On the automorphism of some classes of groups
We study two classes of 2-generated nilpotent groups of nilpotency class 2 and compute the order of their automorphism groups.
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| Datum: | 2009 |
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2009
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| author | Hashemi, M. Хашемі, М. |
| author_facet | Hashemi, M. Хашемі, М. |
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| description | We study two classes of 2-generated nilpotent groups of nilpotency class 2 and compute the order of their automorphism groups. |
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UDC 512.5
M. Hashemi (Univ. Guilan, Iran)
ON THE AUTOMORPHISM OF SOME CLASSES OF GROUPS
PRO AVTOMORFIZM DEQKYX KLASIV HRUP
We study two classes of 2-generated nilpotent groups of nilpotency class 2 and compute the order of
their automorphism groups.
DoslidΩeno dva klasy 2-porodΩenyx nil\potentnyx hrup klasu nil\potentnosti 2 ta obçysleno
porqdok ]x hrup avtomorfizmiv.
1. Introduction. Many authors, have studied automorphism groups, of course most of
these are devoted to p-groups. In [1] Jamali presents some non-abelian 2-groups with
abelian automorphism groups. Bidwell and Curran [2] studied the automorphism
group of a split metacyclic p-group. By a program in [3], one can calculate the order
of small p-groups. Our purpose in the present paper is to calculate the order of the
automorphism groups of two classes of groups. Let G be a group. Z ( G ) denotes the
center of G; G ′ the commutator subgroup of G; Aut ( G ) the automorphism of G
and ϕ ( m ) the Euler function.
First, we state a lemma without proof that establishes some properties of groups of
nilpotency class 2.
Lemma 1. If G is a group and G ′ ⊆ Z ( G ), then the following hold for every
integer k and u, v, w ∈ G:
(i) [ u v, w ] = [ u, w ] [ v, w ] and [ u, v w ] = [ u, v ] [ u, w ];
(ii) [ uk, v ] = [ u, vk
] = [ u, v ]
k;
(iii) ( u v )
k = uk
vk
[ v, u ]
k
(
k
–
1)
/
2.
Theorem 1 ([4, p. 44], Proposition 3). Suppose that we are given a presentation
〈 X | R 〉 for a group G , and a map θ : X → G . Then θ extends to an endo-
morphism of G if and only if for all x ∈ X and all r ∈ R the result of substituting
( x ) θ for x in r yield the identity of G . Furthermore if, in addition ( )X θ
generates G then θ extends to an epimorphism of G.
We consider the finitely presented groups,
K ( n, l ) = 〈 a, b | a bn = bl
a, b an = al
b 〉, where ( n, l ) = 1,
and
Gn = 〈 a, b | an = bn = 1, [ a, b ]
a = [ a, b ], [ a, b ]
b = [ a, b ] 〉, n ≥ 1.
In Section 2, we investigate the automorphism group of K ( n, l ) and compute the
order of its automorphisms group. In Section 3 we solve a system and by using it, find
an explicit formula for the | Aut ( Gn ) |.
Most of theorems of this paper were suggested by data from a computer program
written in the computational algebra system GAP [3].
2. The order of Aut (((( K (((( n, l )))) )))). In this section, we consider the metacyclic Fox
groups K ( n, l ) defined by K ( n, l ) = 〈 a, b | a bn = bl
a, b an = al
b 〉, where ( n, l ) = 1.
We state some known results concerning K ( n, l ), the proofs of which can be found
in [5, 6].
Theorem 2. The groups K ( n, l ) defined by the above presentation have the
following properties:
© M. HASHEMI, 2009
1704 ISSN 1027-3190. Ukr. mat. Ωurn., 2009, t. 61, # 12
ON THE AUTOMORPHISM OF SOME CLASSES OF GROUPS 1705
(i) | K ( n, l ) | = | l – n | 3, if ( l, n ) = 1 and is infinite otherwise;
(ii) if ( l, n ) = 1, then | a | = | b | = ( l – n )
2;
(iii) if ( l, n ) = 1, then a bl n n l− −= .
Lemma 2. (i) For every l ≥ 3, K ( n, l ) ≅ K ( 1, 2 – l ).
(ii) For every i ≥ 2 and ( n, i ) = 1, K ( n, n + i ) ≅ K ( 1, i + 1 ).
Note. If ( m, n ) = 1, then K ( n, m ) ≅ K ( 1, m – n + 1 ) which we may write as
Km n− +1. Hence we only calculate Aut ( Kl ).
Before we present the main result of this section we need to develop some results
concerning Kl .
Lemma 3. Every element of Kl may be uniquely presented by x = a b a lβ γ δ( − )1 ,
where 1 ≤ β, γ, δ ≤ l – 1.
Proof. By parts (ii) and (iii) of Theorem 2, every element of Kl can be written in
this form. Since | Kl | = | l – 1 | 3, that expression is unique.
The lemma is proved.
Lemma 4. In Kl , [ a, b ] = bl−1 ∈ Z ( Kl ).
Proof. Since a bl l− −=1 1 then al−1 ∈ Z ( Kl ). By the relations of Kl we have
[ a, b ] = a b ab a b b a a b a bl l l− − − − − − −= = =1 1 1 1 1 1 1 ∈ Z ( Kl ),
as desired.
The lemma is proved.
Proposition 1. Let l ≥ 3 be an integer and f ∈ Aut ( Kl ). Then there exist 1 ≤
≤ βi , γi , δi ≤ l – 1 for 1 ≤ i ≤ 2 such that f a a b a l( ) = ( − )β γ δ1 1 11 , f b( ) =
= a b a lβ γ δ2 2 21( − ) when βi and γi are solutions of the following system:
γ β β γ β γ γ β2 1 2 1 1 1 1 1
1
2
1− ≡ − + ( − ) ( − )l l
lmod ,
γ γ γ β β β γ β1 2 2 2 1 2 1 1
1
2
1
2
1+ − ( − ) ≡ + + ( − ) ( − )l l l l
lmod ,
(1)
β γ β γ1 1 1 1
21 2
2
1− + ( − )( − ) −
l l l
l, = 1,
β γ β γ2 2 2 2
21 2
2
1− + ( − )( − ) −
l l l
l, = 1.
Proof. Let f ∈ Aut ( Kl ) and f a a b a l( ) = ( − )β γ δ1 1 11 , f b( ) = a b a lβ γ δ2 2 21( − ) ,
where 1 ≤ βi , γi , δi ≤ l – 1 and 1 ≤ i ≤ 2. Since b a = al
b, we have f b f a( ) ( ) =
= f a f bl( ) ( ) . By setting the values f a( ) and f b( ) in the recent relation, we get
a b a a b a a b al l lβ γ δ β γ δ β γ δ2 2 2 1 1 1 1 11 1 1( − ) ( − ) ( − )= ( 11 2 2 21) ( )( − )l la b aβ γ δ .
After some routine calculations and Lemma 3 we see,
γ β β γ β γ γ β2 1 2 1 1 1 1 1
1
2
1− ≡ − + ( − ) ( − )l l
lmod . (2)
ISSN 1027-3190. Ukr. mat. Ωurn., 2009, t. 61, # 12
1706 M. HASHEMI
Also a b = bl
a then we have
γ β β γ β γ γ β1 2 1 2 2 2 2 2
1
2
1− ≡ − + ( − ) ( − )l l
lmod . (3)
By the relations (2) and (3), we have
γ γ γ β β β γ β1 2 2 2 1 2 1 1
1
2
1
2
1+ − ( − ) ≡ + + ( − ) ( − )l l l l
lmod .
Since | a | = | f ( a ) | = ( l – 1 )
2, consequently ( )( − ) ( − )a b a l lβ γ δ1 1 1
21 1 = 1. Thus
a
l
l l l( − ) − + ( − )( − )
1
2 1
2
2
1 1 1 1
2
β γ γ β
= 1. This further implies
β γ β γ1 1 1 1
21 2
2
1− + ( − )( − ) −
l l l
l, = 1.
For | b | = | f ( b ) | = ( l – 1 )
2, by a similar argument we see that
β γ β γ2 2 2 2
21 2
2
1− + ( − )( − ) −
l l l
l, = 1.
Thus the assertions hold.
The proposition is proved.
The following proposition is the main result of this section.
Proposition 2. Let l ≥ 3 be an integer. Then
| Aut ( Kl ) | =
( − ) ( − ) −
( − ) ( − )
l l if l or
l
is even
l l
1 1
1
2
3 1 1
3
3
ϕ
ϕ
, ,
, iif
l
is odd
−
1
2
.
Proof. First, let l be even. Then the system (1) reduces to the following
equivalent system:
γ β β γ β γ2 1 2 1 1 1 1− ≡ − ( − )mod l ,
γ γ β β1 2 1 2 1+ ≡ + ( − )mod l ,
(4)
( − − )β γ1 1 1, l = 1,
( − − )β γ2 2 1, l = 1.
By the second congruence in (4), we get γ β β γ2 1 2 1 1≡ + − ( − )mod l .
Substituting γ2 in the first congruence gives β β β β γ β γ β1
2
1 2 1 1 2 1 1+ − − ≡ –
– γ1 1( − )mod l , or
β β γ β β γ β γ1 1 1 2 1 1 1 1 1( − ) + ( − ) ≡ − ( − )mod l .
Now ( − − )β γ1 1 1, l = 1 implies that β β1 2 1 1+ ≡ ( − )mod l . A consequence of the
last congruence and 1 ≤ β1 + β2 – 1 ≤ 2 l – 3 is that β β2 1= −l . This and the second
congruence in (4) follows that γ γ2 1= −l . Now let ( t, l – 1 ) = 1. Then for every
ISSN 1027-3190. Ukr. mat. Ωurn., 2009, t. 61, # 12
ON THE AUTOMORPHISM OF SOME CLASSES OF GROUPS 1707
β1 ∈ { 1, 2, … , l – 1 } there exist unique integer γ1 ∈ { 1, 2, … , l – 1 } such that β1 –
– γ1 = t (for selection γ1 = t + β1). Combining all these facts, we see that for every f ∈
∈ Aut ( Kl ) there are β1 , γ1 , δ1 , δ2 such that
f a a b at l( ) = + ( − )β β δ1 1 11 ,
f b a b al l t l( ) = − −( + ) ( − )β β δ1 1 21 ,
where 1 ≤ β1 , δ1 , δ2 ≤ l – 1, ( t, l – 1 ) = 1. Now if we denote f by f tβ δ δ1 1 2, , , , then the
assertion yields.
Lastly, let l be odd. Since
l l l
l
( − ) ≡ − ( − )1
2
1
2
1mod , the system (1) simplifies to
the following system:
γ β β γ β γ γ β2 1 2 1 1 1 1 1
1
2
1− ≡ − + − ( − )l
lmod ,
γ γ γ β β β γ β1 2 2 2 1 2 1 1
1
2
1
2
1+ − − ≡ + + − ( − )l l
lmod ,
(5)
β γ β γ1 1 1 1
1
2
1− + − −
l
l, = 1,
β γ β γ2 2 2 2
1
2
1− + − −
l
l, = 1.
Now suppose that
l − 1
2
is even. By the third condition in (5), one of β1 and γ1
is even and the other is odd. Also, it is true about β2 and γ2 . Combining of all these
and (5), we have
γ β β γ β γ2 1 2 1 1 1 1− ≡ − ( − )mod l ,
γ γ β β1 2 1 2 1+ ≡ + ( − )mod l ,
( − − )β γ1 1 1, l = 1,
( − − )β γ2 2 1, l = 1.
The result follows in a similar way as for the first case.
To complete the proof, let
l − 1
2
be odd. Then by (5) we get
γ β β γ β γ2 1 2 1 1 1
1
2
− ≡ − −
mod
l
,
γ γ β β1 2 1 2
1
2
+ ≡ + −
mod
l
,
β γ1 1
1
2
− −
,
l
= 1,
β γ2 2
1
2
− −
,
l
= 1.
ISSN 1027-3190. Ukr. mat. Ωurn., 2009, t. 61, # 12
1708 M. HASHEMI
In a similar way as for the first case, we get β β1 2 1
1
2
+ ≡ −
mod
l
. Since 1 ≤
≤ β1 + β2 – 1 ≤ 2 l – 3, we have β2 =
l
t
−
+ −1
2
1 1β , where t ∈ { 1, 2, 3 }.
Similarly, γ2 =
l
s
−
+ −1
2
1 1γ , where s ∈ { 1, 2, 3 }. By setting the values β2 and
γ2 in the first and second congruences of (5), we have
s tβ γ β γ1 1 1 1 2− ≡ ( )mod ,
s t
l
st
l
s− − −
− −
+
1
2
1
2
1
2
1β +
+
l
s t
l
t
−
( + ) − −
+
+ ≡ (1
2
1
2
1 1 01γ modd 2) .
Moreover, since one of β1 and γ1 is even and the other is odd then
s tβ γ1 1 0 2− ≡ ( )mod ,
(6)
s t
l
st
l
s− − −
− −
+
1
2
1
2
1
2
1β +
+
l
s t
l
t
−
( + ) − −
+
+ ≡ (1
2
1
2
1 1 01γ modd 2) .
We now count the solutions of (6). To do this, we must consider three cases as the
following:
1. Let s and t be odd. Utilizing the first congruence of (6), we have β1 – γ1 ≡
≡ 0 2( )mod and which is a contradiction (for, one of β1 and γ1 is even and the other
is odd).
2. Let s and t be even, then
0 0 2≡ ( )mod ,
β γ1 1 1 2+ ≡ ( )mod .
So that β2 = l – β1 and γ 2 = l – γ1 are solutions of (1), where 1 ≤ β1 , γ1 ≤ l – 1
and β γ1 1
1
2
− −
,
l
= 1. Hence the number solutions of (6) (in this case) is ( l –
– 1 ) ϕ ( l – 1 ).
3. Suppose one of s and t is even and other is odd. First let s be even, then
γ1 0 2≡ ( )mod ,
β1 1 2≡ ( )mod .
Now let s be odd, then
β1 0 2≡ ( )mod ,
γ1 1 2≡ ( )mod .
ISSN 1027-3190. Ukr. mat. Ωurn., 2009, t. 61, # 12
ON THE AUTOMORPHISM OF SOME CLASSES OF GROUPS 1709
So that the number solutions of (6) (in this case) is 2 ( l – 1 ) ϕ ( l – 1 ). Therefore, by
the above considerations, the assertion is established.
The proposition is proved.
3. The order of Aut (((( Gn )))). The goal of this section is to calculate the | Aut ( Gn ) |,
where
Gn = 〈 a, b | an = bn = 1, [ a, b ]
a = [ a, b ], [ a, b ]
b = [ a, b ] 〉, n ≥ 1.
First, we recall the following lemma from [7].
Lemma 5. Let G = Gn
, then | Gn | = n3, | Z ( G ) | = n and Z ( G ) = G′ =
= 〈 x | xn = 1 〉.
We now show that every element in the Gn , where n ∈ N, has standard form:
Lemma 6. Every element of the group G = Gn can be written uniquely in the
form a b b ai j k[ ], , where 0 ≤ i, s, k ≤ n – 1.
Proof. Since [ a, b ]
a = [ a, b ], [ a, b ]
b = [ a, b ], then [ a, b ] ∈ Z ( G ) and
[ ] = [ ]− −( )
−
a b a b b, ,1 11
∈ Z ( G ),
[ ] = [ ] = [ ]− − −( )
−
a b a b a ba1 1 11
, , , ∈ Z ( G ).
Moreover, for every x = x x xs s
k
sk
1 2
1 2 … in Gn , where xi ∈ { a, b } and s1 , s2 , … , sk
are integers, using the relations b a a b b aj i i j j i= [ ], , we may easily prove that every
element of G is in the form a b gi j , where 0 ≤ i < m – 1, 0 ≤ j ≤ n – 1 and g ∈ G′
(by induction method on the length of the word x ). Suppose x = a b gi j = e then
a bi j ∈ Z ( G ) and [ a, bj
] = [ a, b ]
j = 1, thus n | j. Similarly n | i, that is i = j = 0
and g = e. The result is now immediate.
The lemma is proved.
The following proposition is the main result of this section.
Proposition 3. Let n ≥ 2 be an integer. Then f ∈ Aut ( Gn ) if and only if there
exist 0 ≤ si , ti , ki ≤ n – 1 for 1 ≤ i ≤ 2 such that f a a b a bt s k( ) = [ ]1 1 1, , f b( ) =
= a b a bt s k2 2 2[ ], and s1 , s2 , t1 , t2 are solutions of the following system:
s t
n n
n1 1
1
2
0
( − ) ≡ ( )mod ,
s t
n n
n2 2
1
2
0
( − ) ≡ ( )mod , (7)
( − )s t s t n1 2 2 1, = 1.
Proof. Let f ∈ Aut ( Gn ) and f a a b a bt s k( ) = [ ]1 1 1, , f b( ) = a b a bt s k2 2 2[ ], ,
where 1 ≤ si , ti , ki ≤ n , 1 ≤ i ≤ 2. Since | a | = | ( a ) f | = n and ( )a f n =
= a b a bnt ns nk s t
n n
1 1
1 1 1
1
2[ ]
− ( − )
, = [ ]
( − )
a b
s t
n n
,
1 1
1
2 , we get n s t
n n
1 1
1
2
( − )
. Also | b | =
= | ( b ) f | = n so that n s t
n n
2 2
1
2
( − )
. Finally, | [ a, b ] | = n hence [ ] ( − )a b n t s s t, 1 2 1 2 =
= e. This yields that ( − )t s s t n1 2 1 2, = 1.
ISSN 1027-3190. Ukr. mat. Ωurn., 2009, t. 61, # 12
1710 M. HASHEMI
Now it is sufficient to prove that f, with the above conditions, is an isomorphism.
Let u = a b a bs t k[ ], and ( u ) f = e, then by the Lemma 6, we have
t t t s n1 2 0+ ≡ ( )mod ,
s t s s n1 2 0+ ≡ ( )mod , (8)
k t k s t s s t k s t ts1 2 1 2 1 2 1 2+ + ( − ) − –
s t t t s t s s
n1 1 2 21
2
1
2
0
( − )
−
( − )
≡ ( )mod .
By adding s1 times the first congruence of (8) to ( – t1 ) times the second
congruence, we get s t s t s s n1 2 1 2 0− ≡ ( )mod or ( − ) ≡ ( )s t t s s n1 2 1 2 0 mod . Since
( − )t s s t n1 2 1 2, = 1, we have n | s. An identical argument shows that n | t. Using these
in the third congruence of (8), yield that ( − ) ≡ ( )t s s t k n1 2 1 2 0 mod . Hence n | k. That
is u = e and f is an isomorphism.
The proposition is proved.
In order to give an expression for the | Aut ( Gn ) |, we need the following key
lemma.
Lemma 7. Let n = pii
k iα
=∏ 1
, where pi is prime number and α i ≥ 1. Then
the number of solutions of the system
1 ≤ s1 , s2 , t1 , t2 ≤ n – 1,
( − )s t s t n1 2 2 1, = 1,
is n n p pi ii
k iϕ α( ) ( + )−
=∏2 1
1
1 .
Proof. Without loss of generality, we assume k = 2. We know that, the number
of { m | 0 ≤ m ≤ n – 1 and p1 | m } is p p1
1
2
1 2α α− . Also, for p2 and p p1 2 , it is
p p1 2
11 2α α − and p p1
1
2
11 2α α− − respectively. Since ( s1 , s2 ) when s1 and s2 are
multiple of p1 or p2 not being allowed, we may choose ( s1 , s2 ) in t ways, where
t =
p p p p p p p1
2
2
2
1
2 2
2
2
1
2
2
2 2
1
21 2 1 2 1 2 1α α α α α α α− ( − +− − −− −
− −
) =
= ( − − +
2
2
2 2
1
2 2
2
2 2
1
2
2
2
1
2
2
2
2
1 2
p
p p p p p p
α
α α 11
1 1 11
1
1 2
1
2 1
1
1 2
1 2 1
)
( − ) ( − ) ( + )− − −
,
p p p p p p pα α α α22
1 2
1
2
1
1
1 2
1
2
1
1 1
−
− −
( + ) =
= ( ) ( + ) ( + )
p
n p p p pϕ α α .
Now, we select ( t1 , t2 ) such that ( − )s t s t n1 2 2 1, = 1. To do this, we find the
number of ( x, y ) such that ( − )s y s x n1 2 , ≠ 1. In other words, we find the number of
( x, y ) such that
s y s x p1 2 10− ≡ ( )mod or s y s x p1 2 20− ≡ ( )mod .
Let s y s x p1 2 10− ≡ ( )mod , then for every 0 ≤ x ≤ n – 1 there is a unique 0 ≤ y0 ≤
≤ p1 – 1 such that s y s x p1 0 2 1≡ ( )mod (for y0 ≡ 0 or s s x p1 2 1
* mod( ) , where s1
* is
the arithmetic inverse of s1 respect to p1 ). Hence for every 0 ≤ x ≤ n – 1 the number
solutions of s y s x p1 2 10− ≡ ( )mod in Zn is p p1
1
2
1 2α α− (for yi = y0 + p1 k, 0 ≤ k ≤
ISSN 1027-3190. Ukr. mat. Ωurn., 2009, t. 61, # 12
ON THE AUTOMORPHISM OF SOME CLASSES OF GROUPS 1711
≤ p p1
1
2
1 2α α− are solutions). By a similar argument, for every 0 ≤ x ≤ n – 1 the
number solutions of s y s x p1 2 20− ≡ ( )mod in Zn is p p1 2
11 2α α − . Also, we know that
p p1
1
2
11 2α α− − solutions are common in two sets of solutions. Consequently, when ( s1 ,
s2 ) select, we may choose ( t1 , t2 ) in l ways where
l =
p p p p p p p p p1
2
2
2
1 2 1
1
2 1 2
11 2 1 2 1 2 1 2α α α α α α α α− ( + −− −
11
1
2
1
1
2 1
2
2 1
1 2 1 2
1 2
1 2 1
α α
α α
− −
− −
) =
= ( − − + )
p
p p p p p p ,,
.np p p p n n1
1
1 2
1
2
1 21 1α α ϕ− −( − ) ( − ) = ( )
Multiplying the number t and l together we obtain the assertion.
The lemma is proved.
With the previous notations, we prove that the important result of this section.
Proposition 4. Let n = pii
k iα
=∏ 1
be an integer. Then
| Aut ( Gn ) | =
n n p p if n is odd
n
n p
i ii
k i3 2 1
1
3
2
1
3
ϕ
ϕ
α( ) ( + )
( )
−
=∏ , ,
ii ii
k i p if n is evenα −
= ( + )
∏ 1
1
1 , .
Proof. First, let n be odd. Then the system (7) reduces to the equivalent system
0 ≤ s1 , s2 , t1 , t2 ≤ n – 1,
( − )s t s t n1 2 2 1, = 1.
Since the number of solutions of this system is n n p pi ii
k iϕ α( ) ( + )−
=∏2 1
1
1 and k1 ,
k2 ≤ n – 1, the assertion follows from the Proposition 3.
Finally, let n be even. Now s1 , s2 , t1 , t2 are solutions of the system (7) if and
only if for every 1 ≤ i ≤ k – 1 they are solutions of the following system:
s t
n n
pi
i
1 1
1
2
0
( − ) ≡ ( )mod α ,
s t
n n
pi
i
2 2
1
2
0
( − ) ≡ ( )mod α ,
( − )s t s t pi
i
1 2 2 1, α = 1.
When pi is odd number, then this system reduces to
0 ≤ s1 , s2 , t1 , t2 ≤ pi
iα – 1,
( − )s t s t pi
i
1 2 2 1, α = 1,
which was investigated in the Lemma 7.
Now it is sufficient to compute the solutions of the system
ISSN 1027-3190. Ukr. mat. Ωurn., 2009, t. 61, # 12
1712 M. HASHEMI
s t
n n
1 1
1
2
0 2
( − ) ≡ ( )mod α ,
s t
n n
2 2
1
2
0 2
( − ) ≡ ( )mod α ,
( − )s t s t1 2 2 1 2, α = 1.
This is equivalent to:
s t1 1 0 2≡ ( )mod ,
s t2 2 0 2≡ ( )mod , (9)
( − )s t s t1 2 2 1 2, = 1.
From first and third conditions of (9), we note that exactly one of s1 an t1 should
be odd. Then we may choose ( s1 , t1 ) in 22 1α− ways. Now, we select ( s2 , t2 ) such
that ( − )s t s t1 2 2 1 2, = 1. If t1 is even then t2 and s1 are odd. This together with
2 | s2 t2 yields that s2 is even. Therefore, the number of solutions of system (9) for this
case is 24 4α− . Similarly, it is true if t1 is odd. By the above argument, the number
solutions of (9) is 2
2
3
2 2 2 14 3 2 1α
α
α αϕ− −= ( ) ( + )( ) . This completes the proof.
Corollary. Let G be a non-abelian group of order p3 , where p is odd
prime. Then | Aut ( G ) | = p p3 1( − ) or p p p3 21 1( − ) ( + ) .
Proof. By [8], G is isomorphic to one of K p+1 or Gp . Then the result now
follows from Propositions 2 and 4.
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3. The GAP Group, GAP Groups, Algorithms, and programming version 4.4 packages AutPGroup
and Small Groups http. A. R.
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Cambridge: Cambridge Univ. Press, 1977.
5. Campbell C. M., Campel P. P., Doostie H., Robertson E. F. Fibonacci length for metacyclic
groups // Algebra Colloq. – 2004. – P. 215 – 222.
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19. – P. 247 – 248.
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Comput. – 2006. – 20, # 1 – 2. – P. 171 – 180.
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Received 05.06.09
ISSN 1027-3190. Ukr. mat. Ωurn., 2009, t. 61, # 12
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| id | umjimathkievua-article-3131 |
| institution | Ukrains’kyi Matematychnyi Zhurnal |
| keywords_txt_mv | keywords |
| language | Ukrainian English |
| last_indexed | 2026-03-24T02:36:48Z |
| publishDate | 2009 |
| publisher | Institute of Mathematics, NAS of Ukraine |
| record_format | ojs |
| resource_txt_mv | umjimathkievua/2e/ac139f8b88e5b9df60a156204c7b792e.pdf |
| spelling | umjimathkievua-article-31312020-03-18T19:45:55Z On the automorphism of some classes of groups Про автоморфізм деяких класів груп Hashemi, M. Хашемі, М. We study two classes of 2-generated nilpotent groups of nilpotency class 2 and compute the order of their automorphism groups. Досліджено два класи 2-породжених нільпотентних груп класу нільпотентності 2 та обчислено порядок їх груп автоморфізмів. Institute of Mathematics, NAS of Ukraine 2009-12-25 Article Article application/pdf https://umj.imath.kiev.ua/index.php/umj/article/view/3131 Ukrains’kyi Matematychnyi Zhurnal; Vol. 61 No. 12 (2009); 1704-1712 Український математичний журнал; Том 61 № 12 (2009); 1704-1712 1027-3190 uk en https://umj.imath.kiev.ua/index.php/umj/article/view/3131/3010 https://umj.imath.kiev.ua/index.php/umj/article/view/3131/3011 Copyright (c) 2009 Hashemi M. |
| spellingShingle | Hashemi, M. Хашемі, М. On the automorphism of some classes of groups |
| title | On the automorphism of some classes of groups |
| title_alt | Про автоморфізм деяких класів груп |
| title_full | On the automorphism of some classes of groups |
| title_fullStr | On the automorphism of some classes of groups |
| title_full_unstemmed | On the automorphism of some classes of groups |
| title_short | On the automorphism of some classes of groups |
| title_sort | on the automorphism of some classes of groups |
| url | https://umj.imath.kiev.ua/index.php/umj/article/view/3131 |
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