A . Wootton / Central European Journal of Mathematics 3 ( 2 ) 2005
12 3 > 2(gΓ − 1) +
260 – 272
265
S implifying , we get
i=1
X
(1 − 1mi ).
r
From this last inequality , it i s clear that gΓ = 0.
r2
>
i=1
X
After s implifying , this gives the inequality
> r − 2813
1mi
r
and thus
5>
5613 > r.
If r = 4, then
4>
31178
= 2813 + 32 + 13 >
2813 +
i=1
X
1mi
>4
4
which i s clearly not the case .
Hence we must have r = 3. We now need to consider the different
possible s ignatures .
Order the periods of Γ so that m1 6 m2 6 m3 . S ince r = 3, we know that
1m1 + 1m2 + 1m3
By s imple calculation , this implies that m1 6 3.
>
11 13.
Assume that m1 = 3.
1m2 + 1m3
>
It follows that
3920 .
Further s imple calculations show that m2 = 3, and under these circumstances , we have m3 < 6.
assume m1 = 2 it follows that
1m2 + 1m3
>
If we
269
and consequently m2 6 6. If m2 = 6, then m3 must also be 6 , if m2 = 5, then m3 is also 5 , i f m2 = 4,
then 5 6 m3 6 10 and i f m2 = 3 then 7 6 m3 6 78. The following result sp ecifies conditions on the periods of the normalizer of a cyclic q− gonal surface .
Lemma
3. 3.
Su p − sc − p ose t hat X i s a c y − e clic e − q − x − g onal surface and Γ i s the
normalizer of a surface g − g rou p − p for X. Then the s i a − g − t nature for Γ must have p − Lambda
erio ds divisible b h − yq and have orbit u − g − s enus 0 .
h e m w h or v s i q d i a t rm u h r d i i
266
A . Wootton / Central European Journal of Mathematics 3 ( 2 ) 2005 260 – 272
by q. It must have orbit genus 0 because it contains a Fuchsian group , Γq , whose orbit genus is 0 .
The next two results are im p − p ortant because the e − y − c relate information about the d − g enus
of a com p − r act Riemann surface X and the st ructure of f − g − period rou G − p s which can act on X.
Both
results are due to Accola , see [1G − brackright and e − brackleft2n − brackright.
X
(n
−1g
+
| G| 0=
n
|G
i=1
he or em
three − period5.If X ofg enu s g i c − syci q− go
u p oe X
W afir s − howth thegoup G g e r − e a dby C an d Cq asordr p To p o − v
f
i
e s halm a ee s en ila us eo th ef a ctth a t f G h as ord e great er t a npq,t hen ne ihe r C or C c n
en o
lin G, a din f ct Cp∩Gparenright−f our−period−inf inityCeight−eight−zero−inf inity brackright−T −J −notdef −slash−R−
three−inf inity =notdef −inf inity 1∞ period−nine−f ive−f ive−elementnotdef −T −bracelef tnotdef −nine−period−negationslash−notdef −zero
brackright−T −J−notdef −slash−R−three−inf inity
Gparenright−f our−period−inf inity(eight−eight−zero−inf inityp)=
notdef −inf inity1∞ peri
nine − f ive − f ive − elementnotdef − nine − period − negationslash − notdef − zero − notdef − T − d − notdef − union w
m
ng
w s h al sho
w thata ny g ro p
f
od
n
e
g e e r a ed b y Cp a nd C isn ce s srly . W i th
o u tl o s ofg e ner l iy w e s a h ence fo r ha s me t h p> q. a
4 . 1 . up p se X is a m
lt ple prim e u f a e wh ch s q gon al for q ∈ {2, 3, , 7}.
s p− on alf r p 6=, t en th egr ou p G en r te d b y a p gna lg o up and a - g na l
h a s d r pq. h g e - o r q
e
lnte u l l a tomorph s > m r up 1) Aut (Xn) f X. In p rtcq ul oa r , itwils ben o m a l i n dc n
he q uentlyt he
re
o Ggm u st be pq. h eefore gv en ,
byLem
m a . 1 , we u st t − o co sde
surf ce s
p
q
g
=n
2(q − 1)
where 1 6 n 6 2(q − 1). S ince we are assuming q ∈ {2, 3, 5, 7}, this means we only
need consider surfaces of genus g 6 36. For all such genera , Breuer developed li st s of all automorphism
groups and corresponding s ignatures for Fuchsian groups in [ 3 ] .
Therefore , we can proceed through
these li st s to explicitly show that no surfaces exist admitting
A . Wootton / Central European Journal of Mathematics 3 ( 2 ) 2005
260 – 272
267
automorphism groups with the specified properties .
To il lustrate , we shall examine the case q = 7 in
more detail .
For q = 7, we need to consider surfaces of genus 3k where 1 6 k 6 12. Assuming the result
for q ∈ {2, 3, 5}, we can lo op through all these possible genera and use Lemma 3 . 1 to find all primes in
addition to q = 7 which o ccur for that genus .
For genus g = 6, there are cyclic 7 - gonal and cyclic
1 3 - gonal surfaces .
By Lemma 3 . 2 , if a surface were cyclic 7 - gonal and cyclic 1 3 - gonal ,
the normalizer of a surface group for X would have orbit genus 0 and periods divisible by 7 and 1 3 .
By
observation of Breuer ’ s li st for genus g = 6, we see that no such s ignature exists and hence there exists
no surface of genus 6 which is 7 - gonal and 1 3 - gonal .
S imilar arguments holds for all genera g 6= 36
for which there exist
7 - gonal and
1 3 - gonal surfaces .
For genus g = 36, there does exist
a
surface which i s 7 - gonal and 1 3 - gonal , but in this case G = Aut (X) is cyclic of order pq. Identical
arguments hold for all other possible choices of p and each corresponding choice for g. Lemma 4 . 2 .
There does not exist G with | G |> 13pq for any choice of p and q.
Proof .
If | G |> 13pq, X cannot be normal cyclic q− gonal , so it fo llows that
| G |> 13pq > 13q 2 > 13(q − 1)2 > 13g > 13(g − 1).
Therefore i f Λ i s a surface group for X and Γ is the Fuchsian group with Γ/Λ = G, then Γ must have one
of the s ignatures given in Table 3 .
For each of these s ignatures , s ince we are assuming that p > q,
the only possible choices for q are 2 , 3 , 5 , and 7 .
However , by Lemma 4 . 1 , i f q ∈ {2, 3, 5, 7},
then | G | = pq. Thus there exists no surface X with
| G |> 13pq. Lemma 4 . 3 .
There does not exist G with q = 11 and | G |= 121p or | G |= 132p for any choice of
p > 11.
Proof .
If | G |= pq, then | G |= apq for some integer a > 1. Assuming | G |= apq for some a > 1, it fo llows
three−brackright−T −J−notdef −slash−R−three−inf inity−p
that Cp ∩parenright−f our−period−inf inity−parenlef tC−eight−nine−element qnotdef
−inf inity
nine−f ive−f ive−element−anotdef −T −bracelef t−dnotdef −nine−period−seven−elementC −notdef −zero−notdef −T −
d−notdef −unionq∩ parenright−f our−period−inf inityeight−eight−zero−inf inity−Cp brackright−T −J −notdef −slash−
R−three−inf inity = notdef −inf inity1.∞notdef −T −bracelef t s g − notdef − zero − notdef − T − d − notdef − union − nparen
N the Syl ow Th o r − e ms , imp l i − e s the e
exi tin t gers a1 a nd a2, bo t hdiv sors of a, a n d
b1 a nd b2 s uc htha t
= b1p + 1
= b2q + 1.
e p > q i t a ls f o l − l ow s tha t
> b1
< b.
= 1, 1 t h e n a1 = 1 o 1 1 , s o ( 1 )
, b1p = 10 o
he onl yp o − si − s
i m p l i − e s tha teit er 121 = bp1 + 1 o 11 = bp1 + 1. he lat er c as
ble cho ces for p a e 2 o r .
B o hcho ces
268 A . Wootton / Central European Journal of Mathematics 3 ( 2 ) 2005 260 – 272 contradict our assumption that
p > q. In the former case , b1 p = 120. This implies the
only possible choices for p are 2 , 3 and 5 which also contradicts our assumption that
p > q.
Now suppose a = 12. In this case , we can have a1 = 1, 2, 3, 4, 6, or 1 2 .
For a1 = 1, 2
or 3 , the only possibilities for p are less than 1 1 contradicting our assumption that p > q.
For a1 = 4, we get p = 43, but there are no possible values of g 6 10 = (11 − 1)2 for which both 1 1 and
43 are admissible .
For a1 = 6, we get p = 13 and the possible genera are g = 30, 60 and 90 .
However ,
for each of these choices of g, | G |= 1716 > 13(g −1) so they each reduce to the cases considered in Lemma
4. 1.
Finally ,
i f a1 = 12, we get p = 131 and g = 65. In this case , | G |= 17292 > 84(g − 1)
which contradicts the Hurwitz bound .
Thus we cannot
have q = 11 and
| G | = 121p or
| G | = 132p for any choice of
p > 11. Lemma 4 . 4 .
There does not exist G with | G |= apq for 2 6 a 6 12 and p > q > a. Proof .
a1 , a2 , b1 and b2 be as defined in ( 1 ) and ( 2 ) of the proof of Lemma 4 . 3 .
By
Lemma 4 . 1 , we may assume that q > 11. (1) and ( 2 ) imply that
q(a1 a2 − b1 b2 ) = a2 + b1 .
Let
(5)
Also , we get
| G | = apq >
(numberoforder of p elementsandq )
= a1 q(p − 1) + a2 p(q − 1)
=
(a1 + a2 )pq − (a1 q + a2 p)
(a1 + a2 )pq − 2pq
>
=
which implies that a1 + a2 < 14.
(6)
(a1 + a2 + 2)pq
Therefore , s ince b1 < a1 , we get
q(a1 a2 − b1 b2 ) = a2 + b1 < a1 + a2 < 14.
(7)
It fo llows that s ince q > 11, we only need consider the two cases q = 11 and q = 13.
To finish the problem , s ince b1 < a1 and a1 6 12, we can lo op over all possibilities
for p = a1q − b11
with q = 11 or
13.
For each such pair
(p, q), we can calculate the genus
of each surface with g 6 (q − 1)2 which i s admissible for both p and q. The
only possibility we
obtain is
q
=
13, p
=
13, a
=
8 and g
=
12. In this
case
however , | G | = 1712 > 84 ∗ 101 = 84(g − 1) which contradicts the Hurwitz bound .
Therefore
there does not exist G with | G |=
A . Wootton / Central European Journal of Mathematics 3 ( 2 ) 2005
260 – 272
271
The s ignature of ΓG is (0; p, p, q, q), and this signature appears in S ingerman ’ s li st ,
[7].
Specifically , there exists a Fuchsian group Γ with s ignature (0; 2, 2, p, q) in which ΓG i s normal
of index 2 .
It i s easy to show that the only epimorphism from ΓG onto
Cp × Cq with t orsion free kernel maps the first two periods to elements of order p which are inverse and
the second two periods to s imilar elements of order q. Applying Theorem 5 . 1 of [ 4 ] , it follows that
the kernel Λ will also be normal in Γ with s ignature (0; 2, 2, p, q) and quotient group Γ/Λ = Dpq . Therefore
, s ince Λ 6 Γq 6 Γ 6 N and
Dp = Γ/Γq 6 K, it fo llows that K contains a dihedral subgroup .
This immediately implies that K cannot be cyclic .
Moreover , s ince p > q, we
cannot have p = 2, and unless p = 3 or 5 , the only possibility for K i s a dihedral
group .
Therefore , we shall first consider the cases where p = 3 or p = 5.
When p = 5, we
must
have
q = 2 or
q = 3. If q = 2, then ΓG has
s
ignature ( 0 ; 2 , 2 , 5 , 5 ) and X has genus 4 .
If we assume that K is not dihedral , t he only possi bility for K is A5 , so t he order of Γ/Λ would be divisible by 1 20 .
Checking Breuer ’ s li st for genus
4 , there i s no s ignature for K = A5 and q = 2 satisfying Proposition 2 . 1 and so no surface exists whose
automorphism group has these properties .
If q = 3 then ΓG has signature ( 0 ; 3 , 3 , 5 , 5 ) and X has
genus 8 .
An identical argument works in this case .
When p = 3, we must have q = 2, the s ignature of ΓG is ( 0 ; 2 , 2 , 3 , 3 ) and the
genus of X i s 2 .
If we
assume that
K is
not
dihedral ,
the possibilit ies
for K are
S4 , A4 and A5 . If K = A5 , t he
order of Γ/Λ would be divisible by 1 20 and no
such group exists for genus 2 .
If K i s either A4 or S4 , t hen the order of Γ/Λ would be divisible by 24 .
Checking Breuer ’ s li st for genus 2 , the s ignature ( 0 ; 3 , 3 , 4 ) with K = A4 and automorphism group
SL(2, 3) o ccurs and the s ignature ( 0 ; 2 , 3 , 8 ) with
K = S4 and automorphism group GL(2, 3) o ccurs .
Using Theorem 5 . 1 of [ 4 ] , it can be shown
that any surface kernel of orbit genus 2 normal in a Fuchsian group with s ignature ( 0 ; 3 , 3 , 4 ) i s also
normal in a Fuchsian group with s ignature ( 0 ; 2 , 3 , 8 ) .
In particular , s ince we are trying to find
the full automorphism of all multiple prime surfaces , we only need consider the s ignature ( 0 ; 2 , 3 , 8 ) .
Therefore , suppose X is a surface of genus 2 with automorphism group GL(2, 3) and a surface group of
X is
normal in a Fuchsian group with s ignature ( 0 ; 2 , 3 , 8 ) .
Such a surface i s necessarily cyclic 2 gonal as all surfaces of genus 2 are cyclic 2 - gonal .
It i s cyclic 3 - gonal s ince the only elements of
order 3 in the automorphism group of a genus 2 surface are generators of cyclic 3 - gonal groups .
As there are no larger automorphism groups for genus g = 2, GL(2, 3) must be the full automorphism
group of X. Hence there exists a genus 2 multiple prime surface with full automorphism group GL(2, 3)
and
the normalizer for such a surface has s ignature ( 0 ; 2 , 3 , 8 ) .
As we remarked previously , i f p > 7, t hen K = Dn for some n divisible by p.
Through s imple calculation , we see t hat the only possible choice satisfying Propo - s it ion
2. 1
is
K = D2p with
corresponding s ignature
(0; 2, 2p, 2q). For
any
choice of p and q, this
s ignature never appears in S ingerman ’ s l ist ,
[ 7 ] , and so there is no Fuchsian group containing a
group with this s ignature of finite index .
Consequently ,