2 Basic definition and results
For the basic concepts of loop theory, we refer to Bruck [1, 2].
Here we review some definitions, notation and results.
Let 𝑄 be a loop.
Set
A
=
{
L
c
∣
c
∈
Q
}
,
B
=
{
R
d
∣
d
∈
Q
}
.
Then 𝐴 and 𝐵 are left (and right) transversals to
Inn
Q
in
Mlt
Q
,
⟨
A
,
B
⟩
=
Mlt
Q
,
[
A
,
B
]
≤
Inn
Q
and
core
Mlt
Q
Inn
Q
=
1
(i.e. the largest normal subgroup of
Mlt
Q
in
Inn
Q
is trivial).
As a consequence,
A
∩
Inn
Q
=
B
∩
Inn
Q
=
1
holds.
Conversely, consider a group 𝐺 with the following properties: 𝐻 is a subgroup of 𝐺, and
A
,
B
are left transversals to 𝐻 in 𝐺.
Then 𝐴 and 𝐵 are 𝐻-connected transversals by definition if
[
A
,
B
]
≤
H
.
By [10], the above two situations are equivalent.
Theorem 2.1
A group 𝐺 is isomorphic to the multiplication group of a loop if and only if there is a subgroup 𝐻, for which there exist 𝐻-connected transversals 𝐴 and 𝐵 such that
⟨
A
,
B
⟩
=
G
and
core
G
H
=
1
.
A subloop 𝑆 of a loop 𝑄 is normal if and only if
Inn
Q
acts upon 𝑆, which is equivalent to the following:
M
(
S
)
Inn
Q
is a subgroup of
Mlt
Q
,
where
M
(
S
)
=
⟨
L
c
,
R
c
∣
c
∈
S
⟩
.
(This is a standard fact.)
Let 𝑄 be a loop and 𝑆 a normal subloop of 𝑄.
Put
M
(
S
)
=
⟨
L
c
,
R
c
∣
c
∈
S
⟩
.
Put
K
(
S
)
=
core
Mlt
Q
(
M
(
S
)
Inn
Q
)
.
Denote by 𝑓 the natural homomorphism of
Mlt
Q
onto
Mlt
Q
/
K
(
S
)
.
Then
f
(
A
)
and
f
(
B
)
are
f
(
Inn
Q
)
-connected transversals in
Mlt
Q
/
K
(
S
)
, and the multiplication group of the factor-loop
Q
/
S
,
Mlt
(
Q
/
S
)
≅
Mlt
Q
/
K
(
S
)
.
The permutation group generated by all left translations is called the left multiplication group, and we shall denote it by
L
=
L
(
Q
)
=
⟨
A
⟩
.
In a similar way, the right multiplication group
R
=
R
(
Q
)
=
⟨
B
⟩
is generated by all right translations.
Let
L
1
=
L
∩
Inn
Q
and
R
1
=
R
∩
Inn
Q
.
Proposition 2.2
We have
L
1
=
⟨
L
x
y
-
1
L
x
L
y
∣
x
,
y
∈
Q
⟩
,
R
1
=
⟨
R
y
x
-
1
R
x
R
y
∣
x
,
y
∈
Q
⟩
,
and
Inn
Q
is generated by
L
1
∪
R
1
∪
{
T
x
∣
x
∈
Q
}
, where
T
x
=
R
x
-
1
L
x
for all
x
∈
Q
.
The left, middle and right nucleus of a loop 𝑄 are defined, respectively, by
N
λ
=
N
λ
(
Q
)
:=
{
a
∈
Q
∣
a
(
x
y
)
=
(
a
x
)
y
for all
x
,
y
∈
Q
}
,
N
μ
=
N
μ
(
Q
)
:=
{
a
∈
Q
∣
x
(
a
y
)
=
(
x
a
)
y
for all
x
,
y
∈
Q
}
,
N
ϱ
=
N
ϱ
(
Q
)
:=
{
a
∈
Q
∣
x
(
y
a
)
=
(
x
y
)
a
for all
x
,
y
∈
Q
}
.
The intersection
N
=
N
(
Q
)
=
N
λ
∩
N
μ
∩
N
ϱ
is called the nucleus of 𝑄.
The center of 𝑄 is defined by
Z
(
Q
)
=
{
a
∈
N
∣
x
a
=
a
x
for every element
x
∈
Q
}
.
Denote by
C
(
Q
)
the commutant of 𝑄, i.e.
C
(
Q
)
=
{
x
∈
Q
∣
L
x
=
R
x
}
.
Proposition 2.3
Let 𝑄 be a loop.
Then
C
Mlt
Q
(
R
)
=
{
L
c
∣
c
∈
N
λ
}
,
C
Mlt
Q
(
L
)
=
{
R
d
∣
d
∈
N
ϱ
}
.
If
R
⊴
Mlt
Q
, then
C
Mlt
Q
(
R
)
⊴
Mlt
Q
and
N
λ
⊴
Q
.
If
L
⊴
Mlt
Q
, then
C
Mlt
Q
(
L
)
⊴
Mlt
Q
and
N
ϱ
⊴
Q
.
Proof
See [4, Lemma 1.7].
∎
3 Moufang loops
Let 𝑄 be a Moufang loop.
Let
G
=
Mlt
Q
,
H
=
Inn
Q
,
A
=
{
L
x
∣
x
∈
Q
}
,
B
=
{
R
x
∣
x
∈
Q
}
.
In the language of H-connected transversals, the definition of Moufang loops is the following:
a
A
a
=
A
,
b
B
b
=
B
for every
a
∈
A
,
b
∈
B
.
These conditions are equivalent to the Moufang law (giving the left and right Bol identities, respectively).
The consequences of these are
a
A
b
-
1
=
A
,
b
B
a
-
1
=
B
,
A
b
a
=
A
,
B
a
b
=
B
(
A
b
=
{
α
b
∣
α
∈
A
,
b
∈
B
}
)
for every
a
∈
A
,
b
∈
B
∩
a
H
, i.e.
a
=
L
x
and
b
=
R
x
with
x
∈
Q
.
As usual,
[
A
,
B
]
≤
H
,
⟨
A
,
B
⟩
=
G
,
core
G
H
=
1
and
A
∩
H
=
B
∩
H
=
1
.
Let
A
*
=
C
G
(
B
)
,
B
*
=
C
G
(
A
)
.
Proposition 3.1
The following statements hold.
For arbitrary
a
∈
A
,
b
∈
B
∩
a
H
and for all integers 𝑘,
a
b
=
b
a
,
a
k
∈
A
,
b
k
∈
B
∩
a
k
H
and
b
∈
B
∩
H
a
,
o
(
a
)
=
o
(
b
)
,
i.e. the order of
a
is equal to the order of
b
.
⟨
A
⟩
∩
H
=
⟨
B
⟩
∩
H
=
[
A
,
B
]
and
H
=
⟨
[
A
,
B
]
,
a
-
1
b
∣
a
∈
A
,
b
∈
B
∩
a
H
⟩
.
⟨
A
⟩
,
⟨
B
⟩
,
A
*
=
C
G
(
B
)
and
B
*
=
C
G
(
A
)
are normal subgroups in 𝐺.
N
λ
=
N
ϱ
=
N
μ
=
N
and
N
⊴
Q
(i.e. 𝑁 is a normal subloop in 𝑄).
B
*
=
C
G
(
A
)
=
{
R
x
∣
x
∈
N
}
⊆
B
,
B
*
⊴
⟨
B
⟩
,
B
*
B
=
B
B
*
=
B
,
A
*
=
C
G
(
B
)
=
{
L
x
∣
x
∈
N
}
⊆
A
,
A
*
⊴
⟨
A
⟩
,
A
*
A
=
A
A
*
=
A
.
Proof
(i) From the definition of Moufang loops.
(ii) See Proposition 2.2 and the definition.
(iii) From the definition.
(iv) See (iii) and Proposition 2.3.
∎
Proposition 3.2
Proposition 3.2 ([3, Lemma 3.2])
The following statements hold.
[
a
,
b
]
a
j
∈
H
,
[
a
,
b
]
b
j
∈
H
for arbitrary
a
∈
A
,
b
∈
B
and for all integers 𝑗.
Let
a
∈
A
,
b
∈
B
such that
[
a
,
b
]
∈
H
∩
N
G
(
A
)
.
Then
a
∈
C
G
(
[
a
,
b
]
)
and
b
∈
C
G
(
[
a
,
b
]
)
.
Proposition 3.3
Proposition 3.3 ([3, Proposition 3.4])
The following statements are true for every
a
,
a
1
∈
A
,
b
∈
B
∩
a
H
,
b
1
∈
B
∩
a
1
H
:
a
a
1
b
-
1
=
a
b
1
a
1
∈
A
,
b
b
1
a
-
1
=
b
a
1
b
1
∈
B
.
Proposition 3.4
Proposition 3.4 (see [3, Proposition 3.5])
For every
a
∈
A
,
b
∈
B
∩
a
H
, the following statements are true.
Proposition 3.5
Proposition 3.5 (see [3, Proposition 3.6])
For every
a
1
,
a
2
∈
A
,
b
1
∈
B
∩
a
1
H
and
b
2
∈
B
∩
a
2
H
, the following statements are true.
(
a
1
a
2
)
-
1
b
2
b
1
∈
H
.
Let
a
=
a
1
a
2
b
1
-
1
∈
A
,
b
=
b
2
b
1
a
2
-
1
∈
B
.
Then
b
-
1
a
∈
H
.
Proposition 3.6
Proposition 3.6 (see [3, Proposition 3.7])
For every
a
1
,
a
2
∈
A
,
b
1
∈
B
∩
a
1
H
,
b
2
∈
B
∩
a
2
H
, the following statements are true.
b
1
b
2
b
1
-
1
(
b
2
-
1
)
a
1
∈
N
G
(
A
)
.
a
1
a
2
a
1
-
1
(
a
2
-
1
)
b
1
∈
N
G
(
B
)
.
Proposition 3.7
The following statements hold.
Let
h
∈
H
∩
N
G
(
A
)
,
a
∈
A
,
b
∈
B
∩
a
H
.
Then
h
a
=
h
α
b
-
1
with
α
∈
A
.
Let
h
∈
H
∩
N
G
(
B
)
,
b
∈
B
,
a
∈
A
∩
b
H
.
Then
h
b
=
h
β
a
-
1
with
β
∈
B
.
Proof
(i) Clearly,
h
-
1
h
a
=
(
a
-
1
)
h
a
holds, but using
h
∈
N
G
(
A
)
, it follows that
(
a
-
1
)
h
=
a
*
∈
A
.
Hence
h
-
1
h
a
=
a
*
a
.
By Proposition 3.3,
(
a
*
)
b
a
=
α
∈
A
.
Since
(
a
*
)
b
a
=
(
a
*
a
)
b
, we get
a
*
a
=
α
b
-
1
, consequently,
h
a
=
h
α
b
-
1
.
(ii) Similarly.
∎
Proposition 3.8
The following statements hold.
Let
h
∈
H
∩
N
G
(
A
)
,
a
∈
A
such that
h
a
∈
H
.
Then
h
a
=
h
.
Let
h
∈
H
∩
N
G
(
B
)
,
b
∈
B
such that
h
b
∈
H
.
Then
h
b
=
h
.
Proof
(i) By Proposition 3.7, we have
h
a
=
h
a
*
b
-
1
with
a
*
∈
A
,
b
∈
B
∩
a
H
.
Since
[
A
,
B
]
≤
H
, then clearly
(
a
*
)
b
-
1
=
h
0
a
*
with
h
0
∈
H
, whence
h
a
=
h
h
0
a
*
.
(𝐴 is a right transversal to 𝐻.)
But we have
h
a
∈
H
, whence
a
*
∈
H
, i.e.
a
*
=
e
, consequently,
h
a
=
h
.
(ii) Similarly.
∎
4 Main results
In this section, 𝑄 is a Moufang loop.
Let
G
=
Mlt
Q
,
H
=
Inn
Q
,
A
=
{
L
x
∣
x
∈
Q
}
,
B
=
{
R
x
∣
x
∈
Q
}
.
We have
[
A
,
B
]
≤
H
,
⟨
A
,
B
⟩
=
G
and
core
G
H
=
1
.
The commutant of 𝑄 is
C
(
Q
)
=
{
x
∈
Q
∣
L
x
=
R
x
}
.
We denote
A
0
=
{
L
x
∣
x
∈
C
(
Q
)
}
.
Clearly,
A
0
=
A
∩
B
.
Proposition 4.1
The following statements hold.
There exists an automorphism 𝜎 of the multiplication group of 𝑄 such that
σ
(
L
x
)
=
R
x
-
1
,
σ
(
R
x
)
=
L
x
-
1
.
σ
(
h
)
=
h
for every
h
∈
H
(
=
Inn
Q
).
Proof
(i) This is a well-known fact (see [7]).
(ii) By Proposition 3.1 (ii), we have
H
=
⟨
[
A
,
B
]
,
a
-
1
b
∣
a
∈
A
,
b
∈
B
∩
a
H
⟩
.
Thus we can assume
a
=
L
x
,
b
=
R
x
for some
x
∈
Q
.
Clearly, by the definition of 𝜎,
σ
(
a
-
1
b
)
=
a
-
1
b
.
Let
a
1
∈
A
,
b
2
∈
B
be arbitrary, and let
b
1
∈
B
∩
a
1
H
.
Consider
a
1
-
1
(
a
1
)
b
2
.
We have to show
σ
(
a
1
-
1
(
a
1
)
b
2
)
=
a
1
-
1
(
a
1
)
b
2
.
By the definition,
σ
(
a
1
-
1
(
a
1
)
b
2
)
=
b
1
(
b
1
-
1
)
a
2
-
1
with
a
2
∈
A
∩
b
2
H
.
Thus our aim is to prove
b
1
(
b
1
-
1
)
a
2
-
1
=
a
1
-
1
(
a
1
)
b
2
, which is equivalent to
a
1
(
a
2
)
b
1
-
1
=
(
a
1
)
b
2
a
2
.
By Proposition 3.3, this is true.
∎
Proposition 4.2
Let
a
0
=
b
0
∈
A
0
(
=
A
∩
B
).
Then the following statements are true.
a
0
3
=
b
0
3
∈
C
G
(
A
)
∩
C
G
(
B
)
(
=
Z
(
G
)
).
Let
b
∈
B
be arbitrary.
Then
a
0
b
=
a
0
h
with
h
∈
H
∩
N
G
(
A
)
∩
N
G
(
B
)
.
Proof
(i) By Proposition 3.4,
a
0
b
0
2
∈
N
G
(
A
)
.
Since
a
0
=
b
0
, it follows that
a
b
0
3
∈
A
for all
a
∈
A
.
As
a
b
0
3
∈
a
H
(
[
A
,
B
]
≤
H
), we get
b
0
3
∈
C
G
(
a
)
, consequently,
b
0
3
∈
C
G
(
A
)
.
As
b
0
∈
A
∩
B
, similarly, we get
b
0
3
∈
C
G
(
B
)
, consequently,
b
0
3
∈
Z
(
G
)
because of
G
=
⟨
A
,
B
⟩
.
(ii) Applying Proposition 3.6 (i) instead of
a
1
and
b
1
for
a
0
(
=
b
0
) and instead of
b
2
for
b
-
1
, we get
(
b
-
1
)
a
0
-
1
b
a
0
∈
N
G
(
A
)
.
We have
a
0
b
=
a
0
h
with
h
∈
H
(
[
A
,
B
]
≤
H
), whence
b
a
0
=
b
h
-
1
and
(
b
-
1
)
a
0
-
1
=
(
h
-
1
)
a
0
-
1
b
-
1
.
So
(
b
-
1
)
a
0
-
1
b
a
0
=
(
h
-
1
)
a
0
-
1
h
-
1
∈
N
G
(
A
)
.
Let
t
=
(
h
-
1
)
a
0
-
1
h
-
1
.
Clearly, we have
b
a
0
-
1
=
b
(
h
)
a
0
-
1
.
Then
[
A
,
B
]
≤
H
gives
(
h
)
a
0
-
1
∈
H
, and hence
t
∈
H
∩
N
G
(
A
)
.
Clearly,
b
a
0
2
=
b
h
-
1
(
h
-
1
)
a
0
=
b
t
a
0
.
Then
[
A
,
B
]
≤
H
gives
t
a
0
∈
H
.
Proposition 3.8 implies
t
a
0
=
t
, whence
b
-
1
b
a
0
2
∈
H
∩
N
G
(
A
)
.
Since
a
0
3
=
Z
(
G
)
(see Proposition 4.2 (i)), it follows that
t
a
0
=
t
=
b
-
1
b
a
0
-
1
∈
H
∩
N
G
(
A
)
.
Clearly,
t
a
0
2
=
(
b
-
1
)
a
0
b
and
t
a
0
2
=
t
a
0
=
t
∈
H
∩
N
G
(
A
)
holds, whence
t
a
0
2
=
a
0
-
1
a
0
b
=
h
∈
H
∩
N
G
(
A
)
.
As
σ
(
h
)
=
h
,
σ
(
A
)
=
B
(Proposition 4.1), we can conclude that
h
∈
H
∩
N
G
(
B
)
is true.
∎
Proposition 4.3
Let
a
0
∈
A
0
(
=
A
∩
B
),
b
∈
B
be arbitrary.
We have
a
0
b
=
a
0
h
with
h
∈
H
.
Then
h
a
=
h
b
=
h
a
0
=
h
,
a
0
a
=
a
0
b
,
a
0
b
-
1
=
a
0
a
-
1
=
a
0
h
-
1
with
a
∈
A
∩
b
H
.
Proof
By Proposition 4.2 (ii),
h
∈
H
∩
N
G
(
A
)
∩
N
G
(
B
)
.
Proposition 3.2 gives
h
b
∈
H
,
h
a
0
∈
H
, whence, by Proposition 3.8,
h
b
=
h
a
0
=
h
holds.
Applying 𝜎 for both sides of
a
0
b
=
a
0
h
, we get
(
b
0
-
1
)
a
-
1
=
b
0
-
1
h
with
b
0
=
a
0
(
∈
A
∩
B
).
So we have
a
0
a
-
1
=
h
-
1
a
0
.
Using
h
a
0
=
h
, it follows that
a
0
a
-
1
=
a
0
h
-
1
.
From
a
0
b
=
a
0
h
, it holds that
a
0
b
-
1
=
a
0
(
h
-
1
)
b
-
1
.
As
h
b
=
h
, we can conclude that
a
0
b
-
1
=
a
0
h
-
1
.
So
a
0
a
-
1
=
a
0
b
-
1
, and since 𝑎 is an arbitrary element of 𝐴, we get
a
0
a
=
a
0
b
=
a
0
h
.
Clearly,
a
0
a
2
=
a
0
b
2
is true.
Then we have
a
0
a
2
=
a
0
h
h
a
and
a
0
b
2
=
a
0
h
h
b
, whence
h
b
=
h
a
=
h
holds.
∎
Proposition 4.4
Let
a
0
∈
A
0
,
b
∈
B
be arbitrary.
Assume
a
0
b
=
a
0
h
with
h
∈
H
.
Then
h
3
=
e
.
Proof
By Proposition 4.2 (ii),
h
∈
H
∩
N
G
(
A
)
∩
N
G
(
B
)
.
From
a
0
b
=
a
0
h
, it follows that
b
a
0
=
b
h
-
1
.
By Proposition 4.2 (i), we have
a
0
3
∈
Z
(
G
)
, whence
b
=
b
a
0
3
=
b
(
h
-
1
)
(
h
-
1
)
a
0
(
h
-
1
)
a
0
2
.
Proposition 4.3 gives
h
a
0
=
h
, so we get
b
=
b
(
h
-
1
)
3
, consequently,
h
3
=
e
.
∎
Proposition 4.5
Let
a
0
∈
A
0
,
b
1
,
b
2
∈
B
,
a
1
∈
A
∩
b
1
H
,
a
2
∈
A
∩
b
2
H
.
Assume
a
0
b
1
=
a
0
h
1
,
a
0
b
2
=
a
0
h
2
with
h
1
,
h
2
∈
H
.
Then
h
2
-
1
h
2
b
1
=
β
*
with
β
*
∈
B
,
(
β
*
)
a
1
=
β
*
,
(
β
*
)
a
2
=
β
*
,
(
h
1
h
2
)
b
1
b
2
=
h
1
h
2
.
Proof
By Proposition 4.2 (ii), we have
h
1
,
h
2
∈
H
∩
N
G
(
A
)
∩
N
G
(
B
)
.
Proposition 4.3 gives
h
1
b
1
=
h
1
,
h
2
b
2
=
h
2
.
Hence
a
0
b
2
b
1
b
2
=
a
0
h
2
(
h
1
h
2
)
b
1
b
2
.
Using
b
2
b
1
b
2
∈
B
(by the definition), Proposition 4.2 (ii) results in
(
h
1
h
2
)
b
1
b
2
∈
H
∩
N
G
(
A
)
∩
N
G
(
B
)
.
Consider
a
0
a
2
a
1
a
2
.
By using
a
0
b
2
=
a
0
a
2
,
a
0
b
1
=
a
0
a
1
,
h
1
a
1
=
h
1
,
h
2
a
2
=
h
2
(see Proposition 4.3), we get
a
0
a
2
a
1
a
2
=
a
0
h
2
(
h
1
h
2
)
a
1
a
2
.
We have
σ
(
b
2
b
1
b
2
)
=
a
2
-
1
a
1
-
1
a
2
-
1
(see Proposition 4.1), from which it holds that
a
2
a
1
a
2
∈
A
∩
b
2
b
1
b
2
H
.
By Proposition 4.3,
a
0
b
2
b
1
b
2
=
a
0
a
2
a
1
a
2
is true, consequently,
(4.1)
(
h
1
h
2
)
b
1
b
2
=
(
h
1
h
2
)
a
1
a
2
∈
H
∩
N
G
(
A
)
∩
N
G
(
B
)
.
Applying 𝜎 for (4.1) (see Proposition 4.1),
(
h
1
h
2
)
b
1
b
2
=
(
h
1
h
2
)
a
1
-
1
a
2
-
1
=
(
h
1
h
2
)
a
1
a
2
=
(
h
1
h
2
)
b
1
-
1
b
2
-
1
.
Since
(
h
1
h
2
)
b
1
b
2
∈
H
∩
N
G
(
B
)
, conjugating this by
b
2
-
1
(
∈
B
), Proposition 3.7 gives
(4.2)
(
h
1
h
2
)
b
1
=
(
h
1
h
2
)
b
1
b
2
(
β
*
)
a
2
with
β
*
∈
B
.
Using
h
1
h
2
∈
H
∩
N
G
(
B
)
, Proposition 3.7 results in
(
h
1
h
2
)
b
1
=
h
1
h
2
(
β
△
)
a
1
-
1
with
β
△
∈
B
.
Then
[
A
,
B
]
≤
H
implies
(
β
*
)
a
2
∈
H
β
*
and
(
β
△
)
a
1
-
1
∈
H
β
△
.
As
(
h
1
h
2
)
b
1
b
2
,
h
1
h
2
∈
H
and 𝐵 is a right transversal to 𝐻, we get
β
*
=
β
△
.
Thus
(4.3)
(
h
1
h
2
)
b
1
=
h
1
h
2
(
β
*
)
a
1
-
1
.
By the definition,
b
*
=
b
2
-
1
(
b
1
)
a
2
∈
B
.
Using Proposition 4.3, it follows that
a
0
b
*
=
a
0
b
2
-
1
a
2
-
1
b
1
a
2
=
(
a
0
h
1
(
h
2
-
2
)
b
1
)
a
2
=
a
0
h
2
(
h
1
h
2
-
2
)
b
1
a
2
.
By Proposition 4.4,
h
2
-
2
=
h
2
.
Thus
a
0
b
*
=
a
0
h
2
(
h
1
h
2
)
b
1
a
2
.
Proposition 4.2 (ii) implies
(
h
1
h
2
)
b
1
a
2
∈
H
∩
N
G
(
A
)
∩
N
G
(
B
)
.
Conjugating this by
a
2
-
1
, Proposition 3.7 gives
(
h
1
h
2
)
b
1
=
(
h
1
h
2
)
b
1
a
2
α
¯
b
2
-
1
with
α
¯
∈
A
.
By (4.3),
(
h
1
h
2
)
b
1
=
h
1
h
2
(
β
*
)
a
1
-
1
.
Then
[
A
,
B
]
≤
H
implies
(
β
*
)
a
1
-
1
∈
H
β
*
(
(
β
*
)
-
1
∈
B
)
,
α
¯
b
2
-
1
∈
H
α
¯
(
(
α
¯
)
-
1
∈
A
,
b
2
-
1
∈
B
)
.
We have
h
1
,
h
2
,
(
h
1
h
2
)
b
1
a
2
∈
H
.
Then 𝐴 and 𝐵 are right transversals to 𝐻, consequently,
α
¯
∈
H
β
*
.
Hence
α
¯
(
β
*
)
-
1
∈
H
.
By Proposition 3.1 (i),
(
β
*
)
-
1
α
¯
∈
H
.
Let
α
*
=
α
¯
, i.e.
(
β
*
)
-
1
α
*
∈
H
.
Thus
(
h
1
h
2
)
b
1
=
(
h
1
h
2
)
b
1
a
2
(
α
*
)
b
2
-
1
.
Conjugating both sides of this by
b
2
and using
a
2
b
2
=
b
2
a
2
, we get
(
h
1
h
2
)
b
1
b
2
=
(
h
1
h
2
)
b
1
b
2
a
2
α
*
,
whence
(4.4)
(
h
1
h
2
)
b
1
b
2
a
2
=
(
h
1
h
2
)
b
1
b
2
(
α
*
)
-
1
.
By (4.1), we have
(
h
1
h
2
)
b
1
b
2
∈
H
.
Conjugating this by
a
2
, using Proposition 3.7, we get
(4.5)
(
h
1
h
2
)
b
1
b
2
a
2
=
(
h
1
h
2
)
b
1
b
2
(
α
¯
¯
)
b
2
-
1
with
α
¯
¯
∈
A
.
Our (4.4) and (4.5) give
(
α
*
)
-
1
=
(
α
¯
¯
)
b
2
-
1
.
Since
(
α
*
)
-
1
∈
A
,
(
α
¯
¯
)
b
2
-
1
∈
α
¯
¯
H
and 𝐴 is a left transversal to 𝐻, it follows that
(
α
¯
¯
)
b
2
-
1
=
α
¯
¯
.
Hence
(
α
*
)
-
1
=
α
¯
¯
and
(
(
α
*
)
-
1
)
b
2
-
1
=
(
α
*
)
-
1
.
Applying 𝜎, we get
(4.6)
(
β
*
)
a
2
=
β
*
.
From (4.2), we get
(4.7)
(
h
1
h
2
)
b
1
=
(
h
1
h
2
)
b
1
b
2
β
*
,
but
(
h
1
h
2
)
b
1
=
(
h
1
h
2
)
(
β
*
)
a
1
-
1
(see (
4.3
))
.
Hence
(4.8)
(
β
*
)
a
1
-
1
(
β
*
)
-
1
=
(
h
1
h
2
)
-
1
(
h
1
h
2
)
b
1
b
2
.
We have
(4.9)
(
h
1
h
2
)
-
1
(
h
1
h
2
)
b
1
b
2
∈
H
∩
N
G
(
A
)
∩
N
G
(
B
)
(see Proposition 4.2 and (4.1)).
By Proposition 3.7,
(4.10)
(
h
1
)
b
2
=
h
1
β
1
a
2
-
1
with
β
1
∈
B
.
As
(
h
1
h
2
)
b
1
=
(
h
1
h
2
)
(
β
*
)
a
1
-
1
(see (4.3)), using
(
h
1
)
b
1
=
h
1
(see Proposition 4.3) we get
(4.11)
(
h
2
)
b
1
=
h
2
(
β
*
)
a
1
-
1
.
Consequently,
(4.12)
(
h
1
h
2
)
-
1
(
h
1
h
2
)
b
1
b
2
=
h
2
-
1
β
1
a
2
-
1
h
2
(
β
*
)
a
1
-
1
b
2
=
β
1
a
2
-
1
h
2
(
β
*
)
a
1
-
1
b
2
=
(
β
*
)
a
1
-
1
(
β
*
)
-
1
∈
H
(see Proposition 4.3 and (4.8), (4.11), (4.10), (4.9)).
We have
(
β
*
)
a
1
-
1
=
β
*
h
¯
with
h
¯
∈
H
, whence
(
β
*
)
a
1
-
1
(
β
*
)
-
1
=
h
¯
(
β
*
)
-
1
∈
H
∩
N
G
(
A
)
∩
N
G
(
B
)
(see (4.9) and (4.12)).
As
(
h
¯
(
β
*
)
-
1
)
β
*
=
h
¯
∈
H
, Proposition 3.8 gives
h
¯
=
h
¯
β
*
∈
H
∩
N
G
(
A
)
∩
N
G
(
B
)
,
whence
(
β
*
)
a
1
-
1
(
β
*
)
-
1
=
h
¯
=
(
β
*
)
-
1
(
β
*
)
a
1
-
1
.
Thus
(4.13)
β
1
a
2
-
1
h
2
(
β
*
)
a
1
-
1
b
2
=
h
¯
=
(
β
*
)
-
1
(
β
*
)
a
1
-
1
.
We show
h
¯
b
2
-
1
=
h
¯
.
Consider
(
β
*
)
a
2
a
1
-
1
a
2
.
Clearly,
a
2
a
1
-
1
a
2
∈
A
by definition.
By using
(
β
*
)
a
2
=
β
*
(see (4.6)), we get
(
β
*
)
a
2
a
1
-
1
a
2
=
β
*
h
¯
a
2
.
Since
h
¯
a
2
∈
H
(
[
A
,
B
]
≤
H
), applying 𝜎, it follows that
h
¯
a
2
=
h
¯
b
2
-
1
∈
H
.
Using
h
¯
∈
H
∩
N
G
(
B
)
, Proposition 3.8 gives
h
¯
b
2
-
1
=
h
¯
.
By using this, we get
β
1
a
2
-
1
h
2
b
2
-
1
(
β
*
)
a
1
-
1
=
h
¯
=
(
β
*
)
-
1
(
β
*
)
a
1
-
1
(see (
4.13
))
.
Hence
β
1
a
2
-
1
h
2
b
2
-
1
=
(
β
*
)
-
1
.
Proposition 3.4 gives
(
β
*
)
b
2
a
2
2
∈
B
.
As
(
β
*
)
a
2
=
β
*
(see (4.6)), it follows that
(
β
*
)
b
2
∈
B
.
By using
h
2
∈
N
G
(
B
)
, we get
β
1
a
2
-
1
∈
B
, whence
β
1
a
2
-
1
=
β
1
.
We have
(
h
1
)
b
2
=
h
1
β
1
a
2
-
1
(see (4.10)), consequently,
h
1
-
1
(
h
1
)
b
2
∈
B
.
Since
b
2
∈
B
is arbitrary,
h
1
=
a
0
-
1
(
a
0
)
b
1
and
b
1
∈
B
are arbitrary, and using
h
2
=
a
0
-
1
a
0
b
2
,
we can conclude that
h
2
-
1
h
2
b
1
∈
B
.
As
h
2
-
1
h
2
b
1
=
(
β
*
)
a
1
-
1
(see (4.11)), it follows that
(4.14)
(
β
*
)
a
1
=
β
*
,
h
2
-
1
h
2
b
1
=
β
*
,
Applying 𝜎, we get
h
2
-
1
h
2
a
1
-
1
=
(
α
*
)
-
1
with
α
*
∈
A
∩
β
*
H
and
(
α
*
)
b
1
=
α
*
.
From (4.3), we get
(4.15)
(
h
1
h
2
)
b
1
=
h
1
h
2
β
*
.
From (4.7) and (4.15), we can conclude that
(4.16)
(
h
1
h
2
)
b
1
b
2
=
h
1
h
2
.
Summarizing, (4.6), (4.14), (4.16) give our statement.
∎
Theorem 4.6
Let 𝑄 be a Moufang loop;
C
(
Q
)
is the commutant of 𝑄.
Let
A
0
=
{
L
x
∣
x
∈
C
(
Q
)
}
(
=
A
∩
B
)
.
Let
x
0
∈
C
(
Q
)
,
y
,
z
∈
Q
be arbitrary.
Then
C
(
Q
)
is a normal subloop in 𝑄 if and only if
(4.17)
[
L
x
0
,
R
y
]
L
z
=
[
L
x
0
,
R
y
]
R
z
∈
[
L
x
0
,
R
y
]
(
A
∩
B
)
,
[
[
L
x
0
,
R
y
]
,
L
z
]
=
[
[
L
x
0
,
R
y
]
,
R
z
]
=
R
w
=
L
w
∈
A
∩
B
with
w
∈
C
(
Q
)
.
Proof
Let
a
0
=
L
x
0
,
b
2
=
R
y
,
h
2
=
[
a
0
,
b
2
]
∈
H
,
a
1
=
L
z
,
b
1
=
R
z
.
First assume
C
(
Q
)
is a normal subloop in 𝑄.
The normality of the commutant gives that
A
0
H
is a subgroup in 𝐺 (
=
Mlt
Q
).
We have to show
h
2
a
1
=
h
2
b
1
∈
h
2
A
0
,
i.e.
h
2
-
1
h
2
a
1
=
h
2
-
1
h
2
b
1
∈
A
0
.
We need the following statement.
Claim 1
Claim 1 (see [10, Lemma 2.5])
Let 𝐺 be a group; 𝐻 is a subgroup of 𝐺, and
A
,
B
are 𝐻-connected transversals (
[
A
,
B
]
≤
H
).
Let
C
⊆
A
∪
B
,
K
=
⟨
H
,
C
⟩
.
Then
C
⊆
core
G
H
.
Applying this for our 𝐺 (
=
Mlt
Q
), for 𝐻 (
=
Inn
Q
),
C
=
A
0
,
K
=
⟨
A
0
,
H
⟩
.
Clearly,
K
=
A
0
H
, and we get
A
0
⊆
core
G
A
0
H
.
Let
M
=
core
G
A
0
H
.
Clearly,
(
a
0
)
b
2
=
a
0
h
2
∈
M
,
h
2
∈
M
,
(
h
2
)
b
1
∈
M
.
Proposition 4.5 gives
h
2
-
1
(
h
2
)
b
1
=
β
*
∈
B
∩
M
.
Since
B
∩
M
=
A
0
, it follows that
β
*
∈
A
0
(
=
A
∩
B
).
By Proposition 4.5, we have
(
β
*
)
a
1
=
β
*
, and applying 𝜎, we get
(
α
*
)
b
1
-
1
=
α
*
with
α
*
∈
A
∩
β
*
H
.
In our case,
α
*
=
β
*
, thus
(
h
2
-
1
)
b
1
-
1
h
2
=
α
*
=
β
*
, and applying 𝜎 again, we can conclude that
(
h
2
-
1
)
a
1
h
2
=
(
α
*
)
-
1
, consequently,
h
2
-
1
(
h
2
)
a
1
=
h
2
-
1
(
h
2
)
b
1
=
α
*
=
β
*
∈
A
∩
B
and
(
h
2
)
a
1
=
(
h
2
)
b
1
=
h
2
α
*
.
Conversely, assume 𝐺 satisfies our condition (4.17).
We have to show
C
(
Q
)
is a normal subloop in 𝑄, i.e. we need to prove
A
0
H
is a subgroup of 𝐺 (
=
Mlt
Q
).
Let
H
0
=
⟨
a
0
-
1
a
0
b
∣
a
0
∈
A
0
,
b
∈
B
⟩
.
We show that
A
0
H
0
is a normal subgroup in 𝐺 (
=
Mlt
Q
).
Claim 2
Let
γ
0
∈
A
0
,
h
0
=
a
0
-
1
a
0
b
with
a
0
∈
A
0
,
b
∈
B
.
Then the following statements are true.
h
0
γ
0
=
γ
h
0
with
γ
∈
A
0
, i.e.
h
0
γ
0
∈
A
0
H
0
.
(
γ
0
h
0
)
-
1
∈
A
0
H
0
.
Proof of Claim 2
(i) By Proposition 4.2,
h
0
∈
H
∩
N
G
(
A
)
∩
N
G
(
B
)
, and using
γ
0
∈
A
∩
B
, it follows that
γ
0
h
0
-
1
=
γ
∈
A
∩
B
=
A
0
.
(ii)
(
γ
0
h
0
)
-
1
=
h
0
-
1
γ
0
-
1
,
γ
0
-
1
∈
A
0
,
h
0
-
1
∈
H
0
, and (i) gives our statement.
∎
Claim 3
Let
γ
i
,
γ
j
∈
A
0
be arbitrary.
Then
γ
i
⋅
γ
j
∈
A
0
H
0
.
Proof of Claim 3
By definition, we have
γ
i
γ
j
β
i
-
1
∈
A
with
β
i
=
γ
i
.
As
γ
j
=
β
j
, we get
γ
i
γ
j
β
i
-
1
=
β
i
β
j
γ
i
-
1
,
but by definition,
β
i
β
j
γ
i
-
1
∈
B
, so
γ
i
γ
j
β
i
-
1
∈
A
0
(
=
A
∩
B
).
As
[
A
,
B
]
≤
H
, it follows that
γ
j
-
1
γ
j
β
i
-
1
∈
H
0
(see the definition of
H
0
).
Thus we can conclude that
γ
i
γ
j
∈
A
0
H
0
.
∎
Claim 4
Let
γ
0
∈
A
0
,
a
∈
A
,
b
∈
B
∩
a
H
.
Assume
γ
0
-
1
γ
0
b
=
h
*
∈
H
.
Then the following statements are true.
h
*
∈
H
0
and
γ
0
a
=
γ
0
b
=
γ
0
h
*
.
(
h
*
)
a
*
∈
A
0
H
0
,
(
h
*
)
b
*
∈
A
0
H
0
with
a
*
∈
A
,
b
*
∈
B
.
Proof of Claim 4
(i) See the definition of
H
0
and Proposition 4.3.
(ii) By using our condition (4.17),
(
h
*
)
-
1
(
h
*
)
b
*
∈
A
0
, whence
(
h
*
)
b
*
=
h
*
δ
with
δ
∈
A
0
.
Using Claim 2, we can conclude that
(
h
*
)
b
*
∈
A
0
H
0
.
Similarly,
(
h
*
)
a
*
∈
A
0
H
0
.
∎
Claim 5
A
0
H
0
is a normal subgroup of 𝐺.
Proof of Claim 5
Using
G
=
A
⋅
H
=
⟨
A
,
B
⟩
, Claims 2, 3, 4 imply our proof.
∎
Clearly,
A
0
H
=
A
0
H
0
H
.
From Claim 5, it follows that
A
0
H
is a subgroup of 𝐺.
Thus the commutant
C
(
Q
)
is a normal subloop of 𝑄.
∎
Proposition 4.7
Let 𝑄 be a finite Moufang loop, and assume the commutant
C
(
Q
)
is a normal subloop in 𝑄.
Let
A
0
=
{
L
x
∣
x
∈
C
(
Q
)
}
.
Assume 𝑄 is of odd order or the nucleus
N
(
Q
)
is trivial.
Then
Inn
Q
normalizes
A
0
, i.e.
A
0
h
=
A
0
for every
h
∈
H
(
=
Inn
Q
)
(
A
0
h
=
{
a
0
h
(
=
h
-
1
a
0
h
)
∣
h
∈
H
}
)
.
Proof
Let
a
0
∈
A
,
a
0
≠
e
,
h
∈
H
.
Let
d
=
a
0
h
.
Applying 𝜎 (see Proposition 4.1),
σ
(
d
)
=
(
a
0
-
1
)
h
=
d
-
1
.
The normality of
C
(
Q
)
implies
A
0
H
is a subgroup in 𝐺.
Thus
d
=
a
0
h
=
a
*
h
*
with
a
*
∈
A
0
,
h
*
∈
H
.
Then
σ
(
d
)
=
(
a
*
)
-
1
h
*
=
(
h
*
)
-
1
(
a
*
)
-
1
=
d
-
1
.
Hence
(
h
*
)
a
*
=
(
h
*
)
-
1
, consequently,
(
h
*
)
(
a
*
)
2
=
h
*
.
By using Proposition 4.2,
(
a
*
)
3
∈
C
G
(
A
)
∩
C
G
(
B
)
, i.e.
(
a
*
)
3
∈
Z
(
G
)
, which implies
(
h
*
)
a
*
=
h
*
, whence
(
h
*
)
2
=
e
holds.
If 𝑄 is of odd order, we can conclude that
h
*
=
e
, which implies
a
0
h
=
a
*
∈
A
0
.
If
N
(
Q
)
is trivial, then we have
C
G
(
A
)
=
1
(see Proposition 3.1), and we get that
a
0
and
a
*
are of order 3 (see Proposition 4.2).
Consequently,
a
0
h
=
a
*
h
*
is of order 3, too.
Since
a
*
h
*
=
h
*
a
*
, then
h
*
=
e
.
In this case,
a
0
h
=
a
*
∈
A
0
.
∎
We give a proof for the general case of the preceding statement, as a characterization.
Theorem 4.8
Let 𝑄 be a Moufang loop,
A
0
=
{
L
x
∣
x
∈
C
(
Q
)
}
.
Then the commutant
C
(
Q
)
is normal in 𝑄 if and only if
Inn
Q
normalizes
A
0
, i.e.
A
0
h
=
A
0
for every
h
∈
H
(
=
Inn
Q
)
.
Proof
Assume
C
(
Q
)
is a normal subloop in 𝑄.
Let
a
0
∈
A
0
,
b
1
,
b
2
∈
B
,
a
1
∈
A
∩
b
1
H
,
a
2
∈
A
∩
b
2
H
,
(
a
0
)
b
1
=
a
0
h
1
,
(
a
0
)
b
2
=
a
0
h
2
with
h
1
,
h
2
∈
H
.
By Theorem 4.6, we have
(
h
2
)
a
1
=
(
h
2
)
b
1
=
h
2
β
*
with
β
*
∈
A
0
, i.e.
β
*
=
α
*
(
α
*
∈
A
∩
β
*
H
).
We need the following statements.
Claim 6
We have
(
β
*
)
a
1
=
β
*
,
(
β
*
)
b
1
=
β
*
,
(
β
*
)
a
2
=
β
*
,
(
β
*
)
b
2
=
β
*
,
h
2
b
1
=
h
2
β
*
,
h
2
b
1
-
1
=
h
2
(
β
*
)
-
1
,
h
2
a
1
=
h
2
β
*
,
h
2
a
1
-
1
=
h
2
(
β
*
)
-
1
.
Proof of Claim 6
By Proposition 4.5,
(
β
*
)
a
1
=
(
β
*
)
a
2
=
β
*
.
Since
β
*
∈
A
0
, Proposition 4.3 gives
(
β
*
)
b
1
=
(
β
*
)
b
2
=
β
*
.
We have
(
h
2
)
b
1
=
(
h
2
)
a
1
=
h
2
β
*
.
Hence we get
(
h
2
)
a
1
-
1
=
(
h
2
)
b
1
-
1
=
h
2
(
β
*
)
-
1
.
∎
Claim 7
a
0
β
*
=
a
0
(
a
0
=
b
0
∈
A
∩
B
).
Proof of Claim 7
From Proposition 3.3,
b
0
b
1
a
0
-
1
∈
B
.
Consider
h
2
b
0
b
1
a
0
-
1
.
By using
b
0
∈
C
G
(
h
2
)
(see Proposition 4.3), it follows that
h
2
b
0
b
1
a
0
-
1
=
h
2
b
1
a
0
-
1
=
(
h
2
β
*
)
a
0
-
1
=
h
2
(
β
*
)
a
0
-
1
(see Claim
6
)
.
By using
b
0
b
1
a
0
-
1
∈
B
and that
C
(
Q
)
is a normal subloop, Theorem 4.6 implies
h
2
-
1
h
2
b
0
b
1
a
0
-
1
∈
B
,
whence
(
β
*
)
a
0
-
1
=
β
*
.
∎
Claim 8
(i)
h
1
β
*
=
h
1
,
(ii)
h
2
β
*
=
h
2
.
Proof of Claim 8
(i) We have
a
0
b
1
=
a
0
h
1
.
By definition, we have
β
*
b
1
(
α
*
)
-
1
∈
B
(
α
*
=
β
*
).
Using
a
0
β
*
=
a
0
α
*
=
a
0
(see Claim 7), we get
a
0
β
*
b
1
(
α
*
)
-
1
=
a
0
(
h
1
)
(
α
*
)
-
1
.
As
h
1
∈
H
∩
N
G
(
A
)
(see Proposition 4.2) and
h
1
(
α
*
)
-
1
∈
H
(
[
A
,
B
]
≤
H
),
Proposition 3.8 gives
h
1
(
α
*
)
-
1
=
h
1
,
i.e.
h
1
β
*
=
h
1
.
(ii) Similarly.
∎
Claim 9
h
1
h
2
=
h
2
h
1
.
Proof of Claim 9
We have
h
2
-
1
(
h
2
)
b
1
=
β
*
(see Claim 6).
Using Proposition 4.3, we get
a
0
β
*
=
a
0
h
2
-
1
(
h
2
)
b
1
=
(
a
0
h
1
-
1
)
h
2
b
1
=
a
0
h
1
(
h
1
-
1
)
h
2
b
1
.
Then
[
A
,
B
]
≤
H
implies
(
h
1
-
1
)
h
2
b
1
∈
H
.
As
(
h
1
-
1
)
h
2
∈
H
∩
N
G
(
B
)
(see Proposition 4.2), Proposition 3.8 gives
(
h
1
-
1
)
h
2
b
1
=
(
h
1
-
1
)
h
2
.
By Claim 7,
a
0
β
*
=
a
0
, consequently,
h
1
h
2
=
h
1
.
∎
Claim 10
The following statements hold.
h
1
b
2
-
1
=
h
1
β
*
,
h
1
b
2
=
h
1
(
β
*
)
-
1
.
h
1
a
2
=
h
1
(
β
*
)
-
1
,
h
1
a
2
-
1
=
h
1
β
*
.
Proof of Claim 10
(i) We have seen
(
h
1
h
2
)
b
1
=
(
h
1
h
2
)
b
2
-
1
(see Proposition 4.5).
Using
h
1
h
2
=
h
2
h
1
(see Claim 9), we get
(
h
1
h
2
)
b
1
=
(
h
2
h
1
)
b
2
-
1
.
As
(
h
1
)
b
1
=
h
1
,
(
h
2
)
b
2
=
h
2
by Proposition 4.3, it follows that
h
1
(
h
2
)
b
1
=
h
2
(
h
1
)
b
2
-
1
.
As
(
h
2
)
b
1
=
h
2
β
*
(see Claim 6),
h
1
(
h
2
)
b
1
=
h
1
h
2
β
*
=
h
2
h
1
β
*
holds.
Hence
h
1
b
2
-
1
=
h
1
β
*
.
Using
(
β
*
)
b
2
=
β
*
(see Claim 6), it follows that
(
h
1
)
b
2
=
h
1
(
β
*
)
-
1
.
Applying 𝜎, we get
h
1
a
2
-
1
=
h
1
β
*
(
β
*
=
α
*
).
∎
Consider
t
=
a
0
b
1
-
1
b
1
a
2
.
Then
t
=
(
a
0
h
1
-
1
)
a
2
-
1
b
1
a
2
=
(
a
0
h
2
-
1
(
h
1
-
1
)
a
2
-
1
)
b
1
a
2
(see Proposition 4.3).
As
(
h
1
-
1
)
a
2
-
1
=
h
1
-
1
(
β
*
)
-
1
(see Claims 10 and 8), hence
t
=
(
a
0
h
2
-
1
h
1
-
1
(
β
*
)
-
1
)
b
1
a
2
=
(
a
0
h
1
(
h
2
-
1
)
b
1
h
1
-
1
(
β
*
)
-
1
)
a
2
(see Proposition 4.3 and Claim 6).
As
(
h
2
-
1
)
b
1
=
h
2
-
1
(
β
*
)
-
1
(see Claims 6 and 8), hence
t
=
(
a
0
h
1
h
2
-
1
(
β
*
)
-
1
h
1
-
1
(
β
*
)
-
1
)
a
2
=
(
a
0
h
2
-
1
(
β
*
)
-
2
)
a
2
(see Claims 8 and 9).
Finally, we get
t
=
a
0
(
β
*
)
-
2
(
(
β
*
)
a
2
=
β
*
) (see Proposition 4.3 and Claim 6).
We have
(
β
*
)
-
2
∈
A
0
.
By definition,
a
0
(
(
β
*
)
-
2
)
b
0
-
1
∈
A
(
a
0
=
b
0
), Claim 7 gives
a
0
β
*
=
a
0
, consequently,
a
0
(
β
*
)
-
2
∈
A
0
.
Thus
a
0
b
1
-
1
b
1
a
2
∈
A
0
is true.
As
a
0
∈
A
0
,
b
1
∈
B
,
a
2
∈
A
are arbitrary, we can conclude that
[
A
,
B
]
≤
N
G
(
A
0
)
.
We have
Inn
Q
=
H
=
⟨
[
A
,
B
]
,
a
i
-
1
b
i
∣
a
i
∈
A
,
b
i
∈
B
∩
a
i
H
⟩
.
By using Proposition 4.3,
a
0
a
i
=
a
0
b
i
, i.e.
a
0
a
i
-
1
b
i
=
a
0
holds.
Thus
H
≤
N
G
(
A
0
)
.
Conversely, assume
Inn
Q
=
H
≤
N
G
(
A
0
)
.
We show
A
0
H
is a subgroup in 𝐺 (
=
Mlt
Q
).
Let
α
1
,
α
2
∈
A
0
be arbitrary.
It is enough to show
α
1
α
2
∈
A
0
H
.
By definition,
α
1
α
2
β
1
-
1
∈
A
and
β
1
β
2
α
1
-
1
∈
B
with
β
1
∈
B
∩
α
1
H
,
β
2
∈
B
∩
α
2
H
.
Clearly,
α
1
=
β
1
,
α
2
=
β
2
, whence
α
1
α
2
β
1
-
1
∈
A
0
.
As
[
A
,
B
]
≤
H
, it follows that
α
2
-
1
α
2
β
1
-
1
∈
H
,
consequently,
α
1
α
2
∈
A
0
H
.
Using
H
≤
N
G
(
A
0
)
, we can conclude that
A
0
H
is a subgroup in 𝐺.
Thus the commutant
C
(
Q
)
is a normal subloop in 𝑄.
∎
