148

A sufficient condition for a finite group
to be a Borsuk‐Ulam group
Toshio Sumi

Faculty of Arts and Science, Kyushu University

1

Introduction

In this paper, we always assume that a group means a finite group. A G‐map f:Xarrow Y
is said to be a G‐isovariant map if G_{x}=G_{f(x)} for any x\in X , where G_{x} is the isotropy

subgroup, that is, G_{x}=\{g\in G|g\cdot x=x\} . We call a group G is a BUG (Borsuk‐Ulamgroup) [5] if\dim V-\dim V^{G}\leq\dim W-\dim W^{G}

for any isovariant G ‐map f:Varrow W between G‐representation spaces V and W.
Let C_{2} be a cyclic group of order 2 and let f:Varrow W be an isovariant C_{2}|‐map between
C_{2}|‐representation spaces V and W . Fixing a G‐invariant inner product, f induces a free
C_{2} ‐map

S(f|_{V-V^{C_{2}}}):S(V-V^{C_{2}})arrow S(W-W^{C_{2}}) between C_{2} ‐representation spheres,

where V-V^{C_{2}} is an orthogonal vector subspace of V^{C_{2}} inV.

By Borsuk‐Ulam the‐orem, this map gives \dim S(V-V^{C_{2}})\leq\dim S(W-W^{c_{2}}) . Since \dim S(V-V^{C_{2}})=\dim V-\dim V^{c^{2}}-1, C_{2} is a BUG. For a cyclic group C_{p} of prime order p , Kobayashi[2] showed that \dim S(V)\leq\dim S(W) for a free C_{p} ‐map S(f'):S(V)arrow S(W) between
representation spheres and thus C_{p} is a BUG.
Let G be a group extension of K by H:1arrow Harrow Garrow Karrow 1 and f:Varrow W be an
isovariant G ‐map. Since the equality\dim W-\dim W^{G}-(\dim V-\dim V^{G})=(\dim W-\dim W^{H}-(\dim V-\dim V^{H}))+(\dim W^{H}-\dim W^{G}-(\dim V^{H}-\dim V^{G}))holds, if K and H are BUGs then G is a BUG [5]. Therefore any solvable group is a
BUG. Then it is natural to ask whether a group is a BUG or not.

Wasserman [5] proposed a prime condition which implies a sufficient condition for agroup to be a BUG. A positive integer n satisfies the prime condition ifp_{1}^{-1}+p_{2}^{-1}+ +P_{r}^{-1}<1,e_{r} are positive integers such that n=p_{1}^{e_{1}}p_{2}^{e_{2}}\cdots p_{r}^{e_{r}}.
p_{r} are primes and e_{1},where p_{1},A group G satisfies the prime condition if the order of any cyclic subgroup of G satisfies
the prime condition.

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Theorem 1 ([5]) If a group G satisfies the prime condition, then G is a BUG.

Let Cyc1 (G) be the set of all cyclic subgroups of G . Nagasaki and Ushitaki [3] proposeda Möbius condition: A group G satisfies the Möbius condition if

D\inCyc{\imath}(G)\sum_{C\leqD}\mu(\frac{|D}{|C})\geq0
for any cyclic subgroup C of G , where \mu:\mathbb{N}arrow\{0, \pm 1\} is the Möbius function, that is,\mu(n)=\{ begin{ar ay}{l} 1 n=1 0 ifp^{2}|nforsomeprimep (-1)^{r} n=p_{1}p_{2}\cdotsp_{r}fordistnctprimesp_{1},p_{2},p_{r}. \end{ar ay}

Theorem 2 ([3]) If a group G satisfies the Möbius condition, then G is a BUG.
Since if K and H are BUGs then a group extension of H by K is a BUG, if we obtainthat every simple group is a BUG, then any group is a BUG. By the above theorem,
Nagasaki and Ushitaki showed that projective linear groups PSL(2, q) are BUGs. In thispaper, we give a sufficient condition for a group to be a BUG and apply projective linear
groups PSL(3, q) and alternating groups A_{n}.
2A sufficient condition

Let V and W be G ‐representation spaces and let f:Varrow W be an isovariant G‐map. For
a subgroup H of G , let

g_{f}(H)=(\dim W-\dim W^{H})-(\dim V-\dim V^{H}).

Note that If G is a cyclic group then g_{f}(G)\geq 0.
Proposition 3 Let H_{1} and H_{2} be a subgroups of G with H_{1}\triangleleft H_{2} and f an isovariant
G ‐map between representation spaces.

g_{f}(H_{2})-g_{f}(H_{1})=g_{f^{H_{1}}}(H_{2}/H_{1})holds. In particular, if H_{2}/H_{1} is a BUG, g_{f}(H_{2})\geq g_{f}(H_{1}) holds.

Let S(G) denote the set of all subgroups of G . It is made into a poset by defining

H\leq K in

S(G) if H is a subgroup of K . Let Cyc1 (G) be the full subposet of S(G) whichcontains all cyclic subgroups of G.
We put\mu(C,D)=\{ begin{ar ay}{l} \mu(\frac{|D|}{C|}), C\leqD 0, otherwise. \end{ar ay}

Nagasaki and Ushitaki [3] showed that PSL(2, q) satisfies the Möbius condition by using

the following equation.

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Theorem 4 ([3]) Let f:Varrow W be a G ‐map between representation spaces.

|G|g_{f}(G)= \sum_{C\in Cyc1(G)}(\sum_{D\in Cyc{\imath}(G)}\mu(C, D))|C|g_{f} (C)holds. If G satisfies the Möbius condition then G is a BUG.

Let RCyc1 (G) be the set of representatives of conjugacy classes of all cyclic subgroupsof G and let RCyc1_{1}(G) be the set of representatives of conjugacy classes of all nontrivial
cyclic subgroups of G . Recall that g_{f}(\{e\})=0.
Let\tilde{\mu}(C,D)=\{ begin{ar ay}{l} \mu(\frac{|D}{|C}), (C)\leq(D) 0, otherwise. \end{ar ay}

where (C) denotes the conjugacy class of C.
Lemma 5 For

C \in RCyc1(G),\sum_{D\in Cyc{\imath}(G)}\mu(C, D)=\sum_{D\in RCyc1(G)} \frac{|N_{G}(C)|}{|N_{G}(D)|}\tilde{\mu}(C, D).

C<D\in Cyc1(C_{G}(C)), S=\{E\in Cyc1(G)|(D)=(E), E\geq C\} and let
f:N_{G}(C)arrow S be a map which sends g to g^{-1}Dg . Let E\in S . Then E=g^{-1}Dg for someg\in G. C and g^{-1}Cg is a subgroup of E with same index. Since for any k>0 dividingfoasis=g^{-1}\in N_{G}(C)andf(g)=ET
If D_{1} and D_{2} are conjugate, then \tilde{\mu}(C, D_{1})=\tilde{\mu}(C, D_{2}) . Therefore we see that
Proof

Let

\dot{ \imath} cgroup D ufniısquseu’rjwecetisveee. thatthusthemap|D|,g ubgroup o rder kof t .hecyclherefore,dThus,
C \# S=\frac{Cgan|N_{G}(C)|}{|N_{G}(D)|}.
\sum \mu(C, D)= \sum \sum \mu(C, E)

D\in Cyc1(G)D\in RCyc1(G) E \in Cycı(G)(E)=(D)\sum \sum\mu(C, D)

D\in RCyc1(G)E\in S= \sum \frac{|N_{G}(C)|}{|N_{G}(D)|}\tilde{\mu}(C, D).

D \in RCycı(G)

I

Let

and

\beta_{G}(C, D)=\frac{|C|\tilde{\mu}(C,D)}{|N_{G}(D)|}\beta_{G}(C)=\sum_{D\in RCyc1(G)}\beta_{G}(C, D)

We abbreviate to write.

g_{f}(G) as g(G) if f is clear.

Proposition 6

g(G)= \sum_{C\in RCyc{\imath}(G)}\beta_{G}(C)g(C) .

(1)

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151
Proof

By Theorem 4 and Lemma 5, we see thatg(G)= \frac{1}{|G|}\sum_{C\in Cyc{\imath}(G)}(\sum_{D\in Cyc1(G)}\mu(C, D))|C|g (C)= \frac{1}{|G|}\sum_{C\in RCyc{\imath}(G)}\sum_{c'\in c_{yc{\imath}(G)} (\sum_{D\in Cyc1(G)}\mu(C', D))|C'|g(C')= \sum_{C\in RCyc1(G)}\frac{|C|}{|N_{G}(C)|} \sum_{D\in Cyc{\imath}(G)}\mu(C, D))g(C)= \sum_{C\in RCyc{\imath}(G)}\frac{|C|}{|N_{G}(C)|} \sum_{D\in RCyc{\imath}(G)} \frac{|N_{G}(C)|}{|N_{G}(D)|}\tilde{\mu}(C, D) g(C)= \sum_{C\in RCyc1(G)}(\sum_{D\in RCyc1(G)}\frac{|C|}{|N_{G}(D)|}\tilde{\mu}(C, D))g(C)(c)=(c').

1

We write \beta(C)=\beta_{G}(C) for simple. If G is a cyclic group, then\beta(C)=\sum_{D\inRCyc{\imath}(G)}\frac{|C}{|G}\tilde{\mu}(C,D)= \{ begin{ar ay}{l 0,C\neqG 1,C=G. \end{ar ay}
Lemma 7

Proof

(2)

\sum_{C\in RCyc1(G)}\frac{1}{|N_{G}(C)|}\leq 1.

By the class equation for G , we see that

1= \sum_{(x)}\frac{1}{|C_{G}(x)|}\geq\sum_{C\in RCyc1(G)}\frac{1}{|C_{G}(C)|} \geq\sum_{C\in RCyc1(G)}\frac{1}{|N_{G}(C)|}.
1

Lemma 8

|G|= \sum_{C,D\in Cyc1(G)}\mu(C, D)|C| and \sum_{C\in RCyc1(G)}\beta(C)=1 . If
\sum_{C\in RCyc1_{1}(G)}\beta(C)>0.

Proof

G is nontrivial thenLet u:Cyc1(G)arrow \mathbb{Q} be a map defined asu(C)=\{\begin{ar ay}{l } |genC| C\neq\{e\} 1 C=\{e\} \end{ar ay}
and putseev(G)= \sum_{D\in Cyc1(G)}u(D) . Then v(G)=|G| . By the Möbuis inversion formula, weu(D)= \sum_{C\leq D\in Cyc{\imath}(G)}\mu(|D|/|C|)v(C)=\sum_{C\in Cyc{\imath} (G)}\mu(C, D)v(C)152
and then|G|=v(G)= \sum_{C,D\in Cyc1(G)}\mu(C, D)|C|.
Therefore, we see

1= \frac{1}{|G|}\sum_{C\in Cyc1(G)}|C|(\sum_{D\in Cyc1(G)}\mu(C, D))=\frac{1}{|G}\sum\sum_{(c)=(c')}|C'|C\inRCyc{\imath}(G)c'\inc_{yc{\imath} (G)}(\sum_{D\inCyc{\imath}(G)}\mu(C',D)= \sum_{C\in RCyc1(G)}\frac{|C|}{|N_{G}(C)|}\sum_{D\in Cyc{\imath}(G)}\mu(C, D)= \sum_{C\in RCyc{\imath}(G)}\sum_{D\in RCyc{\imath}(G)}|N_{G}(C) |C|\frac{|N_{G}(C)|}{|N_{G}(D)|}\tilde{\mu}(C, D)= \sum_{C\in RCyc1(G)}\sum_{D\in RCyc1(G)}\beta(C, D)= \sum_{C\in RCyc1(G)}\beta(C).

If G is nontrivial then, by Lemma 7, we see\beta(\{e\})=\sum_{D\in RCyc{\imath}(G)}\frac{\mu(|D|)}{|N_{G}(D)|}<\sum_{D\in RCyc{\imath}(G)}\frac{1}{|N_{G}(D)|}\leq 1
and thusG\sum_{C\in RCyc{\imath}_{1}(G)}\beta(C)>0.1
Now we consider\sum_{C\in RCyc{\imath}_{1}(G)}\beta_{G}(C)\gamma(C) for a map \gamma:RCyc1_{1}(G)arrow \mathbb{Q}\geq 0 . We note thatsatisfies the Möbius condition if and only if\sum_{C\in RCyc{\imath}_{1}(G)}\beta_{G}(C)\gamma(C)\geq 0 for an arbitrary

map \gamma:RCyc1_{1}(G)arrow \mathbb{Q}\geq 0 . By (2), we see

Proposition 9 A cyclic group satisfies the Möbius condition.
We recall Proposition 3. For an isovariant

G ‐mapf and subgroups C\triangleleft D of G suchthat D/C is a BUG, g_{f}(C)\leq g_{f}(D) . We say G is a CCG (cyclic condition group),
if for an arbitrary map \gamma:RCyc1_{1}(G)arrow \mathbb{Q}\geq 0 such that \gamma(C)\leq\gamma(D) if (C) \leq(D) ,\sum_{C\in RCyc1_{1}(G)}\beta_{G}(C)\gamma(C)\geq 0. A CCG is a BUG.

Proposition 10 A group satisfying the prime condition is a CCG.
Proof

Let \gamma:RCyc1_{1}(G)arrow \mathbb{Q}\geq 0 be a map such that \gamma(C)\leq\gamma(D) if (C)\leq(D) . Since\sum_{C\in RCyc1_{1} \beta(C)\gamma(C)=\sum_{D(\in RCyc{\imath}_{1}\sum \beta(C, D)\gamma(C)C \in RCycıl (G),153
we show that\sum_{C\in RCyc1_{1}(G)}\beta(C, D)\gamma(C)\geq 0 for each D\in RCyc1_{1}(G) . For an positive

integer n , let \pi(n) be the set of all primes dividing n . Put r=\#\pi(|D|) , the number of
elements of \pi(|D|) , and let D_{0} be the subgroup of D with index. Since\prod_{p\in\pi(|D|)}p(\begin{ar y}{l r 2s \end{ar y}) (r-2s)=(\begin{ar ay}{l } r 2s +1 \end{ar ay}) (2s+1)

we see that,\sum_{C\in RCyc1_{1}(G)}\beta(C, D)\gamma(C)=(D_{0}) \leq(C)\leq(D)\sum_{C\in RCyc1_{1}(G)}\beta(C, D)\gamma(C)(\begin{ar y}{l \Sigma \beta(E,D)\gam (E)+ \Sigma \beta(F,D)\gam (F) |\pi(D|/E)=2s(D_{0})<(E\leq(D)E\inRCyc{\imath}_{1(G) |\pi(D|/F)= 2s+1(D_{0})\leq(F)<DF\inRCyc{\imath}_{1(G) \end{ar y})=\sum_{s=0}^{\lfo r(-1)/2\rflo r}|\pi(|D/E|)=2s(D_{0})<(E)\leq(D) \sum_{E\inRCyc{\imath}_{1(G)}\beta(E,D)\gam a(E)+\frac{1}2s+1}|\pi(|D/F|) =2s+1(D_{0})\leq(F)<(E)\sum_{F\inRCyc{\imath}_{1(G)}\beta(F,D)\gam a(F)=\sum_{s=0}^{\lfor(-1)/2\rflor}|\pi(|D/E|)=2s(D_{0})<(E)\leq(D) \sum_{E\inRCyc1_{ }(G) \frac{|E}N_{G}(D)|}\gam a(E)-\frac{1}2s+1} |\pi(|D/F|)=2s+1(D_{0})\leq(F)<E)\sum_{F\inRCyc{\imath}_{1(G)}\frac{|F} {|N_G}(D)|}\gam a(F)\geq\sum_{s=0}^{\lfor(-1)/2\rflor}|\pi(|D/E|)=2s(D_{0})<(E)\leq(D)\sum_ {E RCyc1_{ }(G) \frac{|E}N_{G}(D)|}\gam a(E)-\frac{1}2s+1}|\pi(|D/F|)=2s +1(D_{0})\leq(F)<E)\sum_{F\inRCyc{\imath}_{1(G)}\frac{|F}N_{G}(D)|} \gam a(E)\geq\sum_{s=0}^{\lfor(-1)/2\rflor}|\pi(|D/E|)=2s(D_{0})<(E)\leq(D)\sum_ {E\inRCyc1_{ }(G) 1-\frac{1}2s+1}|\pi(|E/F|)=1F\inRCyc{\imath}_{1(G) \sum_{(F)<E)}\frac{|F}E|)\frac{|E}N_{G}(D)|}\gam a(E)\geq\sum_{s=0}^{\lfo r(-1)/2\rflo r}\geq\sum_{s=0}^{\lfo r(-1)/2\rflo r}|\pi(|D/|E)=2s(D_{0})<(E)\leq(D)\sum_ {E RCyc{\imath}_{1}(G)}(1-\sum_{p\in pi(|E)}\frac{1}p)\frac{|E}{|N_{G}(D)|} \gam a(E)154
and thus if G satisfies the prime condition, then it is nonnegative.

3

I

Through linear programming

Let RCyc1_{1}^{+}(G) and RCyc1_{1}^{-}(G) be the subsets of RCyc1_{1}(G) consisting of C with \beta_{G}(C)>
0 and \beta_{G}(C)<0 , respectively. We consider a map\psi:RCyc1_{1}^{-}(G)\cross RCyc1_{1}^{+}(G)arrow \mathbb{Q}\leq 0such thatto D then\beta_{G}(C)=D\in RCyc1_{1}^{+}\sum_{(G)}\psi(C, D)

for C\in RCyc1_{1}^{-}(G) and if C is not subconjugate\psi(C, D)=0 . Theng(G) = \sum_{C\in RCyc1_{1}(G)}\beta_{G}(C)g(C)= \sum_{C\in RCyc1_{1}^{+}(G)}\beta_{G}(C)g(C)+ \sum \beta_{G}(C)g(C)C\in RCyc1_{1}^{-}(G)=\sum_{D\in RCyc{\imath}_{1}^{+}(G)}\beta_{G}(D)g(D)+C\in RCyc{\imat\sumh}_{1}^{-} (G) (\begin{ar y}{l } \Sigma \psi(C D) D\in RCyc{\imath}_{1}^{+}(G) \end{ar y}) g(C)\geq \sum_{D\in RCyc{\imath}_{1}^{+}(G)} (\beta_{G}(D)+ \sum \psi(C, D))g(D).

C\in RCyc1_{1}^{-}(G)By Lemma 8,

c \in RCyc{\imath}_{1}^{+}\sum_{(G)}\beta_{G}(C)+\sum_{C\in RCycl_{1}i} \beta_{G}(C)>0. Thus we may expect the exis‐

tence of \psi . We determine whether there exist \psi such thatfor D\in RCyc1_{1}^{+}(G) by linear programming.

\beta(D)+\sum_{C\in RCyc1_{1}^{-}(G)}\psi(C, D)\geq 0\{beginary}{l\psi(C,D)leq0\psi(C,D)=0f(C\notleq(D) \inRCyc{math}_1^{+\sum_(G)}\psiC,D)leq\bta_{G}(C)for\inRCyc1_{}^-(G)\sum_{CinRyc{\math}_1^{-(G)}\psiC,D)geq-\bta_{G}(D)for\inRCyc1_{}^+(G)\end{ary}We can check the existence of \psi for the following groups by using the software GAP[1]:Theorem 11 (1) Alternating groups A_{n}, 5\leq n\leq 11 satisfy the prime condition.
(2)

A_{n}, 12\leq n\leq 21 are CCGs.

(3) Symmetric groups S_{n}, 5\leq n\leq 9 satisfy the prime condition.

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(4)

S_{n}, 10\leq n\leq 22 are CCGs.

(5) All sporadic groups are CCGs.

(6) Automorphism groups of all sporadic groups are CCGs.

4

Projective special linear group

If any simple groups are BUGs, then every group is a BUG. It is important to study
simple groups. Projective special linear groups PSL(3, q) are simple groups.
Lemma 12 Let C be a cyclic subgroup of a group G. Suppose that there is a uniquemaximal cyclic subgroup D of G with C<D. Then N_{G}(C)=N_{G}(D), \beta_{G}(C)=0 and\beta_{G}(D)=\frac{|D|}{|N_{G}(D)|}>0.

Proof Since C<D , for g\in N_{G}(D), 9^{-1}Cg is a subgroup of the cyclic group D withindex |D/C| and thus g^{-1}Cg=C . Therefore, N_{G}(D)\leq N_{G}(C) . If g\in N_{G}(C)\backslash N_{G}(D)exists, then C<g^{-1}Dg\neq D . This is contradiction. Therefore the equality N_{G}(C)=

N_{G}(D) holds.

We see that\beta_{G}(C)=\sum_{E\in RCyc1(G)}\frac{|C|}{|N_{G}(E)|}\tilde{\mu}(C, E)=\frac {|D|}{|N_{G}(D)|}\sum_{E\in RCyc{\imath}(D)}\frac{|C|}{|D|}\tilde{\mu}(C, E)= \frac{|D|}{|N_{G}(D)|}\beta_{D}(C)=0

by (2), and clearly \beta_{G}(D)=\frac{|D|}{|N_{G}(D)|}. I
Let p be a prime and q a power of p . Let C_{q-1}, C_{p} , and C_{q+1} be cyclic subgroups ofSL(2, q) of order q-1, p , and q+1 respectively, and let \pi:SL(2, q)arrow G be a naturalprojection. Then \pi(C_{q\pm 1}) has order (q\pm 1)/gcd(q-1,2) .
Proposition 13

Proofg( PSL(2, q))=\frac{1}{2}g(\pi(C_{q-1}))+\frac{1}{2}g(\pi(C_{q+1}))+\frac{p} {|N_{PSL(2,q)}(C_{p})|}g(C_{p}) .

We may put

RCyc1_{1}(PSL(2, q))=\{H|\{1\}<H\leq C_{r}, r=p, q\pm 1\}.
The cyclic groups \pi(C_{r}), r=p, q\pm 1 are maximal and the orders are p, (q\pm 1)/d , respec‐tively, which are coprime each other. Therefore, any nontrivial cyclic subgroup of G hasa unique maximal cyclic subgroup of G . Thus by Lemma 12, we see

g(G)= \sum_{r\in\{q\pm 1,p\} \frac{|C_{r}| {|N_{PSL(2,q)}(C_{r})|}g(C_{r}).

I

Nonsolvable groups PSL(2, q), SL(2, q), PGL(2, q) , and GL(2, q) are all BUGs (see[3]). Furthermore, a group which does not have a simple group except PSL(2, q) as asubquotient group is a BUG.

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Example 14 The simple group PSL(3,11) does not satisfy the prime condition, since

it has an element of order 120. We confuse order (with type) with the cyclic subgroup

generated by the corresponding element. For example, PSL(3,11) has a unique cyclic
group of order 110 up to conjugate, and two cyclic subgroups of order 5 up to conjugate,
denoted by 5a and 5b , whose are not conjugate. We have

RCyc1_{1}^{+}(PSL(3,11))=\{10b, 10c, 10d, 11a, 110,120,133\},
RCyc1_{1}^{-}(PSL(3,11))=\{2,5a, 5b, 10a, 11b\}.
Since \beta(133)+\beta(11b)=587/1815>0, \beta(110)+\beta(5b)+\beta(10a)=0 , and \beta(10b)+\beta(2)+\beta(5a)=0 , we see thatg(PSL(3,11))\geq\beta(10c)\gamma(10c)+\beta(10d)\gamma(10d)+\beta(11a) \gamma(11a)+ \beta(120)\gamma(120)+\frac{587}{1815}\gamma(133)\geq 0.
The group PSL(3,11) is a CCG. See the following table corresponds with \beta_{PSL(3,11)}(-, -) .
The first columns and first rows are all cyclic subgroups C and D of RCyc1_{1}(PSL(3,11))
respectively, and the last columns are the values \beta_{PSL(3,11)}(-) .

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Table 1:\beta_{PSL(3,11)}(-, -)The group SL(3, q) is of order q^{3}(q-1)^{2}(q+1)(q^{2}+q+1) . Put q=p^{u}, G=SL(3, q), \delta=
1, d=gcd(3, q-\delta), \rho^{r}=1, r=q-\delta, r'=r/d, s=q+\delta, s'=s/gcd(3, s), t=q^{2}+\delta q+1,
t'=t/d, \sigma^{s}=\rho, \tau^{t}=1, \omega=\rho^{(q-1)/d} . A maximal cyclic subgroup of SL(3, q) is conjugate
to one of the followings: C_{pr}= \langle(B \rho^{-1})(\sigma^{\delta} \sigma^{q})a\leq b<r, (r, a, b)=1), C_{rs}= \langle

C_{t} , where

B is conjugate to

conjugate to(\tau \tau^{\deltaq} \tau^{q 2})(\begin{ar y}{l \rho1 \rho\rho^{-2} \end{ar y})\rangle, C_{r}^{(a,b)}= \langle\rangle, C_{dp}^{(c)}=\langle\omega(\rho^{a} \rho^{b} \rho^{-ab})(\begin{ar y}{l 1 \thea^{c} 1 \thea^{c} 1 \end{ar y})\(0\leqc\leqd-1)\rangle(0\leq a<r',,in GL(2, q^{2}) and C_{t} is generated by an element

q-1

in GL(3, q^{3}) [ 4 , Table la]. Let \psi:SL(3, q)arrow PSL(3, q)

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D_{r(a,b)}^{(a,b)}=\psi(C_{r}^{(a,b)}), D_{r's}=\psi(C_{rs}) ,be a canonical projection and put D_{pr'}=\psi(C_{pr}),ra/d\equiv rb/d\equiv-r(a+b)/d
D_{p}^{(c)}=\psi(C_{dp}^{(c)}), D_{t'}=\psi(C_{t}) , where r(a, b)=r/d if
modulo r , and r(a, b)=r otherwise. We may assume that RCyc1 (G) consists of subgroups
d=3 andof the above cyclic subgroups.
Proposition 15 If r satisfies the prime condition, then PSL(3, q) is a CCG.

Proof

The order of a maximal cyclic subgroup is r, r', p,

r's , pr’, or t' .

We see that

For C\in RCyc1_{1}(G) , if D_{p}^{(0)}\leq C, D_{r}^{(1,1)}<C or C<D_{t'} , then amaximal cyclic subgroup containing C is unique. We see that(p, r's)= (pr’s, t' ) =1 .

\sum_{C\in RCyc1_{1}(G)}\beta_{G}(C)\gamma(C)=\sum_{\Leftar ow 0}^{d-1}\beta_{G}(D_{p}^{(c)} \gamma(D_{p}^{(c)} +\beta_{G} (D_{t'})\gamma(D_{t'})+\beta_{G}(D_{r's})\gamma(D_{r's})+ \beta_{G}(D_{pr'})\gamma(D_{pr'})+\sum_{(a,b)_{C} \sum_{\leq D_{r(a,b)}^{(ab) } ,\beta_{G}(C)\gamma(C).

Note that \beta_{G}(D_{pr'})=1/r, \beta_{G}(D_{sr'})=1/2, \beta_{G}(D_{t'})=1/3 and \beta_{G}(D_{p})=-s/p^{2}r , whereD_{p} is a subgroup of D_{pr'} of order p . Then \beta_{G}(D_{pr'})+\beta_{G}(D_{p})>0 and\sum_{C\in RCyc1_{1}(G)}\beta_{G}(C)\gamma(C)\geq\sum_{(a,b)_{C} \sum_{\leq D_ {r(a,b)}^{(ab)} ,\beta_{G}(C)\gamma(C).

By the proof of Proposition 10, if r satisfies the prime condition, then for anyof orderRCyc1_{1}(G) oforderr, \sum_{C}\beta_{G}(C,D)\gamma(C)\geq 0

and thus\sum_{(a,b)}\sum_{C\leq D_{r(a,b)}^{(\alpha,b)} 6_{G}(C)\gamma(C)\geq 0.

D\in

I

Example 16 The number 30 does not satisfy the prime condition. The following tablecorresponds with \beta_{PSL(3,31)}(C, D) such that \beta_{PSL(3,31)}(C)\neq 0.

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Table 2:

\beta_{PSL(3,31)}(-, -)Since \beta(6)+\beta(30a)=0, \beta(10b)+\beta(30d)>0, \beta(15a)+\beta(30c)>0, \beta(15b)+\beta(30b)>0,\beta(31a)+\beta(310)>0 , we get g(PSL(3,31))\geq\beta(3)\gamma(3)+\beta(31b)\gamma(31b)+\beta(31c)\gamma(31c)+\beta(31d)\gamma(31d)+\beta(320)\gamma(320)+\beta(331)\gamma(331)\geq 0 . Thus PSL(3,31) is a CCG.

5

Future work

It is not true that for an extension 1arrow Harrow Garrow Karrow 1 , if H and

K are CCGs, then

G is a CCG.

Proposition 17 A_{22} is not a CCG.

Proof

Let x=(1,2,3,4,5,6,7)(8,9,10,11, 12) (13, 14, 15, 16)(17, 18, 19) (20, 21)\in A_{22}and S=\{\{x^{3}\rangle, \langle x^{4}\rangle, \langle x^{5}\rangle, \langle x^{7} \rangle\} . Let \tilde{S} be the subset of RCyc1_{1}(A_{22}) consisting of C suchthat some element of S is subconjugate to C . Then \tilde{S}=S\cup\{\langle x\rangle, \langle x^{2}\rangle\}.

Let \gamma:RCyc1(G)arrow \mathbb{Q}\geq 0 by
\gamma(C)=0 otherwise. We see that
A_{22} is not a CCG. 1\gam a(C)_{\sum^{=1}\dot{\imath}fC\inRCyc{\imath}_{1}(G)C

is conjugate to some element of \tilde{S} and\beta_{G}(C)\gamma(C)=-61/1935360<0 . Therefore

Suppose that RCyc1 (A_{22})\supset\tilde{S} . There exists representation A_{22} ‐spaces V and W
such that \dim V^{C}=\dim W^{C} for C\in RCyc1(A_{22})\backslash \tilde{S} including \dim V=\dim W , and\dim W^{C}-\dim V^{C} is constant positive number for C\in\tilde{S} . By these condition, we have\dim W<\dim V (see (1)). Suppose that there is an isovariant map f:Varrow W . Let C
be a cyclic subgroup generated by x , and H a solvable subgroup of N_{A_{22}}(C) of order 840

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generated by x and (14, 16)(18, 19). Then H normalizes C and g_{f}(C)=1 and g_{f}(H)= \frac{1}{2},which is a contradiction. Therefore, there is no isovariant map Varrow W.
Thus we consider a new condition. For a map \gamma:RCyc1_{1}(G)arrow \mathbb{Q}\geq 0 and a subgroup
H of G , we put\hat{\gamma}(H)=\sum_{C\in RCyc{\imath}_{1}(H)}\beta_{H}(C)\overline{\gamma} (C),where \overline{\gamma}:Cyc1_{1}(G)arrow \mathbb{Q} is a class function which sends a cyclic subgroup C of G to \gamma(C')
such that C'\in RCyc1_{1}(G) is conjugate to C in G . If H\in RCyc1_{1}(G) then \hat{\gamma}(H)=\gamma(C) ,and if H_{1} and H_{2} are conjugate in G then \hat{\gamma}(H_{1})=\hat{\gamma}(H_{2}) . Recall that g_{f}(H_{2})-g_{f}(H_{1})=g_{f^{H_{1}}}(H_{2}/H_{1}) and if H_{2}/H_{1} is a BUG then g_{f^{H_{1}}}(H_{2}/H_{1})\geq 0 for an isovariant G‐map
f:Varrow W . A group G is a SCG (subgroup condition group) if for an arbitrary map\gamma:RCyc1_{1}(G)arrow \mathbb{Q}\geq 0 such that \hat{\gamma}(H_{1})\leq\hat{\gamma}(H_{2}) for subgroups H_{1}\underline{\triangleleft}H_{2}\leq G with H_{2}/H_{1}a CCG, \hat{\gamma}(G)\geq 0 . A SCG is a BUG.
Question 18 Is the group A_{22}

a

SCG ?
l

A sufficient condition to be a BUG is that the minimizing value of the following linear
programming is zero.
Minimizesubject to\sum_{V\in Ir _{1}(G)}x_{V}\dim V\{ begin{ar ay}{l -1\leqx_{V}\leq1,V\inIr _{1}(G) \sum_{V\inIr _{1}(G)}x_{V}(\dimV^{H_{1}-\dimV^{H_{2})\geq0,H_{1} \triangle ftH_{2}\leqG,H_{2}/H_{1}solvable \end{ar ay}

where Irr_{1}(G) is the set of all irreducible nontrivial representation G ‐spaces.
Since there are many inequalities, we could not check whether the minimizing value iszero or not for A_{22} . We must reduce partial condition to compute.

Acknowledgement
The author was partially supported by JSPS KAKENHI, Grant number JS16K05151.
References[1] The GAP Group, GAP — Groups, Algorithms, and Programming, Version 4.9.2;
2018, (https://www.gap‐system.org).

[2] T. Kobayashi, The Borsuk‐ Ulam theorem for a Z_{q} ‐map from a Z_{q} ‐space to S^{2n+1},
Proc. Amer. Math. Soc. 97 (1986), 714‐716.
[3] I. Nagasaki and F. Ushitaki, New examples of the Borsuk‐Ulam groups, RIMSKôkyûroku Bessatsu, B39 (2013), 109‐119.

[4] W. A. Simpson and J.S. Frame, The character table for SL(3, q), SU(3, q^{2}), PSL(3, q) ,PSU(3, q^{2}) , Can. J. Math. 25 (1973), 486‐494.

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161
[5] A. G. Wasserman, Isovariant maps and the Borsuk‐Ulam theorem, Topology Appl.
38 (1991), 155‐161.
Faculty of Arts and Science
Kyushu University
Motooka 744, Nishi‐ku, Fukuoka, 819‐0395
JAPAN

E‐mail address: sumi@artsci.kyushu‐u.ac.jp