ON THE STRUCTURE OF INVOLUTIONS AND SYMMETRIC
SPACES OF QUASI DIHEDRAL GROUP

arXiv:1401.0112v3 [] 4 Jul 2017

ZAHID RAZA, IMRAN AND BIJAN DAVVAZ
Abstract. Let G = QD8k be the quasi-dihedral group of order 8n and θ be an
automorphism of QD8k of finite order. The fixed-point set H of θ is defined as
Hθ = Gθ = {x ∈ G | θ(x) = x} and generalized symmetric space Q of θ is given
by Qθ = {g ∈ G | g = xθ(x)−1 for some x ∈ G}.
The characteristics of the sets H and Q have been calculated. It is shown that
for any H and Q, H.Q 6= QD8k . The H-orbits on Q are obtained under different
conditions. Moreover, the formula to find the order of v-th root of unity in Z2k
for QD8k has been calculated. The criteria to find the number of equivalence
classes denoted by C4k of the involution automorphism has also been constructed.
Finally, the set of twisted involutions R = Rθ = { x ∈ G | θ(x) = x−1 } has been
explored.

1. Introduction
Serious work on groups generated by reflections began in the 19th century. The
finite subgroups of O(3) generated by isometric reflections on the 2-sphere (or equivalently, by orthogonal linear reflections on R3 ) had been determined by Möbius in
1852. In the second half of the 19th century, work also began on finite reflection
groups on S n for n > 2 (or equivalently, finite linear reflection groups on Rn+1 ).
By the late 19′ s other groups generated by isometric reflections across the edges of
polygons (with 3 or more edges) in the hyperbolic plane had been studied by Klein
and Poincare. The group of symmetries of such a root system was a finite group
generated by reflections had been shown by Weyl in 1925. This intimate connection
with the classification of semi-simple Lie groups cemented reflection groups into a
central place in mathematics. The two lines of research were united in (Coxeter,
1930). He classified discrete groups generated by reflections on the n-dimensional
sphere or Euclidean space. The notion of an abstract reflection group introduced
by Jacques Tits which he called a coxeter group. He considered pairs (W, S) with
W a group and S a set of involutions which generate W so that the group has a
presentation of the following form: the set of generators is S and the set of relations
2010 Mathematics Subject Classification. 20F28 .
Key words and phrases. Automorphisms, involutions, fixed-point set, symmetric spaces, quasidihedral group.
1

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ZAHID RAZA, IMRAN AND BIJAN DAVVAZ

is {(st)m(s,t) }, where m(s, t) denotes the order of st and the relations range over all
unordered pairs s, t ∈ S with m(s, t) 6= ∞. The pair (W, S) is a coxeter system
and W is a coxeter group. This was demonstrated in (Bourbaki, 1968; Bjorner and
Francesco, 2005).
The early work by Helminck on symmetric spaces concerned some algebraic results
about their structure(Helminck, 1988). Classification of the real symmetric spaces
and their fine structure had been included in (Helminck, 1994). By considering an
open subgroup H of the fixed point group Gθ of an involution θ of G, the result was
proved that ‘‘ For k a field of characteristic not two, let G be a connected semi-simple
algebraic k-group and θ an involution of G. Then θ is defined over k if and only if
H θ is defined over k” (Helminck and Wang, 1993). They also proved that ‘‘ if G is a
connected semi-simple algebraic group, θi involution of G and Hi = Gθi , (i = 1, 2).
Let H1θ1 and H2θ2 be the identity components of H1 and H2 respectively. If H1θ1 = H2θ2
then θ1 = θ2 .”
A mini-workshop on generalizations of symmetric spaces was organized by Ralf
Kohl at.el in 2012. Mathematicians in this area brought together to present their current research projects or trigger new collaboration via comparison, analogy, transfer,
generalization and unification of method in this workshop.
The structure of automorphism group of dihedral group has been found by Cunningham at.el (2012). They also characterized the automorphism group with θ2 = id
and calculated H and Q for each automorphism for the family of finite dihedral
groups. They also proved that the set of involutions is partitioned in such a way
that there are at most two distinct equivalence classes of involutions in each of the
partition. In involution case, Q is always a subgroup of G. Moreover, the infinite
dihedral group is also discussed.
Let G = Dcn be the dicyclic group of order 4n. Let φ be an automorphism of G
of order k. Bishop at.el (2013) describe φ and the generalized symmetric space Q of
G associated with φ. When φ is an involution, they describe its fixed point group H
along with the H-orbits and G-orbits of Q corresponding to the action of φ-twisted
conjugation.
In this paper, we extent the study to quasi-dihedral group QD8k .

2. Preliminaries
The quasi-dihedral group is denoted by QD8k , where QD8k = ha, b : a4k = b2 =
1, ba = a2k−1 bi. In reset of the whole paper, we denote it by QD2m , where m = 4k
and n = m2 = 2k. That is QD2m = ha, b : am = b2 = 1, ba = an−1 bi. It is
clear from the presentation that ai bj ∈ QD2m is in normal form for some integer
0 ≤ i < m and j ∈ {0, 1}. In the whole study, Zn denotes the additive group
of integers modulo n and Um denotes the multiplicative group of units of Zn and

ON THE STRUCTURE OF INVOLUTIONS AND SYMMETRIC SPACES OF QUASI DIHEDRAL GROUP
3

r ∈ Um if and only if gcd(r, m) = 1. The v-th root of unity in Zn which can be
v
described by Rm
= {r ∈ Um | r v = 1 (mod n)}.
Let G be a group, then an element a is a conjugate with b in G if and only
if there exists an element x such that a = xbx−1 . For a ∈ G, the function
φa (x) = axa−1 ∀ x ∈ G is called an inner automorphism of G. Let θ1 , θ2 be
two automorphisms, then θ1 ∼ θ2 if and only if they are conjugate to each other. If
there exists another automorphism σ such that σθ1 σ −1 = θ2 .
Definition 3. For any c ∈ Zn , we define ZDiv (c) = {y ∈ Zn | cy ≡ 0 (mod n)} and
easy to see that | ZDiv (c) |= gcd(c, n)
Notation 4. Let us fixed the following notations, which will be used in the whole
paper.
v
• For fixed r ∈ Rm
. The number of equivalence classes with fixed leading
co-efficient r is denoted by Nr =| {θ̄ | θ = rx + s ∈ Autv (D2m )} |
• Let us consider the subset of automorphism denoted by Autv (QD2m ) and is
defined as Autv (QD2m ) = {θ ∈ Aut(QD2m ) | θv = id }.
The following lemma is easy to prove.
Lemma 5. If QD2m be the quasi-dihedral group, then
(1) bai = ai(n−1) b for all i = 0, 1, . . . , (m − 1),
(2) | a2i b |= 2 and | a2i+1 b |= 4 for all i = 0, 1, . . . , (n − 1).
Lemma 6. The automorphism group of QD2m is isomorphic to the group of affine
linear transformations of Zn . In particular, we have
Aut(QD2m ) ∼
= Af f (Zn ) = {rx + s : Zn → Zn | r ∈ Um , s ∈ Zn }
and the action of rx + s on elements of QD2m in normal form is given by
(6.1)

(rx + s)(aj bt ) = arj+[(t−1)n−(t−2)]s bt

Proof. Let θ be an automorphism, then θ(a) = ar and θ(b) = (as b), where gcd(r, m) =
1, s = 0, 2, 4, . . . , (m − 2). Consider arbitrary element aj bt ∈ QD2m , and θ being
homomorphism,
(6.2)

θ(aj bt ) = θ(aj )θ(bt ),

then θ(aj ) = arj and θ(bt ) = (as b)(as b)(as b) . . . (as b) = as (bas )(bas )b . . . (as b).
By Lemma 5, we have
θ(bt ) = as [as(n−1) b](bas )(bas )b . . . (as b) = as+s(n−1) (b2 as )(bas )b . . . (as b)
θ(bt ) = asn (b2 as )(bas )b . . . (as b) = asn (as(n−1) b2 )(bas )b . . . (as b)
θ(bt ) = asn+sn−sb3 (bas )b . . . (as b) = a2sn−s b3 (bas )b . . . (as b) after t-steps, we have
θ(bt ) = a(t−1)ns−(t−2)s bt . Put values in equation 6.2, we get
θ(aj bt ) = arj .a(t−1)ns−(t−2)s bt = arj+(t−1)ns−(t−2)s bt = arj+[(t−1)n−(t−2)]s bt . Since b2 =

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ZAHID RAZA, IMRAN AND BIJAN DAVVAZ

1, so t ∈ {0, 1}. If t = 0, then θ(aj ) = arj+(2−n)s =⇒ rx + (2 − n)s ∈ Af f (Zn ). If
t = 1, then θ(aj b) = arj+s b =⇒ rx + s ∈ Af f (Zn ).
Conversely, For f ∈ Af f (Zn ), that is f = rx + s where r ∈ Um , s ∈ Zn and it is
given as a group action f (aj bt ) = arj+[(t−1)n−(t−2)]s bt for all aj bt ∈ QD2m .
Now, we will show that f is an automorphism. For this let us define θ : QD2m −→
QD2m by θ(aj bt ) = f (aj bt ) = arj+[(t−1)n−(t−2)]s bt . We will show that θ is a bijective
homomorphism. For this we will show that θ is one to one and onto.
one to one: Suppose that θ(aj1 bt1 ) = θ(aj2 bt2 ), then arj1 +[(t1 −1)n−(t1 −2)]s bt1 =
arj2 +[(t2 −1)n−(t2 −2)]s bt2 ⇒ arj1 +[(t1 −1)n−(t1 −2)]s−rj2−[(t2 −1)n−(t2 −2)]s bt1 −t2 = e.
Case 1 :
rj1 + [(t1 − 1)n − (t1 − 2)]s − rj2 − [(t2 − 1)n − (t2 − 2)]s = 0 and t1 − t2 = 2, since,
t ∈ {0, 1} therefore t1 − t2 6= 2 not possible.
Case 2 :
rj1 + [(t1 − 1)n − (t1 − 2)]s − rj2 − [(t2 − 1)n − (t2 − 2)]s = m and t1 − t2 = 0
⇒ r(j1 −j2 )+ns(t1 −t2 )−s(t1 −t2 ) = m ⇒ r(j1 −j2 ) = m ⇒ r | m but gcd(r, m) = 1,
not possible. Now r(j1 − j2 ) = m will be only possible when j1 − j2 = 0, r 6= 0
j1 = j2 ⇒ aj1 = aj2 ⇒ aj1 bt1 = aj2 bt2 ⇒ θ is one to one.
onto: It is cleared from the definition of the group action that for every arj+[(t−1)n−(t−2)]s bt
there exists aj bt ∈ QD2m such that θ(aj bt ) = arj+[(t−1)n−(t−2)]s bt ⇒ θ is onto.
Hence Aut(QD2m ) ∼
= Af f (Zn ). 
7. Structure of Automorphisms of QD2m
Proposition 7.1. For any integer v ≥ 1 and θ = rx + s be an automorphism of
v
QD2m then θv =id if and only if r ∈ Rm
and s ∈ ZDiv(r v−1 + r v−2 + · · · + r + 1).
Proof. We will use the mathematical induction on v to show first the following
statement: θv (x) = r v x + (r v−1 + r v−2 + · · · + r + 1)s.
For v = 2, 3 θ2 (x) = θ[θ(x)] = θ(rx + s) = r(rx + s) + s = r 2 x + rs + s =
r 2 x + (r + 1)s. Similarly, θ3 (x) = r 3 x + r 2 s + rs + s = r 3 x + (r 2 + r + 1)s.
Suppose that the statement is true for v = i that is
i
θ (x) = r i x + (r i−1 + r i−2 + · · · + r + 1)s. We will show that the statement is
true for v = i + 1. Take θi+1 (x) = θ[θi (x)] = θ[r i x + (r i−1 + r i−2 + · · · + r + 1)s]
θi+1 (x) = r[r i x + (r i−1 + r i−2 + · · · + r + 1)s] + s = r i+1 x + (r i + r i−1 + · · · + r + 1)s].
Hence , the statement is true for all positive values of v. We have identified the
identity automorphism with x ∈ Af f (Zn ), thus r v ≡ 1 (mod n) and (r v−1 + r v−2 +
v
· · · + r + 1)s ≡ 0 (mod n) ⇔ r ∈ Rm
and s ∈ ZDiv(r v−1 + r v−2 + · · · + r + 1) as
required. 
3
Example 7.2. Let G = QD56 and v = 3. In this case R28
= {1, 9, 11, 15, 23, 25},
take r = 9, then ZDiv(92 + 9 + 1) = ZDiv(91) = {0, 2, 4, 6, 8, 10, 12}. Now, consider

ON THE STRUCTURE OF INVOLUTIONS AND SYMMETRIC SPACES OF QUASI DIHEDRAL GROUP
5

θ1 = 9x+6 ∈ Aut(QD56 ), then after simple calculation we get θ13 = x =id. Similarly,
for θ2 = 9x + 12 ∈ Aut(QD56 ), we have θ23 = x =id.
Proposition 7.3. For any m and v ≥ 1, we have
X
gcd(r v−1 + r v−2 + · · · + r + 1, n)
|Autv (QD2m )| =
r∈Rvm

v
Proof. This follows from above Proposition: for any r ∈ Rm
there are

|ZDiv(r v−1 + r v−2 + · · · + r + 1)| = gcd(r v−1 + r v−2 + · · · + r + 1, n)
element s such that (r v−1 +r v−2 +· · ·+r +1)s ≡ 0 (mod n) and every automorphism
θ ∈ Autv (QD2m ) must be of this form. 
Proposition 7.4. Let θ1 = r1 x+s1 and θ2 = r2 x+s2 be two different automorphism
of QD2m . Then θ1 ∼ θ2 if and only if r1 = r2 and f s1 − s2 ∈ hr1 − 1i ≤ Zn for some
f ∈ Um .
Proof. Given that θ1 = r1 x + s1 and θ2 = r2 x + s2 ∈ Aut(QD2m ). Suppose that
σ = f x + g be any arbitrary automorphism of QD2m and define σ −1 = f −1 x − f −1 g.
We will check that σ ·σ −1 = x. For this take σ ·σ −1 = σ(σ −1 (x)) = σ(f −1 x−f −1 g) =
f (f −1 x − f −1 g) + g = x − g + g = x. Now, we calculate
(7.1)

σθ1 σ −1 = σ · (θ1 · σ −1 )

Take θ1 · σ −1 = θ1 [σ −1 (x)] = θ1 [f −1 x − f −1 g] = r1 (f −1 x − f −1 g) + s1 = r1 f −1 x −
r1 f −1 g + s1 put in 7.1, we get σθ1 σ −1 = σ(r1 f −1 x − r1 f −1 g + s1 ) = f (r1 f −1 x −
r1 f −1 g+s1 )+g σθ1 σ −1 = r1 x−r1 g+f s1 +g = r1 x+f s1 −g(r1 −1). Now, by definition
σθ1 σ −1 = θ2 if and only if r1 = r2 and f s1 − g(r1 − 1) = s2 ⇒ f s1 − s2 = g(r1 − 1)
⇒ f s1 − s2 ∈ hr1 − 1i. 
Example 7.5. Let G = QD64 and consider θ1 = 7x + 3, θ2 = 7x + 14, θ3 = 7x + 11
three automorphisms of QD64 . Then it is easy to check that θ1 ≁ θ2 because there
does not exist such f ∈ U32 for which 3f − 14 ∈ h6i. And θ1 ∼ θ3 because there
exist some f ∈ U32 for which 3f − 14 ∈ h6i.
Proposition 7.6. Suppose m is fixed and let v ≥ 1.
v
(1) For any r ∈ Rm
; hr − 1i ≤ ZDiv(r v−1 + r v−2 + · · · + r + 1).
v−1 +r v−2 +···+r+1)
v
(2) For any r ∈ Rm
; Um acts on the cosets ZDiv(r hr−1i
.
(3) The set Autv (QD2m ) is partitioned into equivalence classes indexed by pair
v−1 +r v−2 +···+r+1)
v
(r, A), where r ∈ Rm
and A is an orbit of Um on ZDiv(r hr−1i
.
Proof.

(1) We know that (r − 1).(r v−1 + r v−2 + · · · + r + 1) = r v − 1 ≡ 0 (mod n)
⇒ (r − 1).(r v−1 + r v−2 + · · · + r + 1) ≡ 0 (mod n) ⇒ (r − 1) ∈ ZDiv(r v−1 +
r v−2 + · · · + r + 1) ⇒ hr − 1i ≤ ZDiv(r v−1 + r v−2 + · · · + r + 1).

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ZAHID RAZA, IMRAN AND BIJAN DAVVAZ

(2) We simply recognize that Um = Aut(Zn ) acts on Zn by multiplication and
since every subgroup is cyclic, Um must stabilize the subgroups of Zn .
v−1 +r v−2 +···+r+1)
.
(3) It is simply Proposition 7.4 in terms of the Um -action on ZDiv(r hr−1i

Example 7.7. Let G = QD40 ,and v = 2. Then U20 = {1, 3, 7, 9, 11, 13, 17, 19} and
Z10 = {0, 1, 2, . . . , 9}.
2
(1) Since v = 2, so R20
= {1, 9, 11, 19}. Take r = 9 then h9 − 1i = h8i =
{0, 2, 4, 6, 8} and ZDiv(10) = {0, 1, 2, . . . , 9}. So, {0, 2, 4, 6, 8} ⊆ {0, 1, 2, . . . , 9}
(2) For U20 action on the cosets ZDiv(r+1)
. Take r = 9 then ZDiv(r+1)
= ZDiv(10)
=
hr−1i
hr−1i
h8i
E={0,1,2,...,9}
F ={0,2,4,6,8}

= x + F ∀ x ∈ E provide two sets F and F1 = {1, 3, 5, 7, 9}.

Now, the action of U20 on the cosets
(3)

ZDiv(10)
h8i

has two sets.

ZDiv(10)
h8i

= y.F =

F ∀ y ∈ U20 and ZDiv(10)
= y.F1 = F1 ∀ y ∈ U20
h8i
ZDiv(10)
The action of U20 on h8i = F, F1 . Then the number of equivalence classes
2
of Aut(QD40 ) indexed by the pair (r, A) where r ∈ R20
and A is an orbit of
ZDiv(10)
U20 on h8i is 8.

v
Theorem 7.8. If m is a fixed positive integer and r ∈ Rm
, then the number
v−1
v−2
ZD iv(r
+r
+···+r+1)
of orbits of Um on
is equal to the number of divisors of
hr−1i
gcd(r−1,n). gcd(r v−1 +r v−2 +···+r+1,n)
.
n

Proof. The Um -orbit on Zn are indexed by the subgroups of Zn . Thus the Um -orbit
v−1 +r v−2 +···+r+1)
on ZDiv(r hr−1i
are indexed by subgroups L ≤ Zn such that
hr − 1i ≤ L ≤ ZDiv(r v−1 + r v−2 + · · · + r + 1).
It is well known that the subgroup lattice of Zn is isomorphic to the divisor lattice
of n. The subgroup hr − 1i corresponds to the divisor gcd(r − 1, n) and the subgroup
n
,
ZDiv(r v−1 + r v−2 + · · · + r + 1) correspond to the divisor gcd(rv−1 +rv−2
+···+r+1,n)
and the subgroups between these groups correspond to the divisor of n between
n
gcd(r − 1, n) and gcd(rv−1 +rv−2
in the divisor lattice. Finally, it is known
+···+r+1,n)
that this sub-lattice of the divisor of n is isomorphic to the divisor lattice of
gcd(r − 1, n)
gcd(r − 1, n). gcd(r v−1 + r v−2 + · · · + r + 1, n)
=
.
n
n
gcd(r v−1 +r v−2 +···+r+1,n)

8. Fixed Ssts and Symmetric Spaces of Automorphisms
In this section if θ is an automorphism the fixed-point set Hθ and the symmetric
space Qθ will be discussed, where Hθ = {x ∈ G | θ(x) = x} and Qθ = {g ∈ G | g =

ON THE STRUCTURE OF INVOLUTIONS AND SYMMETRIC SPACES OF QUASI DIHEDRAL GROUP
7

xθ(x)−1 for some x ∈ G}. When θ is understood to be fixed the subscript from our
notation has been dropped.
The following theorem characterizes the sets H and Q in the case of quasi-dihedral
groups.
Theorem 8.1. If θ = rx + s is an automorphism of G = QD2m of finite order, then
′
′
H = {aj | j(r + 2s − 1) ≡ 0 (mod n)} ∪ {aj b | j (r + 2s − 1) ≡ −s (mod n)} and
′
Q = {aj | j ∈ hr + 2s − 1i ∪ [l(1 − r) − s] (mod n)} where, j and l are even numbers.
Proof. Since, θ is an automorphism and θ(aj bt ) = arj+[(t−1)n−(t−2)]s bt . First we will
find set the H. Let us consider aj ∈ G, then by definition of H, we have θ(aj ) = aj ,
θ(aj ) = a[r+(2−n)s]j = aj , ⇒ n | j(r + 2s − 1).
′
′
′
′
′
Now consider aj b ∈ G, then θ(aj b) = aj b so, aj b = a[r+(2−n)s]j as b, thus n |
′
′
′
(r+2s−1)j +s. Hence, H = {aj | j(r+2s−1) ≡ 0 (mod n)}∪{aj b | j (r+2s−1) ≡
′
−s (mod n)}, where j is even number.
Now we will find the set Q. L aj ∈ G be an arbitrary element and let ai be a fixed
element of G, where j 6= i, then by definition of Q, aj = ai θ(ai )−1 = ai θ(a−i ) =
ai .a−[r+(2−n)s]i = ai[1−r−(2−n)s] ⇒ j ≡ i[1 − (r + 2s)] (mod n) ⇒ j ∈ hr + 2s − 1i.
Now consider al b ∈ G, where l is even number, then aj = (al b)θ(al b)−1 =
(al b)θ(al b) = (al b)(arl+s b) = al (barl+s )b = al a(rl+s)(n−1) bb = al+(rl+s)(n−1) ⇒ j =
l + (rl + s)(n − 1) ⇒ j = l[1 + r(n − 1)] + s(n − 1) ⇒ j ≡ l(1 − r) − s (mod n).
So, Q = {aj | j ∈ hr + 2s − 1i ∪ [l(1 − r) − s] (mod n)}, where l is an even
numbers. 
Example 8.2. Let us consider G = QD64 and θ = 3x + 12 ∈ Aut(D64 ), then
′
′
H = {aj | 10j ≡ 0 (mod 16)} ∪ {aj b | 10j ≡ −4 (mod 16)} = {1, a8 , a2 b, a10 b} and
′
Q = {aj | j ∈ h10i ∪ (14j + 4) (mod 16)} = {1, a2 , a4 , . . . , a14 }
Using the descriptions of H and Q , we obtain further results as well.
Corollary 8.3. If θ = rx + s is an automorphism of QD2m of finite order, where
m = 4k, n = 2k and k is even, then H will be cyclic or non-cyclic according to the
following conditions
(1) Let gcd(r + 2s − 1, n) = d1 . If s = 2w + 1 is odd and d1 ∤ w then H will be
cyclic that is H ∼
= ZDiv(r + 2s − 1) and | H |= d1
(2) Let gcd(r + 2s − 1, k) = d2 . If s = 2w is even and d2 | w then we have two
cases
a): If w is odd, then H will be cyclic | H |= d1 and H ∼
= ZDiv(r +2s−1).
b): If w is even
i): If d2 ∤ w then H will be cyclic | H |= d1 and H ∼
= ZDiv(r +
2s − 1).
ii): If d2 | w then H will be non-cyclic. We have two cases

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ZAHID RAZA, IMRAN AND BIJAN DAVVAZ
k

I-): if 4 ∤ k then | H |= 23 d1 H = {ha 2 i, a2l b, a2q b | l 6=
q for some l, q ∈ {1, 2, ...k}}. It is not a subgroup of G.
II-): if 4 | k then | H |= 2d1 and H ∼
= ZDiv(r + 2s − 1) × Z2 .
Example 8.4. Let G = QD48 , then gcd(r, m) = 1 ⇒ r ∈ U24 = {1, 5, 7, 11, 13, 17, 19, 23}
and s ∈ Z12 = {0, 1, 2, . . . , 11}.
(1) Consider θ1 = 7x + 11 ∈ Aut(QD48 ) then r = 7, s = 2(5) + 1 = 11
⇒ r + 2s − 1 = 4 and d1 = gcd(28, 12) = 4 ⇒ 4 ∤ 5 then H1 = {1, a3, a6 , a9 }
is cyclic. and ZDiv4 = {0, 3, 6, 9} ⇒ H1 ∼
= ZDiv4 and O(H1) = d1 = 4
(2) a): Take θ2 = 11x + 10 ∈ Aut(QD48 ) then r = 11, s = 2(5) = 10
⇒ r + 2s − 1 = 6 and d1 = gcd(6, 12) = 6, d2 = gcd(6, 6) = 6 then
H2 = {1, a2 , a4 , . . . , a10 } is cyclic and ZDiv6 = {0, 2, 4, 6, 8, 10} ⇒ H2 ∼
=
ZDiv6 and O(H2) = d1 = 6
b):
i): For θ3 = 11x + 4 ∈ Aut(QD48 ) then r = 11, s = 2(2) = 4
⇒ r + 2s − 1 = 6 and d1 = gcd(6, 12) = 6, d2 = gcd(6, 6) = 6
⇒ 6 ∤ −2 then H3 = {1, a2 , a4 , . . . , a10 } is cyclic and ZDiv6 =
{0, 2, 4, 6, 8, 10} ⇒ H3 ∼
= ZDiv6 and O(H3 ) = d1 = 6
ii): I-: For θ4 = 5x + 8 ∈ Aut(QD48 ) then r = 5, s = 2(4) = 8
⇒ r + 2s − 1 = 8 and d1 = gcd(8, 12) = 4, d2 = gcd(8, 6) = 2
⇒ 2 | −4 then H4 = {1, a3 , a6 , a9 , a2 b, a8 b} is non-cyclic and
O(H4 ) = 32 .d1 = 6
II-: Take θ5 = 7x + 4 ∈ Aut(QD48 ) then r = 7, s = 2(2) = 4
⇒ r + 2s − 1 = 2 and d1 = gcd(2, 12) = 2, d2 = gcd(2, 6) =
2 ⇒ 2 | −2 then H5 = {1, a6 , a4 b, a10 b} is non-cyclic and
O(H5 ) = 2.d1 = 4
Corollary 8.5. If m = 4k, n = 2k and k is odd, then H will be cyclic or non-cyclic
according to the following conditions
(1) Let gcd(r + 2s − 1, n) = d1 . If s = 2w + 1 is odd then H will be cyclic
| H |= d1 and H ∼
= ZDiv(r + 2s − 1).
(2) Let gcd(r + 2s − 1, n) = d1 . If s = 2w is even then
a): If d1 = n then H will be cyclic | H |= d1 = n and H ∼
= ZDiv(r +
2s − 1).
b): If d1 = 2 then H will be non-cyclic | H |= 23 d1 = 23 (2) = 3 and
H = {1, ak , a2l b} for some l ∈ {1, 2, ..., k} is not a subgroup of G.
Example 8.6. Let G = QD40 where m = 20, n = 10, k = 5. Since gcd(r, m) = 1
⇒ r ∈ U20 = {1, 3, 7, 9, 11, 13, 17, 19} and s ∈ Z10 = {0, 1, 2, . . . , 9}.
(1) Consider θ1 = 3x + 5 ∈ Aut(QD40 ) then r = 3, s = 5 ⇒ r + 2s − 1 = 2
and gcd(2, 10) = 2 then H1 = {1, a5 } is cyclic and ZDiv2 = {0, 5} ⇒ H1 ∼
=
ZDiv2 and | H1 |= 2

ON THE STRUCTURE OF INVOLUTIONS AND SYMMETRIC SPACES OF QUASI DIHEDRAL GROUP
9

(2)

a): For θ2 = 7x + 2 ∈ Aut(QD40 ) then r = 7, s = 2 ⇒ r + 2s − 1 = 0
and gcd(0, 10) = 10 then H2 = {1, a2 , a3 , . . . , a9 } is cyclic and ZDiv0 =
{0, 1, 2, . . . , 9} ⇒ H2 ∼
= ZDiv0 and | H2 |= 10
b): Take θ3 = 7x+4 ∈ Aut(QD40 ) then r = 7, s = 4 ⇒ r+2s−1 = 4 and
gcd(4, 10) = 2 then H3 = {1, a5 , a4 b} is non-cyclic and | H3 |= 32 .2 = 3

Corollary 8.7. If θ = rx + s ∈ Autv (QD2m ) be of finite order, where m = 4k, n =
2k and k is even number, then Q will be cyclic or non-cyclic according to the
following conditions
(1) If s = 2w + 1 is odd, then Q will be non-cyclic.
(2) Let gcd(r + 2s − 1, n) = d1 . If s = 2w is even, then
′
a): If d1 | (1 − r)j − s and
i): if r + 2s − 1 6≡ k (mod n) then Q will be cyclic.
ii): If r + 2s − 1 ≡ k (mod n) we have two cases
• if 1 − r ≡ k (mod n) then Q will be cyclic.
• If 1 − r 6≡ k (mod n) then Q will be non-cyclic.
′
b): If d1 ∤ (1 − r)j − s and
i): if −s 6≡ k (mod n) then Q will be non-cyclic.
ii): If −s ≡ k (mod n) then Q will be cyclic.
Example 8.8. Let G = QD48 where m = 24, n = 12, k = 6. Since gcd(r, m) = 1
⇒ r ∈ U24 = {1, 5, 7, 11, 13, 17, 19, 23} and s ∈ Z12 = {0, 1, 2, . . . , 11}.
(1) Consider θ1 = 7x + 11 ∈ Aut(QD48 ) then r = 7, s = 11 ⇒ r + 2s − 1 = 4
then Q1 = {1, a, a4 , a8 } is non-cyclic.
(2) a):
i): For θ2 = 5x + 8 ∈ Aut(QD48 ) then r = 5, s = 8 ⇒ r + 2s − 1 =
′
′
8 6≡ k and d1 = gcd(8, 12) = 4, (1 − r)j − s = −4j − 8 at
′
j = 2 yeilds 4 | −16 then Q2 = {1, a4 , a8 } is cyclic.
ii): Take θ3 = 7x + 6 ∈ Aut(QD48 ) then r = 7, s = 6 ⇒ r + 2s − 1 =
′
′
6 ≡ k and d1 = gcd(6, 12) = 6, (1 − r)j − s = −6j − 6 at
′
j = 2 yeilds 6 | −18 and 1 − r = −6 ≡ 6 ≡ k then Q3 = {1, a6 }
is cyclic. And for θ4 = 11x + 4 ∈ Aut(QD48 ) then r = 11, s = 4
′
⇒ r + 2s − 1 = 6 ≡ k and d1 = gcd(6, 12) = 6, (1 − r)j − s =
′
′
−10j − 4 at j = 2 yeilds 6 | −24 and 1 − r = −10 ≡ 2 6≡ k then
Q4 = {1, a4 , a6 , a8 } is non-cyclic.
b):
i): Consider θ5 = x + 10 ∈ Aut(QD48 ) then r = 1, s = 10 ⇒
′
′
r + 2s − 1 = 8 and d1 = gcd(8, 12) = 4, (1 − r)j − s = 0j − 10 ≡
2 ⇒ 4 ∤ 2 and −s = −10 ≡ 2 6≡ k then Q5 = {1, a2 , a4 , a8 } is
non-cyclic.
ii): For θ6 = 5x+ 6 ∈ Aut(QD48 ) then r = 5, s = 6 ⇒ r + 2s −1 = 4
′
′
′
and d1 = gcd(4, 12) = 4, (1 − r)j − s = −4j − 6 for any j ,

10

ZAHID RAZA, IMRAN AND BIJAN DAVVAZ
′

d1 ∤ −4j −6 and −s = −6 ≡ 6 ≡ k then Q6 = {1, a2 , a4 , . . . , a10 }
is cyclic.
Corollary 8.9. If m = 4k, n = 2k and k is odd number, then Q will be cyclic or
non-cyclic according to the following conditions
(1) If s = 2w is even then Q will be cyclic.
(2) If s = 2w + 1 is odd
a): Q will be non-cyclic if either r + 2s − 1 ≡ 0 (mod n) or 1 − r ≡
0 (mod n).
b): Q will be cyclic
i): if 1 − r ≡ 0 and s ≡ k (mod n).
ii): if r + 2s − 1 6≡ 0 and 1 − r 6≡ 0 (mod n).
Example 8.10. Let G = QD40 where m = 20, n = 10, k = 5. Since gcd(r, m) = 1
⇒ r ∈ U20 = {1, 3, 7, 9, 11, 13, 17, 19} and s ∈ Z10 = {0, 1, 2, . . . , 9}.
(1) For θ1 = 7x + 6 ∈ Aut(QD40 ) then r = 7, s = 6 ⇒ r + 2s − 1 = 8 then
Q1 = {1, a2 , a4 , a6 , a8 } is cyclic.
(2) a): Take θ2 = x + 7 ∈ Aut(QD40 ) then r = 1, s = 7 ⇒ r + 2s − 1 =
4 and 1 − r = 0 then Q2 = {1, a2 , a3 , a4 , a6 , a8 } is non-cyclic.
b):
i): Consider θ3 = 11x + 5 ∈ Aut(QD40 ) then r = 11, s = 5
⇒ r + 2s − 1 = 0, 1 − r = 0, s = 5 ≡ k then Q3 = {1, a5 } is
cyclic.
ii): For θ4 = 7x + 5 ∈ Aut(QD40 ) then r = 7, s = 5 ⇒ r + 2s − 1 =
6, 1 − r = −6 6≡ 0 then Q4 = {1, a, a2 , . . . , a9 } is cyclic.
Corollary 8.11. Let θ = rx + s ∈ Autv (QD2m ) be of finite order, where m =
4k, n = 2k and k is even number. Then Q is isomorphic to a cyclic group as
follows:
(1) If | Q |= 2 then Q = hak : an = 1i and Q ∼
= hki.
k
(2) If | Q |= 2 then
a): Let gcd(r + 2s − 1, n) = d1 . If n ∤ r + 2s − 1 then Q = had1 : an = 1i
and Q ∼
= hd1 i.
b): If n | r + 2s − 1 and gcd(r + 2s − 1, n) = n then Q = ha4 : an = 1i
and Q ∼
= h4i.
(3) If | Q |= k and gcd(r + 2s − 1, n) = d1 then
′
a): if (r + 2s − 1) ∩ (1 − r)j − s = ∅ then Q = had1 /2 : an = 1i
and Q ∼
= h d21 i.
′
b): if (r+2s−1)∩(1−r)j −s 6= ∅ then Q = had1 : an = 1i and Q ∼
= hd1 i.
Corollary 8.12. Let θ = rx + s ∈ Autv (QD2m ) be of finite order, if k is odd
number. Then the conditions for Q to be cyclic and isomorphism relationship are:
(1) If s is even, then

ON THE STRUCTURE OF INVOLUTIONS AND SYMMETRIC SPACES OF QUASI DIHEDRAL GROUP
11

a): if r = 1, s = 0 | Q |= 1 then Q = {1} and Q ∼
= h0i.
2
n
∼
b): if | Q |= k then Q = ha : a = 1i = h2i.
(2) If s is odd, then
a): Let n | 1 − r, n | s − k. If | Q |= 2 then Q = hak : an = 1i ∼
= hki.
b): if r + 2s − 1 6≡ 0 and 1 − r 6≡ 0 (mod n) and | Q |= n then Q =
ha : an = 1i ∼
= h1i.
Corollary 8.13. Let θ = rx + s ∈ Aut(QD2m ) be a fixed automorphism. Then for
any H and Q, HQ =
6 QD2m .
Example 8.13.1. Consider the involution θ1 = 17x + 8 of QD48 . Then H1 =
{1, a3 , a6 , a9 , a2 b, a8 b} and Q1 = {1, a4 , a8 } ⇒ H1 Q1 6= QD48 . Now, consider
the automorphism θ2 = 5x + 3 of order 4. Then H2 = {1, a6 b} and Q2 =
{1, a2 , a4 , . . . , a10 , a, a5 , a9 } ⇒ H2 Q2 6= QD48 .
Proposition 8.14. Let θ = rx + s ∈ Autv (QD2m ) and for any al ∈ Q.
′
i): If l is odd, then H\Q = {{aj } | j ∈ hr + 2s − 1i ∪ [(1 − r)j − s] (mod n)}
′
ii): If l is even then H\Q = {{aj , a−j } | j ∈ hr+2s−1i∪[(1−r)j −s] (mod n)}
′
where, j is any even number. In either case G\Q = {Q}, that is there is only a
single G-orbit on Q.
Proof. Since, H is fixed by θ thus the action of H on Q is simply by conjugation.
Notice that Q ⊆ hai ≤ QD2m so, the action of H on general element al of Q. Let
ai ∈ H then, ai .al .a−i = ai+l−i = al . So, hai fixes Q point wise.
′
′
′
′
′
′
′
Suppose ai b ∈ H, then (ai b).al .(ai b)−1 = ai (bal )b−1 a−i = ai (al(n−1) b)b−1 a−i =
′
′
′
l
l
ai +l(n−1)−i = al(n−1) = aln .a−l = (a2n ) 2 .a−l = (am ) 2 .a−l = a−l so, aj b takes
′
al to a−l ⇒ {aj , a−j } is an orbit ⇔ aj b ∈ H which is true ⇔ l is even.
Now, for every element of Q is in G-orbit of 1 ∈ Q. Since, every element of Q are
of the form al ∈ Q. and G has two type of elements ai and ai b. Now let us supposed
that q1 = al = ai .θ(ai )−1 for some ai ∈ G, thus q1 = ai .θ(a−i ) = ai .a−ri+[2−n]y =
ai−ri+[2−n]y = a−i(r−1)+(2−n)y = av(r−1)+(2−n)y , where v = −i.
Also, al = (ai b).θ(ai b)−1 for some ai b ∈ G ⇒ al = (ai b).θ(ai b) = (ai b).(ari+y b) =
i
a (bari+y )b = ai .a(ri+y)(n−1) b.b = ai+(ri+y)(n−1) b2 Let us denote this element q2 =
al = a−v+(y−vr)(n−1) , where v = −i.
For a−v ∈ G, we have a−v .1.θ(a−v )−1 = a−v .θ(av ) = a−v .arv+(2−n)y = av(r−1)+(2−n)y =
q1 . Now for a−v b ∈ G, we have (a−v b).1.θ(a−v b)−1 = (a−v b).θ(a−v b) = (a−v b).a−rv+y b =
a−v (ba−rv+y )b = a−v .a(−rv+y)(n−1) b.b = a−v+(y−rv)(n−1) = q2 .
Therefore, for any q ∈ Q there exist g ∈ G such that g.1.θ(g)−1 = q. Hence
G\Q = {Q}, there is a single G-orbit on Q. 
Example 8.14.1. Revisiting θ1 , θ2 of Example 8.13.1 applying Proposition 8.14 to
obtain that for θ1 . Since all l are even, therefore H1 \Q1 = {{1}, {a4 , a−4 }, {a8, a−8 }},

12

ZAHID RAZA, IMRAN AND BIJAN DAVVAZ

for θ2 there exists even and odd powers of elements of Q. For all even powers
H2 \Q2 = {{1}, {a2 , a−2 }, . . . , {a10 , a−10 }, {a, a−1 }, {a5 , a−5 }, {a9 , a−9 }} and odd
powers H2 \Q2 = {{1}, {a2 }, . . . , {a10 }, {a}, {a5 }, {a9 }}.
9. Involutions in Aut(QD2m )
If θ = rx + s is an automorphism such that θ2 =id, then Proposition 7.1 gives
2
r ∈ Rm
and s ∈ ZDiv(r + 1).
Proposition 9.1. Suppose that m = 4pt11 pt22 , where the pi are distinct odd primes.
2
Then | Rm
|= 8
Proof. Since, m = 4pt11 pt22 ⇒ n = 2pt11 pt22 . Suppose that r ∈ Zn with r 2 ≡ 1 (mod n)
∵ gcd(r, m) = 1 ⇒ gcd(r, 4pt11 pt22 ) = 1. But m is even therefore, r is odd. Assume
that r = 2q + 1 for some 0 ≤ q < (n − 1). Now, r 2 = (2q + 1)2 = 4q 2 + 4q + 1
⇒ 4q 2 +4q ≡ 0 (mod n) ⇒ 4q(q+1) = l.(2pt11 pt22 ) ∵ gcd(4, 2pt11 pt22 ) = 2 ⇒ 2q(q+1) =
l.(pt11 pt22 ) (mod pt11 pt22 ). Either q or q + 1 is odd.
l
.pt11 pt22 ⇒ q = h.pt11 pt22 where 1 ≤ h ≤ 4. Now,
(1) If q is even, then q = 2(q+1)
r = 2q + 1 ⇒ r = 2h.pt11 pt22 + 1 ⇒ r ≡ 1(mod n)
(2) If q + 1 is even, then q + 1 = 2ql .pt11 .pt22 ⇒ q + 1 = h.pt11 pt22 ⇒ q = h.pt11 pt22 − 1
where 1 ≤ h ≤ 4. Now, r = 2q + 1 ⇒ r = 2h.pt11 pt22 − 1 ⇒ r ≡ −1(mod n)
⇒ r ≡ ±1(mod n),

t

Proposition 9.2. Suppose that m = 2α pt11 pt22 . . . pqq where the pi are distinct odd
primes. Then

 2q+1 α = 2
2
2q+2 α = 3
| Rm
|=
 q+3
2
α ≥ 4.

Proof. Similar to the Propositions 10.1. 

Example 9.3. Let G = QD192 , then U96 = {1, 5, 7, 11, 13, 17, 19, 23, 25, 29, 31, 35, 37,
2
41, 43, 47, 49, 53, 55, 59, 61, 65, 67, 71, 73, 77, 79, 83, 85, 89, 91, 95} and R96
= {1, 7, 17,
2
4
23, 25, 31, 41, 47, 49, 55, 65, 71, 73, 79, 89, 95} thus, | R96 |= 2 = 16
Corollary 9.4.
|Aut2 (QD2m )| =

X

gcd(r + 1, n).

r∈R2m
2
Theorem 9.5. Let r ∈ Rm
, then the followings hold:
(1) hr − 1i ≤ ZDiv(r + 1)
iv(r+1) |≤ 2
(2) Nr =| ZDhr−1i

ON THE STRUCTURE OF INVOLUTIONS AND SYMMETRIC SPACES OF QUASI DIHEDRAL GROUP
13

(3) Nr = 2 if and only if m = 2α l ; α ≥ 2, and l is odd, r ≡ ±1 (mod 2α−1 ).
2
(1) Given that r ∈ Rm
, then (r − 1).(r + 1) = r 2 − 1 ≡ 0 ⇒ (r − 1) ≤
ZDiv(r + 1) ⇒ hr − 1i ≤ ZDiv(r + 1).
(2) Suppose

Proof.

|

(9.1)

ZDiv(r + 1)
|= j
hr − 1i

n
It is well known that | hr − 1i |= gcd(r−1,n)
and | ZDiv(r + 1) |= gcd(r + 1, n)
put values in 9.1
gcd(r + 1, n)
=j
n
gcd(r−1,n)

(9.2)

⇒ gcd(r + 1, n). gcd(r − 1, n) = n.j
t

Now, suppose that m = 2α pt11 pt22 . . . pqq where α ≥ 2 ⇒ α − 1 ≥ 1
t
⇒ n = 2α−1 pt11 pt22 . . . pqq where pi are distinct odd primes and ti ≥ 1. Then
gcd(r + 1, n) = 2α1 pa11 pa22 . . . paq q
with α1 ≤ (α − 1) and 0 ≤ ai ≤ ti ∀ i and
gcd(r − 1, n) = 2α2 pb11 pb22 . . . pbqq
with α2 ≤ (α − 1) and 0 ≤ bi ≤ ti . Now, for all i ∈ {1, 2, . . . q} either ai = 0
or bi = 0. Indeed, if ai > 0 and bi > 0 then pi divides both (r + 1) and
(r − 1) which is impossible. ∵ (pi > 2)
Similarly, either α1 ≤ 1 or α2 ≤ 1. Otherwise 2min{α1 ,α2 } divides (r + 1)
and (r − 1) which is impossible.
Since, from 9.2
gcd(r + 1, n). gcd(r − 1, n) = n.j
(2α1 pa11 pa22 . . . paq q ).(2α2 pb11 pb22 . . . pbqq ) = j.2α−1 pt11 pt22 . . . ptqq
(2α1 +α2 pa11 +b1 pa22 +b2 . . . paq q +bq ) = j.2α−1 pt11 pt22 . . . ptqq
Now, since ∀ i either ai = 0 or bi = 0 ⇒ ai + bi = ti
t

2
(9.3)

α1 +α2

= j.2

α−1

pt11 pt22 . . . pqq
a +bq

pa11 +b1 pa22 +b2 . . . pq q

2α1 +α2 = j.2α−1 .
As α1 ≤ (α − 1) and α2 ≤ (α − 1) and since either α1 ≤ 1 or α2 ≤ 1,
⇒ (α − 1) ≤ (α1 + α2 ) ≤ α
equation 9.3 ⇒ j = 2α1 +α2 −α+1 ⇒ 1 ≤ j ≤ 2.

14

ZAHID RAZA, IMRAN AND BIJAN DAVVAZ

(3) By Theorem 7.8, Nr is the number of divisors of
gcd(r − 1, n). gcd(r + 1, n)
n
and Nr is the number of divisor of j computed above. Therefore,
Nr = 1 if j = 1 and Nr = 2 if j = 2.
In either case,
ZDiv(r + 1)
|
|≤ 2.
hr − 1i
Finally according to the proof of (2), j = 2 if and only if
α1 = α − 1, α2 = 1 or α1 = 1, α2 = α − 1.
If α1 = α − 1, α2 = 1 then 2α−1 divides (r + 1).
If α1 = 1, α2 = α − 1 then 2α−1 divides (r − 1).
⇒ r ≡ ±1 (mod 2α−1 ).

Example 9.6. Let G = QD48 where m = 23 .3, α = 3, then
2
U24 = {1, 5, 7, 11, 13, 17, 19, 23} and R24
= {1, 5, 7, 11, 13, 17, 19, 23}.
2
(1) Let r = 5 ∈ R24 then hr − 1i = h4i = {0, 4, 8} = B1 and
ZDiv(r + 1) = ZDiv6 = {0, 2, 4, 6, 8, 10} = B2 clearly, h4i ≤ ZDiv6.
iv(6) = {0,2,4,6,8,10} = B2 = x + B ∀ x ∈ B
(2) For r = 5, ZDh4i
1
2
{0,4,8}
B1
0 + B1 = {0, 4, 8} = B1 , 2 + B1 = {2, 6, 10} = B3
4 + B1 = {4, 8, 0} = B1 , 6 + B1 = {6, 10, 2} = B3
8 + B1 = {8, 0, 4} = B1 , 10 + B1 = {10, 2, 6} = B3
= {B1 , B2 } ⇒ Nr = N5 = 2.
U24 action on ZDiv6
h4i
3
(3) Since, m = 2 .3 ⇒ l = 3 > 1 is odd and r = 5 ≡ 1 (mod 22 ).
Thus N5 = 2 as calculated in (2).
t

Corollary 9.7. Suppose that m = 2α pt11 pt22 . . . pqq , where the pi are distinct odd
primes and α ≥ 2. Then the number of equivalence classes C4k of Aut2 (QD8k ) is
given by
 q+1
2
α=2
C4k =
q+2
2
α ≥ 3.
2
Example 9.8. Let G = QD384 . Thus R192
= {1, 17, 31, 47, 49, 65, 79, 95, 97, 113, 127,
2
|= 24 . It is easy to check that the values r =
143, 145, 161, 175, 191} ⇒| R192
1, 65, 97, 161 satisfy r ≡ 1 (mod 25 ) and the values r = 31, 95, 127, 191 satisfy
r ≡ −1 (mod 25 ). So, Cm =| {1, 31, 65, 95, 97, 127, 161, 191} |= 23 .

Now for θ an involution, the set of twisted involutions is given in the followings
porosities:

ON THE STRUCTURE OF INVOLUTIONS AND SYMMETRIC SPACES OF QUASI DIHEDRAL GROUP
15

Proposition 9.9. If θ2 =id, then the set of twisted involutions is given as
R = {ai | (r + 1)i ≡ −2s (mod n)} ∪ {ai b | (r − 1)i ≡ −s (mod n)}
2
Proof. Since, θ is an involution therefore, r ∈ Rm
and s ∈ ZDiv(r + 1). To find the
i
i
i −1
ri+(2−n)s
set R, , let a ∈ G, then θ(a ) = (a ) ⇒ a
= a−i ⇒ ri + (2 − n)s = −i
⇒ (r + 1)i + (2 − n)s = 0 ⇒ (r + 1)i ≡ −2s (mod n).
Now for ai b ∈ G, we have, θ(ai b) = (ai b)−1 ⇒ ari+s b = ai b ⇒ ri + s = i
⇒ (r − 1)i ≡ −s (mod n), hence R = {ai | (r + 1)i ≡ −2s (mod n)} ∪ {ai b |
(r − 1)i ≡ −s (mod n)}. 

Corollary 9.10. If θ2 =id, m = 4k, n = 2k and k is even number, then Q is a
subgroup of hai and the natural isomorphism ψ : hai −→ Zn gives using Theorem
8.1 and Corollary 8.7 and 8.11
(1) If | Q |= 2, then ψ(Q) = hki.
(2) If | Q |= k2 then

hd1 i If n ∤ r + 2s − 1 and d1 = gcd(r + 2s − 1, n)
ψ(Q) =
h4i If n | r + 2s − 1 and n = gcd(r + 2s − 1, n).
(3) If | Q |= k and gcd(r + 2s − 1, n) = d1 then
ψ(Q) =



h d21 i If {(r + 2s − 1) ∩ [l(1 − r) − s]} = ∅
hd1 i If {(r + 2s − 1) ∩ [l(1 − r) − s]} =
6 ∅,

where l is an even number.
Corollary 9.11. If θ2 =id, m = 4k, n = 2k and k is even number, then Q is a
subgroup of hai and the natural isomorphism ψ : hai −→ Zn gives using Theorem
8.1 and Corollary 8.9 and 8.12
(1) If s is even, then

h0i If r = 1, s = 0, | Q |= 1
ψ(Q) =
h2i If | Q |= k.
(2) If s is odd, then

hki | Q |= 2 If n | 1 − r and n | s − k
ψ(Q) =
h1i | Q |= n If n ∤ r + 2s − 1 and n ∤ 1 − r.
Corollary 9.12. In any Case R 6= Q.
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in Mathematics, vol.231, Springer, New York.

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ZAHID RAZA, IMRAN AND BIJAN DAVVAZ

[3] Nicolas Bourbaki (2002). ”Lie groups and Lie algebras”. Chapter 4-6, Elements of Mathematics
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[4] K. K. A. Cunningham, T. J. Edgar, A. G. Helminck, B. F. Jones, OH, R. Schwell , J.F.
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Department of Mathematics, College of Sciences,University of Sharjah, UAE.
E-mail address: zraza@sharjah.ac.ae
Department of Mathematics, National University of Computer and Emerging
Sciences, Lahore, Pakistan.
E-mail address: imran aimphd@yahoo.com
Department of Mathematics, Yazd University, Yazd, Iran
E-mail address: davvaz@.yazd.ac.ir

