ss="fs6 ls8e ws60">21mn, define
the action


421
:Dih Diff
n
K
by



π
21
41, 0
i
n
ru rue




and



4
1
0,1 ru u
r
.
The quotient 4
K
is the projective plane
22, 2P
containing two cone points of order two. This action ex-
tends to

,
M
bd and we obtain

421
:Dih n


Diff ,
M
bd . The orbifold quotient is
4,


4
Bk W
where

4,,1WIztzt  is the
twisted I-bundle over

22, 2P and
is the homeo-
morphism of induced by
. The orbifold quotient
 
44
,(21),
M
bdb nd
.
5) Quotient type

5,
. Define the following ac-
tions:

521
:Dih Diff
n
K
where



π
21
51,0
i
n
ru rue




and

50,1 ru ru
;


542
:Dih Diff
n
K
where



π
21
5
1
1, 0
i
n
ru ue
r



and


50,1 ru ru
;


542
:Dih Diff
n
K
where



2π
42
51, 0
i
n
ru rue



and



5
1
0,1 ru u
r
;
and


54
:Dih Diff
n
K
where



π
2
51, 0
i
n
ru rue



and



5
1
0,1 ru u
r



. The
orbifold quotient for all these actions is the mirrored disk

22, 2. All these actions extend to

Diff ,
M
bd . If
 

5,,1WIztrzt , where r is a reflection
exchanging a pair of cone points, is the twisted I-bundle
over

22, 2, then the orbifold quotient for these ex-
tended actions is

55
,Bk W

. We obtain
 
55
,21,
M
bdb nd
,
 
35
,21,2
M
bdb nd
,


55
,42,
M
bdb nd

and

55
,4,
M
bd nbd
.
6) Quotient type
6,
. Define actions 66
,:

2
Dih Diff
n
K
as follows:



π
6
1
1, 0
i
n
ru ue
r
and


60,1 ru ru
if
n is even, and



2π
6
1
1, 0
i
n
ru ue
r



and



6
1
0,1 ru u
r
if n is odd. The quotients 5
K
and 5
K
are both a mirrored disk

22 containig
a cone point of order two and two cone points of order
two on the mirror. These actions extend to the prism
manifold
,
M
bd and we obtain
66 2
,:Dih n
Diff ,
M
bd . The orbifold quotient for these actions is
denoted by
66
,Bk W

 where
6,,1WIztrzt
  the twisted I-bundle over
the mirrored disk
22, and r is a reflection leaving
two cone points fixed and exchanging the other two cone
points. The orbifold quotients

5
,Mbd
and
6
,Mbd
are both
6,bnd bnd.
7) Quotient type
7,
. Define
72
:Dih n
Diff
K
and

722
:Dih Diff
n
K
 as fol-
lows:



π
71, 0
i
n
ru rue
,


70,1 ru ru
, and



7
1
1, 0,0ru u
r

,



π
70,1,0
i
n
ru rue



, and
70, 0,1ru ru
. The quotients 7
K
and 7
K
are both a mirrored disk

20 containig four cone
points of order two on the mirror. These actions extend to
the prism manifold
,
M
bd and we obtain


72
:DihDiff ,
n
M
bd
and


722
:DihDiff ,
n
M
bd
 .
If
7,,1WIztrzt
 , where r is a reflection
leaving each cone points fixed, then 7
W is a twisted
I-bundle over the mirrored disk

20. The orbifold
quotient for these extended actions is
6,

6
Bk W
. We obtain
 
77
,2,
M
bd nbd
and
77
,2,2
M
bdnb d
.
3. Prism Manifold Covers of Orbifolds
In this section we give necessary and sufficient condi-
J. E. KALLIONGIS, R. OHASHI
Copyright © 2012 SciRes. APM
152
tions for when the orbifold

,
i
, 17i, is cov-
ered by a prism manifold. The proofs rely on Section 2.
Proposition 1. For the orbifold

1,
, there exists
a prism manifold cover if and only if either
is odd or
0
(mod 4).
Proof. Suppose
is odd. Then there exists a
-ac-
tion on

1,M
such that
 
1
1, 1,,MM

.
If
is even, write 0
2l
where 0
is odd. If
is odd, then there exists a 0
-action on
2,
l
M
such that
 

01
2, 2,,
ll
MM
.
Suppose now that
and
are both even where 0
(mod 4). Write 0
2
and 0
2m
where 0
and
0
are both odd. Then there exists a 0
2
-action on

0
2,
m
M
such that

0
002100
2,2,2 ,2
mm m
MM
.
For the converse, suppose that
and
are both even
and there is a covering

,,
i
Mbd
. Then ei-
ther

21bn
 and d
, or

21bn
 and
2d
. Since
is even, it follows that 2 divides b. In
the first case, 2 would also divide d, contradicting the
fact that b and d are relatively prime. If 0
(mod 4),
then again 2 divides d giving a contradiction.
Proposition 2. For the orbifold

2,
, there exists
a prism manifold cover if and only if 0
(mod 2).
Proof. Suppose that 0
(mod 2). Write 0
2

.
Then there exists a 0
2
-action on

1,M
such that
 
0
22
1, 1,,MM
.
For the converse, suppose that

2
,,Mbd
. Then
either 2nb
and d
, or 2nb
and 2d
.
Proposition 3. For the orbifold

3,
, there exists
a prism manifold cover if and only if

(mod 2).
Proof. Suppose that

(mod 2) and let 2
d
.
Suppose that 0
 (mod 4), and thus there exists
an integer n such that 4n
 . There exists a 4n
-
action on

1,
M
d such that
 

43
3
1,1,2, 2
,.
n
M
dMd ndnd


If 0
 (mod 4), then write 21
2n


for some
n. There exists a

22 1n
-action on

1,
M
d such that




3
22 1
3
1,1,21,21
,.
n
M
dMdn dn d

 
For the converse, suppose that

3
,,Mbd
.
Then either
21bnd
 and

21bnd
, or
2nb d
and 2nb d
for some n. Subtracting
the two equations in both cases, we obtain 2d
.
Propo sitio n 4. For the orbifold

4,
, there exists
a prism manifold cover if and only if either
is odd or
is odd.
Proof. Since
1
,Mbd
always double covers
4
,Mbd
, using a proof similar to that in Proposition
1 shows that there is a prism manifold covering of
4,
if and only if
or
is odd by [4].
Proposition 5. A prism manifold covering for the orbi-
fold
5,
always exists.
Proof. Suppose
is an odd number. Then
1,M
admits a
Dih
-action whose quotient is
5,
.
If β is even, we write 0
2m
where m 1, 0
2n
where 0n, and 0
and 0
are both odd numbers. If
n = 0 or 1n
, then
2,
m
M
and

0
2,
m
M
admit
0
Dih
and
0
2
Dih
-actions respectively, whose
quotient space is
5,
. If n and m are both greater
than 1, or if 1m
and 2n, then

1,M
admits a

10
42
Dih m



or a

0
2
Dih
-action respectively,
whose quotient space is
5,
.
Proposition 6. For the orbifold
6,
, there exists
a prism manifold cover if and only if
(mod 2).
Proof. Since
3
,Mbd
double covers
6
,Mbd
and
3
,Mbd
double covers
6
,Mbd
, the re-
sult follows by Proposition 3.
Proposition 7. For the orbifold

7,
, there exists
a prism manifold cover if and only if 0
(mod 2).
Proof. Since
2
,Mbd
double covers
7
,Mbd
and
2
,Mbd
double covers

7
,Mbd
, the re-
sult follows by Proposition 2.
4. Poset of Actions on Prism Manifolds
Recall that two group actions
:DiffGM
and
:DiffGM

are equivalent if there is a homeo-
morphism :hM M
such that
1
:Gh Gh



.
If
:DiffGM
is an action, let :MM

be the orbifold covering map.
Let be the set of equivalence classes of actions on
prisim manifolds which leave a Heegaard Klein bottle
invariant. Now is partially ordered by setting
if there is a covering :
M
M
such that


. Note that the covering :
M
M
is also
a regular covering.
For a pair of positive integers
and
let
1,
denote the equivalence classes of those actions whose
quotient type is
1,
. Note that by Proposition 1 the
set
1,
is nonempty if and only if either
is odd,
or 0
(mod 4). Unless otherwise stated, we assume
from now on that
and
are integers where either
J. E. KALLIONGIS, R. OHASHI
Copyright © 2012 SciRes. APM
153
is odd or 0
(mod 4).
Let
 
1,, :..,1,
divides,1 (2)and,or2.
bdZZgcd bd
bmoddd
b



 

It follows that

1,
is a partially ordered set under
the ordering

22 11
,,bdbd if 21
bb and 21
dd. Let

1,
be the subset of

1,
consisting of all
ordered pairs
 
1
,,bd
where d
. Note that
 
11
,,
 if
is odd. Moreover, if 0
2
(mod 2) and β is even, then

11
,,2
 
.
Proposition 8. Let

11
,bd and

22
,bd be elements
of the poset

1,
. There exists elements

,bd and

,bd

in

1,
, such that
 
11
,,bdbd and
 
22
,,bdb d, and

11
,,bd bd

and
,bd

22
,bd .
Proof. Let

12
.. ,b gcdbb. Note that since i
d
or 2
for 1,2i, it follows that if
12
min ,ddd,
then i
dd. Thus b divides
and d is
or 2
. If
b
is even, then it follows that 2 divides both 1
bb
and 2
bb, contradicting

12
.. ,bgcdbb. Thus b
is
odd showing
 
1
,,bd
. Moeover
11
,,bdb d
and
 
22
,,bdb d. Let

12
.. ,blcmbb
and d

12
max ,dd. It follows that b
is odd, and hence
 
1
,,bd

. Furthermore

11
,,bd bd

and
 
22
,,bdb d

.
Corollary 9.

1,
is a lattice where for
11
,bd
and

22
,bd in

1,
the join



112 21212
,,..,,min,bdb dgcdbbdd ,
and the meet



112 21212
,,..,,max,bdb dlcmbbd d .
Furthermore,

1,
is a sublattice of
1,
.
Proposition 10. Let

11
,bd and

22
,bd be elements
of

1,
such that

22 11
,,bdbd. Then there
exists either a standard m
-action 1
on
22
,
M
bd ,
or a standard 2m
-action 1
on

22
,
M
bd , which
we denote by
, and a regular covering
 
22 2211
:, ,,
M
bd MbdMbd

.
Proof. If

22 11
,,bdbd, then 21
bb and 21
dd.
Furthermore 1
d
, or 1
2d
and 2
d
, or 2
2d
.
Now 1
2
bm
b,

11
21bn
, and

21
21bn
 for
some integers m, n1 and n2. Since

1122
21 21bnb n ,
it follows that

21
21 21nmn , and therefore m
must be odd. Since 21
dd
, the only possibilities are
12
dd or 21
2dd. If 12
dd
, then there exists a
m
-action 1
on
22
,
M
bd such that

2222 111
,, ,
M
bd MbdMbd

.
If 21
2dd
, then there exists a 2m
-action 1
on
22
,
M
bd such that

2222 111
,,,
M
bd MbdMbd
.
Proposition 11. Let
,bd be an element of
1,
.
Then there exists either a standard 21n
-action 1
on
,
M
bd , or a standard

22 1n
-action 1
on
,
M
bd
which we denote by
, and a regular covering
1
:, ,,Mbd Mbd

.
Proof. Write 21n
b
. If d
, then there is a
21n
-action 1
such that
11
,, ,Mbd Mbd

.
If 2d
, then there is a

22 1n
-action 1
on
,
M
bd
such that
11
,, ,Mbd Mbd

.
Theorem 12. For each pair of positive integers
and
, the poset
1,
is isomorphic to the poset
1,
.
Proof. Define a function

11
:, ,f
 as
follows: let
1
,,bd
. There exists either a stan-
dard
b
-action if d
, or a standard
2b
-action if
2d
on
,
M
bd , which we denote by
, such that

1
,, ,Mbd Mbd

.
Define
1
,,fbd

.
Suppose

1112 22
,,
f
bd fbd

 . Since 1
and 2
are equivalent, there exists a homeomorphism
1122
:, ,hMbdMbd such that
1
12
Gh Gh


. Since

11
,
M
bd and
22
,
M
bd
are homeomorphic, it follows that 12
bb and 12
dd
,
showing f is one-to-one.
Let
1,

. Then there exist a prism manifold
,
M
bd such that

1
:, ,,Mbd Mbd

.
We may assume that b and d are both positive. By [4], η
is equivalent to one of the standard actions 1
or 1
,
and
1
,21,
M
bdb nd
 or


121,2bn d
respectively, for some positive integer n. Therefore
1
b
(mod 2) and d
or 2d
. If
is either
1
or 1
, then
,fbd
, showing f is onto.
Suppose now that
 
22 11
,,bdbd. Let

111
,fbd
and
22 2
,fbd
where 1
and 2
are the standard
J. E. KALLIONGIS, R. OHASHI
Copyright © 2012 SciRes. APM
154

11
b

and

22
b

-actions respectively on
11
,
M
bd
and

22
,
M
bd and 1
i
or 2. We have the coverings
 
1111111
:, ,,MbdMbd


and
 
11111 11
:, ,,Mbd Mbd

.
By Proposition 10 there is a standard

12
bb
-action
on

22
,
M
bd where 1
or 2, and a regular covering
covering
 
22 2211
:, ,,
M
bd MbdMbd


.
Since these are standard actions and 2
12 1
b
bb b
it fol-
lows that 21


. This shows that
21
.
Corollary 13.

1,
is a lattice.
We will now consider maximal and minimal elements
in

1,
. Write 0
2n
where 0
is odd. Then
the maximal element in
1,
is

2,
n
if
is
odd, and

2,2
n
if
is even. Note that if
,bd

1,
, then 2n divides b and 2n
b is odd. In de-
scribing the minimal elements let 0
b be the largest odd
divisor of 0
such that

0
.. ,1gcdb
. If
is odd
or if 0n, then the minimal element in
1,
is

0
2,
nb
, otherwise the minimal element is

0
2,2
nb
.
We say an element

11
,bd is directly below
22
,bd
or that

22
,bd is directly above

11
,bd if whenever
 
112 2
,,,bdbdb d, then either

11
,,bd bd
or
 
22
,,bdb d.
Theorem 14. Let

0
011
2,
mb
and

0
022
2,
nc
be the minimal elements in
1
11
,
and
1
22
,
respectively where 1
i
or 2, and let 01
i
km
i
i
bp
and 01
i
sn
i
i
cq
be the prime decompositions. Sup-
pose one of the following holds:
1) 1
and 2
are both odd.
2) 1
and 2
are both even and 00 0mn.
3) 1
and 2
are both even and 00
0mn.
4) 1
even with 00m and 2
odd.
Then

1
11
,
is isomorphic to

1
22
,
if
and only if ks and after reordering ii
mn
for
1, ,ik.
If 1
is odd and 2
is even with 00n, then

1
11
,
is isomorphic to

1
22
,
if and only if
1
ks, after reordering ii
mn
for 1, 2,,1ik,
and 1
k
m.
Proof. We will first assume that 1
and 2
are both
odd. Suppose

11
112 2
:, ,f

 is an iso-
morphism. Now
0
1
2,
m
and

0
2
2,
n
are the maxi-
mal elements of

1
11
,
and

1
22
,
respec-
tively, and

00
12
2, 2,
mn
f
. The elements directly
below
0
1
2,
m
in
1
11
,
are

0
11
2,,,
mp
0
1
2,
m
k
p
and the elements directly below
0
2
2,
n
in
1
22
,
are
 
00
12 2
2,,,2,
nn
s
qq
. Since f
must take the elements directly below

0
1
2,
m
to the
elements directly below
0
2
2,
n
, it follows that ks
.
The elements directly above

01
,b
in
1
11
,
are listed as

00
12
0
11
112 1
12
1
1
2,,2 ,,,
2,
ii
ki
mmm m
mm
ii
ii
mmm
ki
ik
pppp
pp




Similarly the elements directly above

02
,c
are

00
12
0
11
122 2
12
1
2
2,,2 ,,,
2,.
ii
ki
nnnn
nn
ii
ii
nn n
ki
ik
qqq q
qq




By reordering we may assume
 
00
11
12
2,2,
jj
ii
mn
mmnn
ji ji
ij ij
fppq q



for 1jk
. The number of elements in
1
11
,
is

11
k
i
im
, and this equals the number of elements in
1
22
,
which is

11
k
i
in
. Let



0
11
11 111
1
11
,, ,:
2,,.
ji
j
m
mm
ji
ij
b
ppb




Similarly let



0
11
22 222
1
22
,, ,:
2,,.
ji
j
n
nn
ji
ij
c
qqc
 



It follows that

11
112 2
,,
jj
f

. Thus the
number of elements in
1
11
,
j
which is
1
ji
ij
mm
is equal to

1
ji
ij
nn
the num-
ber of elements in
1
22
,
j
. Using the equations
 
1
11
kk
ii
ij i
mn



and

11
jiji
ij ij
mmnn


,
we obtain

11,
11
i
jij j
jij
ij
n
mm
nmn


and this implies that
j
j
mn
for 1jk.
We now suppose that
0
01
2,
mb
and

0
02
2,
nc
are
the minimal elements in

1
11
,
and
1
22
,
respectively, and 01
i
km
i
i
bp
and 01
i
km
i
i
cq
are
the prime decompositions. If
  
1
111
,,b

, then
J. E. KALLIONGIS, R. OHASHI
Copyright © 2012 SciRes. APM
155
0
1
2i
k
ms
i
i
bp
where 0ii
s
m . Define
  
11
112 2
:, ,f

by


0
12
1
,2 ,
i
k
ns
i
i
fb q
.
It is not hard to check that f is an isomorphism. The proof
in cases (2) - (4) is similar.
We now assume that 1
is odd and 2
is even with
n0 = 0. Since the argument is similar to the previous case,
we will sketch the proof. The elements directly below

0
1
2,
m
in

1
11
,
are
 
00
11 1
2,,,2 ,
mm
k
pp
and the elements directly below
2
1, 2
in
1
22
,
are

122 2
,2,,,2,1,
s
qq

. It follows that k = s +
1. The elements in

1
11
,
directly above

0
01
2,
mb
are
 

00
12
0
11
112 1
12
1
1
2,,2 ,,,
2,,
ii
ki
mmm m
mm
ii
ii
mmm
ki
ik
pppp
pp




and the elements directly above

02
,c
are
 

12
1
11
1222
12
1
12
1
,, ,,,
,,
ii
ki
nn
nn
ii
ii
nn
ki
ik
qqq q
qq






02
,2c
. By relabeling we may assume that

011
12
2, ,
jj
ii
mn
mm n
ji ji
ij ij
fppq q



for 11jk and


01
102
2,,2
ki
mm m
ki
ik
fppc

.
Now
 
1
11
12 1
kk
ii
ii
mn

 

, and for 1jk
Lattice of Finite Group Actions on Prism Manifolds
Advances in Pure Mathematics, 2012, 2, 149-168
http://dx.doi.org/10.4236/apm.2012.23022 Published Online May 2012 (http://www.SciRP.org/journal/apm)
Copyright © 2012 SciRes. APM
Lattice of Finite Group Actions on
Prism Manifolds
John E. Kalliongis1, Ryo Ohashi2
1Department of Mathematics and Computer Science, Saint Louis University, St. Louis, USA
2Department of Mathematics, King’s College, Wilkes-Barre, USA
Email: kalliongisje@slu.edu, ryoohashi@kings.edu
Received October 12, 2011; revised December 11, 2011; accepted December 22, 2011
ABSTRACT
The set of finite group actions (up to equivalence) which operate on a prism manifold M, preserve a Heegaard Klein
bottle and have a fixed orbifold quotient type, form a partially ordered set. We describe the partial ordering of these
actions by relating them to certain sets of ordered pairs of integers. There are seven possible orbifold quotient types, and
for any fixed quotient type we show that the partially ordered set is isomorphic to a union of distributive lattices of a
certain type. We give necessary and sufficent conditions, for these partially ordered sets to be isomorphic and to be a
union of Boolean algebras.
Keywords: Finite Group Action; Prism 3-Manifold; Equivalence of Actions; Orbifold; Partially Ordered Set;
Distributive Lattice
1. Introduction
This paper examines the partially ordered sets consisting
of equivalence classes of finite group actions acting on
prism manifolds and having a fixed orbifold quotient
type. For a fixed quotient type, we show that the partially
ordered set is a union of distributive lattices of a certain
type. These lattices have the structure of factorization
lattices. The results in this paper relate to those in [1],
where those authors study a family of orientation revers-
ing actions on lens spaces which is partially ordered in
terms of a subset of the lattice of Gaussian integers or-
dered by divisibility (see also [2]). Finite group actions
on prism manifolds were also studied in [3].
Let M be a prism manifold and let G be a finite group.
A G-action on M is a monomorphism
:DiffGM
where

Diff
M
is the group of self-diffeomorphisms of
M. Two group actions

:DiffGM
and :G

Diff
M
are equivalent if there is a homeo-morphism
:hM M
such that
 
1
GhGh



, and we
let
denote the equivalence class. If
:DiffGM
is an action, let :MM

be the orbifold covering
map. The set of equivalence classes of actions on prism
manifolds forms a partially ordered set by defining
if there is a covering :
M
M
such that


.
A prism manifold is defined as follows: Let 11
TSS
be a torus where

1:1Sz z  is viewed as the
set of complex numbers of norm 1 and
0, 1I. The
twisted I-bundle over a Klein bottle is the quotient space
,, ,,1WTIuvtuv t
. Let D2 be a unit disk
with 21
DS
and let 12
VSD be a solid torus.
Then the boundary of both V and W is a torus 11
SS
.
For relatively prime integers b and d, there exist integers
a and b such that 1ad bc
 . The prism manifold
,
M
bd is obtained by identifying the boundary of V to
the boundary of W by the homeomorphism :VW

defined by
,,
ab cd
uvuv uv
for

1
,uvV S 
1
S. The integers b and d determine

,
M
bd , up to ho-
meomorphism. An embedded Klein bottle K in
,
M
bd
is called a Heegaard Klein bottle if for any regular
neighborhood
NK
of K,
NK
is a twisted I-bundle
over K and the closure of
 
,
M
bdN K is a solid
torus. Any G-action which leaves a Heegaard Klein bot-
tle invariant is said to split.
We describe in Section 2, the G-actions (up to equiva-
lence) which can act on a prism manifold and the seven
possible quotient orbifolds

,
i
for 17i where
and
are some positive integers. For example, the
orbifold
1,
is an orbifold whose underlying
space is a prism manifold with a simple closed curve as
an exceptional set of type

.. ,kgcd
. The closure
of the complement of the exceptional set is a twisted
I-bundle over a Klein bottle. Section 3 gives necessary
and sufficient conditions for an orbifold of type
,
i
to be regularly covered by a prism manifold.
Let
,
i
be the partially ordered set of equiva-
lence classes of G-actions with orbifold quotient
J. E. KALLIONGIS, R. OHASHI
Copyright © 2012 SciRes. APM
150

,
i
. Define a set
 
1,, :..,1,
divides,1 (mod2),andor2.
bdgcd bdb
dd
b



 


We show in Sections 4 and 6 that the set

1,
is a
distributive lattice which is isomorphic as a partially or-
dered set to

1,
, and this implies that
1,
is
also a distributive lattice. For 27i , we show that

,
i
is isomorphic as a partially ordered set to a
union of lattices of type

1,
x
y. In addition, we give
necessary and sufficient conditions for two lattices of
type

1,
x
y to be isomorphic.
A G-action is primitive if it does not contain a non-
trivial normal subgroup which acts freely. These actions
determine minimal elements in the partially ordered sets.
We determine the primitive actions for each possible
orbifold quotient in Section 5.
In Section 6 we compute the maximum length of a
chain in the partially ordered sets

,
i
. Further-
more, if 0
b is the largest odd divisor of
such that

0
.. ,1gcd b
and 01
i
kl
i
i
bp
is the prime decom-
position, then we show that

1,
is a Boolean al-
gebra if and only if 1
i
l for all 1ik.
When
1,mn is a prism manifold, we consider in
Section 7 a partially ordered set of non-cyclic subgroups

,mn of


11
π,mn. We show that

,mn is a
lattice isomorphic to

1,mn where the partial order-
ing on the groups is given by 21
GG if 2
G is a sub-
group of 1
G. The meet 1212
,GG GG and the join
1212
GGG G. Moreover we show that there exists a
sublattice of

,mn which is a Boolean algebra,
and a lattice homomorphism

,mn which re-
stricts to the identity on .
Section 8 is devoted to several examples which illus-
trate some of the main results.
2. Actions on Prism Manifolds
In this section we describe a set of G-actions on a prism
manifold
,
M
bd which leave a Heegaard Klein bottle
invariant and their quotient spaces

,
M
bd G. We
obtain seven quotient types
,
i
for 17i where
and
are some positive integers. It follows by [4]
that any G-action which leaves a Heegaard Klein bottle
invariant is equivalent to one of the actions in Quotient
type [i] for some 17i, and

,,
i
Mbd G
.
By [4] these actions are completely determined by their
restriction to a Heegaard Klein bottle K. We begin by
describing G-actions on K and note that these actions
extend to all of
,
M
bd . We will list the actions by
their quotient type.
Let
Vk be the orbifold solid torus with exceptional
set the core of type k and let
 
BkV k
be the
Conway ball, where

:Vk Vk
is the involution
defined by
,,uvuv
. The Conway sphere
Bk has 4 cone points, each of order 2.
It is convenient to view the Klein bottle as the set of
equivalence classes

1
:1 22,,
11
1,, 22.
22
Kru ruS
uuuuu



1) Quotient type
1,
. For 21mn, define
actions
1:Diff
m
K
and
12
:Diff
m
K
by



2π
21
1
i
n
ru rue
and



π
21
1
1
1
i
n
ru ue
r



where
1 represents a generator of the group. The quotients
1
K
and 1
K
are both Klein bottles. These actions
extend to the prism manifold

,
M
bd and we denote
these extensions using the same letters to obtain
1:Diff,
m
M
bd
and
12
:Diff,
m
M
bd
.
The orbifold quotient for these actions is denoted by
11
,Vk W

, where 1
W is a twisted I-bundle
over the Klein bottle and

:i
Vk W

is defined
by

,,
kk
uvuvuv

where
and
are inte-
gers, and
,kgcd
 . It follows that the quotient
1
,Mbd
is the orbifold denoted by


121,bn d
and the qoutient
1
,Mbd
is the orbifold denoted by
121,2bn d.
2) Quotient type
2,
. For 2mn define ac-
tions
2:Diff
m
K
and
22
:Diff
m
K

by


π
21
i
n
ru rue
, and



2
1
1, 0ru u
r
and



π
21, 0
i
n
ru rue
. The quotients 2
K
and 2
K
are both mirrored annuli. These actions extend to all of
,
M
bd and we obtain

22
:Diff,
n
M
bd
and
22 2
:Diff,
n
M
bd
 . The orbifold quotient for
these actions is denoted by

22
,Vk W

, where
2,,, ,1WTIuvtuvt
 is a twisted I-bundle
over the mirrored annulus mA and ψ is defined as in Case
1. The orbifold quotient
 
22
,2,
M
bd nbd
and
22
,2,2
M
bdnb d
.
3) Quotient type
3,
. Define 32 21
:n

Diff
K
and

34
:Diff
n
K
by



3
1
1, 0ru u
r
,



π
21
31, 0
i
n
ru rue




, and
J. E. KALLIONGIS, R. OHASHI
Copyright © 2012 SciRes. APM
151



π
2
3
1
1
i
n
ru ue
r



. The quotients 3
K
and 3
K
are mirrored Möbius bands. These actions extend to the
prism manifold
,
M
bd and we obtain 3221
:n



Diff ,
M
bd and


34
:Diff,
n
M
bd
. The or-
bofold quotient for these actions is

33
,Vk W



3
VkW
where

3,, ,,1WTIuvtvut  is
a twisted I-bundle over the mirrored Möbius band mM.
The orbifold quotient
 

33
,21,21
M
bdb ndbnd

and
33
,2,2
M
bdnb dnb d
.
4) Quotient type

4,
. For we
have
 
12 1
ji ji
ij ij
mm nn

 

.
Using these two equations we obtain
j
j
mn
for
11jk. The number of elements greater than or
equal to

01
1
2,
ki
mm m
ki
ik
pp
and

02
,2c
is

1
ki
ik
mm
and

1
11
k
i
in
respectively. Since
these two numbers must be equal, it follows that 1
k
m
.
For the converse suppose that 01
i
km
i
i
bp
where
1
k
m and 1
01
i
km
i
i
cq
. If

1
,b
is any element in
1
11
,
, then 0
1
2i
k
ms
i
i
bp
where 0ii
s
m , and
0
k
s or 1. Let


1
1
12
1
,,2
ks
i
i
fb q

if 0
k
s
,
and

1
1
2
=1 ,
ks
i
iq
if 1
k
s. It follows that f is an
isomorphism.
For a pair of positive integers
and
, let
2,
denote the equivalence classes of those actions whose
quotient type is
2,
. Let
 
2,, :..,1,
divides,0 (2),andor2.
bdgcd bd
bmoddd
b






It follows that
2,
is a partially ordered set under
the ordering
22 11
,,bdbd if 21
bb
, 1
2
1
b
b
(mod
2), and 21
dd.
The proof of the following theorem is similar to that of
Theorem 12.
Theorem 15. For each pair of positive integers
and
, the poset
2,
is isomorphic to the poset
2,
.
We will now consider the structure of the partially or-
dered set
2,
. Write 01
2mbb
where 0
b and
1
b are both odd and 0
b is the largest odd divisor of
which is relatively prime to
.
Theorem 16. For each pair of positive integers
and
, the poset
2,
is a disjoint union of lat-
tices given by



2
1
10
1
0
111
10 0
,
2, if0(2)
,if0(4)
(2,2),if0 (2)
2
mmj
j
mmj
j
bmod
bmod
bb mod









Proof. We first assume that
is odd. Note that for
1jm
, we have

2
0
2, ,
mj
b

. It suffices
to show that if

0
2, ,
mj
bbd
, then

0
2,
mj
b
,bd and hence
0
2,
mj
b
is a minimal element,
and if
2
,,bd
, then


0
,2,
mj
bd b
for
some unique j where 1jm
. Suppose

0
2,
mj
b
,bd , and thus 0
2mj
b
divides b,
divides d, and
0
2mj
b
b
is odd. Now d divides
, which implies d
.
Since b divides 01
2mbb
and

.. ,1gcdb
, it fol-
lows that b must divide 0
2mb. Write 2w
bb
where
b
is odd. Now
0
2
2
w
mj
b
b
being odd implies wmj
and b0 divides b
. Note that 2mj
bb
divides 0
2mb.
Thus b
divides 0
b showing that 0
bb
, and there-
fore 0
2mj
bb
. Let

2
,,bd
. Since
is odd,
it follows that d
. As above we have that b divides
0
2mb. Furthermore, 0
2mb
b must be even. We may write
2r
bb
where 01rm
, b is odd, and b
di-
vides 0
b. Therefore

0
,2,
r
bd b
.
We now assume that
is even and we write 2n
where
is odd. There are two cases to consider: n 2
and n = 1. Suppose first that n 2. Now
0,b
2,
. We will show that

0,b
is the minimal
element in
2,
. Suppose that

2
,,bd
and
0,,bbd
. In this case d
. Also since
0
b
b
J. E. KALLIONGIS, R. OHASHI
Copyright © 2012 SciRes. APM
156
and 0
b are both odd, it follows that b is odd. Since

.. ,1gcdb
and b is odd, it follows that b divides 0
b.
Thus 0
bb showing that

0,b
is the minimal ele-
ment. Now let
 
2
,,bd
. Recall that d
or
2d
. If d
, then since b and d are relatively
prime, b must be odd. Furthermore, b must divide 0
b,
and thus

0
,,bd b
. If 2d
, then 1
2
2
n
d

and d is even since n 2. Now b and 1
2n
d
are
relatively prime, which implies that b is odd. Again we
have that b must divide 0
b, so that


0
,,bd b
. This
shows that

0,b
is the minimal element and


21
0
,,b

. We now consider the case when
n = 1. Note that


2
0
2,2,
mj
b

for 1 j m
– 1 and


2
0,,b

. We need to show that these
are the minimal elements in

2,
, and if
,bd

2,
then either


0
,2,2
mj
bd b
for some
unique j or


0
,,bd b
. The proof to this is similar
to case 1.
Remark 17. Note that by Theorem 7 each

1
10
2,
mmj
jb
and

11
10
2,2
mmj
jb

is a dis-
joint union of isomorphic lattices.
For a pair of positive integers
and
with >
and
even, let

3,
denote the equivalence
classes of those actions whose quotient type is
3,
.
Let


3,,: ,
2
. .,1,divides.
2
bd d
gcdb db





It follows that

3,
is a partially ordered set under
the ordering

21
,,bdbd if 21
bb and 1
2
1
b
b
(mod
2).
Theorem 18. For each pair of positive integers
and
with >
and
even, the poset
3,
is isomorphic to the poset
3,
.
Proof. Let
 
3
,,bd
and let 2
mb
. Ob-
serve that bmd
 and bm d
. There exists a
standard 2m
-action


2
:Diff,
m
M
bd
such
that
33
,,,Mbdbmdbm d


where 3
if m is odd and 3
if m is even. Define
 
33
:, ,f
 by

3
,,fbd

.
Suppose that

21
,,bd bd. Let 1
1
2m
b

and
2
2
2m
b

and let


1
12 1
:Diff,
m
M
bd
and
2
22 2
:Diff,
m
M
bd
be the standard actions. It
follows that 1
21
2
b
mm
b
, and therefore 1
2m
is a sub-
group of 2
2m
. Furthermore,



1
2
21
11
b
b

, which
implies that
21
and thus f is order preserving.
The proof that f is one-to-one and onto is similar to that
in Theorem 12.
We will now consider the structure of the partially or-
dered set
3,
. Write 2
2
m

where
is
odd. Let 0
b be the largest positive odd divisor of
which is relatively prime to 2
d
.
Theorem 19. For each pair of positive integers
and
with >
,
even, and 2
d
, the
poset
3,
is a disjoint union of isomorphic lat-
tices given by


1
00
3
1
0
2,if 0(4)
,
,if0(4)
mmj
jbd mod
bd mod

 


Proof. Suppose first that 0
 (mod 4) (equiva-
lently d is odd). Note that 0
2jb divides 2
for
0jm
, and since d is odd we have

0
.. 2,1
j
gcdbd
.
It follows that

3
0
2, ,
jbd
and

0
2,
jbd is a
minimal element of
1
0
2,
jbd. Let

3
,,bd
.
Write 2k
bb
where b
is odd. Since b divides
2
2
m

, it follows that 0km and b
divides
. Furthermore, b
and d are relatively prime. Since
0
b is the largest positive odd divisor of
which is
relatively prime to d, it follows that b divides 0
b.
Hence
1
0
,2,
k
bdb d for a unique k. Now sup-
pose
1
0
,2,
k
bdb d for some k. By assumption
2
d
. Since b divides 0
2kb and 0
2kb divides
22
m
, it follows that b divides 2
, and hence
3
,,bd
.
The proof for 0
(mod 4) is similar.
Corollary 20.


1
00
3
1
0
2,if 0(4)
,
,if0(4)
mmj
jbd mod
bd mod

 


For each pair of positive integers
and
, let
4,
denote the set of equivalence classes of actions
J. E. KALLIONGIS, R. OHASHI
Copyright © 2012 SciRes. APM
157
on prisim manifolds whose quotient space is
4,
.
Theorem 21.
 
41
,,


Proof. If

4,

, then
is equivalent to



521
:DihDiff,
n
M
bd
for some integers n,
b, and d where

21bn
 and d
. Since
,bd

1,
, define a function

41
:, ,f


by


,
f
bd
. It follows easily that f is an order
preserving surjection.
For each pair of positive integers
and
, let

5,
denote the set of equivalence classes of actions
on prisim manifolds whose quotient space is
5,
.
We now consider the structure of the partially ordered set

5,
. Write 01
2mbb
where 1
b is odd and 0
b
is the largest odd divisor of
that is relatively prime to
.
Theorem 22.






5
1
00
1
0
11
0
1
0
,
2,if0(2)and1
,if0(4)and1
2,2,if0(2)and 1
2
,if1
mj
j
m
bmodm
bmodm
bmodm
bm






 

Proof. Suppose that 0
(mod 2). Let
5,

.
We have a covering

5
,,,Mbd Mbd


for some positive integers b and d. Now
is equivalent
to either of the standard actions 5
, 5
or 5
. The
action 5
is impossible since
is odd. We will de-
fine a function


51
00
:, 2,
mj
j
fb

 as fol-
lows: if
is equivalent to 5
, then

01
212
m
bn bb
for some n and d
. Since b and d are relatively prime
and 0
b is the largest odd divisor of
that is relatively
prime to
, it follows that b divides 0
2mb. Thus


1
0
,2,
m
bd b
and we let



1
0
,2,
m
fbdb
.
If
is equivalent to 5
, then

01
422
m
bn bb , and
this implies that b must divide 1
0
2mb
. Thus
,bd

11
0
2,
mb
and we define



11
0
,2,
m
fbdb
.
If
is equivalent to 5
, then 01
42
m
nbb b for some
n. Write 0
2k
nn where 0
n is odd. This implies that b
divides 2
0
2mkb
 where 02km . This shows that


12
0
,2,
mk
bd b

and we define



12
0
,2,
mk
fbdb

.
We now show that f is an order preserving bijection.
Note that there do not exist integers n, n, b, and b
,
such that

21221bn bn

 , or

214bn nb

 ,
or

22 14bn nb

 , if either b divides b or b
di-
vides b with odd quotient. This implies that f is one-to-
one. Furthermore if
21
, then 2
and 1
are
both equivalent to either 5
, 5
or 5
. From this it
can be shown that

21
ff
. To show f is onto,
suppose
1
0
,2,
j
bd b
for 0jm. Let
0
221
jbn
b
for some positive integer n. If mj
,
then
01 1
221
mbbbnb

, and since 1
b is odd we
may write
21bn
. Hence there is an action


521
:DihDiff ,
n
M
bd
such that

55
,,Mbd

, and thus
5
f
,bd . Similarly, if 1jm
or if 1jm
, we ob-
tain actions 5
and 5
-actions respectively. This shows
that f is onto.
If 0
(mod 4), then if m = 1 there exist only 5
-
actions, and if m > 1 there exist only 5
-actions. For
0
2
(mod 2), if m = 1 there exist only 5
and 5
-
actions, and if m > 1 there exist only 5
and 5
-actions.
If
is odd and
is even, then there exist only 5
and 5
-actions. The proof in all these cases is similar to
the above.
For each pair of positive integers
and
with
>
and
even, let

6,
denote the set
of equivalence classes of actions on prisim manifolds
whose quotient space is

6,
.
Theorem 23. For each pair of positive integers
and
, the poset
6,
is isomorphic to the poset
3,
.
Proof. Let
6,1

. Now
is a
Dih n
-
action on a prisim manifold

,
M
bd and is equivalent
to 6
if n is even or 6
if n is odd. Furthermore,

6
,,
M
bdbn dbn d
,
and therefore bn d
and bn d
. It follows
that
3
,,bd
. Define a function
6
:,f

3,
by
,
f
bd
.
Let
3
,,bd
. Therefore 2
bn
and
2
d
for some n
. This implies bnd
and bn d
. If n is even there exists an 6
-action,
and if n is odd there exists an 6
-action. Therefore f is
onto.
Let
111
,
f
bd
and


222
,
f
bd
and
suppose
12
ff
. Then
 
112 2
,,bd bd and
hence 12
bb
and 12
dd
. We also have 11 1
bnd
and 22 2
bn d
so that 12
nn. Recall that 1
and
2
are equivalent to either an 6
or 6
-action. If
12
nn
is even, then both of them are equivalent to an
6
-action, otherwise they are both equivalent to a 6
-
J. E. KALLIONGIS, R. OHASHI
Copyright © 2012 SciRes. APM
158
action. Hence 1
and 2
are equivalent showing f is
one-to-one.
If
12
, then there is a covering map
 
22 11
:, ,
M
bd Mbd
where 12
dd. Hence,

21
21nbb
for some n
which shows 21
bb
is an odd number. Therefore, we
conclude
 
1122
,,bdb d showing f is order preserv-
ing.
For each pair of positive integers
and
, let

7,
denote the set of equivalence classes of ac-
tions on prisim manifolds whose quotient space is

7,
. The proof of the following theorem is similar
to that of Theorem 23.
Theorem 24. For each pair of positive integers
and
, the poset

7,
is isomorphic to the poset

2,
.
5. Primitive Actions on Prism Manifolds
Let


:Diff ,GMbd
be a G-action on a prism
manifold

,
M
bd with orbifold covering map
 
:, ,Mbd Mbd
.
We say that
is primitive if G does not contain a non-
trivial normal subgroup which acts freely on
,
M
bd .
Therefore for any nontrivial normal subgroup H of G, if


0:Diff ,
H
H
Mbd

 ,
then

0
,Mbd
is not a manifold. In this section we
determine when an action is primitive.
Theorem 25. Let


:Diff,
m
M
bd
be a m
-
action on the prism manifold

,
M
bd .
1) If
is equivalent to 1
, then
is primitive if
and only if for every prime divisor p of m, 0d
(mod
p).
2) If
is equivalent to 1
, then
is primitive if
and only if b is even and for every odd prime divisor p of
m, 0d (mod p).
3) If
is equivalent to 2
or 3
, then
is primi-
tive if and only if either 2n
m, or if p is any odd prime
divisor of m, then 0d (mod p).
4) If
is equivalent to 3
, then
is primitive if
and only if either 2m, or if p is any odd prime divisor
of m, then 0d (mod p).
Proof. We may suppose 1
. Then 21mn
and if l
is a subgroup of m
and 0:
ll




Diff ,
M
bd , then
 
10
,,,Mbdbld Mbd
.
Furthermore

1,bl d is a manifold if and only if

.. ,1gcdbl d. Assume that
is primitive and let p
be a prime divisor of m. Consider the subgroup
p
.
Since

.. ,1gcdb d and
is primitive, it follows that
 
..,..,
g
cdbp dgcdpdp. Thus p divides d. Now
suppose that every prime divisor of m also divides d and
let l
be a subgroup of m
. Let p be a prime divisor
of l. Since l divides m, it follows that p divides m. Hence
by assumption
.. ,1gcdbl d, showing that
is
primitive.
For part 2), suppose that 1
. Then
22 1mn
and if l
is a subgroup of m
and 0:
ll


Diff ,
M
bd , then either 21ls
and
10
,,,Mbdbld Mbd
,
or
22 1ls
and

10
,21,2,Mbdb sdMbd
.
Furthermore
1,bl d is a manifold if and only if
.. ,1gcdbl d
; and
121,2bs d is a manifold if
and only if
..21,21gcdbsd
.
Assume first that
is primitive. If p is an odd prime
divisor of m, then the same argument used in the 1
case shows p divides d. Now 2
is a subgroup of m
and we have a covering
12
,,2,MbdbdMbd.
Since
is primitive,
1,2bd is not a manifold, and
since
.. ,1gcdb d
it follows that 2 divides b. For the
converse suppose that b is even, and if p is any odd prime
divisor of m, then 0d
(mod p). Let l
be any sub-
group of m
. If l is odd, the proof that
is primitive is
identical to the 1
case. If l is even, then
..21,21gcdbsd
, showing that
is primitive.
To prove part 3), suppose that
is equivalent to 2
.
Therefore 2ms
and

22
:Diff,
s
M
bd
is
defined by



π
21
i
s
ru rue
where
ru is any
point in the Heegaard Klein bottle K and 1 denotes a
generator of 2
s
. If l
is any subgroup of 2
s
and
2:Diff,
ll
M
bd


, then



2π
1
i
l
ru rue
where 1 denotes a generator of l
.
We now suppose that 2
is primitive and p be an odd
prime divisor of 2ms
. Letting lp, we obtain a
covering
1
,, ,
M
bdMbdbpd
. Since 2
is primitive
.. , 1gcdbp d
, and since
..,..,
g
cdbp dgcdpd, it follows that p divides d.
Suppose 2n
m
. Then 2t
l and



2ππ
1
22
1
ii
tt
ru ruerue
 
 

 
 
. Now K
is a mir-
rored annulus, showing that

,Mbd
is not a mani-
fold, and thus 2
is primitive. Suppose now that 2n
m
,
and if p is an odd prime divisor of m, then 0d
(mod
p). Assume there is a covering

1
,, ,
M
bdMbdbld
 (Note that the quo-
tient space cannot be
1,2bl d by definition of 2
J. E. KALLIONGIS, R. OHASHI
Copyright © 2012 SciRes. APM
159
and 1
). It follows that l is odd. Let p be any odd prime
divisor of l. It follows that p divides m, and hence by
assumption p divides d. Thus

.. ,1gcdbl d, showing
2
is primitive.
We now suppose that 3
. In this case 4ms
and


34
:Diff,
s
M
bd
where



π
2
3
1
1
i
s
ru ue
r



. If l
is a subgroup of m
, let


3:Diff,
ll
M
bd


. Assume first that
is
primitive. Suppose p is an odd prime divisor of m and
consider the primitive subgroup
p
of m
. Here lp
,
and since m
p is even we have



2π
1
i
p
ru rue
.
Since
is a primitive 1
-action on

,
M
bd , it fol-
lows by the above that 0d (mod p). Now suppose
2n
m and let l
be a subgroup of m
. Then 2t
l
and



π
1
2
1
i
t
ru rue




. In this case
is an 2
-
action and

,Mbd
is not a manifold.
Now suppose that every odd prime divisor of m also
divides d. Let l
be a subgroup of m
. If l is odd, then
4
s
l is even and



2π
1
i
l
ru rue
which is an 1
-
action. The result follows by the above that
is primi-
tive, and thus
is primitive. Now suppose l is even. If
m
l is even then



2ππ
1
ii
ll
ru ruerue
 

 
 
, which
is an 2
-action; and if m
l is odd then



2ππ
11
1
ii
ll
ru ueue
rr




, which is a 3
-action.
In either case the quotient is not a manifold. Hence
is
primitive.
We now prove part 4) and suppose 3
. Therefore

22 1mn
and



322 1
:Diff,
n
M
bd
where



2π
21
3
1
1
i
n
ru ue
r



. If 2m, then 3
K
is a
mirrored Möbius band and the action is primitive. So
suppose that 2m. Assume first that 3
is primitive
and let p be an odd divisor of m, and hence p divides
21n. Consider the subgroup
p
and let


3:Diff,
pp
M
bd


.
Then



4π
1
i
p
ru rue




, and we have a covering
 
1
,, ,
M
bdMbdbpd
. Again as above since
3
is primitive, p must divide d.
We now suppose that for each odd prime divisor p of
m, 0d
(mod p). Let l
be a subgroup of m
. If l
is odd we obtain a covering

1
,, ,
l
M
bdMbdbld, and as above if p is
a prime divisor of l we obtain

.. ,1gcdbl d. Thus the
action is primitive. If l is even then
22 1ls
, and
if 3l

then



2π
21
1
1
i
s
ru ue
r



. In this case
3
,,
l
M
bd bld, which is not a manifold show-
ing that the action is primitive.
Proposition 26. Let
22
:Diff,
m
M
bd

be an action on the prism manifold

,
M
bd where
2m. Then
is primitive if and only if either 2n
m
,
or if p is any odd prime divisor of m, then 0d
(mod
p).
Proof. We may assume that 2
, and therefore
2
22
m
. Suppose that 2
is primitive. This implies
that 2
is primitive and the result follows from Theo-
rem 25. Now suppose that either 2n
m, or if p is an
odd prime divisor of m, then 0d (mod p). Note that
this implies by Theorem 25 that 2
is primitive. Let H
be a subgroup of 22m
 and 2
H
. If
211HZ
, then

,Mbd
is not a manifold.
So we may assume that


211HZ , and hence H
is a subgroup of 2m
. Since 2
is primitive,
,Mbd
is not a manifold showing that 2
is primitive.
Proposition 27. Let

22
:Diff,
M
bd
 be
an action on the prism manifold

,
M
bd . Then
is
primitive if
is equivalent to either 5
, 6
or 7
. If
is equivalent to 2
or 5
, then
is primitive if
and only if 0b
(mod 2).
Proof. If
is either 5
, 6
, 7
, 2
or 5
, then
any subgroup of 22
 restricted to a Heegaard Klein
bottle K is a product of the following homeomorphisms
where
ru K
:
πi
ru rue
,
ru ru, and

1
ru u
r
. The only fixed-point free action on K is
the homeomorphism

π
1i
ru ue
r



. By definition of
5
, 6
or 7
, they do not contain this homeomor-
phism, and hence they are primitive. But the actions 2
and 5
do contain this 2
Z
subgroup and we obtain a
covering

21
,, ,2.
M
bdMbdbd Since
1,2bd is a manifold if and only if 2 does not divides
b, the result follows.
Proposition 28. Let

:DihDiff ,
m
M
bd
be an action on the prism manifold

,
M
bd for m > 2.
1) If
is equivalent to either 4
or 5
, then
is
primitive if and only if for every prime divisor p of m,
0b
(mod p).
2) If
is equivalent to 5
, then
is primitive if
and only if b is even and for every odd prime divisor p of
J. E. KALLIONGIS, R. OHASHI
Copyright © 2012 SciRes. APM
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m, 0d (mod p).
3) If
is equivalent to either 5
, 5
, 6
or 7
,
then
is primitive if and only if either 2n
m, or if p
is an odd prime divisor of m, then 0d (mod p).
4) If
is equivalent to 6
, then
is primitive if
and only if either 2m or, if for each odd prime divi-
sor p of m, 0d (mod p).
Proof. We may assume
is either 4
or 5
, and
thus 21mn and 1
|m
. If
is primitive, then
1
|m
is primitive, and the result follows by Theo-
rem 25. For the converse, suppose that H is a subgroup of

12
Dihmm

. If 21H, then
,
M
bd H
is not a manifold. So we may assume 21H, and
thus H is a subgroup of m
. Hence l
H for some l
with l dividing m. By Theorem 25, 1
is primitive, and
so

,
M
bd H is not a manifold. Thus 4
and 5
are primitive. The proof for the other cases is similar to
Theorem 25.
Proposition 29. Let



22
:DihDiff,
m
M
bd

be an action on the prism manifold

,
M
bd .
1) If 2m, then
is primitive if and only if either
2n
m, or if p is any odd prime divisor of m, then
0d (mod p).
2) If 1m, then
is primitive if and only if 0b
(mod 2).
Proof. We may assume that 7
. Since
22
72
m

 , using Propositions 26 and 27, a proof
similar to that used in Proposition 28 proves the result.
6. Lattice Structure
In this section we compute the maximum length of a
chain in the partially ordered sets

,
i
. In addition,
we give necessary and sufficient conditions for
1,
to be a Boolean algebra.
Theorem 30. Let

11
,bd and

22
,bd be elements
of

1,
, and so ii
d
, where i
is 1 or 2 for
1, 2i. Suppose

22 11
,,bdbd. Then 2
b divides
1
b and 21
.
1) If 21
, then there exists

,bd in
1,
such that

22 11
,,,bdbdbd.
2) If 12
, then 1
2
b
b is prime if and only if there
does not exist
,bd in

1,
such that
22
,bd

11
,,bdb d.
Proof. Since

11
,bd and
22
,bd are elements of

1,
for 1, 2i, it follows that

.. ,1
ii
gcd b d
,
i
b divides
, 1
i
b
(mod 2), and ii
d
where
1
i
or 2. By the definition of
22 11
,,bd bd, we
have 21
bb
and 21
dd
. Now 12
21
d
d
, which implies
21
.
Suppose 21
. Thus 22
and 11
, hence
21
2dd
. Now

1
12
,,bd
and
22 1211
,>,>,bd bd bd showing (1).
We now suppose that 12
, and thus 12
dd
. Sup-
pose there exists
,bd in

1,
such that
22 11
,,,bdbdbd. It follows that 12
ddd
,
2
bb
and 1
bb. Therefore 11
22
bb
b
bb b
 . Note that 1
2
b
b
is prime if and only if either

,bd equals
22
,bd or
11
,bd .
Corollary 31. Let
1
and
2
be elements of
1,
, such that


11
1:Diff,
m
M
bd
and
22
222
:Diff,
m
M
bd
where mi is odd and i
is either 1 or 2. Suppose
21
. Then 2
b divides
1
b and 21
.
1) If 22
and 11
, then there exists
1,

such that
21

 .
2) If 12
and 1
2
b
b is prime, there exists no
1,

such that
21

 .
Recall that the maximal element in
1,
is
2,
n
if
is odd, and

2, 2
n
if
is even. To
obtain the minimal element let 0
b be the largest odd
divisor of
such that

0
.. ,1gcdb
. The minimal
element is
0
2,
nb
if either
is odd or if 0n
,
otherwise the minimal element is

0
2,2
nb
.
Theorem 32. For the partially ordered sets
,
i
where 17i
and 3,6i
, let 12
12
2k
n
nn
n
k
pp p
and 12
12
2k
m
mm
m
k
pp p
be the prime decompositions.
Let 0
b be the largest odd divisor of
relatively prime
to
. Thus, 12
012
k
l
ll
k
bpp p where 0
j
l if and
only if
,0
jj
min nm and
j
j
ln if and only if
,0
jj
min nm
. Then the following chart gives the
length of a maximum chain in each

,
i
.
The maximum length of a chain
Ordered set Conditions Max. length
1,
,
2,
,
7,
0
(mod 2)
11
k
i
il
1,
,
2,
,
7,
0
(mod 2)
12
k
i
il
4,
none
11
k
i
il
5,
0
(mod 2) and n = 0
or n 1
11
k
i
il
5,
0
(mod 2) and n = 0
12
k
i
il
J. E. KALLIONGIS, R. OHASHI
Copyright © 2012 SciRes. APM
161
Proof. We will consider first

1,
. Since
1,
is isomorphic to

1,
we will prove the result for

1,
. Suppose that β and
are both even, and thus

0,2b
and
2,2
n
are the minimal and maximal
elements of

1,
respectively. We may construct a
chain in

1,




12
00101
1
01
,2< ,2<,2<
<,2< <2,2<2,2
lnn
k
bbpbp
bp p



such that dividing any two consecutive first coordinates
yields a single prime in the prime decomposition of 0
b.
Now the length of this chain is

11
k
i
il
. Since any
maximal chain must contain both the minimal and
maximal elements, it follows by the above theorem that
this is a maximal chain. The other cases for
1,
are similar.
For the case

2,
note that

2,
is iso-
morphic to

2,
, which by Theorem 16 is equal to

1
10
2,
mmj
jb
if 0
(mod 2);

1
0,b
if 0
(mod 4); or


11 1
10 0
2,2 ,
mmj
jbb

 if 0
2
(mod 2). By Remark 17 each

1
10
2,
mmj
jb
and

11
10
2,2
mmj
jb

is a disjoint union of isomorphic
lattices. The result now follows by applying the above

1,
case.
By Theorem 21,
 
41
,,

. If
is odd then

11
,,

, and the result follows by the above.
If
is even, then a chain in

1,
can be con-
structed as the above where the second coordinate is re-
placed by
proving this case also.
By Theorem 22,

5,
is isomorphic to

1
00
2,
mj
jb
if 0
(mod 2) and m 1;
1
0,b
if 0
(mod 4) and m 1;

11
0
2,2 ,
mb


if 0
2
(mod 2) and m 1; or

1
0,b
if 0m
.
Using the above results, the first three cases give the
maximum length of

11
k
i
il
. When 0m
, then
there is no restriction on
. Thus the maximum length
is

11
k
i
il
if
is odd and

12
k
i
il
if
is
even.
For the remaining case, it follows by Theorem 24 that
 
72
,,

, which in turn is isomorphic to

2,
. The result now follows by the above.
Theorem 33. For a pair of positive integers
and
with >
,
even, let 2
2
m
where
is odd. Let 0
b be the largest odd divisor of
rela-
tively prime to 2
d
and let 12
12
2k
s
ss
s
k
ppp
and 12
12
2k
t
tt
t
k
pp p
be their prime decompositions.
Thus, 12
012
k
l
ll
k
bpp p where 0
j
l if and only if
,0
jj
min st and
j
j
ls
if and only if
,0
jj
min st
.
For the partially ordered sets

3,
and
6,
the maximum length of a chain is

11
k
i
il
.
Proof. By Theorems 18 and 23 it follows that
3,
and
6,
are both isomorphic to

3,
, which
is equal to
1
10
2,
mmj
jbd
if 0
 (mod 4) or
1,
if 0
(mod 4). In both cases the maxi-
mum length is
11
k
i
il
.
A lattice L is a distributive lattice if for any a, b, c in L,
 
abcab ac
.
Proposition 34.
1,
is a distributive lattice
where for
11
,bd and
22
,bd in

1,
the join

112 21212
,,..,,min,bdb dgcdbbdd
and the meet

112 21212
,,..,,max,bdb dlcmbbd d .
Proof. By Corollary 9,

1,
is a lattice. A com-
putation using the following equation


max ,min,minmax ,,max ,lmnlm ln
for any positive integers l, m and n, shows that
1,
is a distributive lattice.
Remark 35. If we represent the minimal element by
0
2,
nb
and the maximal element by

2,
n
where
is either 1 or 2, then for any element
1
2, ,
nbd
we have
 
0
2,2,2,
nn n
bd bbd


and
 
2, 2,2,
nn n
bd bd

.
A lattice
,,B
is said to be a Boolean algebra if
the following hold:
1) B is a distributive lattice having a minimal element
0 and a maximal element 1.
2) For every aB
, 0aa and 1aa .
3) For every aB
there exists aB
such that
1aa

and 0aa
.
Proposition 36. For the partially ordered set
1,
let 0
b be the largest odd divisor of
such that
0
.. ,1gcdb
and let 01
i
kl
i
i
bp
be the prime
decomposition. Then

1,,,

 is a Boolean
algebra if and only if 1
i
l
for all 1ik.
Proof. Suppose that 1
i
l
for all 1ik. By Remark
35 above and Proposition 34, it remains to show (3) of
the definition. Let

1
11
2, ,
nbd
. Now 10
22
nn
bb
J. E. KALLIONGIS, R. OHASHI
Copyright © 2012 SciRes. APM
162
and so let 0
2
1
b
bb
. Observe that

12
.. ,1gcd bb
and

1212 0
.. ,lcmbbbbb. If either 0n and
is even,
or if
is odd, then all the elements in

1,
have
the same second coordinate—either 2
in the first
case or
in the second case. In this case

11 2101
2,2,2 ,
nnn
bd bdbd
and

11 211
2,2,2,
nnn
bdb dd. The remaining case
is 0n and
even. If 1
d
then let 22d
,
and if 12d
we let 2
d
. It follows that
 
112 2
2, 2,
nn
bdb d gives the minimal element and
 
112 2
2, 2,
nn
bd bd gives the maximal element.
We now suppose that

1,,,

 is a Boolean
algebra. Suppose there exists an >1
j
l. The minimal
element in

1,
is

0
2,
nb
where
is either
1 or 2. Now

1
0
2,
n
j
bp
is an element of
1,
and

1
00
2,>2,
nn
j
bp b

. There exists a com-
plement

2,
nbd such that

1
00
2,2,2,
nn n
j
bd bpb

,
and so

1
00
.. ,
j
lcmbpbb
. It follows that pj divides b.
We also have
 
1
0
2, 2
nn
j
bd bp
equal to the maximal
element

2,
n
, and so

1
0
..2 ,22
nn n
j
gcdbb p
. But
since >1
j
l, it follows that pj divides 1
0
j
bp
, giving a
contradiction.
Proposition 37. Let

00
,Ibd
be an ideal of a
lattice

1,
such that
00
,bd is directly below
2,
nd which denotes the maximum element in the lattice.
If
112 2
,,bdb dI
, then
11
,bd I or

22
,bd I
and I is a maximal ideal.
Proof. Let

112 20 0
,, ,bdb dIb d
. Suppose
both

,
ii
bd I for 1, 2i. Since there is no element
between

00
,bd and
2,
nd, we have
 

00
,,2,
n
ii
bdb dd
, where

0
,2
n
i
g.c.d bb
for
1, 2i. This says that i
b and 0
b do not have a com-
mon odd prime divisor for 1, 2i.
On the other hand,


112 20 0
,, ,bd bdbd so that

012
.. ,blcmbb . Since 0
b and i
b do not have any
common odd prime divisors, this forces 02n
b
. As


00 0
,2,
n
bd d is not the maximum element, 02dd
.
This result is possible only when the second coordinate is
allowed to have an even number, otherwise it would be
contradiction. Note that the second coordinate is an even
number so that we must have 0n, and hence


00
,1,2bdd. In addition, both 1
b and 2
b must be
odd numbers. Now,


112 20 0
,, ,bdb dbd implies
02dd should divide
12
max ,dd. It follows that at
least one of i
d must be equal to 2d. We may assume
12dd
and thus
 
111
,,2bdbd. Since 1
b is an
odd number and 01b
, this shows

1100
,,bd bd
telling us
11
,bdI
, which is a contradiction.
Remark 38. The converse of Proposition 37 is false.
For example, consider
145,11.

9,11 is a prime
ideal but not maximal.
Corollary 39. Let
00
,Ibd
be an ideal of a lattice
1,
such that
00
,bd is directly below
2,
nd
which denotes the maximum element in the lattice. Let
0, 1L be a lattice where the partial ordering on L is
defined by 01
. Then the following are true and
equivalent.
1) I is a prime ideal.
2)
1,
I

is a prime filter.
3) There is a homomorphism

1
:,DI

with
0
I
I
.
Proof. Condition (1) follows by Proposition 37, and
conditions (2) and (3) follow by lattice theory (see for
example [5]).
Proposition 40. Let

00
,Ibd
be an ideal of a
lattice
1,
and
2,
nd denotes the maximum ele-
ment in the lattice. Suppose that if
 
112 2
,,bdbd I,
then
11
,bd I
or
22
,bd I. If

1,
is a Boo-
lean algebra, then I is maximal and

00
,bd is directly
below
2,
nd.
Proof. Since
1,
is a Boolean algebra, I is
maximal. If

00
,,<2,
n
bd bdd

, then
00
,,bd bd
 . This shows

00
,,bd bd
.
7. Group Lattice Structure
Let m and n be relatively prime integers with >1n. De-
fine the group
π,mn to be
112
π,,, 1
mn
m nxyyxyxyx

.
Let V and W denote a solid torus and a twisted I-bundle
over the Klein bottle K respectively. Recall that the prism
manifold
,
M
mn VW
 , where V is identified
to W
by a homeomorphism :VW
 defined
by
,,
s
mtn
uvuv uv
, where s and t are integers
satisfying 1sn tm
 . The fundamental group of
,
M
mn is
π,mn.
Theorem 41. Let H be a normal subgroup of
π,mn.
Then, either H is cyclic or H is isomorphic to
π,bd
for some relatively prime integers b and d satisfying the
following conditions: b divides m, 1
m
b (mod 2), d = n
and
π,mb
mn H, or 2d = n and

2
π,mb
mn H.
Furthermore, there exists a realizable isomorphism
of
π,mn such that if dn then

,
m
b
H
xy
,
J. E. KALLIONGIS, R. OHASHI
Copyright © 2012 SciRes. APM
163
and if 2dn then

2,
m
b
H
xy
.
Proof. Let H be a normal subgroup of

π,mn. Let

:,
M
Mmn
be the regular covering correspond-
ing to H. Choose a component W
of

1W
and let
0:
WWW


. Since W is a twisted I-bundle over a
Klein bottle K and 0:WW
is a covering space, it
follows that W
is either TI
where T is a torus or a
twisted I-bundle over a Klein bottle. Note that each
component of

1V
is a solid torus. If W
is TI
,
then there are two components of
1V
whose
boundaries are being identified with

TI, and thus
M is a lens space. In this case

1
π
M
H is cyclic. If
W
is a twisted I-bundle over the Klein bottle, then there
is only one component of

1V
whose boundary is
being identified with W, and hence M is a prism
manifold. In this case
,
M
Mbd for some relatively
prime integers b and d. Furthermore there is a group ac-
tion G on

,
M
bd such that

,,
M
bdGM mn.
Now

1
K
is a G-invariant Klein bottle. Hence by
[4], the G-action is equivalent, via a homeomorphism h
of

,
M
bd , to either a standard 21r
-action with

21mrb and nd, or a standard

22 1r
-action
with

21mrb and 2nd. These standard actions
arise from the coverings of

,
M
mn corresponding to
the subgroups ,
m
b
x
y and 2,
m
b
x
y respecttively.
Now h projects to a homeomorphism of

,
M
mn real-
izing
.
Theorem 42. Let

11
11
π,,
m
b
bdx y
and

22
22
π,,
m
b
bdx y
be subgroups of

π,mn where
1
i
or 2. Then
 

12
max ,
112 2
π,π,π,,
m
b
bdb dbdxy


where

12
,bgcdbb and

12
min ,ddd. The group
generated by

11
π,bd and

22
π,bd is


12
min ,
π,,
m
b
bd xy


where

12
.. ,blcmbb
and

12
max ,ddd
.
Proof. Let

12
,bgcdbb. Note that we have
bi
mbm
bib
yy




for 1, 2i. This shows that


12
max ,
π,,
m
b
bd xy

is a subgroup of
 
112 2
π,π,bdbd H. Since H
contains
π,bd, it follows that H is not cyclic. By
Theorem 41, H is isomorphic to

π,ln or
π,2ln .
Furthermore, b divides l and l divides i
b, and since
12
,bg.c.dbb, it follows that bl. If 1
d or 2
d is
2n, then since H is a subgroup of

π,
ii
bd , it follows
by the above Theorem 41 that

π,2
H
ln. Since
π,bd is a subgroup of H, we must have 2dn
showing
π,bd H
. We now suppose 12
dd n
,
and thus dn
and

π,
H
ln. It follows that
π,bd H
.
Let J be the group generated by

11
π,bd and
22
π,bd.
Now

12
min ,
x
is clearly a generator of J and
π,bd
.
Since
bm
mbib
bi
yy



 , we have J contained in
π,bd
.
To show
π,bd
is contained in J, we use the easily
verifiable equation

12
12
..,..,
mm
g
cdlcm bbm
bb


 , and
by using
12
.. ,blcmbb
we have
12
.. ,
mm m
gcdbbb



.
Since there exist integers s and t such that
12
mmm
st
bb b
,
we obtain 12
st
mm m
bb b
y
yy




proving the result.
Let
,mn be the collection of subgroups
π,,
m
b
n
bxy


 of
π,mn where 1
or 2.
Theorem 43.
,mn is a lattice of subgroups, and
there exists a lattice isomorphism

1
,,mn mn
which sends an element

π,bd in

,mn to the
element
,bd in
1,mn.
Proof. If
11
π,bd and

22
π,bd are elements in
,mn, define
 
112 2
π,π,bdb d if

22
π,bd is a
subgroup of
11
π,bd . For

11
π,bd and
22
π,bd in
,mn, define
 
112 21122
π,π,π,π,bdb dbdb d
and
112 2
π,π,bdbd to be the group generated by
11
π,bd and
22
π,bd . By the above Theorem 42,
,mn is a lattice. Furthermore, the map which sends
an element
π,bd in
,mn to the element
,bd
in
1,mn is a lattice isomorphism.
Corollary 44.
,mn is a distributive lattice, which
is a Boolean algebra if and only if the prime decomposi-
tion of m is 1
2k
j
i
ip
.
Proof. This follows by Propositions 34 and 36 and
Theorem 43.
J. E. KALLIONGIS, R. OHASHI
Copyright © 2012 SciRes. APM
164
For the following propositions write 0
k
mpm where
p is an odd prime relatively prime to 0
m.
Proposition 45. Let π,
ln
pb



and π,n
pb



be
subgroups of

π,mn where 1l. There exists a sur-
jection :π,π,
l
l
nn
pb pb



.
Proof. Since π,,
l
m
lpb
n
pbx y


 and
π,,
m
pb
n
pbx y


 , define a function
:π,π,
l
l
nn
pb pb



by

l
x
x
and l
mm
p
bpb
lyy




. Clearly l
preserves the first relation in π,
ln
pb



. To show that
l
preserves the second relation, it suffices to show that
1
21
l
pmn
yx
. Write 121
l
ps
, and note that
 
121 2
22 22422
lss s
p
mmm mmmm
yy yyyyy
,
since 41
m
y. Thus 1
22
1
l
pmn mn
yxyx
, showing that
l
is a homomorphism. Since l
takes generators to
generators, it is also a surjection.
Proposition 46. Let 21
21
21
π,π,
ll
nn
pb pb



be
subgroups of
π,mn where 12
1ll. There exist sur-
jections :π,π,
i
i
l
li i
ii
nn
pb pb



for 1, 2i and
a homomorphism 11
11
:π,π,
nn
pb pb



, such
that the following diagram commutes where
and
are inclusions:
2
2
2
π,
ln
pb




2
l
2
2
π,n
pb




1
1
1
π,
ln
pb



1
1
π,n
pb


 1
1
π,n
pb



1
l
Proof. Let
21
2211
21
21
21
π,, ,π,
ll
mm
lpbpbl
nn
pbxyxypb

 

 
 
.
Note that 2
b divides 1
b, 21
and 21
ll
. By
Proposition 45, there exist surjections
:π,π,
i
i
l
li i
ii
nn
pb pb



defined by
ii
i
l
x
x
, =
liii
i
mm
p
bpb
lyy




. Define a
function 11
11
:π,π,
nn
pb pb



by
11
x
x
and 11
l
p
mm
pb pb
yy




where 12
lll. Let 1
0
x
c
and 1
1
m
pb
yc
, and note that the relations in this group
are 11
101 0
ccc c
and 11
2
10
1
n
pb
cc
. Write 21
l
ps
and observe that
 
11 111
21
22 422
11 111
l
pss
p
bpbpbpbpb
cc ccc
 ,
since 1
4
11
pb
c
. Therefore

111 111
22 2
1010 10
1
l
nnn
p
pb pbpb
cccc cc






.
Clearly
preserves the other relation, showing that
is a homomorphism.
Since 2
2
l
pb
divides 1
1
l
pb
and 2
divides 1
, it fol-
lows that the inclusion homomorphisms
and
are
defined as follows:

2
21
1
xx

and
11
22
21
21
l
l
ll
p
b
mm
p
b
pb pb
yy




,

2
21
1
xx

and
1
2
21
b
mm
b
pb pb
yy




.
One can easily check that
 
2
21 2
1
12
ll
x
xx
 
 


and

1
22
22 2
12
()
l
ll
mm
pm
pb bpb
ll
yyy
 


 


,
which verifies that our diagram commutes.
J. E. KALLIONGIS, R. OHASHI
Copyright © 2012 SciRes. APM
165
Proposition 47.

0,pmn is a sublattice of

0,
k
pm n, and there exists a lattice surjection


00
:, ,
k
pm npm n induced by the the family
of group homomorphisms
l
such that restricted
to

0,pm n is the identity.
Proof. It is clear that

0,pm n is a sublattice of

0,
k
pm n. If

0
π,,
k
bdpmn, then l
bpb
for some lk
. By Proposition 45, there exists a surjec-
tion :π,π,
l
l
nn
pb pb




. Define
π,π,
lnn
pb pb







 . By the commutative dia-
gram in Proposition 46, it follows that is order pre-
serving.
Theorem 48. Let 1
2i
km
j
i
i
mp
be the prime de-
composition. Then

1
2,
k
j
i
ipn
is a sublattice of

1
2,
i
km
j
i
ipn
, and there exists a lattice surjection

11
:2,2,
i
kk
m
jj
ii
ii
pn pn




induced by a family of group homomorphisms such that
restricted to

1
2,
k
j
i
ipn
is the identity.
Proof. Apply

11
:2, 2,
i
ir ir
m
mm
jj
i
r
rr iri
ik ik
ppn ppn

 



repeatedly defined in Proposition 47 for 1rk to
obtain the result where is the compositions of those
r
’s.
8. Some Examples
In this section we present several examples which illus-
trate the main theorems.
Example 49.

1315,14. This example illustrates
Theorem 12 that

1315,14 is isomorphic to

1315,14.
Example 50.
11155,11. This is a Boolean lattice/
algebra by Proposition 36 since 1155357 11.
Prime ideals are:
3,11 ,

5,11 and
7,11
Their complements are lters which are:
35,11
,
21, 11
and
15,11
respectively.
Example 51.
2126, 20. Since 200
 (mod 4),
this example illustrates Theorems 15 and 16 that
 
221
126,20126,2063,20 .
Example 52.
25040,20. This example again illus-
trates Theorem 16 and also that

25040,20 is iso-
morphic to
2126, 20.
J. E. KALLIONGIS, R. OHASHI
Copyright © 2012 SciRes. APM
166
Example 53.

25040,5. This illustrates Theorem 16
that

25040,5 is isomorphic to a disjoint union of
isomorphic lattices
 
11 11
63,5126,5252,5504,5 . 
Example 54.

25040,10. This example illustrates Theorem 16 that

31 1
1263,5 63,10
j
j

.
Example 55.

35030,10. This example illustrates
Theorems 18 and 19 that

331
5030,105030,1063,2510
 .
Example 56.

4126,5. This example illustrates
Theorem 21 that

41
126,10 126,5
.
Example 57.
5630,5. This example illustrates
Theorem 22 that

511
630,563,5126,5.
J. E. KALLIONGIS, R. OHASHI
Copyright © 2012 SciRes. APM
167
Example 58.

55040,10. This example illustrates Theorem 22 that
 
511
5040,101008,55040,10 .

Example 59.
663,5. This example illustrates Theo-
rems 19 and 23 that
  
611
63,517, 2934, 29 .
Example 60.

7126, 20. This illustrates Theorems
16 and 24 that

721
126,20126,2063,20 .
J. E. KALLIONGIS, R. OHASHI
Copyright © 2012 SciRes. APM
168
Example 61. This is an example ofcrushto illustrate
Theorem 42.
Apply 3
Apply 5
REFERENCES
[1] J. Kalliongis and A. Miller, “Orientation Reversing Ac-
tions on Lens Spaces and Gaussian Integers,” Journal of
Pure and Applied Algebra, Vol. 212, No. 3, 2008, pp.
652-667. doi:10.1016/j.jpaa.2007.06.022
[2] R. Stanley, “Enumerative Combinatorics Volume 1,” Wads-
worth & Brooks/Cole, New York, 1986.
[3] R. Ohashi, “The Isometry Groups on Prism Manifolds,
Dissertation,” Saint Louis University, Saint Louis, 2005.
[4] J. Kalliongis and R. Ohashi, “Finite Group Actions on
Prism Manifolds Which Preserve a Heegaard Klein Bot-
tle,” Kobe Journal of Math, Vol. 28, No. 1, 2011, pp. 69-
89.
[5] B. A. Davey and H. A. Priestley, “Introduction to Lattices
and Order,” Cambridge Mathematical Textbooks, Cam-
bridge University Press, Cambridge, 1990.