arXiv:1702.03150v1 [] 10 Feb 2017

Autocommuting probability of a finite
group relative to its subgroups
Parama Dutta and Rajat Kanti Nath∗

Department of Mathematical Sciences, Tezpur University,
Napaam-784028, Sonitpur, Assam, India.
Emails: parama@gonitsora.com and rajatkantinath@yahoo.com

Abstract
Let H ⊆ K be two subgroups of a finite group G and Aut(K) the
automorphism group of K. The autocommuting probability of G relative
to its subgroups H and K, denoted by Pr(H, Aut(K)), is the probability
that the autocommutator of a randomly chosen pair of elements, one from
H and the other from Aut(K), is equal to the identity element of G. In
this paper, we study Pr(H, Aut(K)) through a generalization.
Key words: Automorphism group, Autocommuting probability, Autoisoclinism.
2010 Mathematics Subject Classification: 20D60, 20P05, 20F28.

1

Introduction

Let G be a finite group acting on a set Ω. Let Pr(G, Ω) denote the probability that
a randomly chosen element of Ω fixes a randomly chosen element of G. In 1975,
Sherman [11] initiated the study of Pr(G, Ω) considering G to be an abelian group and
Ω = Aut(G), the automorphism group of G. Note that
Pr(G, Aut(G)) =

|{(x, α) ∈ G × Aut(G) : [x, α] = 1}|
|G|| Aut(G)|

where [x, α] is the autocommutator of x and α defined as x−1 α(x). The ratio
Pr(G, Aut(G)) is called autocommuting probability of G.
Let H and K be two subgroups of a finite group G such that H ⊆ K. Motivated
by the works in [2, 6], we define
Prg (H, Aut(K)) =

|{(x, α) ∈ H × Aut(K) : [x, α] = g}|
|H|| Aut(K)|

(1.1)

where g ∈ K. That is, Prg (H, Aut(K)) is the probability that the autocommutator
of a randomly chosen pair of elements, one from H and the other from Aut(K), is
∗ Corresponding

author

1

equal to a given element g ∈ K. The ratio Prg (H, Aut(K)) is called generalized
autocommuting probability of G relative to its subgroups H and K. Clearly, if H = G
and g = 1 then Prg (H, Aut(K)) = Pr(G, Aut(G)). We would like to mention here that
the case when H = G is considered in [3]. In this paper, we study Prg (H, Aut(K))
extensively. In particular, we obtain some computing formulae, various bounds and a
few characterizations of G through a subgroup. We conclude the paper describing an
invariance property of Prg (H, Aut(K)).
We write S(H, Aut(K)) to denote the set {[x, α] : x ∈ H and α ∈ Aut(K)} and
[H, Aut(K)] := hS(H, Aut(K))i. We also write L(H, Aut(K)) := {x ∈ H : [x, α] =
1 for all α ∈ Aut(K)} and L(G) := L(G, Aut(G)), the absolute center of G (see [5]).
Note that L(H, Aut(K)) is a normal subgroup of H contained in H ∩ Z(K). Further,
L(H, Aut(K)) =
∩
CH (α), where CH (α) = {x ∈ H : [x, α] = 1} is a subgroup
α∈Aut(K)

of H. Let CAut(K) (x) := {α ∈ Aut(K) : α(x) = x} for x ∈ H and CAut(K) (H) = {α ∈
Aut(K) : α(x) = x for all x ∈ H}. Then CAut(K) (x) is a subgroup of Aut(K) and
CAut(K) (H) = ∩ CAut(K) (x).
x∈H

Clearly, Prg (H, Aut(K)) = 1 if and only if [H, Aut(K)] = {1} and g = 1
if and only if H = L(H, Aut(K)) and g = 1. Also, Prg (H, Aut(K)) = 0 if
and only if g ∈
/ S(H, Aut(K)). Therefore, we consider H 6= L(H, Aut(K)) and
g ∈ S(H, Aut(K)) throughout the paper.

2

Some computing formulae

For any x ∈ H, let us define the set Tx,g (H, K) = {α ∈ Aut(K) : [x, α] = g},
where g is a fixed element of K. Note that Tx,1 (H, K) = CAut(K) (x). The
following two lemmas play a crucial role in obtaining computing formula for
Prg (H, Aut(K)).
Lemma 2.1. Let H and K be two subgroups of a finite group G such that H ⊆
K. If Tx,g (H, K) 6= φ then Tx,g (H, K) = σCAut(K) (x) for some σ ∈ Tx,g (H, K)
and hence |Tx,g (H, K)| = |CAut(G) (x)|.
Proof. Let σ ∈ Tx,g (H, K) and β ∈ σCAut(K) (x). Then β = σα for some
α ∈ CAut(K) (x). We have
[x, β] = [x, σα] = x−1 σ(α(x)) = [x, σ] = g.
Therefore, β ∈ Tx,g (H, K) and so σCAut(K) (x) ⊆ Tx,g (H, K). Again, let γ ∈
Tx,g (H, K) then γ(x) = xg. We have σ −1 γ(x) = σ −1 (xg) = x and so σ −1 γ ∈
CAut(K) (x). Therefore, γ ∈ σCAut(K) (x) which gives Tx,g (H, K) ⊆ σCAut(K) (x).
Hence, the result follows.
Consider the action of Aut(K) on K given by (α, x) 7→ α(x) where α ∈
Aut(K) and x ∈ K. Let orbK (x) := {α(x) : α ∈ Aut(K)} be the orbit of
x ∈ K. Then by orbit-stabilizer theorem, we have
| orbK (x)| =

| Aut(K)|
.
|CAut(K) (x)|
2

(2.1)

Lemma 2.2. Let H and K be two subgroups of a finite group G such that
H ⊆ K. Then Tx,g (H, K) 6= φ if and only if xg ∈ orbK (x).
Proof. The result follows from the fact that α ∈ Tx,g (H, K) if and only if xg ∈
orbK (x).
The following theorem gives two computing formulae for Prg (H, Aut(K)).
Theorem 2.3. Let H and K be two subgroups of a finite group G such that
H ⊆ K. If g ∈ K then
Prg (H, Aut(K)) =

=

1
|H|| Aut(K)|
1
|H|

X

X

|CAut(K) (x)|

x∈H
xg∈orbK (x)

x∈H
xg∈orbK (x)

1
.
| orbK (x)|

Proof. We have {(x, α) ∈ H × Aut(K) : [x, α] = g} = ⊔ ({x} × Tx,g (H, K)),
x∈H

where ⊔ represents the union of disjoint sets. Therefore, by (1.1), we have
X
|H|| Aut(K)|Prg (H, Aut(K)) = | ⊔ ({x} × Tx,g (H, K))| =
|Tx,g (H, K)|.
x∈H

x∈H

Hence, the result follows from Lemma 2.1, Lemma 2.2 and (2.1).
Considering g = 1 in Theorem 2.3, we get the following computing formulae
for Pr(H, Aut(K)).
Corollary 2.4. Let H and K be two subgroups of a finite group G such that
H ⊆ K. Then
Pr(H, Aut(K)) =

X
1
| orbK (H)|
|CAut(K) (x)| =
|H|| Aut(K)|
|H|
x∈H

where orbK (H) = {orbK (x) : x ∈ H}.
Corollary 2.5. Let H and K be two subgroups of a finite group G such that
H ⊆ K. If CAut(K) (x) = {I} for all x ∈ H \ {1}, where I is the identity element
of Aut(K), then
Pr(H, Aut(K)) =

1
1
1
+
−
.
|H| | Aut(K)| |H|| Aut(K)|

Proof. By Corollary 2.4, we have
|H|| Aut(K)| Pr(H, Aut(K)) =

X

|CAut(K) (x)| = | Aut(K)| + |H| − 1.

x∈H

Hence, the result follows.
3

We also have |{(x, α) ∈ H × Aut(K) : [x, α] = 1}| =

P

|CH (α)| and hence

α∈Aut(K)

Pr(H, Aut(K)) =

1
|H|| Aut(K)|

X

|CH (α)|.

(2.2)

α∈Aut(K)

We conclude this section with the following two results.
Proposition 2.6. Let H and K be two subgroups of a finite group G such that
H ⊆ K. If g ∈ K then
Prg−1 (H, Aut(K)) = Prg (H, Aut(K)).
Proof. Let
A = {(x, α) ∈ H × Aut(K) : [x, α] = g} and
B = {(y, β) ∈ H × Aut(K) : [y, β] = g −1 }.
Then (x, α) 7→ (α(x), α−1 ) gives a bijection between A and B. Therefore |A| =
|B|. Hence, the result follows from (1.1).
Proposition 2.7. Let G1 and G2 be two finite groups. Let H1 , K1 and H2 , K2
be subgroups of G1 and G2 respectively such that H1 ⊆ K1 , H2 ⊆ K2 and
gcd(|K1 |, |K2 |) = 1. If (g1 , g2 ) ∈ K1 × K2 then
Pr(g1 ,g2 ) (H1 × H2 , Aut(K1 × K2 )) = Prg1 (H1 , Aut(K1 ))Prg2 (H2 , Aut(K2 )).
Proof. Let
X = {((x, y), αK1 ×K2 ) ∈ (H1 × H2 ) × Aut(K1 × K2 ) :
[(x, y), αK1 ×K2 ] = (g1 , g2 )},
Y = {(x, αK1 ) ∈ H1 × Aut(K1 ) : [x, αK1 ] = g1 } and
Z = {(y, αK2 ) ∈ H2 × Aut(K2 ) : [y, αK2 ] = g2 }.
Since gcd(|K1 |, |K2 |) = 1, by Lemma 2.1 of [1], we have Aut(K1 × K2 ) =
Aut(K1 ) × Aut(K2 ). Therefore, for every αK1 ×K2 ∈ Aut(K1 × K2 ) there exist
unique αK1 ∈ Aut(K1 ) and αK2 ∈ Aut(K2 ) such that αK1 ×K2 = αK1 × αK2 ,
where αK1 ×αK2 ((x, y)) = (αK1 (x), αK2 (y)) for all (x, y) ∈ H1 ×H2 . Also, for all
(x, y) ∈ H1 × H2 , we have [(x, y), αK1 ×K2 ] = (g1 , g2 ) if and only if [x, αK1 ] = g1
and [y, αK2 ] = g2 . These leads to show that X = Y × Z. Therefore
|Y|
|Z|
|X |
=
·
.
|H1 × H2 || Aut(K1 × K2 )|
|H1 || Aut(K1 )| |H2 || Aut(K2 )|
Hence, the result follows from (1.1).

4

3

Various bounds

In this section, we obtain various bounds for Prg (H, Aut(K)). We begin with
the following lower bounds.
Proposition 3.1. Let H and K be two subgroups of a finite group G such that
H ⊆ K. Then, for g ∈ K, we have
|CAut(K) (H)|(|H|−|L(H,Aut(K))|)
|H|| Aut(K)|

(a) Prg (H, Aut(K)) ≥

|L(H,Aut(K))|
|H|

(b) Prg (H, Aut(K)) ≥

|L(H,Aut(K))||CAut(K) (H)|
|H|| Aut(K)|

+

if g = 1.

if g 6= 1.

Proof. Let C denote the set {(x, α) ∈ H × Aut(K) : [x, α] = g}.
(a) We have (L(H, Aut(K))×Aut(K))∪(H×CAut(K) (H)) is a subset of C and
|(L(H, Aut(K)) × Aut(K)) ∪ (H × CAut(K) (H))| = |L(H, Aut(K))|| Aut(K)| +
|CAut(K) (H)||H| − |L(H, Aut(K))||CAut(K) (H)|. Hence, the result follows from
(1.1).
(b) Since g ∈ S(H, Aut(K)) we have C is non-empty. Let (y, β) ∈ C then
(y, β) ∈
/ L(H, Aut(K)) × CAut(K) (H) otherwise [y, β] = 1. It is easy to see
that the coset (y, β)(L(H, Aut(K)) × CAut(K) (H)) is a subset of C having order
|L(H, Aut(K))||CAut(K) (H)|. Hence, the result follows from (1.1).
Proposition 3.2. Let H and K be two subgroups of a finite group G such that
H ⊆ K. If g ∈ K then
Prg (H, Aut(K)) ≤ Pr(H, Aut(K)).
The equality holds if and only if g = 1.
Proof. By Theorem 2.3, we have
Prg (H, Aut(K)) =

≤

1
|H|| Aut(K)|

X

|CAut(K) (x)|

x∈H
xg∈orbK (x)

X
1
|CAut(K) (x)| = Pr(H, Aut(K)).
|H|| Aut(K)|
x∈H

The equality holds if and only if xg ∈ orbK (x) for all x ∈ H if and only if
g = 1.
Proposition 3.3. Let H and K be two subgroups of a finite group G such that
H ⊆ K. Let g ∈ K and p the smallest prime dividing | Aut(K)|. If g 6= 1 then
Prg (H, Aut(K)) ≤

1
|H| − |L(H, Aut(K))|
< .
p|H|
p

5

Proof. By Theorem 2.3, we have
Prg (H, Aut(K)) =

1
|H|

X

x∈H\L(H,Aut(K))
xg∈orbK (x)

1
| orbK (x)|

(3.1)

noting that for x ∈ L(H, Aut(K)) we have xg ∈
/ orbK (x). Also, for x ∈ H \
L(H, Aut(K)) and xg ∈ orbK (x) we have | orbK (x)| > 1. Since | orbK (x)| is
a divisor of | Aut(K)| we have | orbK (x)| ≥ p. Hence, the result follows from
(3.1).
Proposition 3.4. Let H1 , H2 and K be subgroups of a finite group G such that
H1 ⊆ H2 ⊆ K. Then
Prg (H1 , Aut(K)) ≤ |H2 : H1 |Prg (H2 , Aut(K)).
The equality holds if and only if xg ∈
/ orbK (x) for all x ∈ H2 \ H1 .
Proof. By Theorem 2.3, we have
|H1 || Aut(K)|Prg (H1 , Aut(K)) =

X

|CAut(K) (x)|

X

|CAut(K) (x)|

x∈H1
xg∈orbK (x)

≤

x∈H2
xg∈orbK (x)

= |H2 || Aut(K)|Prg (H2 , Aut(K)).
Hence, the result follows.
Proposition 3.5. Let H and K be two subgroups of a finite group G such that
H ⊆ K. If g ∈ K then
Prg (H, Aut(K)) ≤ |K : H| Pr(K, Aut(K))
with equality if and only if g = 1 and H = K.
Proof. By Proposition 3.2, we have
Prg (H, Aut(K)) ≤ Pr(H, Aut(K))
X
1
=
|CAut(K) (x)|
|H|| Aut(K)|
x∈H
X
1
≤
|CAut(K) (x)|.
|H|| Aut(K)|
x∈K

Hence, the result follows from Corollary 2.4.

6

Note that if we replace Aut(K) by Inn(K), the inner automorphism group
of K, in (1.1) then Prg (H, Inn(K)) = Prg (H, K) where
Prg (H, K) =

|{(x, y) ∈ H × K : x−1 y −1 xy = g}|
.
|H||K|

A detailed study on Prg (H, K) can be found in [2]. The following proposition
gives a relation between Prg (H, Aut(K)) and Prg (H, K) for g = 1.
Proposition 3.6. Let H and K be two subgroups of a finite group G such that
H ⊆ K. If g = 1 then
Prg (H, Aut(K)) ≤ Prg (H, K).
Proof. If g = 1 then by [2, Theorem 2.3], we have
Prg (H, K) =

1
1 X
|H|
| clK (x)|

(3.2)

x∈H

where clK (x) = {α(x) : α ∈ Inn(K)}. Since clK (x) ⊆ orbK (x) for all x ∈ H,
the result follows from (3.2) and Theorem 2.3.
Theorem 3.7. Let H and K be two subgroups of a finite group G such that
H ⊆ K and p the smallest prime dividing | Aut(K)|. Then
Pr(H, Aut(K)) ≥

|L(H, Aut(K))| p(|H| − |XH | − |L(H, Aut(K))|) + |XH |
+
|H|
|H|| Aut(K)|

and
Pr(H, Aut(K)) ≤

(p − 1)|L(H, Aut(K))| + |H| |XH |(| Aut(K)| − p)
−
,
p|H|
p|H|| Aut(K)|

where XH = {x ∈ H : CAut(K) (x) = {I}}.
Proof. We have XH ∩ L(H, Aut(K)) = φ. Therefore
X
|CAut(K) (x)| = |XH | + | Aut(K)||L(H, Aut(K))|
x∈H

X

+

|CAut(K) (x)|.

x∈H\(XH ∪L(H,Aut(K)))

For x ∈ H \ (XH ∪ L(H, Aut(K))) we have {I} =
6 CAut(K) (x) 6= Aut(K)
| Aut(K)|
which implies p ≤ |CAut(K) (x)| ≤
. Therefore
p
X

|CAut(K) (x)| ≥|XH | + | Aut(K)||L(H, Aut(K))|

x∈H

+ p(|H| − |XH | − |L(H, Aut(K))|)
7

(3.3)

and
X

|CAut(K) (x)| ≤|XH | + | Aut(K)||L(H, Aut(K))|

x∈H

+

| Aut(K)|(|H| − |XH | − |L(H, Aut(K))|)
.
p

(3.4)

Hence, the result follows from Corollary 2.4, (3.3) and (3.4).
We have the following two corollaries.
Corollary 3.8. Let H and K be two subgroups of a finite group G such that H ⊆
K. If p and q are the smallest primes dividing | Aut(K)| and |H| respectively
then
p+q−1
.
Pr(H, Aut(K)) ≤
pq
In particular, if p = q then Pr(H, Aut(K)) ≤

2p−1
p2

≤ 43 .

Proof. Since H 6= L(H, Aut(K)) we have |H : L(H, Aut(K))| ≥ q. Therefore,
by Theorem 3.7, we have


1
p−1
p+q−1
Pr(H, Aut(K)) ≤
+1 ≤
.
p |H : L(H, Aut(K))|
pq

Corollary 3.9. Let H and K be two subgroups of a finite group G such that
H ⊆ K and p, q be the smallest primes dividing | Aut(K)| and |H| respectively.
If H is non-abelian then
Pr(H, Aut(K)) ≤

q2 + p − 1
.
pq 2

In particular, if p = q then Pr(H, Aut(K)) ≤

p2 +p−1
p3

≤ 58 .

Proof. Since H is non-abelian we have |H : L(H, Aut(K))| ≥ q 2 . Therefore, by
Theorem 3.7, we have


1
p−1
q2 + p − 1
Pr(H, Aut(K)) ≤
.
+1 ≤
p |H : L(H, Aut(K))|
pq 2

Now we obtain two lower bounds analogous to the lower bounds obtained in
[9, Theorem A] and [8, Theorem 1].
Theorem 3.10. Let H and K be two subgroups of a finite group G such that
H ⊆ K. Then


1
|S(H, Aut(K))| − 1
Pr(H, Aut(K)) ≥
1+
.
|S(H, Aut(K))|
|H : L(H, Aut(K))|
The equality holds if and only if orbK (x) = xS(H, Aut(K)) for all x ∈ H \
L(H, Aut(K)).
8

Proof. For all x ∈ H \ L(H, Aut(K)) we have α(x) = x[x, α] ∈ xS(H, Aut(K)).
Therefore orbK (x) ⊆ xS(H, Aut(K)) and so | orbK (x)| ≤ |S(H, Aut(K))| for
all x ∈ H \ L(H, Aut(K)). Now, by Corollary 2.4, we have


X
X
1 
1
1

Pr(H, Aut(K)) =
+
|H|
| orbK (x)|
| orbK (x)|
x∈L(H,Aut(K))

x∈H\L(H,Aut(K))

1
|L(H, Aut(K))|
+
≥
|H|
|H|

X

x∈H\L(H,Aut(K))

1
.
|S(H, Aut(K))|

Hence, the result follows.
Lemma 3.11. Let H and K be two subgroups of a finite group G such that
H ⊆ K. Then, for any two integers m ≥ n, we have




n−1
1
m−1
1
1+
≥
1+
.
n
|H : L(H, Aut(K))|
m
|H : L(H, Aut(K))|
If L(H, Aut(K)) 6= H then equality holds if and only if m = n.
Proof. The proof is an easy exercise.
Corollary 3.12. Let H and K be two subgroups of a finite group G such that
H ⊆ K. Then


1
|[H, Aut(K)]| − 1
Pr(H, Aut(K)) ≥
1+
.
|[H, Aut(K)]|
|H : L(H, Aut(K))|
If H 6= L(H, Aut(K)) then the equality holds if and only if [H, Aut(K)] =
S(H, Aut(K)) and orbK (x) = x[H, Aut(K)] for all x ∈ H \ L(H, Aut(K)).
Proof. Since |[H, Aut(K)]| ≥ |S(H, Aut(K))|, the result follows from Theorem
3.10 and Lemma 3.11.
Note that the equality holds if and only if equality holds in Theorem 3.10
and Lemma 3.11.
It is worth mentioning that Theorem 3.10 gives better lower bound than the
lower bound given by Corollary 3.12. Also


1
|[H, Aut(K)]| − 1
|L(H, Aut(K))|
1+
≥
|[H, Aut(K)]|
|H : L(H, Aut(K))|
|H|
p(|H| − |L(H, Aut(K))|)
.
+
|H|| Aut(K)|
Hence, Theorem 3.10 gives better lower bound than the lower bound given by
Theorem 3.7.

9

4

A few Characterizations

In this section, we obtain some characterizations of a subgroup H of G if equality
holds in Corollary 3.8 and Corollary 3.9. We begin with the following result.
Theorem 4.1. Let H and K be two subgroups of a finite group G such that
H ⊆ K. If Pr(H, Aut(K)) = p+q−1
for some primes p and q. Then pq divides
pq
|H|| Aut(K)|. Further, if p and q are the smallest primes dividing | Aut(K)|
and |H| respectively, then
H
∼
= Zq .
L(H, Aut(K))
In particular, if H and Aut(K) are of even order and Pr(H, Aut(K)) =
H
∼
L(H,Aut(K)) = Z2 .

3
4

then

Proof. By (1.1), we have (p + q − 1)|H|| Aut(K)| = pq|{(x, α) ∈ H × Aut(K) :
[x, α] = 1}|. Therefore, pq divides |H|| Aut(K)|.
If p and q are the smallest primes dividing | Aut(K)| and |H| respectively
then, by Theorem 3.7, we have


p−1
1
p+q−1
≤
+1
pq
p |H : L(H, Aut(K))|
which gives |H : L(H, Aut(K))| ≤ q. Hence,

H
L(H,Aut(K))

∼
= Zq .

Theorem 4.2. Let H ⊆ K be two subgroups of a finite group G such that H
2
is non-abelian and Pr(H, Aut(K)) = q +p−1
for some primes p and q. Then
pq2
pq divides |H|| Aut(K)|. Further, if p and q are the smallest primes dividing
| Aut(K)| and |H| respectively then
H
∼
= Zq × Zq .
L(H, Aut(K))
In particular, if H and Aut(K) are of even order and Pr(H, Aut(K)) =
H
∼
L(H,Aut(K)) = Z2 × Z2 .

5
8

then

Proof. By (1.1), we have (q 2 + p − 1)|H|| Aut(K)| = pq 2 |{(x, α) ∈ H × Aut(K) :
[x, α] = 1}|. Therefore, pq divides |H|| Aut(K)|.
If p and q are the smallest primes dividing | Aut(K)| and |H| respectively
then, by Theorem 3.7, we have


q2 + p − 1
p−1
1
≤
+
1
pq 2
p |H : L(H, Aut(K))|
which gives |H : L(H, Aut(K))| ≤ q 2 . Since H is non-abelian we have |H :
H
∼
L(H, Aut(K))| 6= 1, q. Hence, L(H,Aut(K))
= Zq × Zq .

10

The following two results give partial converses of Theorem 4.1 and 4.2
respectively.
Proposition 4.3. Let H and K be two subgroups of a finite group G such that
H ⊆ K. Let p, q be the smallest prime divisors of | Aut(K)|, |H| respectively
and | Aut(K) : CAut(K) (x)| = p for all x ∈ H \ L(H, Aut(K)).
(a) If

H
L(H,Aut(K))

∼
= Zq then Pr(H, Aut(K)) =

(b) If

H
L(H,Aut(K))

∼
= Zq × Zq then Pr(H, Aut(K)) =

p+q−1
pq .
q2 +p−1
pq2 .

Proof. Since | Aut(K) : CAut(K) (x)| = p for all x ∈ H \ L(H, Aut(K)) we have
|CAut(K) (x)| = | Aut(K)|
for all x ∈ H \ L(H, Aut(K)). Therefore, by Corollary
p
2.4, we have
Pr(H, Aut(K)) =
=

|L(H, Aut(K))|
1
+
|H|
|H|| Aut(K)|

Pr(H, Aut(K)) =

(b) If

5

H
L(H,Aut(K))
H
L(H,Aut(K))

|CAut(K) (x)|

x∈H\L(H,Aut(K))

|L(H, Aut(K))| |H| − |L(H, Aut(K))|
+
.
|H|
p|H|

Thus

(a) If

X

1
p




p−1
+1 .
|H : L(H, Aut(K))|

(4.1)

∼
= Zq then (4.1) gives Pr(H, Aut(K)) = p+q−1
pq .
2
∼
= Zq × Zq then (4.1) gives Pr(H, Aut(K)) = q +p−1
pq2 .

Autoisoclinic pairs

In the year 1940, Hall [4] introduced the concept of isoclinism between two
groups. Following Hall, Moghaddam et al. [7] have defined autoisoclinism
between two groups, in the year 2013. Recall that two groups G1 and G2
G2
G1
→ L(G
,
are said to be autoisoclinic if there exist isomorphisms ψ : L(G
1)
2)
β : [G1 , Aut(G1 )] → [G2 , Aut(G2 )] and γ : Aut(G1 ) → Aut(G2 ) such that the
following diagram commutes
G1
L(G1 )

ψ×γ

× Aut(G1 ) −−−−→

a(G ,Aut(G ))
1
y 1

[G1 , Aut(G1 )]

β

−−−−→

G2
L(G2 )

× Aut(G2 )

a(G ,Aut(G ))
2
y 2

[G2 , Aut(G2 )]

Gi
× Aut(Gi ) → [Gi , Aut(Gi )], for i = 1, 2,
where the maps a(Gi ,Aut(Gi )) : L(G
i)
are given by
a(Gi ,Aut(Gi )) (xi L(Gi ), αi ) = [xi , αi ].

Such a pair (ψ × γ, β) is called an autoisoclinism between the groups G1 and
G2 . We generalize the notion of autoisoclinism in the following way:
11

Let H1 , K1 and H2 , K2 be subgroups of the groups G1 and G2 respectively.
The pairs of subgroups (H1 , K1 ) and (H2 , K2 ) such that H1 ⊆ K1 and H2 ⊆ K2
H1
are said to be autoisoclinic if there exist isomorphisms ψ : L(H1 ,Aut
K1 ) →
H2
L(H2 ,Aut(K2 )) ,

β : [H1 , Aut(K1 )] → [H2 , Aut(K2 )] and γ : Aut(K1 ) → Aut(K2 )
such that the following diagram commutes
H1
L(H1 ,Aut(K1 ))

ψ×γ

× Aut(K1 ) −−−−→

a(H ,Aut(K ))
1
y 1

[H1 , Aut(K1 )]

where the maps a(Hi ,Aut(Ki )) :
i = 1, 2, are given by

H2
L(H2 ,Aut(K2 ))

β

−−−−→

Hi
L(Hi ,Aut(Ki ))

× Aut(K2 )

a(H ,Aut(K ))
2
y 2

[H2 , Aut(K2 )]

× Aut(Ki ) → (Hi , Aut(Ki )), for

a(Hi ,Aut(Ki )) (xi L(Hi , Aut(Ki )), αi ) = [xi , αi ].
Such a pair (ψ × γ, β) is said to be an autoisoclinism between the pairs of groups
(H1 , K1 ) and (H2 , K2 ). We conclude this paper with the following generalization
of [3, Theorem 5.1] and [10, Lemma 2.5].
Theorem 5.1. Let G1 and G2 be two finite groups with subgroups H1 , K1 and
H2 , K2 respectively such that H1 ⊆ K1 and H2 ⊆ K2 . If (ψ × γ, β) is an
autoisoclinism between the pairs (H1 , K1 ) and (H2 , K2 ) then, for g ∈ K1 ,
Prg (H1 , Aut(K1 )) = Prβ(g) (H2 , Aut(K2 )).
H1
×
Proof. Let us consider the sets Sg = {(x1 L(H1 , Aut(K1 )), α1 ) ∈ L(H1 ,Aut(K
1 ))
Aut(K1 ) : [x1 L(H1 , Aut(K1 )), α1 ] = g} and Tβ(g) = {(x2 L(H2 , Aut(K2 )), α2 ) ∈
H2
L(H2 ,Aut(K2 )) × Aut(K2 ) : [x2 L(H2 , Aut(K2 )), α2 ] = β(g)}. Since (H1 , K1 ) is
autoisoclinic to (H2 , K2 ) we have |Sg | = |Tβ(g) |. Again, it is clear that

|{(x1 , α1 ) ∈ H1 × Aut(K1 ) : [x1 , α1 ] = g}| = |L(H1 , Aut(K1 ))||Sg |

(5.1)

and
|{(x2 , α2 ) ∈ H2 × Aut(K2 ) : [x2 , α2 ] = β(g)}| = |L(H2 , Aut(K2 ))||Tβ(g) |. (5.2)
Hence, the result follows from (1.1), (5.1) and (5.2).

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12

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