1 Introduction
We write
H
⊴
f
G
to indicate that 𝐻 is a normal subgroup of finite index in 𝐺.
The profinite completion of an abstract group 𝐺 is the projective limit of finite quotients
G
^
=
lim
←
H
⊴
f
G
G
/
H
,
endowed with the limit topology.
There is a canonical homomorphism
i
:
G
→
G
^
.
In fact, the profinite completion is a universal object in the following sense: for every homomorphism
φ
:
G
→
H
, where 𝐻 is a profinite group, there is a unique continuous homomorphism
φ
^
:
G
^
→
H
such that
φ
=
φ
^
∘
i
.
The completion
G
^
itself is obviously profinite, namely, it is an inverse limit of a directed system of finite groups.
We say that a group 𝐺 is residually finite if
⋂
H
⊴
f
G
H
=
{
e
}
.
The canonical homomorphism
i
:
G
→
G
^
is injective if and only if 𝐺 is residually finite.
Let 𝐺 be a profinite group.
Regarding 𝐺 as an abstract group, it also has a profinite completion.
Moreover, every profinite group is residually finite, so
i
:
G
→
G
^
is injective.
We say that a profinite group 𝐺 is strongly complete (or rigid in some papers, such as [3]) if 𝐺 is equal to its own profinite completion, i.e., if the natural embedding
i
:
G
→
G
^
is an isomorphism.
One easily sees that the following conditions are equivalent for a profinite group 𝐺.
Every subgroup of finite index in 𝐺 is open.
Every normal subgroup of finite index in 𝐺 is open.
Every homomorphism
φ
:
G
→
H
onto a finite group is continuous.
Strongly complete profinite groups, or more generally, the connection between a profinite group and its profinite completion, have been the focus of many papers, such as [4, 3, 10, 5].
In [7], the authors proved the following useful result.
Proposition 1
The following conditions are equivalent for a profinite group 𝐺.
For every 𝑛, there are finitely many subgroups of index 𝑛.
There are at most countably many subgroups of finite index.
We prove the following.
Theorem 1
Considered as an abstract group, a nonstrongly complete profinite group is never isomorphic to its profinite completion.
In order to prove Theorem 1, we should be familiar with an invariant of profinite groups called the local weight.
Definition 1
Let 𝐺 be an infinite profinite group.
The local weight of 𝐺 (sometimes referred to as the rank), which is denoted by
ω
0
(
G
)
, is the cardinality of the set of all open subgroups of 𝐺.
Let us recall some properties of the local weight.
Proposition 2
Proposition 2 ([2, Chapter 17])
Let 𝐺 be an infinite profinite group.
ω
0
(
G
)
is equal to the cardinality of the set of all open normal subgroups of 𝐺.
If 𝐴 is a profinite group and
φ
:
G
→
A
a continuous epimorphism, then
ω
0
(
A
)
≤
ω
0
(
G
)
.
If
H
≤
c
G
is a closed subgroup, then
ω
0
(
H
)
≤
ω
0
(
G
)
.
In addition, if 𝐻 is open, then
ω
0
(
H
)
=
ω
0
(
G
)
.
There is a strong connection between the number of finite index subgroups of 𝐺 and the local weight of
G
^
, which arises from the following proposition.
Proposition 3
Proposition 3 ([6, Proposition 3.2.2])
Let 𝐺 be a residually finite group.
One may identify 𝐺 with its image in
G
^
and assume
G
⊆
G
^
.
Then there is a one-to-one correspondence
{
O
∣
O
≤
o
G
^
}
⇄
Ψ
Φ
{
N
∣
N
≤
f
G
}
defined by
Φ
(
O
)
=
O
∩
G
and
Ψ
(
N
)
=
N
¯
,
where
N
¯
is the topological closure of 𝑁.
Moreover, for every
N
≤
f
G
, we have
G
/
N
≅
G
^
/
N
¯
.
Corollary 1
Let 𝐺 be a profinite group.
Then
ω
0
(
G
^
)
is the cardinality of the set of all finite index subgroups in 𝐺.
The paper is organized as follows.
In the introduction, we give some background on profinite groups and profinite completions.
In Section 2, we prove Theorem 1.
In Section 3, we present the infinite tower of profinite completions and compute the number of open subgroups in each level.
The computation is based on a computation of the cardinality of a certain set of ultrafilters satisfying some conditions.
In Section 4, we give a simpler proof for the abelian case.
2 The weight of a profinite group
A fundamental example of a nonstrongly complete profinite group is given in [6, Chapter 4].
Example 1
Let 𝑇 be a finite group and 𝐼 an infinite index set.
Define
G
=
Π
I
T
.
Obviously, 𝐺 is a profinite group with the product topology.
Every continuous projection onto a finite quotient splits through some
Π
J
T
, where 𝐽 is a finite subset of 𝐼; therefore,
ω
0
(
G
)
=
|
I
|
.
To every nonprincipal ultrafilter ℱ on 𝐼, one can associate the subgroup
H
F
=
{
(
t
i
)
∣
{
i
∈
I
∣
t
i
=
1
}
∈
F
}
.
It is easily checked that, for every ℱ,
H
F
is a dense subgroup which satisfies
G
/
H
F
≅
T
, where each
t
∈
T
corresponds to its diagonal image
(
t
)
i
∈
I
.
Moreover, by [8, Section 111], there are
2
2
|
I
|
ultrafilters on 𝐼.
So 𝐺 has at least
2
2
ω
0
(
G
)
subgroups of finite index.
This is actually the accurate amount of finite index subgroups, since
|
G
|
=
|
∏
I
T
|
=
2
|
I
|
=
2
ω
0
(
G
)
, and the number of finite index subgroups cannot exceed the number of maps from 𝐺 to a finite set, which is equal to
2
|
G
|
.
We are now ready for the main theorem.
Theorem 2
Let 𝐺 be a nonstrongly complete profinite group with
ω
0
(
G
)
=
m
>
ℵ
0
.
Then
ω
0
(
G
^
)
=
2
2
ω
0
(
G
)
.
Proof
We first prove that
ω
0
(
G
^
)
≥
2
2
ω
0
(
G
)
.
Since there are only countably many finite groups, there is some finite group 𝐾 such that there are 𝐦 distinct open subgroups
H
⊴
o
G
for which
G
/
H
≅
K
.
We choose 𝐾 so that no proper quotient of 𝐾 has this property.
Let
M
=
{
H
⊴
o
G
∣
G
/
H
≅
K
}
and
M
=
⋂
H
∈
M
H
.
Denote by
ℓ
(
K
)
the length of a composition series of 𝐾.
We prove the claim by induction on
ℓ
(
K
)
.
When
ℓ
(
K
)
=
1
, 𝐾 is a simple group.
According to [6, Lemma 8.2.2],
G
/
M
≅
Π
i
∈
I
K
,
the product of
|
I
|
copies of 𝐾 for some index set 𝐼.
Note that
ω
0
(
G
/
M
)
≥
m
since the
H
i
/
M
for all
H
i
∈
M
are distinct open subgroups of
G
/
M
.
On the other hand, by Proposition 2,
ω
0
(
G
/
M
)
≤
ω
0
(
G
)
=
m
, so
ω
0
(
G
/
M
)
=
m
.
Now, by the fundamental example, Example 1,
ω
0
(
Π
i
∈
I
K
)
=
|
I
|
, so
|
I
|
=
m
.
Thus, applying the fundamental example again,
G
/
M
has
2
2
m
subgroups of finite index.
Taking their preimages, we get that 𝐺 has
2
2
m
subgroups of finite index, so
ω
0
(
G
^
)
≥
2
2
m
.
Assume that
ℓ
(
K
)
=
n
>
1
and that the claim holds for all profinite groups 𝐻 which have a finite quotient 𝐴 of shorter length than 𝐾, which appears
ω
0
(
H
)
times.
Choose a simple quotient 𝑆 of 𝐾.
Thus, 𝑆 is a simple quotient of 𝐺 as well.
By minimality of 𝐾, there are less than 𝐦 open normal subgroups
U
⊴
o
G
such that
G
/
U
≅
S
.
Hence, we can choose such a subgroup 𝑈 that contains 𝐦 different subgroups from
M
=
{
H
⊴
o
G
∣
G
/
H
≅
K
}
.
This 𝑈 is a profinite group, with
ω
0
(
U
)
=
m
, by Proposition 2.
In addition, there is some proper subgroup
K
′
◁
K
such that, for all
H
∈
M
,
U
/
H
≅
K
′
.
Note that
ℓ
(
K
′
)
<
ℓ
(
K
)
since
K
′
is a proper normal subgroup of 𝐾.
Choose a quotient
K
′′
of
K
′
which is minimal with respect to the property that it can be realized as a quotient of 𝑈 in 𝐦 different ways.
We get that
ℓ
(
K
′′
)
≤
ℓ
(
K
′
)
<
ℓ
(
K
)
, so we can apply the induction hypothesis to the profinite group 𝑈.
Thus, 𝑈 has
2
2
m
finite index subgroups, and since 𝑈 has finite index in 𝐺, each one of them is of finite index in 𝐺.
In conclusion,
ω
0
(
G
^
)
≥
2
2
m
.
We now prove that
ω
0
(
G
^
)
≤
2
2
ω
0
(
G
)
.
Denote
ω
0
(
G
)
=
m
.
By the explicit construction of the inverse limit, we have
G
⊆
Π
U
⊴
o
G
G
/
U
.
Since 𝐺 has 𝐦 open normal subgroups and, for each subgroup, the quotient is finite, we get
|
G
|
≤
2
m
.
So the number of finite index normal subgroups is equal to the number of epimorphisms from 𝐺 to finite groups.
Since there are countably many finite groups, we get that the number of such epimorphisms is at most
2
|
G
|
≤
2
2
m
.
∎
Theorem 1 is now a direct corollary of the above result.
Proof of Theorem 1
First, assume that
ω
0
(
G
)
=
ℵ
0
.
By Proposition 1, if a profinite group has only countably many subgroups of finite index, then it is strongly complete.
So 𝐺 must have uncountably many subgroups of finite index.
By Corollary 1, this cardinality is
ω
o
(
G
^
)
.
Write
ω
0
(
G
^
)
=
m
.
By Theorem 2, the cardinality of the set of subgroups of finite index in
G
^
is
2
2
m
.
We get that 𝐺 and
G
^
have a different number of subgroups of finite index, so
G
≇
G
^
.
Now assume
ω
0
(
G
)
=
m
and
ω
0
(
G
)
>
ℵ
0
.
Then, by Theorem 2,
ω
0
(
G
^
)
=
2
2
m
and
ω
0
(
G
^
^
)
=
2
2
2
2
m
.
So, again, 𝐺 and
G
^
have a different number of subgroups of finite index and thus cannot be isomorphic.
∎
We also get a result about the connection between the cardinality of 𝐺 and its local weight.
Proposition 4
Let 𝐺 be an infinite profinite group.
Then
|
G
|
=
2
ω
0
(
G
)
.
Proof
Write
ω
0
(
G
)
=
m
.
We already saw that
|
G
|
≤
2
m
in the proof of Theorem 2.
For the other direction, assume first that
ω
0
(
G
)
=
m
>
ℵ
0
.
As we showed in the proof of Theorem 2, 𝐺 admits a subgroup 𝑈 that projects onto
Π
i
∈
I
H
, where 𝐻 is some finite group and 𝐼 is a set of indices of cardinality 𝐦.
Thus,
|
G
|
≥
|
U
|
≥
|
Π
i
∈
I
H
|
=
2
m
.
Now assume that
ω
0
(
G
)
=
ℵ
0
.
Hence, by [6, Corollary 2.6.6], we have that
G
=
lim
←
n
∈
ω
G
n
is the inverse limit of an inverse system of finite groups indexed by 𝜔 such that all the maps in the system are proper projections.
Thus, for every 𝑛 and every
g
∈
G
n
, 𝑔 admits at least two preimages in
G
n
+
1
.
So
|
G
|
≥
2
ℵ
0
.
∎
3 The tower of profinite completions
Now we would like to generalize this result for the whole tower of completions.
Let 𝐺 be a nonstrongly complete profinite group.
In [1], we studied the following construction.
We define an ascending chain over the ordinals by setting
G
0
=
G
.
For a successor ordinal
α
=
β
+
1
, we have
G
α
=
G
β
^
with the natural map
φ
β
α
:
G
β
→
G
α
and such that, for every
γ
<
β
,
φ
γ
α
=
φ
β
α
∘
φ
γ
β
.
For a limit ordinal 𝛼, the system of maps
{
φ
γ
β
}
γ
<
β
<
α
is compatible, so we can define
H
α
=
lim
→
{
G
β
,
φ
γ
β
}
with the natural maps
φ
β
α
′
:
G
β
→
H
α
, and then
G
α
=
H
α
^
with the natural map
ψ
α
:
H
α
→
G
α
; we also define, for every
β
<
α
,
φ
β
α
=
ψ
α
∘
φ
β
α
′
.
We proved that the chain never terminates and, moreover, that all the maps are injective.
So we may assume that, for every
β
<
α
,
G
β
<
G
α
.
Now we would like to calculate
ω
0
(
G
α
)
for all 𝛼.
Theorem 2 gives the computation for the successor case.
Remark 1
For every ordinal 𝛼,
ω
0
(
G
α
+
1
)
=
2
2
ω
0
(
G
α
)
.
We are left with finding
ω
0
(
G
α
)
, where 𝛼 is a limit ordinal.
Recall that, by Proposition 3,
ω
0
(
G
α
)
is the number of subgroups of finite index in
H
α
.
Lemma 1
Let 𝛼 be a limit ordinal.
Then
ω
0
(
G
α
)
≤
Π
β
<
α
ω
0
(
G
β
)
.
Proof
For any group 𝐴, we denote by
FI
(
A
)
the set of finite index normal subgroups in 𝐴.
As we pointed out, if 𝛼 is a limit ordinal, then
ω
0
(
G
α
)
=
|
F
I
(
H
α
)
|
.
Now consider the map
θ
:
F
I
(
H
α
)
→
Π
β
<
α
F
I
(
G
β
)
defined by
θ
(
U
)
=
Π
β
<
α
(
U
∩
G
β
)
.
This map is obviously one-to-one, so
ω
0
(
G
α
)
≤
Π
β
<
α
|
F
I
(
G
β
)
|
=
Π
β
<
α
2
2
ω
0
(
G
β
)
=
Π
β
<
α
ω
0
(
G
β
+
1
)
=
Π
β
<
α
ω
0
(
G
β
)
.
The second equality is due to Corollary 1 and Theorem 2.
∎
Now we are going to prove that, for every nonstrongly complete profinite group,
ω
0
(
G
α
)
=
Π
β
<
α
ω
0
(
G
β
)
.
We will do so in several steps.
Lemma 2
Let 𝐺 be a nonstrongly complete profinite group and
U
≤
o
G
.
Then 𝑈 is nonstrongly complete too, and
ω
0
(
G
α
)
=
ω
0
(
U
α
)
for every ordinal 𝛼.
Proof
First we show that 𝑈 is nonstrongly complete.
Let
H
≤
f
G
be a finite-index subgroup that is not open.
Then
H
∩
U
≤
f
U
.
In addition,
U
∩
H
is not open in 𝐺 since otherwise
int
(
H
)
≠
∅
implies that 𝐻 is open.
Consequently, since 𝑈 is open in 𝐺,
H
∩
U
is not open in 𝑈.
In order to prove the lemma, by Proposition 2, it is enough to prove that, for all 𝛼,
U
α
≤
o
G
α
.
We prove this by transfinite induction.
Successor ordinal:
Assume that
U
α
≤
o
G
α
.
Then
U
α
+
1
=
U
α
^
,
G
α
+
1
=
G
α
^
.
By [6, Proposition 3.2.2],
U
α
¯
≤
o
G
α
+
1
and
[
G
α
:
U
α
]
=
[
G
α
+
1
:
U
α
¯
]
.
By [6, Lemma 3.2.4],
U
α
¯
=
i
^
[
U
α
]
, where
i
^
is the lifting of the inclusion map
i
:
U
α
→
G
α
.
Finally, note that, since by the induction hypothesis
U
α
≤
o
G
α
, then every finite-index subgroup if
U
α
is also a finite-index subgroup of
G
α
, so the profinite topology of
G
α
induces on
U
α
its full profinite topology, and thus, by [6, Lemma 3.6.2],
i
^
[
U
α
]
=
U
α
^
.
In conclusion,
U
α
+
1
≤
o
G
α
+
1
and
[
G
α
:
U
α
]
=
[
G
α
+
1
:
U
α
+
1
]
.
Limit ordinal: Now assume that 𝛽 is a limit ordinal such that, for every
γ
<
α
<
β
,
U
α
≤
o
G
α
and, in addition,
[
G
α
:
U
α
]
=
[
G
γ
:
U
γ
]
.
Then
lim
→
α
<
β
U
α
is a subgroup of
lim
→
α
<
β
G
α
of the same index, and thus, by [6, Proposition 3.2.2],
U
β
=
lim
→
α
<
β
U
α
^
≤
o
lim
→
α
<
β
G
α
^
=
G
β
of the same index.
∎
Remark 2
Let 𝐺 be a nonstrongly complete profinite group.
In order to compute
ω
0
(
G
α
)
for a limit ordinal 𝛼, we may assume 𝐺 has a quotient of the form
∏
ω
0
(
G
)
S
for some finite simple group 𝑆.
Proof
By the proof of Theorem 2, every nonstrongly complete profinite group of
ω
0
(
G
)
=
m
>
ℵ
0
has an open subgroup 𝑈 with a quotient of the form
∏
m
S
, where 𝑆 is some finite simple group.
By Lemma 2, we can replace 𝐺 by 𝑈.
Note that, in order to compute
ω
0
(
G
α
)
, where 𝛼 is a limit ordinal, we may assume that the tower begins with
G
^
.
Thus, in the case
ω
0
(
G
)
=
ℵ
0
, we can replace 𝐺 by
G
^
and get that
ω
0
(
G
)
=
m
>
ℵ
0
.
∎
Proposition 5
Let
ϕ
:
G
→
H
be an epimorphism of nonstrongly complete profinite groups.
Then, for every ordinal 𝛼, there are epimorphisms
ϕ
α
:
G
α
→
H
α
which are compatible with
φ
α
β
.
Proof
We will prove this by transfinite induction.
Let 𝛼 be an ordinal such that, for each
β
<
α
, the claim holds.
First case:
α
=
γ
+
1
.
Then there is an epimorphism
ϕ
γ
:
G
γ
→
H
γ
.
By [6, Proposition 3.2.5], there is an epimorphism
ϕ
^
:
G
γ
+
1
=
G
γ
^
→
H
γ
^
=
H
γ
+
1
which makes the following diagram commutative:
Define
ϕ
γ
+
1
=
ϕ
γ
^
.
Second case: 𝛼 is a limit ordinal.
We can define
ϕ
α
to be the direct limit
lim
→
β
<
α
ϕ
β
.
∎
Corollary 2
Let
G
→
H
be an epimorphism of nonstrongly complete profinite groups.
Then, for every ordinal 𝛼,
ω
0
(
G
α
)
≥
ω
0
(
H
α
)
.
Proof
By Proposition 2, it is enough to prove that
H
α
is a quotient of
G
α
.
So, by Proposition 5, we are done.
∎
Corollary 3
It is enough to prove the computation for nonstrongly complete profinite groups of the form
∏
m
S
for a finite simple group 𝑆.
Proof
By Remark 2, we may assume 𝐺 has a quotient of the form
H
=
∏
m
S
, where
m
=
ω
0
(
G
)
.
Recall that, by Example 1,
ω
0
(
H
)
=
m
.
Assume we have proved that, for every limit ordinal 𝛼,
ω
0
(
H
α
)
=
∏
β
<
α
ω
0
(
H
β
)
.
We will prove by transfinite induction that, under this assumption, for every ordinal 𝛼,
ω
0
(
G
α
)
=
ω
0
(
H
α
)
.
α
=
0
:
ω
0
(
H
)
=
m
=
ω
0
(
G
)
.
α
=
β
+
1
:
ω
0
(
H
β
+
1
)
=
2
2
ω
0
(
H
β
)
=
2
2
ω
0
(
G
β
)
=
ω
0
(
G
β
+
1
)
.
𝛼 is a limit ordinal: By Proposition 5,
ω
0
(
G
α
)
≥
ω
0
(
H
α
)
.
But, by the assumption,
ω
0
(
H
α
)
=
∏
β
<
α
ω
0
(
H
β
)
.
By induction hypothesis,
ω
0
(
H
β
)
=
ω
0
(
G
β
)
for all
β
<
α
.
So
ω
0
(
G
α
)
≥
ω
0
(
H
α
)
=
∏
β
<
α
ω
0
(
H
β
)
=
∏
β
<
α
ω
0
(
G
β
)
≥
ω
0
(
G
α
)
.
∎
In order to prove our claim, we need some propositions about ultrafilters and the subgroups they define.
Proposition 6
Proposition 6 ([9, Theorem 7.6])
Let 𝑋 be a set of cardinality 𝐦.
Denote by 𝒜 the set of all subsets
Y
⊆
X
for which
|
Y
c
|
<
m
.
It has the finite intersection property and thus is contained in a filter.
The number of ultrafilters containing 𝒜 is equal to
2
2
m
.
Let 𝐼 be a set.
We denote the set of all ultrafilters on 𝐼 by
β
I
.
Lemma 3
Let 𝑆 be a finite simple group, 𝐼 an infinite set of cardinality 𝐦, and let 𝒞 be the set of all subgroups of
∏
m
S
, of the form
H
F
for
F
∈
β
I
as defined in Example 1.
Then 𝒞 forms a subbase for a topology on 𝐺, contained in the profinite topology.
Denote the completion of 𝐺 with respect to this topology by
G
C
^
.
Then
G
C
^
≅
∏
β
I
S
is the product of
2
2
m
copies of 𝑆.
Proof
By results of [8, Example 111], there are
2
2
m
nonprincipal ultrafilters on a set of cardinality 𝐦.
The only thing we need to show is that, for finitely many ultrafilters
F
1
,
…
,
F
n
,
G
/
(
H
F
1
∩
⋯
∩
H
F
n
)
≅
∏
n
S
.
We prove this by induction on 𝑛.
Assume the claim holds for 𝑛, and consider
G
/
(
H
F
1
∩
⋯
∩
H
F
n
+
1
)
.
Clearly, the natural morphism
G
/
(
H
F
1
∩
⋯
∩
H
F
n
)
→
G
/
(
H
F
1
∩
⋯
∩
H
F
n
)
×
G
/
H
F
n
+
1
≅
(
∏
n
S
)
×
S
is injective.
So the only thing left to show is that it is onto.
Since
G
/
H
F
n
+
1
is simple, it is enough to show that
H
F
1
∩
⋯
∩
H
F
n
+
1
is a proper subset of
H
F
1
∩
⋯
∩
H
F
n
.
Indeed, since
F
n
+
1
is different from
F
1
,
…
,
F
n
, for every
i
∈
{
1
,
…
,
n
}
, there is some
B
i
⊆
I
such that
B
i
∈
F
i
and
B
i
c
∈
F
n
+
1
.
So
B
=
B
1
∪
⋯
∪
B
n
∈
F
1
∩
⋯
∩
F
n
,
while
B
c
∈
F
n
+
1
.
Hence, taking an element
g
∈
G
which satisfies
g
i
=
e
(for all
i
∈
B
) and
g
j
≠
e
(for all
j
∈
B
c
), we get that
g
∈
H
F
1
∩
⋯
∩
H
F
n
, while
g
∉
F
n
+
1
.
In conclusion, since the epimorphisms between the quotient groups are precisely the natural epimorphisms
∏
I
′
S
→
∏
J
′
S
, where
J
′
⊂
I
′
, we get by [6, Exercise 1.1.4] that
G
C
^
≅
∏
2
2
m
S
.
∎
Proposition 7
Let 𝑋 be a set of cardinality
2
n
, and let
Y
⊆
P
(
X
)
be a set of cardinality 𝐧, closed under finite intersections, such that, for each
B
∈
Y
,
|
B
|
=
2
n
.
Then there are
2
2
n
ultrafilters ℱ on 𝑋 consisting of sets of cardinality
2
n
which contains 𝑌.
Proof
We construct a set of subsets of 𝑋,
{
C
i
,
D
i
}
i
<
2
n
such that, for every
i
<
2
n
,
C
i
∩
D
i
=
∅
and, for every disjoint nonempty finite subsets
I
,
J
⊆
2
n
,
(
⋂
i
∈
I
C
i
)
∩
(
⋂
j
∈
J
D
j
)
∩
B
is of cardinality
2
n
for every
B
∈
Y
.
Let us look at the following set of quarters
(
B
,
I
,
J
,
α
)
such that
B
∈
Y
, 𝐼 and 𝐽 are disjoint nonempty finite subsets of
2
n
, and
α
<
2
n
is an ordinal.
Since this set is of cardinality
2
n
, we can index it by
β
<
2
n
.
We construct the required subsets by recursion.
We build subsets
{
C
i
,
β
,
D
i
,
β
}
i
<
2
n
,
β
<
2
n
and then define for all
i
<
2
n
,
C
i
=
⋃
β
<
2
n
C
i
,
β
and
D
i
=
⋃
β
<
2
n
D
i
,
β
.
Suppose
C
i
,
γ
and
D
i
,
γ
were defined for every
γ
<
β
, and their cardinality is less than or equal to 𝛾, and only less than
2
n
of them are not empty.
Now look at the quarter
(
B
,
I
,
J
,
α
)
corresponding to the ordinal 𝛽.
Choose
x
∈
B
∖
(
(
⋃
i
<
2
n
,
γ
<
β
C
i
,
γ
)
∪
(
⋃
i
<
2
n
,
γ
<
β
D
i
,
γ
)
)
.
Define
C
i
,
β
=
{
(
⋃
γ
<
β
C
i
,
γ
)
∪
{
x
}
if
i
∈
I
,
⋃
γ
<
β
C
i
,
γ
otherwise
.
Likewise, define
D
i
,
β
=
{
(
⋃
γ
<
β
D
i
,
γ
)
∪
{
x
}
if
j
∈
J
,
⋃
γ
<
β
D
i
,
γ
otherwise
.
Note that, since the cardinality of each one of the subsets
C
i
,
γ
,
D
i
,
γ
for all
γ
<
β
is less than or equal to 𝛾, then
(
⋃
i
<
2
n
,
γ
<
β
C
i
,
γ
)
∪
(
⋃
i
<
2
n
,
γ
<
β
D
i
,
γ
)
is of cardinality less than 𝛽 which in turn is less than
2
n
.
Thus, using the fact that
|
B
|
=
2
n
, it is possible to find such an element
x
∈
B
.
In addition, it is easy to see that
C
i
,
β
and
D
i
,
β
satisfy the properties in the recursion hypothesis.
Define
C
i
=
⋃
β
<
2
n
C
i
,
β
and
D
i
=
⋃
β
<
2
n
D
i
,
α
.
By the construction, they satisfy the required properties.
Now denote by 𝒜 the set of all subsets
B
⊆
X
for which
|
B
c
|
<
2
n
.
For each
I
⊆
2
n
(not necessarily finite), define the following sets:
A
I
=
Y
∪
{
C
i
}
i
∈
I
∪
{
D
j
}
j
∉
I
∪
A
.
Since all the finite intersections of subsets from 𝑌,
{
C
i
}
i
∈
I
and
{
D
j
}
j
∉
I
are of cardinality
2
n
, then so are the finite intersections with elements from 𝒜.
Thus, for all
I
⊆
2
n
,
A
I
is contained in some ultrafilter.
In addition, if
I
≠
J
, then without loss of generality, there is some
i
∈
I
∖
J
.
Thus,
C
i
∈
A
I
and
D
i
∈
A
J
are disjoint, so
A
I
and
A
J
cannot be contained in the same ultrafilter.
In conclusion, there are
2
2
n
different ultrafilters containing 𝑌, each of them consisting of subsets of maximal cardinality.
∎
Theorem 3
Let 𝑆 be a finite simple group, and let
G
=
∏
m
S
.
Denote by
{
G
α
}
α
the infinite tower of profinite completions of 𝐺.
Then there exists an infinite chain of profinite groups
{
K
α
}
α
, with compatible morphisms
ϵ
β
α
:
K
β
→
K
α
for all
β
<
α
, which satisfies the following conditions.
For every ordinal 𝛽,
K
β
+
1
=
∏
2
2
ω
0
(
K
β
)
S
For every limit ordinal 𝛼,
K
α
=
∏
n
S
, where
n
=
Π
β
<
α
ω
0
(
K
β
)
.
For every ordinal 𝛼, there are epimorphisms
η
α
:
G
α
→
K
α
which are compatible with the homomorphisms
φ
β
α
:
G
β
→
G
α
and
ϵ
β
α
:
K
β
→
K
α
.
Every
K
α
admits a set of normal subgroups
A
α
such that
for every
N
∈
A
α
, 𝑁 is a subgroup of the form
H
F
, where ℱ is an ultrafilter on
ω
0
(
K
α
)
, all of whose elements have maximal cardinality,
for every
N
∈
A
α
,
K
α
/
N
≅
S
,
for every
β
<
α
and
N
∈
A
α
,
N
∩
K
β
∈
A
β
,
for every ordinal 𝛼 and
N
∈
A
α
, there are
2
ω
0
(
K
α
+
1
)
subgroups
N
′
in
A
α
+
1
which satisfy
N
′
∩
K
α
=
N
,
for every limit ordinal 𝛼 and every series
(
N
β
)
β
<
α
of normal subgroups,
N
β
∈
A
β
such that
N
β
∩
K
γ
=
N
γ
if
γ
<
β
, there are
Π
β
<
α
ω
0
(
K
β
)
subgroups
N
α
∈
A
α
whose intersection with each
K
β
,
β
<
α
, is equal to
N
β
.
Proof
Let 𝛼 be an ordinal, and assume that, for all
β
<
α
, such groups exist.
First case:
α
=
0
.
Define
A
0
to be set of all subgroups of the form
H
F
for ℱ an ultrafilter on 𝐦, all of whose elements have cardinality 𝐦.
By Proposition 6,
|
A
0
|
=
2
2
m
.
Second case:
α
=
β
+
1
for some ordinal 𝛽.
By the induction hypothesis, we have
K
β
≅
∏
n
S
for some cardinal 𝐧.
Let 𝒞 be the set of all subgroups of the form
H
F
, where ℱ is an ultrafilter on
n
=
ω
0
(
K
β
)
.
Define
K
β
+
1
=
K
β
,
C
^
.Then, by Lemma 3,
K
β
+
1
≅
∏
2
2
n
S
.
By definition of the completion with respect to some directed system of subgroups, there is a natural homomorphism
ϵ
β
,
α
:
K
β
→
K
β
,
C
^
=
K
α
.
Define
ϵ
γ
,
α
for all
γ
<
β
to be
ϵ
β
,
α
∘
ϵ
γ
,
β
.
Since 𝒞 is a subtopology of the profinite topology, then there is a natural epimorphism
ξ
β
:
K
β
^
→
K
β
,
C
^
which commutes with the natural homomorphisms
K
β
→
K
β
^
,
K
β
→
K
β
,
C
^
.
By the induction hypothesis, there is a natural epimorphism
η
:
G
β
→
K
β
.
By [6, Proposition 3.2.5], it can be lifted to a compatible epimorphism
η
β
^
;
G
α
=
G
β
^
→
K
β
^
.
So define
η
β
+
1
to be
ξ
β
∘
η
β
^
.
Let
N
∈
A
β
.
By the induction hypothesis, 𝑁 has the form
H
F
for some ultrafilter on 𝐧 such that all of its elements have cardinality 𝐧.
Recall that, for each ultrafilter ℱ, there is a given isomorphism
∏
n
S
/
H
F
≅
S
.
We use this to identify the quotient with 𝑆.
Use this isomorphisms to identify all the quotient groups with the same group.
For all
g
∈
K
β
, define
A
g
to be the set of all ultrafilters
F
′
for which the image of 𝑔 in the natural epimorphism
∏
n
S
→
∏
n
S
/
H
F
′
is equal to the image of 𝑔 in the natural epimorphism
∏
n
S
→
∏
n
S
/
H
F
.
There is a set of indices
A
∈
F
determining the image of 𝑔 in
∏
n
S
/
H
F
, so in fact
A
g
=
{
F
′
∣
A
∈
F
′
}
.
Note that, since ℱ contains only sets of maximal cardinality,
|
A
|
=
n
, so the number of ultrafilters containing 𝐴 is equal to the number of ultrafilters on 𝐴, which is
2
2
n
.
In addition, for each
g
1
≠
g
2
∈
K
β
, there is some
g
3
∈
K
β
for which
A
g
1
∩
A
g
2
=
A
g
3
.
Indeed, let 𝐴 and 𝐵 be the subsets determining the images of
g
1
and
g
2
correspondingly.
Then
A
g
1
∩
A
g
2
is equal to the set of all ultrafilters containing
A
∩
B
.
So define
g
3
to be 𝑒 on
A
∩
B
and 𝑥 on
(
A
∩
B
)
c
for some
e
≠
x
∈
S
.
Let
β
n
denote the set of all ultrafilters on 𝐧.
Then the set of the subsets of the form
A
g
for
g
∈
K
β
is a set of
2
n
subsets of the maximal cardinality, closed under finite intersections.
Thus, by Proposition 7, there are
2
ω
0
(
K
α
)
ultrafilters containing all these sets, which consist of subsets of maximal cardinality.
Denote this collection by
B
F
,
α
.
For each
G
∈
B
F
,
α
, the subgroup
H
G
of
K
α
has the property that its intersection with
K
β
is equal to
H
F
and
K
α
/
H
G
≅
S
.
Define
B
α
=
⋃
{
F
∣
H
F
∈
A
β
}
B
F
,
α
and
A
α
=
{
H
F
∣
F
∈
B
α
}
Third case: Let 𝛼 be a limit ordinal, and define
K
α
′
to be the direct limit of
{
K
β
,
ϵ
γ
,
β
}
.
Denote by
C
β
the set of all subgroups 𝑁 of the form
lim
→
N
β
, where
{
N
β
}
β
<
α
is a compatible system of normal subgroups as described in condition 6.
For every
β
<
α
and
N
β
in the system,
K
β
/
N
β
≅
S
, and those isomorphisms are compatible, so we get that
K
α
′
/
N
≅
S
.
In addition, if we take
N
1
,
…
,
N
l
different subgroups of this form, then they are direct limits of different systems
N
i
=
lim
→
N
i
,
β
.
So there is some 𝛽 from which, for all
β
<
γ
<
α
, the subgroups
N
i
,
γ
are different.
Thus, for every
β
<
γ
<
α
,
K
γ
/
N
1
,
γ
∩
⋯
∩
N
l
,
γ
≅
∏
l
S
.
Hence,
K
α
′
/
N
1
∩
⋯
∩
N
l
≅
∏
l
S
.
Define
K
α
to be the completion of
K
α
′
with respect to this set of subgroups.
Then
K
α
=
∏
n
S
when
n
=
Π
β
<
α
ω
0
(
K
β
)
.
Define
ϵ
β
,
α
to be the compositions of the natural maps to the direct limit, and from the direct limit to its completion,
ϵ
β
,
α
:
K
β
→
lim
→
β
<
α
K
β
→
lim
→
β
<
α
K
β
^
=
K
α
,
and define
η
α
to be the lifting of
lim
→
β
<
α
η
β
to the corresponding completions.
For each
g
∈
K
α
′
and a subgroup 𝑁 of this form, let
A
g
be the set of all such subgroups
N
′
such that the image of 𝑔 in
K
α
′
/
N
′
is equal to its image in
K
α
′
/
N
.
Since 𝑔 is an element in the direct limit, it comes from some
K
β
.
In addition, 𝑁 is the direct limit of a compatible chain
{
N
β
}
β
<
α
.
Thus,
N
′
=
lim
→
β
<
α
N
β
′
satisfies the condition if and only if
N
β
+
1
′
is a lifting of
N
β
in
K
β
+
1
.
So the number of such subgroups is equal to
ω
0
(
K
α
)
.
In addition, by the same argument, we can show that the set of all
A
g
is closed under finite intersections.
Thus, in a similar way to the successor case, we can show there are
2
ω
0
(
K
α
)
subgroups extending any such
N
⊴
K
α
′
and satisfying all the conditions.
Define
A
α
to be the set of all such extensions.
∎
Theorem 4
Let 𝐺 be a nonstrongly complete profinite group.
Then, for every limit ordinal
α
≠
0
,
ω
0
(
G
α
)
=
Π
β
<
α
ω
0
(
G
β
)
.
Proof
By Corollary 3, it is enough to prove the claim for
G
=
∏
m
S
.
By Theorem 3, there is a series of subgroups
K
α
as described in the theorem.
We will prove by transfinite induction that
ω
0
(
G
α
)
=
ω
0
(
K
α
)
.
So assume this equality holds for every
β
<
α
.
For
α
=
0
, the claim is clear.
For
α
=
β
+
1
,
ω
0
(
G
α
)
=
2
2
ω
0
(
G
β
)
.
Since 𝒞 has
2
2
ω
0
(
K
β
)
subgroups, we have that
ω
0
(
K
α
)
=
2
2
ω
0
(
K
β
)
=
2
2
ω
0
(
G
β
)
=
ω
0
(
G
α
)
.
For 𝛼 a limit ordinal, we already know that
ω
0
(
G
α
)
≤
∏
β
<
α
ω
0
(
G
β
)
.
By Proposition 5, we get that
ω
0
(
G
α
)
≥
ω
0
(
K
α
)
.
Since
K
α
is defined as the completion of
K
α
′
with respect to the set 𝒞 of subgroups,
ω
0
(
K
α
)
=
|
C
|
.
By the construction of 𝒞,
|
C
|
=
2
2
m
∏
0
≠
β
<
α
2
ω
0
(
K
β
)
=
2
2
m
∏
0
≠
β
<
α
2
ω
0
(
G
β
)
=
∏
β
<
α
2
2
ω
0
(
G
β
)
=
∏
β
<
α
ω
0
(
G
β
+
1
)
=
∏
β
<
α
ω
0
(
G
β
)
.
∎
In fact, we can calculate
Π
β
<
α
ω
0
(
G
β
)
explicitly.
Lemma 4
Lemma 4 ([9, Lemma 5.9])
If 𝜆 is an infinite cardinal and
⟨
κ
β
:
β
<
λ
⟩
is a nondecreasing sequence of nonzero cardinals, then
Π
β
<
λ
κ
β
=
sup
β
{
κ
β
}
λ
.
Definition 2
Let 𝜅 be a cardinal.
We say that 𝜅 is a strong limit cardinal if, for all
λ
<
κ
,
2
λ
<
κ
.
In our case, the limit ordinal 𝛼 is not necessarily a cardinal, but we may replace it by its cofinality.
So one may take
λ
=
cof
(
α
)
, and
κ
β
=
ω
0
(
G
ψ
(
β
)
)
, where
ψ
:
λ
→
α
is an order-preserving cofinal map.
Note that
sup
β
κ
β
is a strong limit cardinal with cofinality
cof
(
α
)
=
λ
.
Thus, we can use the following proposition.
Proposition 8
Proposition 8 ([9, p. 58, equation 5.23])
Let 𝜅 be a strong limit cardinal.
Then
κ
cof
(
κ
)
=
2
κ
.
Corollary 4
We have that, for a limit ordinal 𝛼,
ω
0
(
G
α
)
=
∏
β
<
α
ω
0
(
G
β
)
=
2
sup
β
<
α
{
ω
0
(
G
β
)
}
.
Remark 3
Let 𝐺 and
G
′
be two nonstrongly complete profinite groups.
Then there exists some ordinal 𝛼 such that, for all
β
≥
α
,
ω
0
(
G
β
)
=
ω
0
(
G
β
′
)
.
Proof
Write
m
0
=
ω
0
(
G
)
and
n
0
=
ω
0
(
G
′
)
.
It is enough to prove that there exists 𝛼 such that
ω
0
(
G
α
)
=
ω
0
(
G
α
′
)
, since each successor level depends only on the previous level, and any limit level does not depend on initial segments.
Assume
n
0
>
m
0
.
Since the series of local weights is a strongly ascending series over the ordinals, it is not bounded.
So denote by
α
0
the minimal ordinal for which
ω
0
(
G
α
0
)
≥
n
0
.
Now continue by recursion.
Assume
α
n
is defined.
Define
α
n
+
1
to be the minimal ordinal for which
ω
0
(
G
α
n
+
1
)
≥
ω
0
(
G
α
n
′
)
.
Eventually,
α
=
sup
{
α
n
}
n
∈
ω
is the desired ordinal.
∎
4 The abelian case
If 𝐺 is a nonstrongly complete profinite abelian group, then we can offer a simpler derivation for the local weight of
G
α
, where 𝛼 is a limit ordinal.
By Remark 2, we may assume 𝐺 has a quotient of the form
∏
ω
0
(
G
)
S
for some finite simple group.
Since 𝐺 is abelian, then so is 𝑆.
In conclusion, there is a finite simple abelian group 𝑆 onto which 𝐺 projects continuously in
ω
0
(
G
)
different ways.
Lemma 5
Let 𝐺 be a nonstrongly complete profinite group and 𝑆 a finite simple group which appears as a quotient of 𝐺 in 𝐧 different ways for some infinite cardinality 𝐧.
Then 𝐺 has
2
2
n
dense normal subgroups
H
i
such that
G
/
H
i
≅
S
.
Proof
By the proof of Theorem 2, since 𝐺 has 𝐧 projections onto 𝑆, then 𝐺 projects onto
∏
n
S
.
By Example 1,
∏
n
S
has
2
2
n
noncontinuous epimorphisms on 𝑆.
Compose them with the epimorphism from 𝐺 to
∏
n
S
to get
2
2
n
noncontinuous epimorphisms from 𝐺 to 𝑆.
For any such epimorphism, the kernel is a subgroup 𝐻 of finite index which is not open, and thus not closed, and for which
G
/
H
≅
S
.
Thus, there are no normal subgroups between 𝐻 and 𝐺.
But, since 𝐻 is normal, so is
H
¯
.
So
H
¯
=
G
.
That is, 𝐻 is dense.
∎
Lemma 6
Let 𝐺 be a nonstrongly complete profinite group.
Then 𝐺 is a semi-directed component of
G
^
.
Proof
Look at the identity map
G
→
G
.
By the universal property of profinite completion, we get the following diagram:
Thus,
G
^
≅
G
⋉
K
for
K
=
ker
i
d
^
.
∎
Remark 4
If 𝐺 is abelian then so are all its finite quotients, and thus
G
^
is abelian as an inverse limit of abelian groups.
Corollary 5
If 𝐺 is abelian, then
G
^
≅
G
×
K
for
K
=
ker
i
d
^
.
Remark 5
Let 𝐺 be a nonstrongly complete profinite group.
Write
K
=
ker
i
d
^
.
Then, since 𝐾 is the kernel of the natural epimorphism
lim
←
U
⊴
f
G
G
/
U
→
lim
←
U
⊴
o
G
G
/
U
and since, for every
U
⊴
f
G
, the closure
U
¯
satisfies
U
¯
⊴
o
G
, we get that
K
=
lim
←
U
⊴
G
U
¯
/
U
.
From now on, we write 𝐾 for
ker
(
i
d
^
:
G
^
→
G
)
, and
K
α
for
ker
(
i
d
^
:
G
α
^
→
G
α
)
.
Corollary 6
If 𝐻 is a normal dense subgroup of finite index of 𝐺, then 𝐾 projects naturally on
G
/
H
.
Moreover, if
H
1
≠
H
2
, the natural projections
K
→
G
/
H
1
,
K
→
G
/
H
2
are different.
Proof
Being the inverse limit of the directed system of finite groups
{
U
¯
/
U
}
U
⊴
f
G
,
every group in the system corresponds to a continuous quotient of 𝐾.
So each dense normal subgroup 𝐻 of 𝐺 corresponds to a continuous epimorphism
K
→
H
¯
/
H
=
G
/
H
.
∎
Proposition 9
Let 𝐧 be an infinite cardinality, and let 𝐺 be a nonstrongly complete profinite abelian group, which projects onto some finite simple group 𝑆 in 𝐧 different ways, for which the kernel is a dense subgroup.
Then each projection
G
→
S
has
2
2
n
different liftings to
G
^
.
Proof
By Corollary 6, 𝐾 has 𝐧 different projections onto 𝑆.
So, by Lemma 5, 𝐾 has
2
2
n
abstract epimorphisms
ψ
i
:
K
→
S
.
Let
f
:
G
→
S
be some projection.
Recall that, since 𝐺 is abelian, then so is 𝑆.
By Corollary 5,
G
^
=
G
×
K
.
Since 𝑆 is abelian, we can define epimorphisms
(
f
,
ψ
i
)
:
G
×
K
→
S
, which clearly extend 𝑓.
∎
Proposition 10
Let 𝐺 be nonstrongly complete profinite abelian group which has
ω
0
(
G
)
projections onto some finite simple group 𝑆.
Consider
{
G
α
}
to be the chain of profinite completions as described in the beginning of the previous section.
Then
ω
0
(
G
α
)
=
∏
β
<
γ
<
α
2
2
ω
0
(
G
γ
)
for all limit ordinals
α
.
For each 𝛼, there is a set
A
α
of not necessarily open subgroups
H
i
which satisfy the following conditions.
For each
H
i
∈
A
α
,
G
α
/
H
i
≅
S
For each
β
<
α
and
H
i
∈
A
α
,
H
i
∩
G
β
∈
A
β
.
For each limit ordinal 𝛼, all the subgroups in
A
α
are open.
For each
β
<
α
and
H
∈
A
β
, there are
n
α
subgroups
H
′
∈
A
α
for which
H
′
∩
G
β
=
H
, where
n
α
=
{
2
2
ω
0
(
G
α
)
,
α
is a successor ordinal
,
ω
0
(
G
α
)
,
α
is a limit ordinal
.
For each 𝛼,
|
A
α
|
=
n
α
.
Proof
We prove the claim by transfinite induction.
For
α
=
0
, take
A
0
=
{
ker
f
i
∣
f
i
:
G
→
S
is an epimorphism
}
.
By assumption, 𝐺 has
ω
0
(
G
)
different continuous projections onto 𝑆, so by Lemma 5,
|
A
0
|
=
2
2
ω
0
(
G
)
.
Let
α
=
β
+
1
, and assume
A
β
is defined.
By induction hypothesis,
|
A
β
|
=
2
2
ω
0
(
G
β
)
or
ω
0
(
G
β
)
.
In the first case, since the number of open subgroups of
G
β
is equal to
ω
0
(
G
β
)
, by definition, then
2
2
ω
0
(
G
β
)
of the subgroups in
A
β
are not open, and thus, since 𝑆 is simple, they are dense.
In the second case, we can use Lemma 5 to conclude that
G
β
has
2
2
ω
0
(
G
β
)
dense subgroups 𝐻 for which
G
/
H
≅
S
.
Thus, by Proposition 9, each subgroup
H
∈
A
β
has a set
A
β
+
1
,
H
of cardinality
2
2
2
2
ω
0
(
G
β
)
=
2
2
ω
0
(
G
α
)
consisting of normal subgroups
H
′
⊴
G
α
for which
G
/
H
′
≅
S
and
H
′
∩
G
β
=
H
.
Define
A
α
=
⋃
H
∈
A
β
A
β
+
1
,
H
.
Let 𝛼 be a limit ordinal.
For each series
(
H
β
)
β
<
α
of normal subgroups from
A
β
such that, for every
γ
<
β
,
H
β
∩
G
γ
=
H
γ
, the union is a normal subgroup
H
⊴
lim
→
β
<
α
G
β
for which the quotient is isomorphic to 𝑆 and the intersection with
G
β
is equal to
H
β
.
By induction hypothesis, there are
∏
β
<
γ
<
α
2
2
ω
0
(
G
γ
)
such series.
Define
A
α
to be the set of closures of all these subgroups in
lim
→
β
<
α
G
β
^
=
G
α
.
We get that
G
α
has at least
lim
→
β
<
α
G
β
open subgroups.
On the other hand, by Lemma 1,
ω
0
(
G
β
)
≤
∏
γ
<
β
ω
0
(
G
β
)
, so
ω
0
(
G
β
)
=
∏
γ
<
β
ω
0
(
G
γ
)
.
∎
