Download Subgroups of Finite Index in Profinite Groups

Subgroups of Finite Index in Profinite Groups
Sara Jensen
May 14, 2013
1
Introduction
In addition to having a group structure, profinite groups have a nontrivial topological
structure. Many results pertaining to profinite groups exploit both structures, and therefore both structures are important in understanding profinite groups.
An amazing result due to Nikolov and Segal is the following theorem.
Theorem 1.1. Suppose that G is a topologically finitely generated profinite group. Then
every subgroup of G of finite index is open.
One way to view Theorem 1.1 is as a statement that the algebraic structure of a
finitely generated profinite group somehow also encodes the topological structure. That
is, if one wishes to know the open subgroups of a profinite group G, a topological property,
one must only consider the subgroups of G of finite index, an algebraic property. As
profinite groups are compact topological spaces, an open subgroup of G necessarily has
finite index. Thus it is also possible to begin with a subgroup of G having a particular
topological property (the subgroup is open) and deduce that this subgroup must also
have a particular algebraic property (the subgroup has finite index).
The proof of Theorem 1.1 is quite extensive, and requires the classification of finite
simple groups. However, if one restricts attention to a smaller class of groups, the result
can be done in a fairly straightforward manner.
Suppose that G is a finite group having a normal series 1 = Gl ⊆ Gl−1 ⊆ . . . ⊆
G1 ⊆ G0 = G such that Gi /Gi+1 is nilpotent for all 0 ≤ i < l. Then we say that G
belongs to the class N l . Note that finite nilpotent groups all belong to N1 , while finite
supersolvable groups all belong to N2 since the commutator subgroup of a supersolvable
group is nilpotent.
The theorem we aim to prove in this document is the following.
Theorem 1.2. Suppose that G is a topologically finitely generated profinite group such
that there exists some fixed l with G/N ∈ N l whenever N is an open normal subgroup of
G. Then every subgroup of G of finite index is open.
1
2
Preliminary Facts
We begin by establishing some notation. Let
w = w(x1 , . . . , xn ) =
s
Y
xjii
i=1
where i ∈ {1, −1} and ji ∈ {1, . . . , n}. That is, w is a finite expression of the variables
x1 , . . . , xn and their inverses. Then if G is any group, we write
w(G) = {w(g1 , . . . , gn ) | g1 , . . . , gn ∈ G}
and we write w(G) for the subgroup of G algebraically generated by w(G). We call
the group w(G) the verbal subgroup of G. Now suppose Y is an arbitrary subset of G
and t a positive integer. We write Prt (Y ) for the collection of all possible words of length
t involving only elements of Y or their inverses.
Example. Suppose that w = w(x, y) = [x, y] = y −1 x−1 yx. Then for an arbitrary group
G, w(G) is the set of commutators in G, while w(G) is the commutator subgroup of G. If
we let t = 4, then w(G) ⊆ Prt (G). If we pick g ∈ [G, G], then we know that there exists
some t so that g is the product of t commutators. For this t, we have that g ∈ Prt (w(G)).
An introductory fact about profinite groups that we will need is the following.
Proposition 1. Suppose G is a profinite group and C1 , C2 , . . . is a countably infinite set
∞
[
of nonempty closed subsets of G having empty interior. Then G 6=
Ci .
i=1
Lemma 2.1. Let G be a profinite group and let X be a closed subset of G such that
X = X −1 and 1 ∈ X. Then G = hXi if and only if G = Prm (X) for some positive
integer m.
Proof. If G = Prm (X) for some positive integer m, then it is clear that G = hXi. Now
suppose that G = hXi. For every g ∈ G, we know that g ∈ Prn (X) for some positive
∞
[
integer n, and thus G =
Pri (X). Note that if i < j, then Pri (X) ⊆ Prj (X) since
i=1
1 ∈ X. Now for each n, we know that Prn (X) is closed, and by Proposition 1, we know
that G cannot be an infinite union of closed subsets with empty interior. This means
that Prt (X) contains[a non-empty open subset U for some positive integer t. Thus we
may also write G =
gU , and by compactness, we know there exists a finite collection
g∈G
g1 , . . . , gr ∈ G for which G =
r
[
gi U . As G = hXi and 1 ∈ X, we know that there exists
i=1
s for which g1 , . . . , gr ∈ Prs (X). Finally, set m = s + t. Then if g ∈ G, we can write
g = gi u for some gi ∈ {g1 , . . . , gr } and u ∈ U . Since gi ∈ Prs (X) and u ∈ Prt (X), we see
that we can write g as a combination of m members of X, as claimed.
2
A crucial lemma which will be used to prove Theorem 1.2, and which is used in
Nikolov and Segal’s proof of Theorem 1.1, is the following.
Lemma 2.2. Let w be a group word. The following are equivalent.
(a) For every finitely generated profinite group G, w(G) is closed.
(b) There exists an integer valued function f such that for each natural number k and
for each finite group H that can be generated by k elements, one has
w(H) = Prf (k) (w(H)) ,
that is, for each k-generated finite group H, the word w has width f (k) in H.
Proof. The image of a compact topological space under a continuous map is compact,
k
Y
and the map from
G → G that sends (g1 , . . . , gk ) → w(g1 , . . . , gk ) or (g1 , . . . , gk ) 7→
i=1
w(g1 , . . . , gk )−1 is continuous; hence w(G) is compact. Likewise, the map from
t
Y
w(G) →
i=1
Prt (w(G)) that sends (w1 , . . . , wt ) →
t
Y
wi is continuous, and therefore Prt (w(G)) is also
i=1
compact for each t.
Write G = lim Gi with projection maps ϕi : G → Gi . Now w(G) = lim w(Gi ), and
←−
←−
i
i
therefore w(G) is also a profinite group. Of course, Prt (w(G)) ⊆ w(G) for every integer
t. As we have just shown that Prt (w(G)) is compact for each t and w(G) is Hausdorff,
it follows that Prt (w(G)) is closed for each t.
It is easy to see that ϕi (Prt (w(G))) = Prt (w(Gi )) for all t and for all i. Assume now
that (b) holds, and let d be the number of topological generators for G. That is, each Gi
is topologically generated by d elements, and so there exists an integer valued function
f for which w(Gi ) = Prf (d) (w(Gi )). Since Prf (d) (w(G)) is closed, we have
Prf (d) (w(G)) = lim Prf (d) (w(Gi ))
←−
i
= lim w(Gi )
←−
i
= w(G)
However, we also clearly have that Prf (d) (w(G)) ⊆ w(G), and therefore w(G) =
Prf (d) (w(G)) ⊆ w(G). This gives us that w(G) is closed, as claimed.
We now assume that (a) holds. Let k be a natural number and F the free profinite group on k generators. By hypothesis, w(F ) = w(F ) and so w(F ) is a profinite
3
group which is algebraically generated by w(F ). It follows by Lemma 2.1 that w(F ) =
Prm (w(F )) for some natural number m. Define f (k) = m. Then if H is an arbitrary profinite group generated by k elements, then we have a surjection ϕ : F → H. As a result,
we have that w(H) = ϕ(w(F )) = ϕ(Prm (w(F ))) = ϕ(Prf (k) (w(F ))) = Prf (k) (w(H)).
We will be particularly interested in the statement of Lemma 2.2 as it applies to the
word w = x−1 y −1 xy. That is, we are interested in what Lemma 2.2 has to say about the
commutator subgroup of a finitely generated profinite group.
In any topological space, one is concerned with the structure of open subgroups of
that space. Since profinite groups are topological structures, we wish to know about the
open subgroups of a profinite group G.
Lemma 2.3. Suppose that G is a profinite group and H is a subgroup of G. Then H is
open in G if and only if H is closed of finite index.
Proof. If H is open, then all cosets of H in G are also open, and the cosets of H in G
form an open cover of G. Since G is compact, we know that finitely many cosets will
suffice to cover G, and therefore H has finite index in G. To see that H is also closed,
note that the complement of H in G is open, since the complement of H in G is the
union of the cosets of H in G that are distinct from H, each of which is open.
Now suppose that H ⊆ G is a closed subgroup of finite index. To see that H is open,
we will show that its complement is closed. The complement of H in G is the union of
the cosets of H in G distinct from H. Since H has finite index, the complement of H in
G is a finite union of closed sets, and is therefore closed. Hence H is also open in G.
3
Solvable, Supersolvable, and Nilpotent Groups and
their Profinite Counterparts
We say that a group G is solvable if there exists some finite integer n for which its derived
series is trivial. That is, if we define G(0) = G and write G(k) = (G(k−1) )0 for every k > 0,
then the solvable groups are those for which there exists some n for which we have
1 = G(n) ⊆ G(n−1) ⊆ . . . ⊆ G(1) ⊆ G .
If G is a solvable group and n is minimal so that G(n) = 1, then we say that n is the
derived length of G.
Note that this initial definition of a solvable group is equivalent to saying that G
admits a (finite) normal series 1 = Nk C Nk−1 . . . C N0 C G for which Ni /Ni+1 is abelian
for all 0 ≤ i < k. We say that a group G is supersolvable if G admits a (finite) normal
series 1 = Nk C Nk−1 . . . C N0 C G for which Ni /Ni+1 is cyclic for all 0 ≤ i < k. We say
that a group G is nilpotent if G admits a (finite) normal series 1 = Nk CNk−1 . . .CN0 CG
for which Ni /Ni+1 is central in G/Ni+1 for all 1 ≤ i < k. Alternatively, we can say that a
4
group G is nilpotent if its lower central series eventually reaches the identity. The lower
central series begins with G = G1 , has G2 := [G, G], and has Gi = [Gi−1 , G]. If G is a
nilpotent group and n is minimal with Gn+1 = 1, then we say that G has nilpotency class
n.
We say that a group G is prosolvable (prosupersolvable, pronilpotent) if G is the inverse
limit of a surjective inverse system of finite solvable (supersolvable, nilpotent) groups.
Example. Suppose that {Gi } is a collection of finite groups, and that G = lim Gi . If the
←−
i
{Gi } are solvable, it need not be the case that G is solvable. To see this, suppose that the
{Gi } are composed of solvable groups of arbitrary derived length; that is, for each integer
n, there exists Gi with derived length n. Then we claim that G cannot be solvable. If G
were solvable, then there would be a finite integer n for which G(n) is trivial. Now if G(n)
is trivial, then this holds for all quotients of G as well, which includes the Gi . However,
we know that there exists a group Gi with derived length n + 1, a contradiction. Thus it
need not be the case that G is solvable. As there exist nilpotent groups of arbitrarily large
nilpotency class, a similar example can be constructed demonstrating that a pronilpotent
group need not be nilpotent.
Lemma 3.1. Suppose that G is a finite group, and that H, K are normal in G with both
G/H and G/K nilpotent. Then G/(H ∩ K) is nilpotent.
Proof. Suppose that K ⊆ Y ⊆ X ⊆ HK with both X and Y normal in G and where
X/Y is central in G/Y ; not that such subgroups exist as G/K is nilpotent. Consider
the subgroups X ∩ H and Y ∩ H, both of which are normal in G. We wish to show that
[G, X ∩ H] ⊆ Y ∩ H. Yet [G, X ∩ H] ⊆ [G, H] ⊆ H since H C G, and [G, X ∩ H] ⊆
[G, X] ⊆ Y since X/Y is central in G/Y . This gives is that [G, X ∩ H] ⊆ Y ∩ H, and we
obtain a central series for G/(H ∩ K) by taking a central series for G/H and extending
it via intersecting a central series for HK/K in G/K with H.
4
The Class N l
In this section, we study the class of groups G belonging to the class N l for some l
more carefully, and study the structure of a profinite group G satisfying the properties
of Theorem 1.2.
Suppose that G = lim Gi is a finitely generated profinite group and that there exists
←−
i
some fixed l with G/N ∈ N l whenever N is an open normal subgroup of G. Since the
maps πi : G → Gi are continuous for all i and each Gi is a finite group, it follows that
Gi ∈ N l for each i. In particular, this means that each Gi is solvable, and therefore G is
a finitely generated prosolvable group.
Recall that a finite group G belongs to the class N l if G admits a normal series
1 = Gl ⊆ Gl−1 ⊆ . . . ⊆ G1 ⊆ G0 = G .
5
That is, G ∈ N l if there are l nilpotent factors in the series for G.
We first note that a finite group G belongs to N l even if the given series is only
subnormal. We do this by induction. Of course, G1 C G, so we assume that G2 is
normal in G1 , but not necessarily normal in G. Then if we replace G2 by coreG (G2 ),
then coreG (G2 ) C G, and G1 /coreG (G2 ) is nilpotent since G1 /G2 , and therefore G1 mod
each conjugate of G2 , is nilpotent, and if G/H and G/K are both nilpotent then so is
G/(H ∩ K) by Lemma 3.1.
Lemma 4.1. Suppose that G is a finite group and that G ∈ N l for l > 0. Then if H, K
are normal subgroups of G with G/H and G/K belonging to N l−1 , then G/(H ∩ K) ∈
N l−1 .
Proof. Suppose that H and K are normal subgroups of G with G/H and G/K belonging
to N l−1 . We will show by induction on l that G/(H ∩ K) belongs to N l−1 . If l = 1, then
G is nilpotent, and therefore G/(H ∩ K) is a quotient of a nilpotent group. As factor
groups of nilpotent groups are nilpotent, our base case holds. We may therefore assume
that l > 1, and that whenever k < l − 1 then our Lemma holds.
Let H = Hl−1 ⊆ . . . ⊆ H1 ⊆ H0 = G be a normal series for G/H with Hi /Hi+1
nilpotent for all 0 ≤ i < l − 1 and letK = Kl−1 ⊆ . . . ⊆ K1 ⊆ K0 = G be a normal series
for G/K with Ki /Ki+1 nilpotent for all 0 ≤ i < l − 1. We work inside G/(H ∩ K).
Now both H1 /H and K1 /K belong to N l−2 . Since K and H are disjoint normal
subgroups modulo H ∩K, we know that H and K commute modulo H ∩K, and therefore
H1 K/H and K1 H/K belong to N l−2 . We wish to have both H and K contained in
H1 ∩ K1 . If, for example, H * K1 , replace K1 by K1 H so that H ⊆ K1 . As we have just
argued, this can be done without affecting that H1 /H ∈ N l−2 .
We now have that H, K ⊆ H1 ∩K1 and that both H and K are such that (H1 ∩K1 )/H
and (H1 ∩K1 )/K belong to N l−2 . By induction, we have that (H1 ∩K1 )/(H ∩K) ∈ N l−2 .
We now argue that G/(H1 ∩ K1 ) is nilpotent, from which it will follow that H ∩ K is
such that G/(H ∩ K) ∈ N l−1 . Yet this follows from Lemma 3.1 since we know that
both G/H1 and G/K1 are nilpotent. Thus G/(H1 ∩ K1 ) is nilpotent and we have that
G/(H ∩ K) ∈ N l−1 , as wanted.
Lemma 4.2. Let K ∈ N l with l > 1. Then K contains a unique smallest normal
subgroup H such that K/H ∈ N l−1 . Moreover, H is nilpotent and [K, H] = H.
Proof. Since K ∈ N l , we know that K admits a normal series 1 = Kl ⊆ Kl−1 ⊆ . . . ⊆
K0 = K. It follows that K/Kl−1 ∈ N l−1 , and therefore there certainly exist subgroups
H ⊆ K for which K/H ∈ N l−1 . Now suppose that H1 and H2 are normal in K and that
both K/H1 and K/H2 belong to N l−1 . By Lemma 4.1, it follows that H1 ∩ H2 is such
that K/(H1 ∩ H2 ) ∈ N l−1 . Since K is finite, it follows that we can take H to be the
intersection of all normal subgroups of K for which the factor group belongs to N l−1 ,
proving the minimality and uniqueness of H. Since one subgroup with this property is
Kl−1 and Kl−1 is nilpotent, it also follows that H is nilpotent. Finally, we must show
6
that [K, H] = H. Certainly [K, H] C G and [K, H] ⊆ H since H C K, so to see that
[K, H] = H we will show that K/[K, H] ∈ N l−1 .
Let H = Hl−1 ⊆ . . . ⊆ H1 ⊆ H0 = K be a normal series with Hi /Hi+1 nilpotent for
0 ≤ i < l − 1. Since each Hi C K, we have that [K, Hi ] ⊆ Hi for all i, and we therefore
have the normal series
[K, H] = [K, Hl−1 ] ⊆ . . . ⊆ [K, H1 ] ⊆ K .
For each i, we have that Hi /[K, Hi+1 ] is nilpotent because Hi /Hi+1 is nilpotent
and Hi+1 /[K, Hi+1 ] is central in K/[K, Hi+1 ]. Then since [K, Hi ] ⊆ Hi , it follows
that [K, Hi ]/[K, Hi+1 ] is nilpotent as it is a subgroup of a nilpotent group. Hence
[K, H] = [K, Hl−1 ] ⊆ . . . ⊆ [K, H1 ] ⊆ K is a normal series where each factor group
is nilpotent, and therefore K/[K, H] ∈ N l−1 . That is, we also have H ⊆ [K, H] by the
minimality of H, and therefore H = [K, H].
5
Easy cases
Lemma 5.1. Suppose A is a finitely generated abelian profinite group. Then every subgroup of finite index is open in A.
Proof. Since A is finitely generated, we know that A is a quotient of
n
Y
Ẑ for some
i=1
positive integer n. Under the quotient topology, we know that a subgroup N ⊆ A is
n
Y
open in A if its preimage is open in
Ẑ. It is therefore sufficient to show that every
finite index subgroup of
n
Y
i=1
Ẑ is open, and so we assume that A has this structure.
i=1
Now Ẑ is procyclic, so we can write A =
Then tA =
n
Y
n
Y
hai i. Suppose that N ⊆ A has index t.
i=1
htai i, and we see by considering the product topology that tA is a closed
i=1
subgroup of finite index in A. By Lemma 2.3, it follows that tA is open in A. However,
tA ⊆ N , and therefore N can be thought of as the union of the cosets of tA in N . That
is, N is a union of open subgroups of A, and therefore N is also open in A.
Lemma 5.2. Suppose that G is a finitely generated prosolvable group with [G, G] closed
in G. Then every subgroup of finite index in G is open.
Proof. Write N for our subgroup of finite index. We first claim that we may assume that
N is normal in G. Now if N has finite index in G, then so does coreG (N ). If coreG (N )
7
is open in G, then N is the union of the cosets of coreG (N ), and therefore N is open in
G. Hence we may assume that N C G, and we proceed by induction on |G : N |.
If |G : N | = 1 then we have nothing to prove, so we may assume that N < G. If there
exists H ⊆ G with N < H < G, then H will be open in G by induction. This means
that H is also closed in G by Lemma 2.3, and therefore H is also a finitely generated
prosolvable group. Also by induction, we have that N is open in H, and therefore
N = N ∩ H is open in G by considering the subgroup topology. We can therefore assume
that no such H exists, so that |G : N | = p for some prime p. In particular, we have that
[G, G] ⊆ N . Since [G, G] is closed in G, we know that G/[G, G] is a finitely generated
abelian profinite group, and therefore N/[G, G] is open in G/[G, G] by Lemma 5.1. Since
N is the preimage of N/[G, G] and N/[G, G] is open in G/[G, G], we have that N is open
in G, as claimed.
Our goal now is to reduce to the case handled in Lemma 5.2. To do this, we need
to establish a few preliminary results. For the next two results, we make frequent use of
the commutator identity
[ab, c] = [a, c]b [b, c] .
(1)
Lemma 5.3. Suppose that K is a group and B C K is a subgroup with [B, K] central in
K.
(a) For each k ∈ K, the map ϕk : B → [B, K] via b 7→ [b, k] is a homomorphism.
(b) For each b ∈ B, the map ϕb : K → [B, K] via k 7→ [b, k] is a homomorphism.
Proof. Let a, b ∈ B, and consider ϕk (ab) = [ab, k]. By Equation (1), we know that
[ab, k] = [a, k]b [b, k]. However, we know that [B, K] is central in K, and therefore [a, k]b =
[a, k]. This gives us that ϕk (ab) = [a, k]b [b, k] = [a, k][b, k] = ϕk (a)ϕk (b), establishing that
ϕk is a homomorphism.
Now suppose that h, k ∈ K, and consider ϕb (hk) = [b, hk]. Now [b, hk] = [hk, b]−1 .
We can again use Equation (1) to rewrite [hk, b] = [h, b]k [k, b]. Now [B, K] = [K, B], and
so we know that [h, b] and [k, b] are also central in K. Thus [h, b]k = [h, b], and we have
that [hk, b] = [h, b][k, b]. This gives us that
ϕb (hk) = [hk, b]−1 = [h, b]−1 [k, b]−1 = [b, h][b, k] = ϕb (h)ϕb (k) ,
as desired.
Lemma 5.4. Let K = hx1 , x2 , . . . , xr i be a finitely generated group.
(a) If A is an abelian normal subgroup of K, then every element of [A, K] can be
expressed in the form
r
Y
[ai , xi ]
i=1
for a1 , a2 , . . . , ar ∈ A.
8
(b) Assume that H is a nilpotent normal subgroup of K. Suppose that H is generated
by y1 , . . . , ys as a normal subgroup. That is, H = hy1 , . . . , ys iK . Then every element
of [H, K] can be written as
r
s
Y
Y
[hi , xi ] [h0j , yj ]
i=1
for
h1 , . . . , hr , h01 , . . . , h0s
j=1
∈ H.
(c) Assume that K is nilpotent. Then every element of [K, K] can be expressed in the
form
r
Y
[ki , xi ]
i=1
for k1 , . . . kr ∈ K.
Proof. We can use Equation (1) to show that for each 1 ≤ i ≤ r, we have that [A, xi ] =
{[a, xi ] | a ∈ A} is actually a subgroup of A. Certainly, 1 ∈ [A, xi ] since 1 ∈ A. Also,
[A, xi ] is closed under taking inverses since
1 = [1, xi ] = [aa−1 , xi ]
−1
= [a, xi ]a [a−1 , xi ]
= [a, xi ][a−1 , xi ] ,
where the second line follows from the first by Equation (1) and the third line follows
from the second since A is abelian and normal in K. This gives us that [a, xi ]−1 = [a−1 , xi ],
and thus [A, xi ] is closed under inverses. Finally, we have that for a, b ∈ A, we have
[ab, xi ] = [a, xi ]b [b, xi ] = [a, xi ][b, xi ] since [a, xi ], b ∈ A and A is abelian. Thus for each
1 ≤ i ≤ r, we have that [A, xi ] is a subgroup of A, as claimed.
r
Y
Let L =
[A, xi ]. Since for each a ∈ A and generator xi we have [a, xi ] ∈ L, we
i=1
have that axi ≡ a (mod L), and therefore K acts trivially on A modulo L. That is, A
is central in K modulo L, and we have that [A, K] ⊆ L. However, it is clear from our
r
Y
definition of L that L ⊆ [A, K], and so we have [A, K] = L =
[A, xi ], proving part (a).
i=1
For part (b), we induct on the nilpotency class of H. If H has class 1, then H is
abelian, and part (a) implies the result. We therefore assume that H has class c > 1 and
that part (b) holds for nilpotent normal subgroups of smaller class. Consider the lower
central series for H, which we write as
1 = Zc+1 ⊆ Zc ⊆ . . . ⊆ Z2 ⊆ Z1 = H .
9
We further abbreviate B = Zc−1 and A = Zc . By definition, we know that [B, H] = A
and [A, H] = 1, so that A is central in H. In particular, A is an abelian normal subgroup
r
Y
of K, and by part (a), we know that [A, K] =
[A, xi ].
i=1
We now claim that [B, yi ] is a subgroup of A for each 1 ≤ i ≤ s. Let us fix yi for some
1 ≤ i ≤ s and write y = yi . Of course, 1 = [1, y] ∈ [B, y], and if a, b ∈ B then we have
[ab, y] = [a, y]b [b, y] by Equation (1). Yet [a, y] ∈ A, b ∈ H, and A is central in H, giving
us that [a, y]b = [a, y]. Hence [ab, yi ] = [a, y][b, y], and therefore [B, y] is closed. Finally,
we must argue that [B, y] is closed under taking inverses. Suppose that b ∈ B. Then
1 = [1, y] = [bb−1 , y]
−1
= [b, y]b [b−1 , y]
= [b, y][b−1 , y] ,
where the second line follows from the first by Equation (1), and the third line follows
from the second as b−1 ∈ H and [B, y] is central in H. Multiplying both sides of this
equation by [b, y]−1 , we have that [b, y]−1 = [b−1 , y], and [b−1 , y] ∈ [B, y]. This shows that
[B, y] is a subgroup, and as y was an arbitrary yi for 1 ≤ i ≤ s, this holds for all yi . Set
s
Y
J=
[B, yj ]. As each [B, yi ] ⊆ A for 1 ≤ i ≤ s, we know that J ⊆ A.
j=1
We now claim that A = [A, K]J. If this holds, then we will have that every element
of A can be expressed in the form
r
s
Y
Y
[ai , xi ] [bj , yj ]
i=1
j=1
for ai ∈ A and bj ∈ B with 1 ≤ i ≤ r and 1 ≤ j ≤ s.
Of course, [A, K]J ⊆ A, so we must show the reverse containment. We now claim
that [A, K]J C K. Now A C K and so is [A, K], and therefore [A, K]K ⊆ [A, K]. However, [J, K] ⊆ [A, K] since J ⊆ A, and therefore J K ⊆ J[A, K]. Thus ([A, K]J)K ⊆
[A, K]J[A, K] = [A, K]J since [A, K] and J are both contained in the abelian subgroup
A. Now for each 1 ≤ i ≤ s, we have that yi centralizes B mod J, and therefore each
yi centralizes B mod [A, K]J. That is, each yi ∈ CK/[A,K]J (B) for 1 ≤ i ≤ s. Since
CK/[A,K]J (B) is normal in K and each yi ∈ CK/[A,K]J (B), we have that H, the subgroup
generated by all conjugates of the yi , is contained in CK/[A,K]J (B). That is, H acts
trivially on B/[A, K]J, and therefore [B, H] ⊆ [A, K]J. But [B, H] = A, and so we have
A = [A, K]J, as claimed.
Finally, let u ∈ [H, K]; we will show that u takes on the desired form for (b). Now
H/A is nilpotent of length < c, and so by induction, we have that
u≡
r
Y
i=1
[hi , xi ]
s
Y
[h0j , yj ]
j=1
10
(mod A)
for h1 , . . . , hr , h01 , . . . , h0s ∈ H. As A = [A, K]J, we know that every element of A can
be written as a specific product of commutators, so we can write
u=
r
s
r
s
Y
Y
Y
Y
[hi , xi ] [h0j yj ] [ai , xi ] [bj , yj ]
i=1
j=1
i=1
j=1
for ai ∈ A with 1 ≤ i ≤ r and for bj ∈ B with 1 ≤ j ≤ s. Finally, since A is central
in H, [A, K] and J are contained in A, and [H, K] ⊆ H, we can rearrange these terms
to align all the xi ’s and yj ’s, giving us that
u=
r
s
Y
Y
[hi , xi ][ai , xi ] [bj , yj ][h0j , yj ] .
i=1
j=1
Using Equation (1) and the fact that A is central in H, we see that [hi , xi ][ai , xi ] =
[hi , xi ]ai [ai , xi ] = [hi ai , xi ] for each 1 ≤ i ≤ r. Similarly for each 1 ≤ j ≤ s, we have that
0
[bj , yj ] is central in H, and therefore [bj , yj ][h0j , yj ] = [bj , yj ]hj [h0j , yj ] = [bj h0j , yj ]. Hence
r
s
Y
Y
u=
[hi ai , xi ] [bj h0j , yj ] ,
i=1
j=1
which proves part (b).
Last, we establish part (c). Again, the proof will proceed by induction on the nilpotency class c of K. If c = 1, then K is abelian, [K, K] = 1, and there is nothing to prove.
This establishes our base case, and we assume that c > 1. Consider the lower central
series for K, which we write as
1 = Zc+1 ⊆ Zc ⊆ . . . ⊆ Z2 ⊆ Z1 = K .
For convenience, we write A = Zc and B = Zc−1 . By definition, we know that
A = [B, K]. Since K is finitely generated and B C K, we know that B is also finitely
generated. Thus B = hy1 , y2 , . . . , ys iK for some collection {y1 , . . . , ys }.
We first argue that we can write every a ∈ A as a product of r commutators of the
form [bi , xi ] for 1 ≤ i ≤ r. Since K is generated by x1 , . . . , xr , we know that each yi
can be written as some product of the xi . Moreover, A is central in K, and by part (b)
of Lemma 5.3, we have that the map ϕb : K → A via k 7→ [b, k] is a homomorphism.
Because each of the ϕb is a homomorphism, we can write each yj in its decomposition
as a combination of xi ’s, and use these observations to write every a ∈ A of the form
n
Y
[bj , zj ], where each zj ∈ {x1 , . . . , xr } and bj ∈ B. However, each [bj , zj ] ∈ A and A
j=1
is central in K, so we can rearrange this product to be of the form
ni
r Y
Y
[bij , xi ], where
i=1 j=1
ni is an integer that depends on i. Finally, we can combine [bij xi ] and [bi(j+1) , xi ] to be
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[bij bi(j+1) , xi ] since ϕxi : B → A is a homomorphism by part (a) of Lemma 5.3. This
r
Y
shows that every a ∈ A can be written as
[bi , xi ] for some bi ∈ B.
i=1
Now let g ∈ [K, K]. Since K/A has class c − 1, we know by induction that
g≡
r
Y
[ki , xi ]
(mod A) ,
i=1
for some ki ∈ K. Using the decomposition we have established for all elements of A,
we see that we can write
r
r
Y
Y
g=
[ki , xi ] [bj , xj ]
i=1
j=1
for some bj ∈ B. Again, [B, K] = A and A is central in K, so we have that
g=
r
Y
[bi , xi ][ki , xi ] .
i=1
Yet the centrality of A in K gives us that [bi , xi ] = [bi , xi ]ki , and so we have that
r
Y
[bi , xi ][ki , xi ] = [bi ki , xi ] for all 1 ≤ i ≤ r. Thus we can write g =
[bi ki , xi ], and the
i=1
result holds.
Lemma 5.5. Let K = H o L. If K can be generated by d elements, then there are d
elements of H that generate it as a normal subgroup of K.
Proof. Write K = hk1 , . . . , kd i. Since K = HL, for all 1 ≤ i ≤ d we can write ki = hi li
for some hi ∈ H and li ∈ L. Now consider the normal subgroup of K generated by the
conjugates of the collection of hi ’s in this decomposition. Since each hi ∈ H and H C G,
we certainly have that N ⊆ H. On the other hand, we have that K = N L since ki
belongs to N L for every 1 ≤ i ≤ d. Thus N = H, and we see that the normal closure of
the d elements h1 , . . . , hd generate H.
The next lemma, we state without proof.
Lemma 5.6. Let K ∈ N l and let H be a normal subgroup of K minimal so that K/H ∈
N l−1 . Suppose further that H is abelian. Then there exists L so that K = H o L.
Finally, we are ready to prove Theorem 1.2.
Proof. By Lemma 5.2, it suffices to show that [G, G] is closed in G. By Lemma 2.2, this
will follow if we can exhibit an integer valued function f such that if K ∈ N l is generated
by k elements, then every element of [K, K] is generated by f (k) elements. We will show
that we can take f (k) = k(2l − 1).
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We argue by induction on l. If l = 1, then K is nilpotent, and part (c) of Lemma
5.4 gives us that every element of [K, K] can be written as a product of k = k(2 · 1 − 1)
commutators. Thus the base case of our induction holds, and we may assume that l > 1.
By Lemma 4.2, we know that there exists a unique normal subgroup H ⊆ K minimal
with the property that K/H ∈ N l−1 . Furthermore, we know that this subgroup H is
nilpotent and that [K, H] = H. As H is nilpotent, we know that [H, H] < H and
[H, H] C K. By the minimality and uniqueness of H, it follows that K/[H, H] ∈ N l
and that H/[H, H] is the unique normal subgroup of K/[H, H] with K/H ∈ N l−1 . Now
H/[H, H] is abelian, so we can appeal to Lemma 5.6 to find a complement L/[H, H] for
H/[H, H] in K/[H, H]. Since K can be generated by k elements, we know by Lemma
5.5 that H/[H, H] is generated as a normal subgroup of K/[H, H] by k elements. That
is, there are elements y1 , . . . , yk ∈ H so that H = [H, H]hy1 , . . . , yk iK . As H is nilpotent,
we know that [H, H] ⊆ Φ(H). Yet elements of Φ(H) are not needed to generate any
element if H, so we deduce that in fact H = hy1 , . . . , yk iK . We can now apply part (b)
of Lemma 5.4 to conclude that every element of H = [H, K] can be written as a product
of 2k commutators. Now let g ∈ [K, K]. By induction, the element gH ∈ K/H can be
written as the product of k(2(l − 1) − 1) = k(2l − 3) commutators in K/H. This means
that g = vh, where v is the product of k(2l − 3) commutators of K and h ∈ H. We have
just shown that h can be written as the product of 2k commutators, and that gives us
that g can in fact be written as a product of k(2l − 3) + 2k = k(2l − 1) commutators.
References
[1] Nikolov, N., and Segal, D. Finite index subgroups in profinite groups. Comptes
Rendus Mathematique 337, 5 (2003), 303–308.
[2] Nikolov, N., Segal, D., and Nikolav, N. On finitely generated profinite groups,
i: strong completeness and uniform bounds. Annals of mathematics (2007), 171–238.
[3] Ribes, L., and Zalesskii, P. Profinite groups. Springer, 2010.
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