta\rangle$

$Q(\beta)$

$Q(\alpha\beta)$

$ST\backslash (S\cup T)$

$Q(\alpha\beta)$

LEMMA 3.8. Suppose relatively prime. Assume $\{1, -1\}$

or

$\beta_{0}\in\{1, -1\}$

$Q(\beta)$

and

$\beta$

are comPlex roots of unity whose orders are

$\alpha_{0}\in\langle\alpha\rangle,$

$\beta_{0}\in\langle\beta\rangle$

, and

$\alpha_{0}\beta_{0}\in Q(\alpha)+Q(\beta)$

$\alpha_{0}\neq\pm 1$

$\beta_{0}\neq\pm 1$

$\{1, \alpha_{0}\}\subseteqq S$

$ST\backslash (S\cup T)$

LEMMA 3.9.

.

Then

$\alpha_{0}\in$

.

, then there are If and , respectively, such that and , and a contradiction is reached.

PROOF.

and

$\alpha$

$Q$

-bases

$S$

$\{1, \beta_{0}\}gT$

and for . Then $T$

$Q(\alpha)$

$\alpha_{0}\beta_{0}\in$

$\square$

SuPPose

relatively prime. Assume Suppose further that 1,

and

$\alpha$

$\alpha_{0},$

$\beta_{0},$

$\beta$

are complex roots of unity whose orders are , and .

$\alpha_{1}\in\langle\alpha\rangle,$

$\beta_{1}$

$\beta_{0},$

$\beta_{1}\in\langle\beta\rangle$

are distinct and

$\alpha_{0}\beta_{0}+\alpha_{1}\beta_{1}\in Q(\alpha)+Q(\beta)$

$\{1, \beta_{0}, \beta_{1}\}$

is linearly independent

579


over Q. Then

First suppose

PROOF. such that

.

$\{\alpha_{0}, \alpha_{1}\}\subseteqq\{1, -1\}$

.

. Let

be a $Q$ -basis for is linearly independent over

$\{\alpha_{0}, \alpha_{1}\}\cap\{1, -1\}=\emptyset$

$T$

$Q(\beta)$

$Q$ . Assume Then there is a $Q$ -basis such that . Then for , and a contradiction is reached. Next, and suppose is linearly dependent over . Renumbering if necessary, we such may suppose . Let be a $Q$ -basis for with , Then and . that , and another contradiction is reached. , so , then . If Therefore if Similarly, . by Lemma 3.8 since $\{1, \beta_{0}, \beta_{1}\}\subseteqq T$

$\{1, \alpha_{0}, \alpha_{1}\}$

$S$

$\in ST\backslash (S\cup T)$

$Q(\alpha)$

$\{1, \alpha_{0}, \alpha_{1}\}\subseteqq S$

$\alpha_{1}\beta_{1}$

$\alpha_{0}\beta_{0},$

$\alpha_{0}\beta_{0}+\alpha_{1}\beta_{1}\in Q(\alpha)+Q(\beta)$

$Q$

$\{1, \alpha_{0}, \alpha_{1}\}$

$\alpha_{1}=a_{0}+a_{1}\alpha_{0}$

$\{1, \alpha_{0}\}\subseteqq S$

$a_{0},$

$\alpha_{0}\beta_{0},$

$S$

$a_{1}\in Q$

$Q(\alpha)$

$\alpha_{0}\beta_{0}+a_{1}\alpha_{0}\beta_{1}=\alpha_{0}\beta_{0}+\alpha_{1}\beta_{1}-$

$\alpha_{0}\beta_{1}\in ST\backslash (S\cup T)$

$a_{0}\beta_{1}\in Q(\alpha)+Q(\beta)$

$\alpha_{0}=\pm 1$

$\{\alpha_{0}, \alpha_{1}\}\cap\{1, -1\}\neq\emptyset$


$\alpha_{1}=\pm 1$

.

For

$\alpha_{1}\beta_{1}\in Q(\alpha)+Q(\beta)$


$\beta_{1}\neq\pm 1$

$\square$

a positive integer,

$k$

$\omega_{k}$

will denote a primitive complex k-th root of

unity.

is a -element in $C_{G}(g)$ for LEMMA 3.10. Suppose is -singular and is the p’-part of . Then some prime . Assume $g’h$ is not -singular, where $\chi(g’h)$ is rational. $h\neq 1$

$\chi$

$g$

$p$

$p$

$g’$

$\chi$

$g$

Suppose to the contrary that $\chi(g’h)$ is irrational. By Lemma 3.5, $\chi(g’h)=x(y)$ for some element of $G$ of order , where $q=3$ or $q=5$ . Lemma 3.7 applies to show $q\neq p$ . Thus divides tbe order of . , and let be the eigenvalue of Let co be a nonidentity eigenvalue of . . Let be the -part of , so on the -eigenspace of $\chi(g’h)=x(h)-\lambda+\lambda\omega’=x(y)$ , is not real. In this case Case 1. Suppose 3.8, by so Lemma Thus . so $(Q(\omega_{o(h)})+Q(\omega_{q}))\cap Q(\omega’)=Q(\omega_{q})$ , and hence we may choose . so that $\chi(g’h)\not\in R$ $\chi(h)\in Q(\omega_{o(h)})\cap Q(\omega_{q})=Q$ $q=3$ , and hence . . Thus Then NOW, $\chi(h)-\lambda\pm\lambda\omega_{3}=x(g’h)=x(y)=n-3^{s}+3^{s}\omega_{3}^{j}$ for some and ] by Lemma 3.4. Comparing imaginary parts, we find $3^{s}=1$ . Then $\in\{n-1, n-2, n-3\}$ since is faithful. Since $n-\chi(h)$ must be divisible by , $p=2$ $\chi(h)=n-2$ , so $\lambda=-1$ and , and hence and divides we have

PROOF.

$y$

$q$

$g’$

$q$

$\rho(g)$

$\rho(h)$

$\omega’$

$\rho(g)$

$\omega$

$\lambda$

$P’$

$\omega’\neq\pm 1$

$\omega$

$\chi(g)$

$\omega’\in$

$\lambda=\pm 1$

$\lambda\omega’\in Q(\omega_{o(h)})+Q(\omega_{q})\subseteqq Q(\omega_{o(h)})+Q(\omega’)$

$\omega’=\pm\omega_{q}$

$\omega_{q}$

$s$

$\chi(h)=n-1+\lambda+\omega_{3}^{j}-\lambda\omega’$

$\chi$

$p$

$\omega’=-\omega_{3}^{j}$

a contradiction. Case 2. Suppose

$o(g’)$

$\overline{\omega}$

,

-eigenspace of

is real. Let In this case

$\chi(g)$

$\rho(g)$

.

$\lambda’$

, to’} is linearly dependent over , so Suppose $q=3$ and can be chosen so that $\omega’$

$\{\lambda, \lambda’\}_{\frac{\not\subset}{arrow}}\{1, -1\}$

$\omega_{3}$

$Q(\omega_{o(h)})+Q(\omega_{3})$

,

so

$\lambda’=\pm 1$

$Q$

be the eigenvalue of

on the , so or else

$\rho(h)$

$\chi(g’h)=x(h)-\lambda-\lambda’+\lambda\omega’+\lambda’\overline{\omega}’=x(y)$

$\lambda\omega’+\lambda’\overline{\omega}’\in Q(\omega_{o(h)})+Q(\omega_{q})\subseteqq Q(\omega_{o(h)})+Q(\omega’)$

{1,

$P$

, and hence

$\{\lambda, \lambda’\}1\{1, -1\}$

by Lemma 3.9. is linearly dependent

$\{1, \omega’,\overline{\omega}’\}$

by Lemma 3.8,

over

$Q$

.

, then . Note that if a contradiction. Thus

$\omega’=\pm\omega_{3}$

$\lambda=\pm 1$

Then $\lambda’\overline{\omega}_{3}\in$

$\lambda\neq\pm 1$

, and

580


similarly $\lambda-\lambda’\in Q$

$\lambda’\neq\pm 1$

.

.

,

$(\lambda-\lambda’)\omega_{3}=\lambda\omega_{3}+\lambda’\overline{\omega}_{3}-\lambda’\in Q(\omega_{o(h)})+Q(\omega_{3})$

.

we must have so $\chi(g’h)=x(y)\in Q(\omega_{o(h)})\cap Q(\omega_{q})=Q$ , a contradiction.

Since

$\lambda\omega_{3}+\lambda’\overline{\omega}_{3}=-\lambda$

Also, $p\neq 3$

and

$\{\lambda, \lambda’\}\cap\{1, -1\}=\emptyset$

,

Therefore . Then

$\lambda=\lambda’$

Therefore . Since and divides the order of , it follows that can be chosen so that . Thus . , then If is real, so $q=5$ . Then is for some and some 7‘ by Lemma 3.4. Since rational, we must have $5^{s}=1$ . Then , and thus $\chi(h)\in\{n-1, n-4, n-5\}$ since is faithful. Since must divide $n-\chi(h)$ , we , which is imposhave $p=2$ and $\chi(h)=n-4$ , so and or Of $o(g’)$ sible since is odd. , so Thus is not real, and hence $q=3$ . Then $\chi(g’h)=x(y)=n-3^{s}+3^{s}\omega_{3}^{j}$ for some . Comparing imaginary parts, we see $\sqrt{3}=3^{s}\sqrt{3}/2$ , which is impossible. This completes the proof of the lemma. $\{\lambda, \lambda’\}\subseteqq\{1, -1\}$

$\omega’\pm\overline{\omega}’\in(Q(\omega_{o(h)})+Q(\omega_{q}))\cap Q(\omega’)=Q(\omega_{q})$

$g’$

$q$

$\omega’=\pm\omega_{q}$

$\omega_{q}$

$\chi(h)\in Q(\omega_{o(h)})\cap Q(\omega_{q})=Q$ $\lambda=\lambda’$

$\chi(h)-2\lambda\pm\lambda(\omega_{5}+\overline{\omega}_{5})=x(g’h)=x(y)$

$\chi(y)$

$s\geqq 0$

$=n-2\cdot 5^{s}+5^{s}\omega_{5}^{j}+5^{s}\overline{\omega}_{5}^{j}$

$\chi(h)$

$\chi(h)=n-2+2\lambda+\omega_{5}^{j}+\overline{\omega}‘-\lambda(\omega^{f}+\overline{\omega}’)$

$\chi$

$p$

$\lambda=-1$

$\lambda\neq\lambda’$

$\omega’=-\omega_{5}^{j}$

$=-\overline{\omega}_{5}^{j}$

$\chi(y)$

$\chi(h)\pm(\omega_{3}-\overline{\omega}_{3})=$

$j$

$\square$

$h$

LEMMA 3.11. Suppose is -singular and $\chi$

is a -singular element of . Then $\chi$

$g$

$\langle g\rangle\subseteqq\langle h\rangle$

.

$G$

and

whenever

$\langle g\rangle=\langle h\rangle$


PROOF. Suppose not. Let be a -element of $C_{G}(g)$ for some prime . Then $gh$ is not -singular by Lemma 3.6. Let be the such that is rational. , so Lemma 3.10 applies to show -part of . Then Similarly, is rational. , and let denote the eigenvalue Let be a nonidentity eigenvalue of -eigenspace of . of on the is not real. Then and Case 1. Suppose $\chi(gh)=x(h)-\lambda+\lambda\omega\in Q$ $0(\omega)=12$ . hence . This is impossible since and eigenvalue of be the on the is real. Let Case 2. Suppose -eigenspace of . or . Then $\chi(gh)-\chi(g^{-1}h)=(\lambda-\lambda’)(\omega-\omega)\in Q$ , so ) Suppose does not divide ( . Since $Q$ , linearly independent applies Lemma 3.9 is over to show and by the above. Then , and hence , is rational, a contradiction. so by an Therefore divides $o(g)$ . Next assume $o(h)=p$ . Replacing appropriate element of if necessary, we may suppose . Then $\chi(h)=x(gh)-\omega-\overline{\omega}+2$ is irrational and real, and hence $p>3$ . We may suppose . Then the nonidentity eigenvalues of is an eigenvalue of multiplicity by Lemma 3.4. There is for some with and are . Then the nonidentity eigenvalues of are some such that , both with multiplicity $p^{s}+1$ , contradicting Lemma 3.4. and has a unique NOW we return to the general case. By the above, $h$

$P$

$\chi$


$P$

$P$

$g_{p}$

$\chi(gh)$

$g_{p}h\neq 1$

$g$

$\chi(g^{-1}h)$

$\rho(g)$

$\omega$

$\rho(h)$

$\lambda$

$\rho(g)$

$\omega$

$\chi(gh)-\chi(g^{-1}h)=\lambda(\omega-\overline{\omega})\in Q)$

$\chi(g)$

$\lambda^{2}=-1$

$\lambda’$

$\chi(g)$

$\rho(h)$


$\lambda’=\lambda$

$\rho(g)$

$\overline{\omega}$

$o$

$P$

$\lambda\omega+\lambda’\overline{\omega}=x(gh)-\chi(h)+\lambda+\lambda’\in Q^{(}\omega_{o(h)}$

$g^{)}$

$\{1, \omega,\overline{\omega}\}$

$\{\lambda, \lambda’\}$

$\lambda=\lambda’=\pm 1$

$\subseteqq\{1, -1\}$

$\omega+\overline{\omega}\in Q(\omega_{o(h)})\cap Q(\omega)=Q$

$\chi(g)$

$h$

$P$

$\lambda=\lambda’=1$

$\langle g_{p}, h\rangle\backslash \langle g_{p}\rangle$

$\rho(h)$

$\omega_{p}$

$\omega_{p}$

$j$

$\rho(h)$

$p^{s}$

$\overline{\omega}_{p}$

$s\geqq 0$

$\rho(g^{j}h)$

$d=\omega_{p}$

$\overline{\omega}_{p}$

$\langle g_{p}, h\rangle$

$\omega_{p}$

SharP characters with irrational

so

581

values

is cyclic. Therefore $o(h)>o(g_{p})$ and are the only . We may as well suppose $h^{p}=g_{p}$ . Then and since eigenvalues of of order greater than . Suppose is irrational. Then and are the only nonidentity eigen, by Lemma 3.4. If , then $gh$ is -singular and values of $p=2$ by Lemma 3.4. Since and so we have a contradiction. Therefore , and therefore , we have , . Then which is a contradiction. is rational. Then $p=2$ and , so Therefore . Note $o(g)=12$ , and therefore . This is and $o(h)=4$ . Also, $o(h)>o(g_{2})$ completes proof This of the . the lemma. since a contradiction

subgroup of order

$p$

,

$g_{p}\in\langle h\rangle$

$\langle g_{p}, h\rangle$

$\lambda’$

$\lambda$


$\rho(h)$

$p$

$\chi(h)$

$\lambda’$

$\lambda$


$\rho(h)$

$\chi$


$\lambda’=\lambda$

$h^{2}=g_{2}$

$\lambda^{2}=-1$

$(\lambda^{f})^{2}=\overline{\lambda}^{2}$

$\chi(gh)=n-2+\lambda\omega+\lambda\overline{\omega}\not\in R$

$\lambda^{2}=-1$


$\lambda’=-\lambda$

$\chi(h)$

$(\lambda-\lambda’)(\omega-\overline{\omega})=2\lambda(\omega-\overline{\omega})\in Q$

$\square$

is an element of LEMMA 3.12. Suppose of prime power order such $C_{G}(g)$ contains no -singular elements. Then is irrational. Assume that that $C_{G}(g)$ is a p-group. $G$

$g$

$p^{\iota}$

$\chi$

$\chi(g)$

be an element of $C_{G}(g)$ of prime order , PROOF. Suppose not. Let $q\neq p$ . , and let be the character of denote the -eigenspace of Let $C_{G}(g)$ acting on . Note that . Let be a nonidentity eigenvalue of $\dim(V_{\omega})=p^{s}$ by Lemma 3.4. Put , for some nonnegative integer , so . We claim that it suffices to prove that . Indeed, $1\leqq j\leqq q-1$ , then , and hence if for since affords a multiple of the regular character of . Thus divides $\dim(V.)=p^{s}$ , a con$h$

$q$

$\rho(g)$

$\xi$

$V_{\xi}$

$V_{\xi}$

$\theta_{\xi}$

$\rho(g)$

$\omega$

$\alpha=\theta_{\omega}(h)$

$s$

$\beta=\theta_{\overline{\omega}}(h)$

$\alpha=0$

$\beta\in Q(\omega_{q})$

$\alpha,$

$\alpha=0$

$\theta_{\omega}(h^{j})=0$

$V_{\omega}$

$p\neq q$

$\langle h\rangle$

$q$

tradiction. Suppose

divide $=Q,$

$j$

.

$\chi(gh)$

is rational. Then

In particular

$andthus\alpha=\beta$

.

$\chi(g^{j}h)$

is rational whenever ,

$\chi(gh)-\chi(g^{-1}h)=(\alpha-\beta)(\omega-\overline{\omega})\in Q$

$Thus\alpha(\omega+\overline{\omega})=x(gh)-\theta_{1}(h)\in Q(\omega_{q})$

.

so

$p$

does not

$\alpha-\beta\in Q(\omega)\cap Q(\omega_{q})$

$If\alpha\neq 0,$

$then\omega+\overline{\omega}\in Q$

,

, so . In either case is not an eigenvalue of or and hence , a contradiction. Hence , and thus and we are done. is irrational. By our For the remainder of the proof we suppose assumptions, $gh$ is not -singular. Therefore Lemma 3.5 gives $\chi(gh)=x(y)$ for some element of order , with $r=3$ if and $r=5$ if $\chi(gh)\in R$ . Note , and also that $r=p$ or $r=q$ . We have since . Then $p^{l}=4$

$p^{\iota}=3$

$\alpha=0$

$\beta=0$

$\rho(g)$

$\overline{\omega}$

$\alpha=0$

$\chi(gh)$

$\chi$

$\chi(gh)\not\in R$

$r$

$y$

$\chi(gh)\in Q(\omega_{r})$

$\chi(g^{-1}h)\in Q(\omega_{r})$

$p\neq q$

$\chi(gh)-\chi(g^{-1}h)=(\alpha-\beta)(\omega-\overline{\omega})\in Q(\omega_{r})\subseteqq Q(\omega)+Q(\omega_{q})$

since

$r\in\{p, q\}$

$\alpha-\beta\in Q$

$\omega+\overline{\omega}\in Q$

.

Since

. Then or else

$\{1, \omega-\overline{\omega}\}$

extends to a

$Q$

-basis for

$Q(\omega)$

,

it follows that

. Therefore either or and ru is not an in all cases, and hence we

$\alpha(\omega+\overline{\omega})=x(gh)-\theta_{1}(h)+(\alpha-\beta)\overline{\omega}\in Q(\omega)+Q(\omega_{q})$

eigenvalue of also have

$\rho(g)$

$\beta\in Q$

Case 1.

$\alpha\in Q$

,

so

.

If

$\beta=0$

$\omega+\overline{\omega}\in Q$

and

,

$\alpha\in Q$

then

$p^{l}=3$

. Thus

$p^{l}=4$

$\alpha\in Q$

.

Suppose

$r=q$

.

Then

$\alpha\omega+\beta\overline{\omega}=x(gh)-\theta_{1}(h)\in Q(\omega)\cap Q(\omega_{q})=Q$

. If

582



is linear dependent over

then $p^{l}=3$ or and since is linearly inde. On the other hand, if in either case, and we are done. . Thus In this case . . We may suppose $1=1$ , for otherwise $Q$

,

$p^{\iota}=4$

$\chi(g)$

$\beta=0$

is irrational, and hence pendent over $Q$ , then $\alpha=\beta=0$ Case 2. Suppose $r=p$ . Thus and the proof is complete. If $r=p=3$ , then $\theta_{1}(h)+\alpha\omega_{3}=x(gh)=x(y)=n-3^{t}+3^{t}oi_{3}$ for some and 7‘ by , then $\theta_{1}(h)=n-3^{t}$ is rational. If since Lemma 3.4, and thus and and $h=1$ since is faithful, a contradiction. Thus $n-\chi(gh)=3^{t+1}+3^{t}\omega_{3}$ $n-\chi(gh)$ is . The product of the algebraic conjugates of so $M$ , is the set of values of on . However, if divisible by $|P|$ by Theorem 2.1, $f_{M}(n)$ $P\in Sy1_{7}(G)$ by is , and divisible then where is sharp of type . is divisible by $7|P|$ , which is impossible since so is real. Then since Suppose $r=p=5$ . Note by Lemma 3.4. Thus since for some and , so $h=1$ , a contradiction. On the , then is rational. If , so , then divides , a other hand, if proof of the lemma. contradiction. This completes the $\alpha=0$


$\alpha=0$

$\theta_{1}(h)=x(gh)-\alpha\omega-\beta\overline{\omega}\in Q(\omega)\cap Q(\omega_{q})=Q$

$\alpha=\beta=0$

$\alpha\omega+\beta\overline{\omega}=x(y)-\theta_{1}(h)\in Q(\omega_{p})$

$t$

$\alpha=\pm 3^{t}$

$\alpha=3^{t}$

$\theta_{1}(h)$

$\alpha=-3^{t},$

$\chi$

$\theta_{1}(h)=n-2\cdot 3^{t}$

$\chi$

$(3+\omega_{3})(3+\overline{\omega}_{3})=7$

$P\backslash \{1\}$

$\chi(gh)\not\in M$

$L$

$(G, \chi)$

$f_{L}(n)$

$\chi(gh)$

$\alpha=\beta$

$n-2\cdot 5^{t}+5^{t}(\omega_{5}^{j}+\overline{\omega}_{5}^{j})$

$\theta_{1}(h)+\alpha(\omega_{6}+\overline{\omega}_{\overline{o}})=$

$\alpha=\pm 5^{t}$

$t$

$J$

$\alpha=5^{t}$

$\theta_{1}(h)$


$\alpha=-5^{t}$


$n-2\cdot 5^{t}-\theta_{1}(h)=5^{t}$

$q$

$\square$

LEMMA 3.13.

of

$G$

be Prime. Suppose Let and orders such that $p$

of different

$\chi(g)$

$g$

and are commuting p-elements are irrational. Then is $h$

$\chi(h)$

$\langle g, h\rangle$

cyclic.

be a nonidentity eigenvalue of , PROOF. We suppose $o(g)$ $o(g)\geqq 8$ nonidentity eigenvalues , is has irrational and Thus , so . A and , contradicting Lemma 3.4. Therefore , if . If and similar argument shows $\rho(g^{O(q)/2}h)$ $p=2$ , and the only nonidentity eigenvalues of are and then $\rho(g)$

$\omega$

$\rho(h)$

$\xi$

$V_{\lambda}$

$W_{\lambda}$

$\rho(h)$

$\lambda$

$\chi(gh^{-})$

$W_{1}$

$\rho(g)$

$\alpha$

$\xi-\overline{\xi}\in Q(\xi^{p})$

$\alpha\neq 0$

$\xi^{2}=1$

$\alpha_{1}$

$\rho(g)$

$W_{\xi}\oplus W_{\xi}^{-}$

$\chi(h)$

$\alpha=0$

$\chi(g)=\alpha_{1}\in Q$

$h^{-1}$

$h$

$V_{\omega}\oplus V_{\overline{\omega}}$

$\omega\xi$

$\subseteqq W_{\xi}\oplus W_{\xi}^{-}$

$V_{\omega}\subseteqq W_{\xi}$

$\xi=\overline{\omega\xi}$

$\xi$

$\xi^{2}=\overline{\omega}$

.

$\xi^{3}$

$\overline{\omega}\xi$

$V_{1}\cap W_{\xi}\neq\{0\}$

$V_{\overline{\omega}}\subseteqq W_{\overline{\xi}}$

$\overline{\xi},$

$\xi$

$V_{\omega}\subseteqq W_{\xi}^{-}$

$\omega\xi$

$V_{\omega}\subseteqq W_{\xi}$

$\omega\overline{\xi}$

$\chi(g^{-1}h)$

$\chi(g)$

$\chi(h)$

$\rho(g^{-1}h)$

$\xi$

$V_{1}\cap W_{\xi}=\{0\}$

$\overline{\xi}^{3}$

$V_{1}\cap W_{\overline{\xi}}=\{0\}$

$W_{\overline{\xi}}\subseteqq V_{1}$

$V_{\overline{\omega}}\neq\{0\}$

$V_{1}\cap W_{\overline{\xi}}\neq\{0\}$

$V_{\omega}=W_{\xi}$

$V_{\overline{\omega}}=\{0\}$

SharP characters with irrational and 6. Hence Therefore $-\xi$

$\overline{\xi}=-\xi$

$V_{1}\cap W_{\overline{\xi}}=\{0\}$

so

,

,

so

and Thus

$\xi^{2}=-1$

$V_{\overline{\omega}}=W_{\overline{\xi}}$

.

$\chi(g)$

$g\in\langle h\rangle$

583

values

is rational, a contradiction. , as required.

$\square$

LEMMA 3.14. Let be prime. Suppose is a -element of such that is irratimal and $C_{G}(g)$ is a p-group. Assume $h\in C_{G}(g),$ $o(h)=o(g)$ and . irratimal. Then $p$

$G$

$p$

$g$

$\chi(g)$

$\chi(h)$

is


PROOF. Let be a nonidentity eigenvalue of . Replacing by a power if necessary, we may suppose is an eigenvalue of . Denote by , respectively. Let the -eigenspaces of , and and $h$

$\rho(g)$

$\omega$

$\rho(h)$

$\omega$

$W_{\xi}$

$\rho(g)$

$\xi$

$p^{t}=\dim(W_{\omega})$

. Put

$V_{\xi}$

$p^{s}=\dim(V_{\omega})$

$\rho(h)$

.

$X=\{x\in G|\chi(x)=x(g)\}$

Case 1. Suppose $o(g)\not\in\{3,4,5,8\}$ . In this case the eigenvalues of $\rho(gh)$ , whicb is linearly independent over $Q$ . are elements of the set $\chi(gh)$ or We claim that is rational. Suppose to the contrary that $\chi(gh)$ and are irrational. Assume both . Then $\rho(gh)$ by or has an eigenvalue or , and therefore $\rho(gh)$ has no eigenvalue Lemma 3.4. Thus . The same argument applied to shows . Then . Thus and has both and as eigenvalues, which is impossible. Therefore . Similarly, . Hence $V_{1}=W_{1}$ , so $g=h$ . If and and are not real, then and we are done. If only one of is real, then $p=2$ and and . Then the nonidentity eigenvalues of or or and are and , contradicting Lemma 3.4. Hence we may suppose $s=t$ . are both real, so . The multiplicity of as an eigenvalue of $\rho(gh)$ is Let is equal to as an eigenvalue of equal to , while the multiplicity of $p^{s}-d$ . by Lemma 3.4. These multiplicities are equal to or a power of $d=2^{s-1}$ $d=p^{s}$ $d=0$ $p=2$ $p=2$ , , or else . However, if and $d=2^{s-1}$ , and Hence is irrational and , so then $p(g^{2}h)$ has nonidentity eigenvalues since $o(g)>8$ , and we have a contradiction to Lemma 3.4. Hence $d=p^{s}$ or $d=0$ , so $g=h$ or $g=h^{-1}$ and we are done. is rational, and thus $g=h^{-1}$ or $g=h$ , so . Therefore $\chi(gh)$ or is not real. In this case the Case 2. Suppose $o(g)=5$ or $o(g)=8$ and , which is linearly eigenvalues of $\rho(gh)$ are elements of the set $Q$ independent over . The argument given in Case 1 shows that $\chi(gh)$ or . is rational. Hence $g=h^{-1}$ or $g=h$ , so Case 3. Suppose $o(g)=3$ . In this case is the only nonidentity eigenvalue and arrive at a contradiction. . We suppose and of $\{1, \omega,\overline{\omega}, \omega^{2},\overline{\omega}^{2}\}$

$\chi(gh^{-1})$

$\chi(gh^{-1})$

$(V_{\omega}\oplus V_{\overline{\omega}})\cap W_{1}\neq\{0\}$

$\omega^{2}$

$\overline{\omega}$

$\omega$

$\overline{\omega}^{2}$

$gh^{-1}$

$V_{\omega}\cap W_{\omega}=V_{\overline{\omega}}\cap W_{\overline{\omega}}=\{0\}$

$V_{\omega}\cap W_{\overline{\omega}}=V_{\overline{\omega}}\cap W_{\omega}=\{0\}$

$W_{\omega}\oplus W_{\overline{\omega}}\subseteqq V_{1}$

$V_{\omega}\oplus V_{\overline{\omega}}\subseteqq W_{1}$

$\omega^{2}$

$\rho(gh^{2})$

$\omega$

$(V_{\omega}\oplus V_{\overline{\omega}})\cap W_{1}=\{0\}$

$\chi(g)$

$V_{\omega}\oplus V_{\overline{\omega}}=W_{\omega}\oplus W_{\overline{\omega}}$

$V_{1}\cap W_{\omega}\oplus W_{\overline{\omega}}=\{0\}$

$V_{\omega}=W_{\omega}$

$\chi(h)$

$\chi(g)$

$W_{\omega}=$

$\chi(h)$

$\rho(g^{2}h)$

$V_{\omega}=W_{\omega}\oplus W_{\overline{\omega}}$

$V_{\omega}\oplus V_{\overline{\omega}}$

$\rho(gh^{2})$

$\chi(g)$

$\omega^{3}$

$\omega$

$d=\dim(V_{\omega}\cap W_{\omega})$

$\chi(h)$

$\omega^{2}$

$d$

$\omega^{2}$

$\rho(gh^{-1})$

$0$

$\omega,$

$P$

$\omega^{3},\overline{\omega}$


$\chi(gh^{-1})$

$\chi(g^{h})$


$\chi(g)$

$\{1, \omega, \omega^{2},\overline{\omega}\}$

$\chi(gh^{-1})$


$\omega$

$\rho(g)$

$\langle g\rangle\neq\langle h\rangle$

$\rho(h)$

CLAIM 1. We have $\dim(V_{\omega}\cap W_{\omega})=2\cdot 3^{S-1}$

.

$s=t\geqq 1$

.

Moreover,

$\dim(V_{1}\cap W_{\omega})=\dim(V_{\omega}\cap W_{1})=3^{s-1}$

and

584


, CLAIM 1. Let $a=\dim(V_{\omega}\cap W_{1}),$ $b=\dim(V_{1}\cap W_{\omega}),$ $c=3^{s}=3^{t}$ , so $a+c=\dim(V_{\omega})=3^{S}$ and $b+c=\dim(W_{\omega})=3^{t}$ . If $a=0$ or $b=0$ , then $b>0$ $a>0$ has both and so $g=h$ , a contradiction. Suppose and . Then as eigenvalues, so is rational by Lemma 3.4, and thus $a=b$ and $s=t$ . Also, $\rho(gh)$ has is rational by as an eigenvalue with multiplicity $2a$ , so . Lemma 3.4. Therefore $a=b=3^{s-1}$ and

PROOF

OF

$c=\dim(V_{\omega}\cap W_{\omega})$

$\rho(gh^{-1})$

$\omega$

$\chi(gh^{-1})$

$\overline{\omega}$

$\chi(gh)$

$\omega$

$c=2\cdot 3^{S-1}$

CLAIM 2.

$X\cap C_{G}(\langle g, h\rangle)E\langle g, h\rangle$

$\square$

.

PROOF OF CLAIM 2. Suppose not. Let be the -eigenspace of . Put $d=\dim(V_{1}\cap W_{1}\cap U_{\omega})$ . the pairs $\{g, h\}$ , $\{g, x\}$ and $\{h, x\}$ , we see $\rho(x)$

$\xi$

$\dim(V_{\omega})=\dim(W_{\omega})=\dim(U_{\omega})=3^{s}$

.

Let Applying Claim 1 to

$x\in X\cap C_{G}(\langle g, h\rangle)\backslash \langle g, h\rangle$

$U_{\xi}$

,

$\dim(V_{1}\cap W_{1}\cap U_{\omega})=\dim(V_{1}\cap W_{\omega}\cap U_{1})=\dim(V_{\omega}\cap W_{1}\cap U_{1})=d$

,

$\dim(V_{1}\cap W_{\omega}\cap U_{\omega})=\dim(V_{\omega}\cap W_{1}\cap U_{\omega})=\dim(V_{\omega}\cap W_{\omega}\cap U_{1})=3^{s- 1}-d$

,

and $\dim(V_{\omega}\cap W_{\omega}\cap U_{\omega})=3^{s-1}+d$

.

are $3^{s-1}+3d$ Therefore the multiplicities of and to as eigenvalues of , respectively. Hence $d=0$ by Lemma 3.4. Thus acts as on and nonidentity eigenvalues of , and the remaining and $\rho(ghx^{-1})$

$\omega$

$3^{s-1}$

$\rho(x)$

$V_{1}\cap W_{\omega}$

occur on

$3^{S-1}$

$V_{\omega}\cap W_{1}$

$V_{\omega}\cap W_{\omega}$

$\omega$

$\rho(x)$

.

.

agree on $V_{1}+W_{1}$ and Then as an eigenvalue of be the multiplicity of on Let . Then $\rho(xy)$ has as an eigenvalue with multiplicity $2(3^{S-1}-a)$ . It follows that $a=0$ or $a=3^{s-1}$ by Lemma and to with multiplicity 3.4. If $a=0$ then $y=(ghx)^{-1}$ , while if $a=3^{s-1}$ then $y=x$ . Put $X_{0}=\{g, h, x, (ghx)^{-1}\}$ . Let $H=C_{G}(\langle g, h, x\rangle)$ , so $H$ is a 3-group. We , then have shown $X\cap H=X_{0}$ . Let $N=N_{G}(H)$ , and let $N_{3}\in Sy1_{3}(N)$ . If has eigenvalues has an orbit of size 3 on the eigenspaces of $\rho(H)$ , so , is rational by Lemma 3.4. Therefore and for some , and tbus $N_{3}\cap X=X_{0}$ , so $N_{3}\in Sy1_{3}(G)$ . , and thus : $H|\leqq 3$ . Note . Suppose to We next show assumes only rational values on is irrational. If $o(z)>3$ , then and the contrary that by Lemma 3.13, a contradiction. Then $o(z)=3$ and $z\in X$ or $z^{-1}\in X$ by , a contradiction. , so or Claim 1. Hence . Then vanishes NOW let be the character of $H$ afforded by . Also, , and hence $|H|$ divides on $\chi(gh^{-1})=n-3^{s}$ $\chi(gh)=n-2\cdot 3^{s}$ , and therefore is divisible by , we have , which is sharp. Therefore is since Suppose by the above.

$y\in X\cap C_{G}(\langle g, h, x\rangle)\backslash \langle g, h\rangle$

$a$

$\rho(x)$

$\rho(y)$

$\rho(y)$

$\omega$

$V_{\omega}\cap W_{\omega}\cap U_{\omega}$

$\omega$

$2\cdot 3^{s-1}+a$

$z\in N_{3}\backslash H$

$\rho(z)$

$\rho(z)$

$\xi$

$\omega\xi$

$\chi(z)$

$\xi$

$\overline{\omega}\xi$

$N_{G}(N_{3})\subseteqq N$

$|N_{3}$

$H\backslash \langle g, h, x\rangle$

$\chi$

$z\in H\backslash \langle g, h, x\rangle$

$\chi(z)$

$\langle g\rangle=$

$\langle h\rangle\subseteqq\langle z\rangle$

$z\in X_{0}$

$z^{-1}\in X_{0}$

$z\in\langle g, h, x\rangle$

$V_{\omega}\cap W_{\omega}\cap U_{\omega}$

$\theta$

$|H|\cdot(\theta, \theta)_{H}=3^{3}(3^{s-1})^{2}=3^{zs+1}$


$|G|_{3}$

$3^{s}(1-\omega)3^{s}(1-\overline{\omega})3^{2s}=3^{4s+1}$

$\theta$

$(G, \chi)$

$3^{4S+1}\leqq 3^{2s+2}$

585


impossible since

$s\geqq 1$

.

This contradiction completes the proof of Claim 2.

$\square$

, so $H$ is a 3-group. Also, $H\cap X=\{g, h\}$ , and NOW put assumes only rational therefore $H\in Sy1_{3}(G)$ . As in the proof of Claim 2, . Then . Let be the character of $H$ afforded by values on vanishes on . Therefore $|H|$ divides . AS in the proof of Claim 2, $|H|=|G|_{3}$ is divisible by , and we have a contradiction since is sharp. Therefore since , as required. . Hence Case 4. Suppose $o(g)=4$ . In this case is the only nonidentity eigenvalue is irrational and of . Let and . If $d=0$ , then has an eigenvalue $-1$ , contradicting Lemma 3.4. Therefore $d>0$ and $\rho(gh)$ has an eigenvalue $-1$ . and $g=h$ . is rational, so Then Case 5. Suppose $o(g)=5$ . The strategy used in this case is similar to that assumes only real values on used in Case 3. By Case 2 we may suppose . We suppose for any . Thus and obtain a contradiction. $H=C_{G}(\langle g, h\rangle)$

$\chi$

$H\backslash \langle g, h\rangle$

$\theta$


$\theta$

$|H|\cdot(\theta, \theta)_{H}=3^{2}(2\cdot 3^{s-1})^{2}=4\cdot 3^{zs}$


$3^{s}(1-\omega)3^{s}(1-\overline{\omega})3^{2\}=3^{4s+1}$

$3^{4s+1}\leqq 3^{2S}$

$(G, \chi)$

$s\geqq 1$


$\omega$

$\rho(g)$

$\chi(g^{2}h)$

$d=\dim(V_{\omega}\cap W_{\omega})$

$\rho(h)$

$\rho(g^{2}h)$

$V_{\omega}=W_{\omega}$

$\chi(gh)$

$\chi$


$\xi,$

$\dim(V_{\xi}\cap W_{\lambda})=\dim(V_{\overline{\xi}}\cap W_{\overline{\lambda}})$

CLAIM 1.

We have

$s=t\geqq 1$

.

Moreover,

$\dim(V_{\omega}\cap W_{\omega})=\dim(V_{\omega}\cap W_{\overline{\omega}})=2\cdot 5^{S-1}$

$\lambda$

$\langle g\rangle\neq\langle h\rangle$

$\dim(V_{1}\cap W_{\omega})=\dim(V_{\omega}\cap W_{1})=5^{s-1}$

and

.

. CLAIM 1. Let Suppose both are irrational. Then by Lemma 3.4, $\rho(gh)$ and , and has no eigenvalue has no eigenvalue since . since $\rho(gh)$ is irrational and . Hence and Therefore , multiplicity contradicting has Lemma 3.4. as an eigenvalue with is rational. Therefore or Suppose is rational and is irrational. We have $b=0$ since . Then the multiplicity of and as eigenvalues of $\rho(gh)$ are $5^{s}+5^{t}-2a$ , and and , respectively, so $5^{s}+5^{t}=3a$ . However, is so we have a contradiction. Similarly, a contradiction is obtained if is rational. irrational and and are both rational. Comparing multiplicities of Therefore , we see $5^{s}+5^{t}-2(a+b)=a=b$ , and as eigenvalues of $\rho(gh)$ and and $a=b=2\cdot 5^{s-1}$ . It follows that $\dim(V_{1}\cap W_{\omega})=\dim(V_{\omega}\cap W_{1})=5^{s-1}$ , hence $s=t$ and and the claim is proved. PROOF

OF

$a=\dim(V_{\omega}\cap W_{\omega}),$

$\chi(gh)$

$b=\dim(V_{\omega}\cap W_{\overline{\omega}})$

$\chi(gh^{-1})$

$g\neq h$

$\omega^{2}$

$g\neq h^{-1}$

$\chi(gh)$

$W_{\omega}\oplus W_{\overline{\omega}}\subseteqq V_{1}$

$V_{\omega}\oplus V_{\overline{\omega}}\subseteqq W_{1}$

$\omega^{2}$

$\rho(gh^{-1})$

$5^{S}+5^{t}$

$\omega$

$\chi(gh)$

$\chi(gh^{-1})$

$\chi(gh^{-1})$

$\chi(gh)$

$\omega^{2}$

$g\neq h^{-1}$

$\omega$

$a\leqq\dim(V_{\omega})=5^{s}$

$a$

$a\leqq\dim(W_{\omega})=5^{t}$ $\chi(gh)$

$\chi(gh^{-1})$

$\chi(gh)$

$\chi(gh^{-1})$

$\omega^{2}$

$\rho(gh^{-1})$

$\omega$

$\square$

Note

$\langle g, h\rangle\cap X=\{g^{\pm 1}, h^{\pm 1}\}$

by Claim

1. Suppose now that

$x\in C_{G}(\langle g, h\rangle)$

is irrational. If $o(x)>5$ , then by Lemma 3.13, a conand is not real, then by Case 2, tradiction. If $o(x)=5$ and real, is is an algebraic and hence another contradiction. Therefore conjugate of . $\chi(x)$

$\langle g\rangle=\langle h\rangle\subseteqq\langle x\rangle$

$\chi(x)$

$\langle g\rangle=\langle x\rangle=\langle h\rangle$

$\chi(x)$

$\chi(g)$

$\chi(x)$

586


CLAIM 2.

$X\cap C_{G}(\langle g, h\rangle)\subseteqq\langle g, h\rangle$

.

CLAIM 2. Suppose to the contrary that . Let be the -eigenspace of $p(x)$ , and put Then since assumes only real values on $\{g, h\},$ pairs 1 and $\{h, x\}$ , we find the to Claim PROOF

OF

$x\not\in\langle g, h\rangle$

$x\in X\cap C_{G}(\langle g, h\rangle)$

and

$d_{a,\beta.\gamma}=\dim(V_{a}\cap W_{\beta}\cap U_{\gamma})$

$\xi$

$U_{\xi}$

$\chi$

$C_{G}(\langle g, h\rangle)$

$d_{\alpha,\beta.\gamma}=d_{\overline{a}.\overline{\beta}.\overline{\lambda}}$

.

.

Applying

$\{g, x\}$

2

if

$d_{\alpha,\beta.\xi}= \sum_{\xi}d_{\alpha.\xi.\beta}=\sum_{\xi}d_{\xi.a,\beta}=5^{s-1}$

or

$\{\alpha, \beta\}=\{1, \omega\}$

,

$\{1, \overline{\omega}\}$

(3.6)

and

2

if

$d_{a.\beta.\xi}= \sum_{\xi}d_{a.\xi,\beta}=\sum_{\xi}d_{\xi.a.\beta}=2\cdot 5^{s-1}$

where

in the sums ranges over

$\xi$


.

It follows that and

$d_{1,1.\omega}+d_{1.\omega.\omega}+d_{1.\overline{\omega}.\omega}=d_{1.\omega.1}+d_{1.\omega,\omega}+d_{1.\omega,\overline{\omega}}=5^{s-1}$

$d_{1.1.\omega}=d_{\omega.1.1}$

or

, {to} ,

$\{\alpha, \beta\}=\{\omega\}$

$\{\omega,\overline{\omega}\}$

,

$d_{1.1.\omega}=d_{1.\omega.1}$

$d_{1.\overline{\omega}.\omega}=d_{1,\omega.\overline{\omega}}$

.

.

(3.7)

since

Similarly,

is irrational. If is an eigenvalue of $p(ghx)$ , then 3.4, by Lemma by (3.7). Similarly so , contradicting (3.7). On the . Then is an eigenvalue of $\rho(ghx)$ , then other hand, if since and . However, , so we have a contradiction. Therefore $\chi(ghx)$ is rational, so Suppose that

$\chi(ghx)$

$\omega$

$d_{\omega,\omega.\overline{\omega}}=2\cdot 5^{s-1}$

$d_{\omega.1.\omega}=d_{\omega,\omega.\omega}=0$

$d_{\omega,\overline{\omega}.\omega}=$

$d_{1,\omega,\overline{\omega}}+d_{\omega,\omega,\overline{\omega}}+d_{\overline{\omega},\omega.\overline{\omega}}\geqq 4\cdot 5^{s-1}$

$d_{\overline{\omega}.\omega.\overline{\omega}}=2\cdot 5^{s-1}$

$\omega^{2}$

$d_{1,\omega.\overline{\omega}}=2\cdot 5^{s-1}$

$d_{\omega.\omega.\overline{\omega}}+d_{\overline{\omega}.\omega.\overline{\omega}}=2\cdot 5^{s-1}$

$d_{1,\omega,\overline{\omega}}+$

$d_{I.\omega.1}+d_{1,\omega,\omega}+d_{\iota,\omega.\overline{\omega}}=$

$d_{\omega.\omega.\overline{\omega}}=d_{\overline{\omega}.\omega.\overline{\omega}}=0$

$5^{s-1}$

$3d_{1.1.\omega}+d_{\omega,\omega,\delta}+d_{cu.\overline{w}.\omega}+d_{\overline{\omega}.\omega.\omega}=d_{1.\omega.\omega}+d_{\omega.1.\omega}+d_{\omega.\omega.\iota}+d_{\overline{\omega}.\overline{\omega},\overline{\omega}}$

.

. From this together with (3.6) and (3.7) it follows that is irrational. If has an eigenvalue to, then a NOW suppose is an eigenvalue of $p(ghx^{2})$ . It follows that contradiction arises. Therefore . Moreover, if is the set of all triples for all or , then for which $d_{1.1.\omega}=0$

$\chi(ghx^{2})$

$\rho(ghx^{2})$

$\omega^{2}$

$d_{a,\beta.\gamma}=0$

$d_{a.\beta.\gamma}=5^{\epsilon-1}$

$\alpha,$

$\beta,$

$I$

$\gamma$

$d_{\alpha.\beta.\gamma}=5^{s-1}$

$(\alpha, \beta, \gamma)$

$I=\{(1, \omega, \omega),$

$(\omega, 1, \omega),$

$(\omega, \omega, 1),$

$(1, \overline{\omega},\overline{\omega}),$

$(\overline{\omega}, 1,\overline{\omega}),$

$(\overline{\omega},\overline{\omega}, 1)$

$(\omega, \omega,\overline{\omega}),$

, (to,

$\overline{\omega},$

$\omega$

),

$(\omega,\overline{\omega}, \omega),$

$(\overline{\omega}, \omega,\overline{\omega})$

(

$\omega,\overline{\omega}$

, (di,

, th),

$\omega,$

$\omega$

)

$\}$

.

Using an argument similar to that used in the proof of Claim 2 of Case 3, it is irrational, then and can be shown that if $y=g^{2}h^{2}x^{4}$ . or either $y=x$ is rat\’ional. Then or for all NOW assume . Moreover, if is defined as above, then one of the following holds. $\chi(ghy^{2})$

$y\in X\cap C_{G}(\langle g, h, x\rangle)\backslash \langle g, h\rangle$

$\chi(ghx^{2})$

$d_{a.\beta.\gamma}=0$

$d_{\alpha.\beta.\gamma}=5^{s-1}$

$\alpha,$

$\beta,$

$\gamma$

$I$

$I=\{(1, \omega,\overline{\omega}),$

$(1, \overline{\omega}, \omega)$

$(\omega, 1, \omega),$

, (hi, 1,

$\overline{\omega}$

$(\omega, \omega, \omega),$

), (di,

$\overline{\omega},\overline{\omega}$


), (di,


( , to, 1), ( , to, w), $\omega$

), (ru,

$\omega$

$\omega$

$\omega,$

$1$

),

$(\overline{\omega}, \omega, \omega)\}$

(3.8)

587

Sharp characters with irrational values $I=\{$ ( $1,$

$\omega$

, to), (

$1$ $\omega,$

, to), (to,

$1$ $\omega,$

),


$(1, \overline{\omega}, \omega),$

$(\overline{\omega}, 1, \omega),$

( , to, 1), (di,

$I=\{(1, \omega, \omega),$

$(\omega, 1,\overline{\omega}),$


$(\overline{\omega}, 1, \omega),$

( , to, to), (to,

$(1, \overline{\omega},\overline{\omega}),$

$\overline{\omega}$

$\overline{\omega},\overline{\omega}$

),

$(\omega,\overline{\omega}, \omega),$

$(\overline{\omega}, \omega,\overline{\omega})$


$\overline{\omega}$


$\omega$

$1$ $\omega,$

$\omega$

, (to,

$(\omega,\overline{\omega}, 1),$

), (di,

( , to, to), $\omega$

$\omega,$

)

$(\omega,\overline{\omega}, \omega)$

), (

$\omega$


, to)

(3.9)

$\}$

, $\}$

.

(3.10)

if (3.9) holds, and if (3.10) is irrational. and is irrational. Let by if necessary, we may suppose $H=C_{G}(\langle g, h, x\rangle)$ , and put $X_{0}=X\cap H$. Then

Define holds. Then Replacing

$x_{0}=g^{3}h^{4}x^{3}$

if (3.8) holds,

$x_{0}\in C_{G}(\langle g, h\rangle)\cap X\backslash \langle g, h\rangle$

$x$

$x_{0}=g^{3}h^{4}x^{2}$

$x_{0}=x^{-1}$

$\chi(ghx_{0}^{2})$

$\chi(ghx^{2})$

$x_{0}$

$X_{0}=\{g^{\pm 1}, h^{\pm 1}, x^{\pm 1}, (g^{4}h^{2}x^{2})^{\pm 1}, (g^{2}h^{4}x^{2})^{\pm 1}, (g^{2}h^{2}x^{4})^{\pm 1}\}$

.

, then be a Sylow 5-subgroup of $N$ . If Let $N=N_{G}(H)$ , and let , of of with number and thus elements commutes an even with commutes : $H|$ S5. Arguing as in the proof of exactly 2 elements of . Therefore $N_{6}\cap X=X_{0}$ , so $N_{6}\in Sy1_{6}(G)$ and $|G|_{6}=|N_{6}|\leqq 5|H|$ . Claim 2 of Case 3, we see Also, $|H|$ divides since assumes only rational values on . is sharp and assumes the values $\chi(gh)=n-2\cdot 5^{S},$ $X(gh^{2})=n-3\cdot 5^{s}$ , Since , and is divisible by . $4s+1\leqq 2s+2$ , which is impossible since . This completes the Therefore proof of Claim 2. $y\in N_{6}\backslash H$

$N_{6}$

$X_{0}$

$y$

$y$

$|N_{g}$

$X_{0}$

$5^{2S+1}$

$\chi$


$\chi$

$(G, \chi)$

$\chi(g)=n-2\cdot 5^{s}+5^{s}(\omega+\overline{\omega})$

$\chi(g^{2})=n-2\cdot 5^{s}+5^{S}(\omega^{2}+\overline{\omega}^{2}),$

$5^{4S+1}$

$|G|_{6}$

$s\geqq 1$

$\square$

Put

$H=C_{G}(\langle g, h\rangle),$

$X_{0}=H\cap X$

.

By Claim 2 we have

.

Thus assumes rational . There-

$X_{0}=\{g^{\pm 1}, h^{\pm 1}\}$

. Arguing as in Case 3, we see since . As in the proof of Claim 2, we have values on , as claimed. fore 4s+l$2s, a contradiction. Thus $o(g)=8$ Suppose strategy . The used in this case is similar to that Case 6. and arrive at a contraused in Case 3 and Case 5. We suppose diction. By Case 2 we may assume assumes only real values on . We may also assume has been chosen so that . $H\in Sy1_{5}(G)$

$|H|\leqq 5^{zs}$

$\chi$

$|G|_{6}\geqq 5^{4\+1}$



$h\in C_{G}(g)\backslash \langle g\rangle$

$\chi$

$\omega+\overline{\omega}=\sqrt{}\overline{2}$

$\omega$

CLAIM 1.

We have

$=\dim(V_{\overline{\omega}}\cap W_{\overline{\omega}})=2^{s-1}$


$s=t\geqq 1$

and

$\dim(V_{\omega}\cap W_{\omega})=\dim(V_{\overline{\omega}}\cap W_{\omega})=\dim(V_{\omega}\cap W_{\overline{\omega}})$

.

. If $a=b=0$ , has an eigenvalue $-1$ , contradicting Lemma $a>0$ , then has $-1$ as an eigenvalue, so $\chi(gh^{3})=n-2^{s+1}-2^{t+1}+2b+(2^{\}-2^{t})\sqrt{2}$ is rational, and hence $s=t$ . Similarly, $s=t$ if $b>0$ . Therefore $s=t$ in both cases. In particular, $X(g)=x(h)=n-2^{\epsilon+1}+2^{S}$ V2. NOW, if $a>0$ , then $\rho(gh)$ has and as eigenvalues, so PROOF OF CLAIM 1. Put is irrational and then 3.4. Thus $a>0$ or $b>0$ . If $\chi(gh^{4})$

$a=\dim(V_{\omega}\cap W_{\omega})$

,

$b=\dim(V_{\omega}\cap W_{\overline{\omega}})$

$\rho(gh^{4})$

$\rho(gh^{3})$

$\omega^{2}$

$-\omega^{2}$

$X(gh)=n-2^{S+2}+2(a+b)+2(2^{s}-a-b)\sqrt{2}+a(\omega^{2}+\overline{\omega}^{2})+2b$

588


is rational, and hence

$a+b=2^{S}$

.

Similarly,

if

$a+b=2^{S}$

$\chi(g^{2}h)=n-2^{S+1}+(b-a)\sqrt{2}$

$b>0$

.

Then

.

is irrational, then $b-a=2^{s}$ or $b-a=-2^{s}$ , so $h=g^{-1}$ or If is rational, so $a=b=2^{s-1}$ . to our assumptions. Therefore

$h=g$ ,

$\chi(g^{2}h)$

$\chi(g^{2}h)$

contrary

$\square$

is irrational. If $o(x)o(g)$

$\langle g\rangle\backslash \langle g^{2}\rangle$

$\langle g\rangle=\langle h\rangle\subseteqq\langle x\rangle$

$\chi(g)$

$\chi(x)$


$\chi$

CLAIM 2.

$C_{G}(\langle g, h\rangle)\cap X\subseteqq\langle g, h\rangle$

.

. PROOF OF CLAIM 2. Suppose to the contrary that $\rho(ghx)$ are elements of . Let By Claim 1, the eigenvalues of $\rho(ghx)$ eigenvalue Applying of an as . Claim 1 to all be the multiplicity of as eigenvalues pairs in $\{g, h, x\}$ , we see that the multiplicities of to, , and $3d$ $2^{s-1}-d$ ), , and , respectively. Hence $\chi(ghx)=n-2^{s+1}+$ of $\rho(ghx)$ are and $d=2^{s-2}$ since $\chi(ghx)\in R$ . Therefore . It follows that $\chi(ghx)=n-2^{S+1}+2^{s-1}\sqrt{}^{-}2^{-}$ , which is impossible since has a single algebraic conjugacy class of irrational values on . $x\in C_{G}(\langle g, h\rangle)\cap X\backslash \langle g, h\rangle$

$d$

$\{1, \omega^{3}, \omega,\overline{\omega},\overline{\omega}^{3}\}$

$\omega^{3}$


$\overline{\omega}$

$3_{\backslash }’2^{s-1}-d$

$s\geqq 2$

$2d\overline{\omega}+2(2^{s-1}-d)\omega$

$\chi$


$\square$

NOW, it follows easily from Claim 1 that has and $K=C_{G}(\langle g, h\rangle)$ , eigenConsidering by 2. so 32. Claim Put order values, we see that $K\cap X=\{g, g^{-1}, h, h^{-1}\}$ . Let $N=N_{G}(K)$ . Then $N$ acts on $K\cap X$ by conjugation, and $N/K$ is isomorphic to a 2-subgroup of . Hence $K$ $|N|$ divides $8|K|$ . Now, let , let be the character of afforded by $K$ afforded by , -eigensPace of of let be and the character the be . Then assumes only rational values on . Suppose by Claim 2, so and Thus $\langle g, h\rangle$

$\langle g\rangle\cap\langle h\rangle=\langle g^{4}\rangle$

$K\cap X\subseteqq\langle g, h\rangle$

$S_{4}$

$\theta$

$V_{\omega}$

$\rho(h)$

$\omega$

$\psi$

$\chi$

$x\in K\backslash \langle g, h\rangle$


$\theta_{\omega}(x)=0$

$x\langle g, h\rangle$

$\theta_{\omega}(ghx)=0$

.

$0=\theta_{\omega}(ghx)=\psi(x)\omega^{2}+(\theta_{\omega}(x)-\psi(x))\cdot 1=\psi(x)(\omega^{2}-1)$

and therefore

$\psi(x)=0$

. Since

$W_{\omega}$

$\psi$

vanishes on

$K\backslash \langle g, h\rangle$

$|K|(\psi, \psi)_{K}=32(2^{s-1})^{2}=2^{2s+3}$

,

,

we have

,

. Hence $|N|$ is a divisor of and therefore $|K|$ is a divisor of . , then has a nontrivial orbit on the set of eigenspaces of If , and thus $X(x)$ is , least has at one pair of eigenvalues so on rational by Lemma 3.4. Hence $N\cap X=K\cap X$ , so $N_{G}(N)=N$ and $N\in Sy1_{2}(G)$ . NOW, assumes the values $\chi(g^{4})=n-2^{S+2},$ $\chi(g^{2})=n-2^{s+1},$ $\chi(g)=n-2^{S+1}+$ $\chi(gh)=n-2^{s}$ , and $\chi(g^{5}h)=n-3\cdot 2^{S}$ on $K$ . Since $2^{2s+3}$

$x\in N\backslash K$

$V$

$\rho(x)$

$\rho(x)$

$\chi$

$2^{s}\sqrt{2},$

$8\cdot 2^{2s+3}=2^{2s+6}$

$\chi(g^{3})=n-2^{S+1}-2^{S}\sqrt{}\overline{2},$


$\zeta,$

$-\zeta$

589


is sharp and

is divisible by $s=0$ , contradicting Claim 1. Therefore and hence the lemma is complete. $(G, \chi)$

$N\in Sy1_{2}(G),$

$2^{6S+4}$

$|N|$

. Therefore 6s+4$2s+6,


the proof of

, and

$\square$

is irrabe prime. Assume is a -element of $G,$ timal, and whenever is a p-group. Suppose further that Moreover, nonidentity any eigenvalue and $X(h)$ is irratimal. Then . occurs with multiplicity1. of LEMMA 3.15.

Let

$p$

$\chi(g)$

$p$

$g$

$C_{G}(g)$

$\langle g\rangle\subseteqq\langle h\rangle$



$\rho(g)$

is not PROOF. Put $H=C_{G}(g),$ $N=N_{G}(H),$ $X=\{x\in G|\chi(x)=x(g)\}$ . If real, then $H\cap X=\{g\}$ by Lemma 3.14, and hence $N=H\in Sy1_{p}(G)$ . On the other hand, if , so $|N:H|\leqq 2$ , and thus $N\in Sy1_{p}(G)$ is real, then $p=2$ $H\in Sy1_{p}(G)$ , and if and if $p>2$ . Let be a nonidentity eigenvalue of multiplicity let be the . of . Put by Lemma Suppose $h\in H$ and is irrational. If $o(h)>o(g)$ , then $o(h)\leqq o(g)$ by 3.13, and we have a contradiction. Therefore , and so . be Let Lemmas 3.13 and 3.14. Hence assumes rational values on the character of $H$ afforded by the -eigenspace of . Then vanishes on , so $|H|$ divides $|H|$ . $(\theta, 0)_{H}=p^{2S+l}$ . Therefore if $s=0$ , then and we are done. . Let $M$ be the set of values of on $\chi(g)$

$H\cap X\subseteqq\{g, g^{-1}\}$

$\chi(g)$

$\rho(g)$

$\omega$

$p^{s}$

$p^{\iota}=o(g)$

$\omega$

$\langle g\rangle\subset\langle h\rangle$

$\chi(h)$

$h\in\langle g\rangle$

$\theta$

$\rho(g)$

$\omega$

$H=\langle g\rangle$

$H\backslash \langle g\rangle$

$\chi$

Case 1.

Suppose

$\chi(g)$

$\langle g\rangle\backslash \{1\}$

In this case we have

is not real.

$f_{M}(n)= \prod_{j=1}^{\iota}p^{s\varphi(p^{j})+1}=p^{S(p^{l}-1)+l}$

.

is sharp, $|H|=|G|_{p}$ is divisible by $f_{M}(n)$ , and therefore $s(p^{l}-1)+$ . Suppose $s>0$ and . Thus either $s=0$ and we are done, or else , and therefore , then . If assumes a rational value on . But then $|G|_{3}=$ is divisible by $3f_{M}(n)$ , a contradiction. Therefore , and bence $s=0$ . Case 2. Suppose $X(g)$ is real. If $p=2$ , then

Since

$(G, \chi)$

$1\leqq 2s+l$

$p^{\iota}=3$

$p^{\iota}=3$

$H\neq\langle g\rangle$

$X$



$|G|_{3}$

$3$

$\theta$


$\chi$

$f_{M}(n)=2^{s+2} \cdot 2^{s+1}\cdot\prod_{j=3}^{l}2^{s\varphi(2^{f})/2+1}=2^{2^{l-1}s+\iota+1}$

Since

so

is divisible by $f_{M}(n)$ and since and we are done.

$|G|_{2}$

$s=0$

$1\geqq 3$

$|G|_{2}\leqq 2|H|$

Suppose

,

we have

$P>2$

.

$2^{\iota-1}s+l+1\leqq 2s+l+1$ ,

. Then

$f_{M}(n)= \prod_{j=1}^{l}p^{s\varphi(p^{j})/z+1}=p^{S(p^{l}-1)/2+l}$

.

Since $|G|_{p}=|H|$ , we have $s(p^{\iota}-1)/2+1\leqq 2s+l$ , so either $s=0$ and we are done, , then . If assumes a rational or else $p^{l}=5$ . Suppose $s>0$ and $p^{\iota}=5$

$H\neq\langle g\rangle$

$\chi$

590


value on

so

$s=0$

.

, so $|G|_{6}\geqq 5f_{M}(n)$ , and we have a contradiction. Thus This completes the proof of the lemma.



,

$\square$

is -srngular or else LEMMA 3.16. Supp $ose\chi(x)$ is irratimal and either and either has prime power order. If is not real or $o(g)>2$ , $C_{G}(g)=C_{G}(x)$ then . $\chi$

$x$

$x$

$\chi(x)$

$g\in\langle x\rangle\backslash \{1\}$

coincide. and PROOF. It suffices to show that the eigenspaces of If has only one nonidentity eigenvalue, and hence we is not real, then has two complex conjugate nonidentity are done. If is real, then eigenvalues, so has two distinct nonidentity eigenvalues since $o(g)>2$ , as required. $\rho(g)$

$\rho(x)$

$\chi(x)$

$\rho(x)$

$\chi(x)$

$\rho(x)$

$\rho(g)$

$\square$

is irrational. We claim PROOF OF THEOREM 2.2. Assume $g\in G$ and prove that it suffices to is irrational such that for some and either is -singular or has prime power order. Indeed, if this is the case then $C_{G}(g)=C_{G}(x)=\langle x\rangle$ by Lemma 3.16, so (i) holds. Also, (ii) holds by has prime Lemma 3.5 if is -singular, and (ii) holds by Lemma 3.15 if power order. Since (iii) is a consequence of (ii), the proof is complete. NOW, suppose be commutes with some -singular element . Let by Lemma maximal such that . Then is -singular and $g\in C_{G}(x_{0})=C_{G}(x)=\langle x\rangle$ by 3.11. Since is -singular, we have and Lemma 3.16, and we are done by the claim. Next, suppose $C_{G}(g)$ contains no -singular element of . Suppose is a $C_{G}(g)$ is prime. Then is a -group by Lemma 3.12. -element of , where $X(x)$ Let is irrational and . Then is a be maximal such that -element, and by Lemma 3.15, so again we are done by the claim. Finally, suppose $C_{G}(g)$ contains no -singular element of and does not of order , have prime power order. Then $\chi(g)=x(y)$ for some element of is real by Lemma 3.5. Since where $q=3$ if $X(g)$ is not real and $q=5$ if has prime order, we may apply the argument above to conclude , -singular power prime is irrational and either is where or else has in , then . Thereorder. If $X$ is the conjugacy class of fore of contains a Sylow q-subgroup of , so we can assume the -part $o(y)=q$ , and hence is an element of . Note since by Lemma 3.16, and the proof of Theorem 2.2 is complete. $\chi(g)$

$g\in C_{G}(x)=\langle x\rangle$

$\chi$

$x$

$\chi(x)$

$x$

$x$

$\chi$

$x$

$x$

$\chi$

$g$

$\chi$

$C_{G}(x)=\langle x\rangle$

$\langle x_{0}\rangle\subseteqq\langle x\rangle$

$\chi$

$x_{0}\in\langle x\rangle$

$x_{0}$

$G$

$\chi$

$G$

$P$

$\langle x\rangle$

$x_{0}$

$X$

$p$

$g$

$P$

$\langle x\rangle$

$x$

$\langle g\rangle\subseteqq\langle x\rangle$

$g\in C_{G}(x)=\langle x\rangle$

$P$

$G$

$\chi$

$y$

$g$

$G$

$q$

$\chi(g)$

$y$

$y\in C_{G}(x)=\langle x\rangle$

$\chi(x)$

$x$

$\chi$

$x$

$G$

$x$

$G$

$\langle x\rangle$

$\langle x\rangle$

$g$

$y\in\langle g_{q}\rangle$

$C_{G}(y)=\langle x\rangle$

$X\cap\langle x\rangle\subseteqq\{x, x^{-1}\}$

$q$

$g_{q}$

$g\in C_{G}(g_{q})\subseteqq$

$\square$

References [1]

D. Alvis, On finite groups admitting certain sharp characters with irrational values, Comm. Algebra, 21 (1993), 535-554.

Sharp characters with irrational values [2] [3]

591

D. Alvis, M. Kiyota, H. Lenstra and S. Nozawa, Sharp characters with only one rational value, Comm. Algebra, 22 (1994), 95-115. H. F. Blichfeldt, A theorem concerning the invariants of linear homogeneous groups with some applications to substitution-groups, Trans. Amer. Math. Soc., 5 (1904),

461-466. P. J. Cameron and M. Kiyota, Sharp characters of finite groups, J. Algebra, 115 (1988), 125-143. [5] T. Ito and M. Kiyota, Sharp permutation groups, J. Math. Soc. Japan, 33 (1981), 435-444. [4]

[6] [7] [8]

[9]

T. Kataoka, Torsion part of (R $(G)/Z\rho_{G})^{*}$ , preprint. T. Kataoka, Virtual characters of a finite group having prescribed values, preprint. T. Kataoka, On sharp characters of a finite group in the sense of Kiyota-Cameron, preprint. M. Kiyota and S. Nozawa, Sharp characters whose values at non-identity elements are 0 and a family of algebraic conjugates, J. Algebra, 161 (1993), 216-229.

Dean ALVIS Department of Mathematics

and Computer Science Indiana University South Bend South Bend Indiana,

USA

46634

S\^ohei NOZAWA Department of Mathematics and Informatics Faculty of Sciences Chiba University 1-33, Yayoicho, Inage-ku Chiba 263 Japan