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Pro-p Groups, Lent 2009
Solutions 1
1. see Prop (2.2). If f, g : X → Y are continuous and Y Hausdorff then
{x ∈ X|f (x) = g(x)} is closed in X.
Proof. Let N = {x ∈ X : f (x) 6= g(x)}. Let y ∈ N and U , V be disjoint
open subsets of Y containing f (y) and g(y). Then f −1 (U ) ∩ g −1 (V ) is an
open neighbourhood of y and is contained in N . Thus N is a union of open
sets, so open, hence result.
2. Let G be a group and L a non-empty family of normal subgroups s.t. if
K1 , K2 ∈ L and K3 is a normal subgroup containing K1 ∩ K2 then K3 ∈ L.
Let T be the family of all unions of sets of cosets Kg with K ∈ L, g ∈ G.
Show that T is a topology on G and that G is a topological group with respect
to this topology. Show also that L is the set of open normal subgroups of G
with respect to this topology.
S
Proof. Clearly G = g∈G Kg for some K ∈ L and ∅ is given by the empty
union. Also, unions lie in T by definition. Need to consider Kg ∩ Lh where
K, L ∈ L and g, h ∈ G. We can write K as a union of cosets of K ∩ L and
hence Kg can be written as a union of cosets of K ∩ L. Similarly we can
write L and hence Lh as a union of cosets of K ∩ L. Thus Kg ∩ Lh can be
written as a union of cosets of K ∩ L ∈ L. So T is a topology on G.
To show that G is a topological group with this topology we need to show
that multiplication m : G × G → G and inversion i : G → G are continuous.
−1
Note that i−1
= Kh−1 , thus inversion is continuous. Also
S(Kh) = h K −1
−1
m (Kg) = h∈G (Kgh × Kh ), hence multiplication continuous.
Finally, suppose N is a normal open subgroup of G. Then N can be
written as a union of cosets of subgroups of L. In particular K ≤ N for some
K ∈ L and thus N ∈ L by property of L.
3. Lemma (1.1)(b). G a topological group. To prove:
Every open subgroup of G is closed.
Every closed subgroup of finite index is open.
If G compact, every open subgroup of G has finite index.
S
Proof. (i) H is open, so Hg is open for all g ∈ G. Thus K = Hg6=H Hg is a
union of open sets so open and H = G \ K is closed.
(ii) As (i)Sbut using that a finite union of closed sets is closed.
(iii) G = Hg so the S
Hg form an open cover, then by compactness we have
a finite subcover G = ni=1 Hgi . Hence result.
4. Prop (1.3)(b). G compact, totally disconnected topological group. To
prove:
A subset of G is both closed and open if and only if it is a union of finitely
many cosets of open normal subgroups.
S
Proof. Suppose P is both closed and open. By (1.3)(a) P = Ki xi where
the Ki are open normal subgroups. By (1.1)(b) Ki open implies Ki xi open,
thus we have an open covering of P . So by compactness of P we have
P = K1 x1 ∪ . . . ∪ Kn xn for some 1 ≤ i ≤ n.
Conversely if P = N1 x2 ∪ . . . ∪ Nm xm where the Ni are open normal
subgroups of G. Then Ni xi is open, and thus P is open. Also by (1.1)(c) Ni
is closed, hence Ni xi is closed and thus P is closed.
5. Recall if S ⊆ G then the centralizer of S is
CG (S) = {g ∈ G|gs = sg ∀s ∈ S}
and the normalizer of a subgroup H is
NG (H) = {g ∈ G|g −1 hg ∈ H & ghg −1 ∈ H ∀h ∈ H}.
Suppose G is a Hausdorff topological group.
(a) Prove that centralizers of subsets and normalizers of closed subgroups are
closed.
(b) Prove that each closed abelian subgroup A is contained in a maximal
closed abelian subgroup (i.e. the family of closed abelian subgroups containing A, partially ordered with respect to inclusion, has a maximal element).
(c) Prove that if there is a series
1 = G0 C G1 C · · · C Gn = G
such that Gi /Gi+1 is abelian for each i, then there is such a series consisting
of closed subgroups. (Consider closures).
Proof. (a) First we consider centralizers. Fix s ∈ S and consider the two
maps rs , ls : G × G → G which send (g, s) 7→ gs and (s, g) 7→ sg respectively.
These maps are both continuous, so by Question 1, the set {g ∈ G : gs = sg}
is closed in G. Take the intersection of these closed sets over s ∈ S to get
that the centralizer is closed.
Now consider normalizers. Fix h ∈ H and define the map fh : G → G by
fh (g) = g −1 hg. Then fh is continuous, so the preimage of H, call it Uh , is
closed. So ∩h∈H Uh = {g ∈ G : g −1 hg ∈ H ∀h ∈ H} is closed. Similarly we
can define kh (g) = ghg −1 , etc.
(b) Let AS
≤ A1 ≤ A2 ≤ · · · be a chain of closed abelian subgroups containing
A. Then Ai is an abelian subgroup, call it X. Further, we claim that the
closure of X, X, is an abelian subgroup. First we need to show X is a
subgroup. Let x, y ∈ X, and U be an open neighbourhood of xy. Then by
continuity of multiplication there exist open neighbourhoods, A and B of x
and y respectively such that AB ⊆ U . But A ∩ X 6= ∅ and B ∩ X 6= ∅
so AB ∩ X 6= ∅ and xy ∈ X as required. Similarly given U an open set
containing x−1 , then i−1 (U ) is an open set containing x so i−1 (U ) intersects
non-trivially with X and hence so does U . So x−1 ∈ X as required. Next
we need to show that X is abelian. Let x, y ∈ X and suppose xy 6= yx.
Then, since G Hausdorff, we can find open, disjoint sets U and V such that
xy ∈ U and yx ∈ V . Since multiplication is continuous we can can find open
neighbourhoods A1 and B1 of x and y in G such that A1 B1 ⊆ U . Similarly
we can find open neighbourhoods A2 and B2 of x and y such that B2 A2 ⊆ V .
Then x ∈ A = A1 ∩ A2 and y ∈ B = B1 ∩ B2 and A and B open so there
exists h ∈ X ∩ A and k ∈ X ∩ B. But then since X abelian hk = kh ∈ U ∩ V
which is a contradiction.
(c) As the question suggests let us consider
1 = G0 ≤ G1 ≤ · · · ≤ Gn = G.
Clearly 1 C G1 and G1 is abelian by proof in (b). We want to show that
G1 C G2 . Similar arguments to those used in (b) show that G1 ≤ G2 . We
now show that G2 lies inside the normaliser of G1 in G2 . Let x ∈ G1 , we need
to consider g −1 xg for g ∈ G2 . Let U be an open neighbourhood of g −1 xg
then gU g −1 is an open neighbourhood of x, since multiplication by a fixed
element is a homeomorphism. So gU g −1 ∩ G1 6= ∅ which implies U ∩ G1 6= ∅.
Thus NG2 (G1 ) ≥ G2 , but by (a) the normaliser of a closed subgroup is closed
so we must have NG2 (G1 ) = G2 as required. And, again by (b), G2 G1 /G1
abelian implies that G2 /G1 abelian, etc.
6. G a profinite group and X ≤ G, then
X = ∩{K|X ≤ K ≤0 G}.
Proof. We know from (1.3) that X = ∩{N X : N Eo G} and clearly this
contains ∩{K : X ≤ K ≤o G}. To go the other way note that an open
subgroup of G contains an open normal subgroup (also (1.3)). So given K
open in G and containing X we can find N normal, open in G and N ≤ K
so N X ≤ K, as required.
7. Let X be a dense subgroup of a topological group G (i.e. X = G). Prove
that N = N ∩ X for each open subgroup N of G.
Proof. Recall, an open subgroup is closed so N ∩ X ≤ N . We want to show
that N ∩ X is dense in N . Thus given any open set U ⊆ N we need to show
U ∩ (N ∩ X) 6= ∅. However X is dense in G so U ∩ X 6= ∅ but since U ⊆ N
it follows that U ∩ N ∩ X = U ∩ X 6= ∅.
8. If G is finite what is the profinite completion of G?
Answer. G.
9. Give an example of a group which has trivial pro-p completions for each
p but a non-trivial profinite completion (or describe the properties such a
group should have).
Q
Example.
n≥5 An . Since An simple for n ≥ 5 it follows that An has no
soluble quotients, so
Q no p-power quotients. But we do have normal subgroups
of finite index, eg n≥6 An .