Dinakar Ramakrishnan Robert J. Valenza
1
Topological Groups
Fourier Analysis on Number Fields
Masarykova univerzita
přírodovědecká fa ulía
Ústřední k.i.hovna Hlav. in v. č.
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Signatura_________7W. TJ^
Springer
Our work begins with the development of a topological framework for the key elements of our subject. The first section introduces the category of topological groups and their fundamental properties. We treat, in particular, uniform continuity, separation properties, and quotient spaces. In the second section we narrow our focus to locally compact groups, which serve as the locale for the most important mathematical phenomena treated subsequently. We establish the essential deep feature of such groups: the existence and uniqueness of Haar measure; this is fundamental to the development of abstract harmonic analysis. The last two sections further specialize to profinite groups, giving a topological characterization, a structure theorem, and a set of results roughly analogous to the Sylow Theorems for finite groups. The prerequisites for this discussion will be found in almost any first-year graduate courses in algebra and analysis.
1.1 Basic Notions
Definition. A topological group is a group G (identity denoted e) together with a topology such that the following conditions hold:
(i) The group operation
GxG ->G (g,h) H* gh
is a continuous mapping. (The domain has the product topology.)
(ii) The inversion map
G ->G
is likewise continuous.
By convention, whenever we speak of a finite topological group, we intend the discrete topology.
2      1. Topological Groups
Clearly the class of topological groups together with continuous homomor-phisms constitutes a category.
It follows at once that translation (on either side) by any given group element is a homeomorphism G->G. Thus the topology is translation invariant in the sense that for all geG and £/cG the following three assertions are equivalent:
(i) U is open.
(ii) gU is open.
(iii) Ug is open.
Moreover, since inversion is likewise a homeomorphism, U is open if and only if U~x — {x '. jc-' e U} is open.
A fundamental aspect of a topological group is its homogeneity. In general, ifXis any topological space, Homeo(X) denotes the set of all homeomorphisms X->X. If S is a subset of Homeo(J0, then one says that X is a homogeneous space under S if for all x,yeX, there exists feS such that f(x)=y. (When S is unspecified or perhaps all of Homeo(A~), one says simply that X is a homogeneous space.) Clearly any topological group G is homogeneous under itself in the sense that given any points g,heG, the homeomorphism defined as left translation by hg~] (i.e., x\-^hg'^x) sends g to h. From this it follows at once that a local base at the identity eeG determines a local base at any point in G, and in consequence the entire topology.
Examples
(1) Any group G is a topological group with respect to the discrete topology.
(2) R*, R*, and C* are topological groups with respect to ordinary multiplication and the Euclidean topology.
(3) R" and C are topological groups with respect to vector addition and the Euclidean topology.
(4) Let k=R or C. Then the general linear group
GL„(*) = {geMJk): det(g)*0}
is a topological group with respect to matrix multiplication and the Euclidean topology. The special linear group
SLn(k) = {geGL„(*): detfe)= 1} («>1)
is a closed subgroup of GLn(k).
1.1. Basic Notions 3
In subsequent discussion, if X is a topological space and xeX, we shall say that C/cI is a neighborhood of x if x lies in the interior of U (i.e., the largest open subset contained in U). Thus a neighborhood need not be open, and it makes sense to speak of a closed or compact neighborhood, as the case may be.
A subset S of G is called symmetric if S=S~l. This is a purely group-theoretic concept that occurs in the following technical proposition.
1-1 Proposition. Let G be a topological group. Then the following assertions hold:
(i) Every neighborhood U of the identity contains a neighborhood V of the identity such that VVc: U.
(ii) Every neighborhood U of the identity contains a symmetric neighborhood V of the identity.
(iii) IfH is a subgroup of G, so is its closure.
(iv) Every open subgroup of G is also closed.
(v) IfK] and K2 are compact subsets of G, so is K}K2.
Proof, (i) Certainly we may assume that U is open. Consider the continuous map (p-.UxU^G defined by the group operation. Certainly p~\U) is open and contains the point (e,e). By definition of the topology on Ux U, there exist open subsets VVV2 of U such that (e,e)eVixVfSetV=V]nV2. Then Vis a neighborhood of e contained in U such that by construction VVc U.
(ii) Clearly g e£/nt/_1 <=> g,g~] e U, so V= t/ntT1 is the required symmetric neighborhood of e.
(iii) Any two points g and h in the closure of H may be exhibited as the limits of convergent nets in H itself. Hence by continuity their product is likewise the limit of a convergent net in H and similarly for inverses.
(iv) If H is any subgroup of G, then G is the disjoint union of the cosets of H, and hence H itself is the complement of the union of its nontrivial translates. If H is open, so are these translates, whence H is the complement of an open set and therefore closed.
(v) KtK2 is the image of the compact set KtxK2 under the continuous map (fc, ,k2) i-> kyk2. It is therefore compact by general topology □
Note that (i) and (ii) together imply that every neighborhood U of the identity contains a symmetric neighborhood Fsuch that VVcU.
4      1. Topological Groups
1.1. Basic Notions 5
Translation of Functions and Uniform Continuity
Given an arbitrary function/on a group, we define its left and right translates by the formulas
Lhm=f{h-'g) and Rhf(g)=f(gh) .
If / is a (real- or complex-valued) continuous function on a topological group, we say that /is left uniformly continuous if for every e>0 there is a neighborhood Vote such that
heV=*\LJ-fl<s
where || ||u denotes the uniform, or sup, norm. Right uniform continuity is defined similarly. Recall that ^.(G) denotes the set of continuous functions on G with compact support.
1-2 Proposition. Let G be a topological group. Then every function f in &C(G) is both left and right uniformly continuous.
Proof. We prove right uniform continuity. Let AT=supp(/) and fix s>0. Then for every geK there exists an open neighborhood Ug of the identity such that
heUg^\f(gh)-f(g)\<s .
Equivalently,/(g') is e-close to/(g) whenever g~'g' lies in Ug. Moreover, by the comment following the previous proposition, each Ug contains an open symmetric neighborhood V of the identity such that V^^Vg Clearly the collection of subsets gVg covers K, and a finite subcollection {gyfjL,, „ suffices. Henceforth we write Vj for Vg and Uj for U . Define V, a symmetric open neighborhood of the identity e, by the formula 1
V = f]VJ .
7=1
If geAT, then gegjVj for some/ For he V we consider the difference f(gh)-f(g): \figh) -fig)I < \f(gh) -/(g,)| + \f(g) -fig)| .
The point is that both gr'g and gj~]gh lie in Uj, so that both terms on the right are bounded by & (Here is where we use that property f^.f^c for all/.) This establishes right uniform continuity for K.
When g does not lie in K, then we must bound \f(gh)\. Iff(gh)*0, then ghe gjVjior some /, and therefore f(gh) is £-close to /(g). Moreover, gj~xg=gf]ghh~^ lies in Uj (here is where we use the symmetry of V), and it follows that \f(g)\<e since gj is close to g and/(g)=0 by assumption. Consequently \f(gh)\<2e, and the argument is complete. □
Separation Properties and Quotient Spaces
Some authors assume as part of the definition of a topological group that the underlying topology is Tv In this case it is also customary to reserve the term subgroup for a closed subset that constitutes a subgroup in the ordinary algebraic sense. Note that in general we accept neither of these assumptions.
The following proposition shows, among other things, that for a topological group the separation axioms 71, and T2 (Hausdorff) have equal strength.
1-3 Proposition. Let G be a topological group. Then the following assertions are equivalent:
(i) GisTx.
(ii) G is Hausdorff.
(iii) The identity e is closed in G.
(iv) Every point of G is closed.
Proof. (i)=>(ii) If G is 7",, then for any distinct g,heG there is an open neighborhood U of the identity lacking g/r1. According to Proposition 1-1, U admits a symmetric open subset V, also containing the identity, such that VVc, U. Then Vg and Vh are disjoint open neighborhoods of g and h, since otherwise g/r1 lies
inV-]V=VVc:U.
(ii) =>(iii) Every point in a Hausdorff (or merely 7",) space is closed.
(iii) =>(iv) This is a consequence of homogeneity: For every point xeG there is a homeomorphism that carries e onto x. Hence if e is closed, so is every point.
(iv) => (i) Obvious by general topology. □
If H is a subgroup of the topological group G, then the set GIH of left cosets of G acquires the quotient topology, defined as the strongest topology such that the canonical projection p.gh-^gH is continuous. Thus U is open in GIH if and only if p~](U) is open in G. Recall from algebra that GIH constitutes a group under coset multiplication if and only if H is moreover normal in G. We shall see shortly that in this case GIH also constitutes a topological group with respect to the quotient topology.
6      1. Topological Groups
The following two propositions summarize some of the most important properties of the quotient construction.
1-4 Proposition. Let G be a topological group and let H be a subgroup of G. Then the following assertions hold:
(i) The quotient space GIH is homogeneous under G.
(ii) The canonical projection p.G-* GIH is an open map.
(iii) The quotient space GIH is    if and only ifH is closed.
(iv) The quotient space GIH is discrete if and only ifH is open. Moreover, ifG is compact, then H is open if and only if GIH is finite.C0^ «6ogW0
(v) IfH is normal in G, then GIH is a topological group with respect to the quotient operation and the quotient topology.
(vi) Let H be the closure of {e} in G. Then H is normal in G, and the quotient group GIH is Hausdorjf with respect to the quotient topology.
Proof, (i) An element xeG acts on GIH by left translation: gHh-> xgH. The inverse map takes the same form, so to show that left translation is a homeo-morphism of GIH, it suffices to show that left translation is an open mapping on the quotient space. Let U be an open subset of GIH. By definition of the quotient topology, the inverse image of U under p is an open subset U of G, and it follows that the inverse image of gU under p is gU, also an open subset of G. Therefore gU is open, and left translation is indeed an open map, as required.
(ii) Let Fbe an open subset of G. We must show that p(V) is open in the quotient. But p(V) is open in GIH if and only if p~](p(V)) is open in G. By elementary group theory, p~l(p(V))=V-H. Letxlie in V-H, so that x=vh for some veV and heH. Since Vis open, given any veV, there is an open neighborhood C/Vc V containing v. Thus Uv-h is an open neighborhood of x contained in V-H, which is accordingly open.
(iii) By general topology, GIH is Tx if and only if every point is closed. Since a coset of H is its own inverse image under projection, each coset is a closed point in GIH if and only if each is likewise a closed subset of G. But by homogeneity this is the case if and only if H itself is closed in G. (Note that we cannot appeal to the previous proposition, since the topological space GIH is not necessarily a topological group with respect to multiplication of cosets.)
(iv) Let Hbz a subgroup of G. Then by part (ii), H is an open subset of G if and only if H is an open point of GIH. Since GIH is homogeneous under G, this
1.1. Basic Notions 7
holds if and only if GIH is discrete. Assume now that G is compact. Then so is GIH, since p is continuous. But then H is open if and only if GIH is both compact and discrete, which is to say, if and only if GIH is finite. (Recall our convention that a finite topological group carries the discrete topology.)
(v) Assume that H is a normal subgroup of G. Then from part (ii) and the commutative diagram
T
G g  ) G
PI IP
GIH       TpU) > GIH
(where T denotes left translation by g), we see at once that translation by any group element is continuous on the quotient. A similar diagram establishes the continuity of the inversion map.
(vi) Since {<?} is a subgroup of G, so is its closure H. Moreover, it is the smallest closed subgroup of G containing e and therefore normal, since each conjugate of H is likewise a closed subgroup containing e. In light of the previous proposition, the full assertion now follows from parts (iii) and (v) above. □
Part (vi) shows that every topological group projects by a continuous homo-morphism onto a topological group with Hausdorff topology. In this sense the assumption that a given group is Hausdorff is not too serious.
1-5 Proposition. Let G be a Hausdorff topological group. Then the following assertions hold:
(i) The product of a closed subset F and a compact subset K is closed.
(ii) IfH is a compact subgroup of G, then p: G^GIH is a closed map.
Proof, (i) Let z lie in the closure of the product FK. Then there exists a net converging to z of the form {xaya} with xaeF and yaeK. Since K is compact, we may replace our given net by a subnet such that {ya} converges to some point y in K. We claim that this forces the convergence of {xa} in F to zyrx, showing that z = zyxy lies in FK, which is therefore closed. To establish the claim, consider an arbitrary open neighborhood U of the identity e. We may choose yet another neighborhood of e contained in U such that Wq U. Then the nets {^xaya} and {y'^y} are both eventually in V, whence the product z~xxa ya y^y = z~lxa y is eventually in U. Thus lim xa=zy~\ as required.
8      1. Topological Groups
1.2. Haar Measure 9
(ii) If X is a closed subset of G, then arguing as the second part of the previous proposition, we are reduced to showing that X-H is likewise a closed subset of G. But if H is compact, this is just a special case of assertion (i). □
Remark. The requirement that H be compact is essential. For example, in the case G=R2, with subgroup H={(0,y) :>>eR}, we have clearly G/H=R, and under this identification, p(x,y) = x. Now let X= {(x,y)eR2: xy = 1}. Then X is closed, but p(X)=R* is not.
Locally Compact Groups
Recall that a topological space is called locally compact if every point admits a compact neighborhood.
Definition. A topological group G that is both locally compact and Hausdorff is called a locally compact group.
Note well the assumption that a locally compact group is Hausdorff. Accordingly, all points are closed.
1-6 Proposition. Let G be a Hausdorff topological group. Then a subgroup H ofG that is locally compact (in the subspace topology) is moreover closed. In particular, every discrete subgroup of G is closed.
Proof. Let K be a compact neighborhood of e in H. Then K is closed in H, since H is likewise Hausdorff, and therefore there exists a closed neighborhood U of e in G such that K- Ur^H. Since Ur>H is compact in H, it is also compact in G, and therefore also closed. By Proposition 1-1, part (i), there exists a neighborhood V of e in G such that VV^U. We shall now show that x eH =>x eH.
First note that H is a subgroup of G by Proposition 1-1, part (iii). Thus if x eH, then every neighborhood of x-1 meets H. In particular, there exists some yeVx~xr\H. We claim that the productyx lies in Ur\H. Granting this, both y and ja lie in the subgroup H, whence so does x, as required.
Proof of Claim. Since Ur\H is closed, it suffices to show that every neighborhood Wofyx meets UnH. Sincey~]tVis a neighborhood ofx, so isy~xWr\xV. Moreover, by assumption x lies in the closure of H, so there exists some element zey~x Wr\xVr\H. Now consider:
(i) the product yz lies both in W and in the subgroup H;
(ii) by construction, ye Vx~x\
(iii) by construction, zexV.
The upshot is that yz lies in Vx~x •xV= VV,a subset of U, and therefore the intersection Wr^(Un,H) is nonempty. This establishes the claim and thus completes the proof. □
1.2 Haar Measure
We first recall a sequence of fundamental definitions from analysis that culminate in the definition of a Haar measure. We shall then establish both its existence and uniqueness for locally compact groups.
A collection Tt of subsets of a set X is called a o-algebra if it satisfies the following conditions:
(i) Xe<SR.
(ii) If yl eTi, then A'elR, where ylc denotes the complement of A inX.
(iii) Suppose thatj4ne2D? (n> 1), and let
A = (jA„ .
Then also^4 e2ft; that is, Tt is closed under countable unions.
It follows from these axioms that the empty set is in IK and that <M is closed under finite and countably infinite intersections.
A set X together with a cr-algebra of subsets is called a measurable space. IfXis moreover a topological space, we may consider the smallest oalgebra 38 containing all of the open sets of X. The elements of 33 are called the Borel subsets ofX.
A positive measure n on an arbitrary measurable space (X^) is a function //:2)T-»R+u{oo} that is countably additive; that is,
for any family {An} of disjoint sets in Tl. In particular, a positive measure defined on the Borel sets of a locally compact Hausdorff space X is called a Borel measure.
Let ju be a Borel measure on a locally compact Hausdorff space X, and let E be a Borel subset of X. We say that ju is outer regular on E if
fj(E) = inf{//(C/): U3E, C/open} .
10     1. Topological Groups
We say that fi is inner regular on E if
/j(E) = sup{fj(K): KcE, K compact} .
A Radon measure on X is a Borel measure that is finite on compact sets, outer regular on all Borel sets, and inner regular on all open sets. One can show that a Radon measure is, moreover, inner regular on definite sets (that is, countable unions of //-measurable sets of finite measure).
Let G be a group and let ju be a Borel measure on G. We say that n is left translation invariant if for all Borel subsets E of G,
»(sE) = ii(E)
for all seG. Right translation invariance is defined similarly.
Definition. Let G be a locally compact topological group. Then a left (respectively, right) Haar measure on G is a nonzero Radon measure fj on G that is left (respectively, right) translation-invariant. A bi-invariant Haar measure is a nonzero Radon measure that is both left and right invariant.
The following proposition shows that the existence of a left Haar measure is equivalent to the existence of a right Haar measure and, in a sense, equates the translation invariance of measure with that of integration. As usual, we let
gf(G) = {/ e^(G):f(s) > 0 Vs e G and ||/||„> 0} .
We often abbreviate this to ^+ when the domain is clear.
1-7 Proposition. Let G be a locally compact group with nonzero Radon measure ju. Then:
(i) The measure n is a left Haar measure on G if and only if the measure Ji defined by ju(E) = //(£""' ) is a right Haar measure on G.
(ii) The measure fi is a left Haar measure on G if and only if
\LJdn = \fdn
G G
for all /e^+ and seG.
(iii) If ft is a left Haar measure on G, then fj is positive on all nonempty open subsets of G and
1.2. Haar Measure 11
jfdM>0
forallfeW*.
(iv) If n is a left Haar measure on G, then /i(G) is finite if and only ifG is compact.
Proof, (i) By definition, we have the equivalence
ji(E) = ji(Es)VseG <=> M^"') = ft^E'1) ^ eG
for all Borel sets E; the assertion follows at once. (For any topological group G, clearly £ is a Borel subset of G if and only if E'1 is.)
(ii) If n is a Haar measure on G, then the stated equality of integrals follows by definition for all simple functions /e^+ (i.e., finite linear combinations of characteristic functions on G), and hence, by taking limits, for arbitrary feWc+. Conversely, from the positive linear functional ia-dfj on ^.(G) we can, by the Riesz representation theorem, explicitly recover the Radon measure n of any open subset C/cG as follows:
//([/) = sup{J/^:/eg?(G), ||/||„<1, andsupp(/)cU} .
G
From this one sees at once that if the integral is left translation invariant, then {i(sU) = /x(U) for all open subsets U of G, since supp(/)c U if and only if supp(Lsf)^sU. The result now extends to all Borel subsets of G because a Radon measure is by definition outer regular.
(iii) Since n is not identically 0, by inner regularity there is a compact set K such that /a{K) is positive. Let [/be any nonempty open subset of G. Then from the inclusion
Kc{JsU
seG
we deduce that K is covered by a finite set of translates of U, all of which must have equal measure. Thus since /j(K) is positive, so is ju(u). If /g^+, then there exists a nonempty open subset U of G on which / exceeds some positive constant R. It then follows that
\fdn^Rfj{U)>Q
as claimed.
12    1. Topological Groups
(iv) If G is compact, then certainly /i(G) is finite by definition of a Radon measure. To establish the converse, assume that G is not compact. Let K be a compact set whose interior contains e. Then no finite set of translates of K covers G (which would otherwise be compact), and there must exist an infinite sequence     in G such that
sHe\JSjK
(1.1)
Now K contains a symmetric neighborhood U of e such that UU^K. We claim that the translates sp (/i 1) are disjoint, from which it follows at once from (iii) that /v(G) is infinite.
Proof of Claim. Suppose that for i<j we have gft**$p where u,ve U. Then Sj= s^#|H eSjK, since U is symmetric and UUcK. But this contradicts Eq. 1.1. □
With these preliminaries completed, we now come to one of the major theorems in analysis.
1-8 Theorem. Let G be a locally compact group. Then G admits a left (hence right) Haar measure. Moreover, this measure is unique up to a scalar multiple.
Via the Riesz representation theorem and statement (ii) of the previous proposition, the existence part of the proof reduces to the construction of a left-invariant linear functional on %{G). The key idea is the introduction of a translation-invariant device for comparing functions in
Preliminaries to the Existence Proof
Let/ <pe ^+. Set U={seG: <p(s) > \\<p\\u/2}, so that a finite number of translates of the open set U suffice to cover supp(/). Then there are n elements s,,...,stte G such that a linear combination of the translates of <p by the s, dominates / in the following sense:
IMI ^
The point is that if sesupp(/), then seSjU for some/, so that srlseU if <p is sufficiently large. Thus it makes sense to define (f:g), the Haar covering number of/with respect to <p, by the formula
1.2. Haar Measure 13
(/:«») = inf|jc> :°<ci.-.c-and/^ZcA^ forsomes"""s« eGJ 1
Note that since ||/||B is assumed positive, the Haar covering number is never zero. We shall see shortly that (f:<p) is almost linear in/for appropriately chosen (p.
1-9 Lemma. The Haar covering numberhas the following properties: (i) <J:(p) = {LJ:<p)forallseG
00 (f)+f2:<P)^(fi-<P) + (f2:<P)
(iii) (cf :<p) = c(f: <p)for any c >0
(iv) (/",:?) 5 (J2-<P) whenever f<f2
(v) tf:*)*||/fl,/|Ml
(vi) (f,:<p)<if,f0)(fo-<P)
Proof, (i) Since left multiplication by any given group element constitutes a permutation of the ambient group, for all seG we have the equivalence
f{t) < Y.CjL.jrt) V/ eG » LJ(t) 1 YcjL«j<P(0 Vr e G which is to say that
Hence precisely the same sets of coefficients cy occur in the calculation of (/: <p)
as for (Lsf. <p).
(ii), (iii), (iv) Obvious.
(v) If the coefficients cj appear in the calculation of (/: <p), then
f(s) < ^Cjpisp) < (2>,)|ML  V* eG
whence Zc^ |L/1|U/|HIU, and the assertion follows.
(vi) We have the implication
/^XcA/oand/0<£<4^  => /^ECAV^
14     1. Topological Groups
1.2. Haar Measure 15
whence
C/»£inf Y,cjdk = i<£c,)inf(£«*,) = WMU<P)
as claimed. This completes the proof. □
The Haar covering number yields an "approximate" functional as follows. Fix^e g£+ and define
(fo-<P)
By (vi) above, we have the inequalities
(f:<p) < (f:f0)(f0-<P) and (fo-9) < (fo-f)(/--<P) ■
Dividing the first by (f0:<p) and the second by (f:<p), we find that 1^ is bounded as follows:
(f0:fT]<W)<(f:fo). (1-2)
f(s) = hisMs)zYcjPiSj'sMs)<YcJ<p(s-Jls)(ht(sJ) + 8)  (/ = 1,2)
j J
and it follows that
£,[//,(*,) + <?]   0 = 1,2).
j
Since hx+h2< 1, this last inequality implies that
(/1:rt+(/2:p)<(l+2^^ • j
But ~Lcj may be made arbitrarily close to (h: <p), and therefore by definition of 1^ and part (ii) of the previous lemma,
I9{fx)+lr{f2)^+^)I,{h)
<(l+2S)[Iy(fl+f2) + SIy(g)]
= y.A+/2) + 2<^(/i+/2) + <%(g)] •
This bound is crucial to the existence of a Haar measure for G.
One would expect that as the support of <p shrinks, 7p will become more nearly linear. This is confirmed by the following lemma.
1-10 Lemma. Given/, andf2 in      for every e>0 there is a neighborhood V of the identity e such that
whenever the support of <p lies in V.
Proof. By Urysohn's lemma for locally compact Hausdorff spaces, there exists a function ge^+ that takes the value 1 on supp(/j+/^) = supp(/,)^supp(/2). Choose 8>0 and let h=fi +f2 + 8g, so that h is continuous. Next let hj=fflh, i=l,2, with the understanding that ht is 0 off the support of/,. Clearly both hi lie in %+, and their sum approaches 1 from below as 8 tends to 0. By uniform continuity, there exists a neighborhood Uof e such that \hi(s)-hi(f)\< 8whenever t-heU.
Assume that supp((p) lies in U and suppose that
J
Then
Finally, Eq. 1.2 asserts that all of the /^-terms on the right are bounded independently of <p, and so for any positive s>0 we can choose 8 sufficiently small that the stated inequality holds. □
Existence of Haar Measure
We now prove the existence of a Haar measure for a locally compact group G. The idea is to construct from our approximate left-invariant functional Ip an exact linear functional. We shall obtain this as a limit in a suitable space.
Let X be the compact topological space defined by the bounds of Ip(f) as follows:
X= nK/o:/r',(/:/o)] •
/esrc+
Then every function 1^ (in the technical sense of a set of ordered pairs in Wc+ x R*) lies in X. For every compact neighborhood U of e, let Ku be the closure of the set {/ :supp(p)c£/} inX. The collection {Ka} satisfies the finite intersection property, since
16     1. Topological Groups
and the right side is nonempty by Urysohn's lemma. Therefore, since X is compact, r\Kv contains an element /, which will in fact extend to the required left-invariant positive linear functional on ^C(G). Note that /, which lies in a product of closed intervals excluding zero, cannot be the zero function on g*c(G), so that the extended functional will likewise be nontrivial.
Since / is in the intersection of the closure of the sets {/ : supp(^)c [/}, it follows that every open neighborhood of / in the product X intersects each of the sets     supp(0c [/}. We may unwind this assertion as follows:
For every open neighborhood U of e, and for every trio of functions fvf2,f3e.^+ and every s>0, there exists a function <pe^+ with supp(p) G [/such that -IJiJ^<eJ=W.
(This statement extends to any finite collection of £, but we shall need only three.) So givenfeWc+ and ceR, we may simultaneously make I(cf) arbitrarily close to IJpf) and cl(j) arbitrarily close to clp(f). Appealing to Lemma 1-9 above, this shows that I(cf) = cl(j). Similarly we have that / is left translation-invariant and at least subadditive. To see that / is in fact additive, we use Lemma 1-10 to choose a neighborhood U of e such that
///i)+V/2)^yy;+/2)+4
whenever supp(p)c U. Then choose (p with supp(^)c U such that /(/,), J(f2), and /(/1+/2) a11 likewise lie within el A of Ip(f{), Ip(f2), and Ip(fy+f2), respectively. Since £ is arbitrary, it follows at once from the inequality above and the general sublinearity of / that /(/,+^) =I{f\) +I(fi)> as required.
Finally, extend / to a positive left translation-invariant linear functional on ^C(G) by setting I(f)=I(f+)-I(f~)- As we remarked above, in view of our general discussion of translation-invariant measures and the Riesz representation theorem, this implies that G admits a left Haar measure // and completes the existence proof. □
Uniqueness of Haar Measure
We now prove that the Haar measure on a locally compact group G is unique up to a positive scalar multiple. Given two Haar measures //and von G, clearly it suffices to show that the ratio of integrals
G_
\f(x)dv
1.2. Haar Measure 17
is independent of/e2g+. To simplify the notation, we shall often write 1(f) and J(f) for the indicated integrals with respect to // and vy respectively. Given two functionsf,ge^, the plan is to produce a function heW* such that the ratios !(/)&(/) and l(g)/J(g) can both be made arbitrarily close to I(h)/J(h).
Let AT be a compact subset of G, the interior of which contains e. Then K contains an open symmetric neighborhood of the identity whose closure AT0 is compact and symmetric. (The symmetry is clearly preserved by closure.) Define compact subsets A^-and Kg of G by
Kf= supp(/) -K0 u AT0-suppCO and Kg = supp(g) -K0 u AT0-supp(g).
(Recall that the group product of compact sets is compact.) For tsK0, define ytf by
Equivalently, we have
rtm=fW-f(ts) .
r,f = Rj-L.j.
Define ytg similarly. Clearly ytf and ytg are supported in Kf and Kg, respectively, and both vanish on the center of G and in particular at e. Let e>0 be given. Then by left and right uniform continuity, K0 contains an open neighborhood U0 of e such that for all seG and teU0, both \ytf(s)\ and IftgCs)] are bounded by ell. Now U0 in turn contains a symmetric open neighborhood [/, of e whose closure AT, is symmetric, compact, and contained in K0. Moreover, by continuity we have that \ylf(s)\<eand \ytg(s)\<efor all seG and all feA",. The point is that as long as t remains in AT,, translation of/and g by t on either side has approximately the same effect.
We now construct h. We claim first that since e lies in the interior of AT,, there exists a second compact neighborhood K2 of e such that K2 is contained in the interior of AT,. Granting this, it follows immediately from Urysohn's lemma for locally compact topological spaces that there exists a continuous function h :G -> R+ that is 1 on K2 and 0 outside of Kv Define h: G-»R+ by
h(s) = h(s)+h(s-]) .
Then certainly he Wc+, supp(/;) lies in AT,, and h is an even function in the sense W\7Ah(s)=h(.s-^).
Proof of Claim. Since G is Hausdorff and the boundary B of AT, is likewise compact, B admits a finite cover by open sets each of which is disjoint from a corresponding open neighborhood of e in AT,. The intersection of these neighborhoods thus constitutes an open neighborhood U2 of e in AT,, and we now set K2 equal to the closure of U2. Then by construction K2 is contained in the interior of A",, as required. □
18     1. Topological Groups
We come to the main calculations. All integrals are implicitly over G and are translation-invariant, since n and v are by assumption Haar measures. First,
I(f)J(h) = fff(s)h(t)dfisdvt = || f(ts)h(t)dftsdvt
The second calculation uses the property that h is even.
I(h)J(f)=\\h(s)f(t)dp.sdvt
= \\h(r's)f(t)dfisdvt = jjh(s-]t)f(t)d{isdvt = \\h(t)f(st)dnsdvt .
From these we can easily estimate the difference:
\I(h)J(f)- 1(f)J(h)\=\||h(t){f(st)- f(ts)}dMsdv, |
=\fJKt)rtf(s)dMsdvt\
<eM(Kf)J(h) .
The point in the last line of the calculation is that supp(/i) lies in a Kx where ytf is small. Similarly,
\I(h)J(g)-l(g)J(h)\=\\\h(t){g(st)-g(ts)}d»sdvt\
=\\\W)rtg{s)d^sdvt\
< £M(Kg)J(h) . Dividing the first inequality by J(h)J(f) yields
w i(f)
J(h) J(f) Dividing the second by J(h)J(g) yields
m Kg)
m j(g)
Af)
SM(Kg) Ag)
1.3. Profinite Groups     19
Since e is arbitrary, this shows that the ratio I(f)U(f) is independent of/as claimed. □
1.3 Profinite Groups
This section introduces a special class of topological groups of utmost importance to our subsequent work. We begin by establishing a categorical framework for the key definition that follows.
Projective Systems and Projective Limits
Let / be a nonempty set, which shall later serve as a set of indices. We say that I is preordered with respect to the relation < if the given relation is reflexive (i.e., /'</' for all iel) and transitive (i.e., i<j and j<k => i<k for all kel). Note that we do not assume antisymmetry (i.e., i<j and j<i need not imply that /' -j); hence a preordering is weaker than a partial ordering. Clearly the elements of a preordered set/constitute the objects of a category for which there is a unique morphism connecting two elements / and j if and only if i<j.
We say that a preordered set / is moreover a directed set if every finite subset oil has an upper bound in /; equivalently, for all ijel there exists kel such that i<k and j<k. (Recall that directed sets are precisely what is needed to define the notion of a net in an abstract topological space.) While most of the specific instances of preordered sets that we meet below will moreover be directed, we shall need only the preordering for the general categorical constructions to follow. Beware, however, that directed sets will play a crucial but subtle role in establishing that the projective limit of nonempty sets is itself nonempty. (See Proposition 1-11.)
Example. The integers Z are preordered (but not partially ordered) with respect to divisibility and in fact constitute a directed set: a finite collection of integers is bounded with respect to divisibility by its least common multiple.
Assume that / is a preordered set of indices and let {G;}IE/ be a family of sets. Assume further that for every pair of indices ijel with i<j we have an associated mapping <p^ G^G,., subject to the following conditions:
(0  <Pii = lGi V/e/
(ii) <pij°<pJk=<pik ViJ,k el,i<j<k
Then the system (G., <ptJ) is called a projective (or inverse) system. Note that if we regard / as a category, then the association /' h-> G,. defines a contravariant functor.
20     1. Topological Groups
1.3. Profinite Groups 21
Definition. Let (G,, <pj}) be a projective system of sets. Then we define the projective limit (or inverse limit) of the system, denoted limG,, by
<
limG, = {(g,) eJjG,: i<Lj =>^(g,) = g,} .
** ie/
Note that as a subset of the direct product, limG, comes naturally equipped with a family of projection maps p.: limG, -> G,, and with regard to these projections, the projective limit manifests the following universal property:
Universal Property. Let H be a nonempty set and let there be given a system of maps ({/,://—> G,)/e/ that is compatible with the projective system (G,, 0>J in the sense that for each pair of indices i,jel with i<j, the following diagram commutes:
H
y \«
Gj-- G,
%
77ie« ^ere ex/'ste a unique map y/:H^> lim G, shc/i f/iar./or eac// iel the dia-gram
H-limG,
also commutes.
The mapping y/ is of course none other than h i-> (y/, (h))jeI, just as for the direct product of sets, but in this case the compatibility of the y/f guarantees that the image falls into the projective limit.
Note carefully that neither the definition of a projective limit nor the associated universal property asserts that a given projective limit of sets is nonempty. In particular, the projection maps may have empty domain. Of course, if a compatible system (i//t:H-> G,.)i€/ exists with nonempty domain H, then one infers from the existence of elements of the form that the projective limit is
likewise nonempty.
The construction of the projective limit works equally well in the category of groups (in which case the set maps are replaced by group homomorphisms, and
the group operation is defined componentwise) or the category of topological spaces (in which case the set maps must be replaced by continuous functions, and the topology on the projective limit is the subspace product topology induced from the direct product). In the case of groups, note that the projective limit is never empty, since the identity element of the direct product clearly lies in the projective limit. It follows from these remarks that the projective limit of a projective system of topological groups is itself a topological group with respect to the componentwise multiplication and the subspace topology.
Remark. A more obvious topology on a product space Y\Xi is the box topology, generated by sets of the form n Ut with Ut open in Xt for all i. But this is a much finer topology than the standard product topology. Moreover, with respect to the box topology the product of compact spaces need not be compact.
In the following subsection we shall be concerned with projective limits of finite groups. In passing we shall require conditions under which the projective limit of finite sets is nonempty. It is here that the notion of a directed set reappears critically.
1-11 Proposition. Assume that I is a directed set, and let (G,., <pt) be a projective system of finite sets. Set G = lim G,. Then:
<—
(i) If each G, is nonempty, G is nonempty.
(ii) For each index iel,
Pi(G) = C\<PviGj) ■
Proof. Our proof is adapted from a more general result in Bourbaki's Theory of Sets, Chapter III, §7.4. Let us call (St)iel a compatible family (with respect to our given projective system) if the following conditions are satisfied:
(a) Forall/e/,S,.cG..
(b) For all ijel with i<j, ^(SpeS,..
(c) Forall/e/,S;*0.
Note well that if (St) is a compatible family of the form S,^*,} for all iel, then in fact (x.)eG, which in this case is ipso facto nonempty.
Henceforth let Z denote the set of all compatible families. We impose an ordering on Z as follows: given compatible families (Sf) and (T,), we shall write (S;)<(7;.) ffSpr, for allIf 2' is a totally ordered subset of E, then clearly Z' admits the upper bound (7)) defined by
22     1. Topological Groups
Conditions (a)-(c) are trivially satisfied, and only the last of these requires fi-niteness. Hence the given ordering is inductive.
Suppose that there exists a maximal compatible system (S^eE. We claim that Sj= PijiSj) for all i<j. To prove this, let (7/.) be defined by
m
Since (5,.) is assumed maximal, our claim is established, provided that we can show that also (T)el.. Again (a) and (b) are routine; (c) is interesting. First observe that if i<j<k, then p^S^QpJSp. Now consider the intersection that defines Tt Each of the factors appearing is a subset of the finite set Sr There are only finitely many such subsets, and consequently we may assume that the intersection is over a finite set of indices.,jr. But I is directed, so there exists an element k in I such that k>j\,...,jr. Thus by our previous observation,
m=l
and therefore Tt is manifestly nonempty.
We continue to assume that (S) is maximal in Z and shall demonstrate next that each S, contains exactly one element. Fix i and let x,eSr Define (TJ) as follows:
T =\SJn<pJ(xi) WttJ J   ]Sj otherwise.
Note in particular that T^ty}, since <pu is the identity on Sr Then (7/;) lies in Z: (a) is obvious, (b) is an easy exercise, and (c) follows from the claim of the previous paragraph, namely that SffJBp for all j>i. Moreover, by construction (S) < (T), whence, since (S) is maximal, we must in fact have equality. This shows that St"*{xf}, Since i was arbitrary, this suffices.
We now address both statements of the proposition. Again fix iel. By definition of a projective system,
A(G)en^(G>).
1.3. Profinite Groups 23
One may argue as above that since all but finitely many factors on the right are redundant, the given intersection is nonempty; thus it contains an element xr Define (TV) as follows:
tpfixt) if/<; G, otherwise.
Note in particular that Ti={xi}. One sees without difficulty that (7})eZ (at last establishing that Z is nonempty!), and so by Zorn's lemma there is a maximal element (Sp of Z with the additional property that (SJ)i(2J). But then (5;)={y;} and G is nonempty, as required by (i). Moreover, x^y^p^G), which in light of the preceding inclusion establishes (ii). □
Profinite Groups
We now come to the principal definition of this section. It may seem at first to be essentially group-theoretic, with the topology as an afterthought, but we shall see shortly that this is not the case.
Consider a projective system of finite groups, each of which we take as having the discrete topology. Their projective limit acquires the relative topology induced by the product topology on the full direct product. This is called the profinite topology, and accordingly the projective limit acquires the structure of a topological group.
Definition. A topological group isomorphic to the projective limit of a projective system of finite groups (endowed with the profinite topology) is called a profinite group.
The following proposition summarizes the most fundamental global properties of a profinite group.
1-12 Proposition. Let G be a profinite group, given as the projective limit of the projective system (G., pu). Then the following assertions hold:
(i) G is Hausdorff with respect to the profinite topology.
(ii) G is a closed subset of the direct product Yl G,.
(iii) G is compact.
Proof, (i) The direct product of Hausdorff spaces is also Hausdorff, and any subset of a Hausdorff space is clearly also Hausdorff in the induced topology.
24    1. Topological Groups
(ii) We may realize the complement of G in FIG. as an open set as follows:
Gc=uu«^)enG*:^)^&} •
>' fit
Therefore G is closed, as claimed.
(iii) Since the direct product FIG. is compact by Tychonoff s theorem, this assertion follows from (ii) on general principles: a closed subset of a compact space is itself compact. □
Examples
(1) Let Gn=Z/nZ, n> 1, the additive group of integers modulo n. Then {GJ is a projective system, since there is a canonical projection
<pmn:ZlnZ ->Z/mZ
whenever m \ n, and these projections are clearly compatible in the required sense. We may thus form their projective limit
Z = limZ/«Z .
Note that Z also admits the structure of a topological ring.
(2) Let Hn = (Z//jZ)*, n> 1, the group of units in ZlnZ. Then {Hn} is a projective system, since a (unital) ring homomorphism maps units to units. Set
Z* = lim(Z/«Z)x .
<-
Then Zx is a topological group under multiplication and in fact is the group of units of Z.
(3) Fix a rational primep and set Gm = Z/pmZ, m^l. Again {Gm} is a projective system, and we form its projective limit to obtain a ring
Z =limZ/pmZ . This is called the ring of p-adic integers.
(4) Let Hm = (Z/pmZ)*, mZl, so that {Hm} is a projective system as in Example 2. Then set
1.3. Profinite Groups 25
Z\;=Iim(Z//>mZ)x .
One checks easily that Zx is the group of units in Zp; this is called the
group of p-adic units.
(5) The set of all finite Galois extensions K/Q within a fixed algebraic closure Q of Q forms a directed set with respect to inclusion. We have a corresponding directed system of finite groups Gal(.K7Q), where if KqzL, the associated homomorphism Gal(I/Q) ->• Gal(A7Q) is just restriction. Moreover, we have an isomorphism
Gal(Q/Q) -^>lim Gal(AT/Q) a h-> (a\K) .
Topological Characterization of Profinite Groups
Recall that a topological space X is called connected if whenever X= Ukj V for nonempty open subsets U and V, then Ur\V*0. (Evidently an equivalent statement results if we substitute nonempty closed subsets for open ones.) Every point xeX is contained in a maximal connected subset of X, which is called the connected component of x. In the special case of a topological group G, the connected component of the identity e is denoted G°.
A topological space X is called totally disconnected if every point in X is its own connected component. Clearly a homogeneous space is totally disconnected if and only if some point is its own connected component. In particular, a topological group G is totally disconnected if and only if G°={e}.
1-13 Lemma. G° is a normal subgroup of G. Moreover, the quotient space GIG° is totally disconnected, whence (G/G°)° is the trivial subgroup of the quotient.
Proof. Let xeG°. Then x_1G° is connected (by homogeneity) and contains e, whencex~]G°cG°. Thus G° is closed under inverses. The same argument now shows that xG°cG°, and that for all-^eG, we have further that j/Cjr'c G°. Consequently G° is indeed a normal subgroup of G°, as claimed. The second statement is immediate: by homogeneity, the connected components of G are precisely the elements of G/G°, and so by general topology (see Exercise 5 below), G/G° is totally disconnected. □
1-14 Theorem. Let G be a topological group. Then G is profinite if and only ifG is compact and totally disconnected.
26    1. Topological Groups
Proof. =>) We have already seen that G is compact. Thus it remains to show that G°={e}. Let Ube any open subgroup of G. Then Ur\G° is open in G° and nonempty. Now consider the subset VofG defined by
V=   (j x-(£/nG°) .
xeG°-U
Then since each x-(Ur^G°) is open in G°, so is V. Moreover, by elementary group theory, Ur\V=0, and G° is the disjoint union of two open sets, namely Ur\G° and V. But by definition G° is connected, so either Ur\G° or ^must be empty. Since the former is not, the latter is, and in fact G°=Un,G°, which is to say that G°c U. Since U is an arbitrary open subgroup of G, we have accordingly,
G°c    p| U .
U an open subgroup of G
We must now make use of the profinite nature of G. Indeed, let
G = limG,
<-
where each G, is a finite group with the discrete topology. Recall that for each index i we have a projection map pi,: G -> G. that is just the restriction of the corresponding map on the full direct product. Let y =()>,) lie in G and assume that.y is not the identity element. Then for some index /0, it must be the case that » &e(. But now consider the set U0 = p^(ej). Since the topology on GIq is discrete and the projections are continuous, U0 is open in G. Since the projections are moreover group homomorphisms, UQ is in fact a subgroup of G. But by construction, U0 excludes y. This shows that the only element in the intersection of all open subgroups of G is the identity. Thus G° is trivial, as required.
The proof of the converse is more delicate and requires three lemmas. We begin with some preliminary analysis.
Let be the family of open, normal subgroups of G. This is clearly a directed set with respect to the relation M<N if Nc,M. (In fact, two subgroups M and Ninyf have a least upper bound Mr^N injf.) Moreover, the following observations are elementary:
(i) For each NeJ^, the quotient group GIN is both compact and discrete, hence finite.
1.3. Profinite Groups 27
(ii) For each pair of subgroups M,NeJf, with M<,N, the kernel of the canonical projection G -> G/M contains N, and hence this map factors through GIN to yield the induced map
<Pm,n'gin ->G/M xN y-*xM .
From this description it is clear that if L<M<N in J^, then
<Pl,m°Pm,n= <Pl,n and {GIN}N&jfr constitutes a projective system of finite groups.
The point, of course, is to show that G is isomorphic to the projective limit of this system.
1-15 Lemma. Let the profinite group G' be given by
G' = lim GIN
<-
n
where N varies overjV, as defined above. Then there exists a surjective, continuous homomorphism a:G->G'.
Proof. For A^e^let denote the canonical projection from G to GIN, which is surjective. Since GIN is homogeneous, we establish that aN is also continuous by noting that a~J(eGm) = N, which by hypothesis is open in G. Arguing as in (ii) above, it is clear that whenever M<N inJ^, the following triangle is commutative:
GIM
Thus by the universal property of projective limits, we have a continuous homo-morphism a: G -> G' such that aN = pN oa for all NeJ^, where pN denotes projection from G' onto GIN, the component of the projective limit corresponding to N.
It remains to show that a is surjective. We claim that a has dense image in G'. Granting this, we conclude the argument as follows: Since G is compact
28     1. Topological Groups
1.3. Profinite Groups 29
and G' is Hausdorff, the image of a is, moreover, closed in G'. Thus Im(a), being dense, must be all of G', as required.
To establish the claim we shall show that no open subset of G' is disjoint from Im(a). Consider the topology of G': this is generated by sets of the form Ph(Sn) , where SN is an arbitrary subset of GIN. Every open set in G' is thus expressible as a union of finite intersections of these pjJ(SN). Such an intersection U consists of elements of the form
where at most only finitely many of the coordinates are constrained to lie in some given proper subset of the corresponding quotient; the rest are unrestricted. Now suppose that the constrained coordinates correspond to the subgroups Nv...,Nr and that
M = fV,.. J"1
Then given (xN)sG', the coordinates xN are all determined as images of the coordinate xM under the associated projection maps. Since a^-.G-^G/M is sur-jective, there is at least one element in feG such that a(t)M=xM, and consequently t also satisfies a(t)N=xN. for j=\,...,r. In particular, if (xN)eU, then certainly a(t)ell, since a(t) agrees with (xN) in all of the constrained coordinates. Thus U manifestly intersects Im(a), and by our previous remarks, so, too, does every open set in G'. This completes the proof. □
1-16 Lemma. Let X be a compact Hausdorff space. For a fixed point PeX, set %?= {K:K is a compact, open neighborhood of/3}. Define Y<zXby
Then Y is connected.
Proof. Note that the collection ^ is nonempty because X itself is compact and open.
Suppose that Y is the disjoint union of closed subsets Yx and Y2. We must show that either F, or Y2 is empty. Recall from general topology that a compact Hausdorff space is normal. Accordingly, there exist disjoint open subsets £/, and U2 containing, respectively, F, and Y2. Now set Z=X-(Ux\j U2), which is closed and therefore compact. Since Yc,UxkjU2, Z and Fare disjoint, which is to say that Z lies in the complement of Y. Thus we have an open cover for Z
Zc\jKc
that admits a finite subcover. Hence there exist KxKre % such that
Zn(C\Kj) = 0 . J
Let W denote the intersection of the C. Then W is a compact, open neighborhood of P, and so Wis itself in W. But also
W = (WnUx) U (WnU2)
since W is disjoint from Z, the complement of C/,u U2. We now make note of the following assertions:
(i) Both Wr\ Ux and Wo U2 are compact, open subsets of X.
(ii) P lies exclusively in one of fVn Ux or (fn U2. Say Pe Wr\ Ux.
From (i) and (ii) it follows that Wn£/,e^and so FcWnC/,. Since F2cFand F2 is disjoint from £/,, it follows that F2 is empty, as required. □
1-17 Lemma. Let G be a compact, totally disconnected topological group. Then every neighborhood of the identity contains an open normal subgroup.
Proof. As a preliminary, note that G is Hausdorff: If x and>» are distinct points in G, then {x,y} is disconnected with respect to the subspace topology. Therefore there exist respective open neighborhoods of x and y that are disjoint. The proof now proceeds in three steps: First, we show that every open neighborhood U of the identity contains a compact, open neighborhood W of the identity. Second, we show that W in turn contains an open, symmetric neighborhood V of the identity such that WVc, W. Third, from V we construct an open subgroup, then an open, normal subgroup of G contained in U, as required.
Let ^denote the set of all compact, open neighborhoods of the group identity e. Applying the previous lemma with P=e, we find that
Y=f]K
is a connected set containing e. But G is totally disconnected, so in fact Y={e}. Now let U denote any open neighborhood of e. Then G-U is closed and therefore compact. Since e is the only element of G common to all of the K in
30     1. Topological Groups
there exist subsets Kx,...,Kre % whose complements cover G-U, and therefore
is a subset of U and a compact, open neighborhood of e. In particular, Welt. This completes the first step.
To begin the second step, consider the continuous map p. WxW^>G defined by restriction of the group operation. We make the following observations:
(i) For every we W, the point (w,e)epr\W).
(ii) Since W is open, the inverse image of W itself under p. is open in Wx W.
(iii) It follows from (i) and (ii) that for every w&W, there exists open neighborhoods Uw of w and Vw of e such that UwxVwcjw\lV). Moreover, by Proposition 1-1, we may assume that each Vw is symmetric.
(iv) The collection of subsets Uw (weW) constitutes an open cover for W. Since Wis compact, a finite subcollection Ul,...,Ur suffices.
Let Vx,...,Vr correspond to Uv...,Ur in (iii) above. Define an open neighborhood Fc Wof the identity as follows:
7=1
By construction WVc^W, and by induction WVqzWfor all n^O. In particular, V"cz Wfor all n>0. This completes the second step.
For the final step, we expand Kto an open subgroup 0 of G contained in W by the formula
0 = (jV .
(Note that 0 is closed under inversion because V is symmetric.) The quotient space GIO is compact and discrete, hence finite, so we can find a finite collection of coset representatives xv...,xs for O in G. It follows that O likewise has only finitely many conjugates in G: all take the form
xpxf    (j = \,...,s).
Thus
1.3. Profinite Groups 31
N = f]XjOx-/
7=1
is an open, normal subgroup of G. Moreover, since one of the conjugates of O is O itself, NcOgzWc U. This completes the proof. □
This brings us at last to the conclusion of the topological characterization of profinite groups.
Proof of Theorem 1-14, Converse. By Lemma 1-15, we have a surjective homomorphism a:G->G', where G' is the projective limit of the finite quotients GIN for N an open, normal subgroup of G (i.e., /Ve ). Appealing to Exercise 9 below, we see that it suffices to show that a has trivial kernel and hence is injective.
Since a simultaneously projects on all of the quotients, it is clear that
Ker(a)=     N .
By the previous lemma, every open neighborhood of eeG contains an open, normal subgroup, which is therefore represented in the intersection above. It follows that Ker(a) is contained in every neighborhood of e and hence in the intersection of all such neighborhoods. But G is Hausdorff: the intersection of all neighborhoods of e consists merely of e itself. Hence Ker(a) is indeed trivial, and the theorem is proved. □
The Structure of Profinite Groups
The following theorem shows in particular that closed subgroups of profinite groups and profinite quotients by closed normal subgroups are likewise pro-finite.
£ -     t &
1-18 Theorem. Let G be a profinite group and let H be a subgroup of G. Then ■ His open if and only if GIH is finite. Moreover, the following three statements are equivalent.
(i) His closed.
(ii) H is profinite.
(iii) H is the intersection of a family of open subgroups.
Finally, //(i)-{iii) are satisfied, then GIH is compact and totally disconnected.
32     1. Topological Groups
Proof. The first statement follows from Proposition 1-4, part (iv), since a profinite group is necessarily compact. We next establish the given equivalences.
(i) =>(ii) H is a closed subset of a compact space and therefore itself compact. Hence it remains to show that H is totally disconnected. But this is trivial: since G°={e}, also H°={e}, and this suffices by homogeneity.
(ii) =>(i) If H is itself profinite, it is a compact subset of a Hausdorff space and hence closed.
(iii) =>(i) Suppose that H is the intersection of some family of open subgroups of G. Then since every open subgroup is also closed [Proposition 1-1, part (iv)], H is also the intersection of a family of closed subgroups of G, and therefore itself closed.
(i)=>(iii) As above, let jV denote the family of all open, normal subgroups of G. IfNesr, then since N is normal, NH is a subgroup of G. By part (i), [G:N] is finite, whence [G:NH] is likewise finite and NH is open. Moreover, clearly
It remains only to demonstrate the opposite inclusion. So let x lie in the indicated intersection, and let U be any neighborhood of x. Then Ux~l is a neigh-4- P borhood of e, and so by Lemma 1-16, t/x-1 contains some N^eJ^. Since x lies in the given intersection, xeN0H. Now by construction, also xeN0x. Hence N0x is equal to NQh for some heH, and consequently heN0xgz U. The upshot is that every neighborhood of x intersects H, and hence x lies in the closure of H. But H is closed by hypothesis, and therefore xeff, as required.
For the final statement, the compactness of the quotient follows at once from the compactness of G. Let p:G->G//f denote the canonical map. To see that KG/H is totally disconnected, assume that p(X) is a connected subset of GIH that properly contains p{H). Then Y=X-H is nonempty, and since we may assume that H is nontrivial, Y contains more than one point. Hence Y is the disjoint union of nonempty open (hence closed) sets F, and F2. One checks easily that since H is closed, F, and F2 are both open (hence closed) in X. Thus X is the disjoint union of the two nonempty closed sets F^H and F2. But then the image of F2 under p is (a) nonempty, (b) not the full image of X, and (c) both open and closed in pQC). Since p(X) is connected, this is a contradiction. Hence the connected component of p(H) is p(H) itself, and the quotient is totally disconnected, as claimed. □
1.3. Profinite Groups 33
A Little Galois Theory
We close this section by showing how profinite groups make a momentous appearance in connection with the Galois theory of infinite extensions. To begin, we recall the following elements of field theory:
(i) Let F be a field. An element a that is algebraic over F is called separable if the irreducible polynomial of a over F has no repeated roots. An algebraic field extension KIF is called separable if every element of K is separable over F.
(ii) Assume that K is an algebraic extension of F contained in an algebraic closure F of F. Then we call KIF & normal extension if every embedding of K into F that restricts to the identity on F is in fact an automorphism of K. (We say that such an automorphism is an automorphism of K over
(iii) A field extension KIF is called a Galois extension if it is both separable    B i and normal. The set of all automorphisms of K over F constitutes a group Ojwij^-under composition; this is called the Galois group of K over F and denoted Gal(KIF). JfFgzLczK is a tower of fields and KIF is Galois, then
KIL is likewise Galois.
Note that these notions do not require that KIF be finite. Our aim now is to extend the fundamental theorem of Galois theory to infinite extensions. This will require the introduction of some topology.
If S is any set of automorphisms of a field F, as usual Fs denotes the fixed field of S in F; that is, the subfield of F consisting of all elements of F left fixed by every automorphism of S.
Suppose that KIF is a Galois extension with Galois group G. Consider the set^-of normal subgroups of G of finite index. If N,MeJ^and Mc^N, we have a projection map pNM:G/M^GIN, and hence a projective system of quotients {G/N}N&/r. This system is certainly compatible with the family of canonical projections : G -> GIN, which corresponds to the restriction map from Gal(A7F) to Gal(A^/F). Thus we have a canonically induced homomorphism p from G into the projective limit of the associated quotients.
1-19 Proposition. Let K, F, G, and J>~be as above. Then the canonical map
p:G^> lim GIN
is in fact an isomorphism of groups. Hence G is a profinite group in the topology induced by p.
34    1. Topological Groups
In this context, we shall simply speak of the Galois group G as having the profinite topology.
Proof. We show first that p is injective. Certainly
Ker(A)= f]N
IVe/
and so we need only demonstrate that this intersection is trivial. Let aeKer(/?)
m i    ?   -an(* 'et xeK. Then by elementary field theory there exists a finite Galois exten-f,.,'.'-   sion F'lF such that F'^K and xeF'. Now the restriction map from
I « G=Gal(K/F) to Ga\(F'IF) has kernel Gal(K/F% which is therefore a normal subgroup of G of finite index. But then creGal(AT/F'), and so o(x)=x. Since x is arbitrary, cr is the identity on K, and Ker(p) is trivial, as required.
We show next that pis also surjective. Fix (aN) in the projective limit. Given an arbitrary element xeK, again we know that x lies in some finite Galois extension F' of F with N=Gal(K/F') normal and of finite index in G and Gal(F'/F)=G/N. Now define aeGal(K/F') by o(x)=aN{x). By construction of the projective limit, a is independent of the choice of extension F', and hence is a well defined automorphism of K. Moreover, it is clear that o~N is pN(o) for all N. □
Note that the isomorphism constructed in the previous proposition is essentially field-theoretic, and not merely group-theoretic. (See Exercise 12 below.)
m g    1-20 Theorem. (The Fundamental Theorem of Galois Theory) Let KIF be a
z^/a , ^Galois extension (not necessarily finite) and let G=GaX(KIF) with the
o <V profinite topology. Then the maps
a.Lh-* H = Gal(A7Z) 0:Hh*L = K"
constitute a mutually inverse pair of order-reversing bijections between the set of intermediate fields L lying between K and F, and the set of closed subgroups of G. Moreover, L is Galois over F if and only if the corresponding subgroup H is normal in G.
Proof. Note that in the case of a finite extension KIF, we may ignore the topological restriction, and the statement amounts to the fundamental theorem of Galois theory for finite extensions, a result that we assume. We proceed in four steps.
Step 1. We must show first that the map a is well-defined; that is, that a indeed yields closed subgroups of G. (The map p is of course well-defined on ar-
1.3. Profinite Groups 35
bitrary subsets of G.) According to the previous proposition, H is profinite as the Galois group of KIL, and Exercise 14 shows that this topology is identical to that induced by G. Thus H is a profinite subgroup of a profinite group and is therefore closed by Theorem 1-18.
Step 2. We claim that po a is the identity map. Let L be an intermediate field. By definition a(L) fixes L, and so clearly P(a(L))^L. Conversely, suppose that z lies in P(a(L)). Then since z lies in K and is therefore separable over L, z also belongs to a finite Galois extensionMof L contained in K. Let aeGal(M/L). Then there exists ere Ga\(K/L) that restricts to a. (The extensibility of automorphisms for infinite extensions follows from the finite case by Zorn's lemma.) By construction, o(z) =z, and hence a(z) = z for all aeGal(M/L). But by the fundamental theorem for finite extensions, we know that zeL. Hence we have also that /?(a(Z,))cZ,, and the claim is established.
Step 3. We shall show now that a°P is likewise the identity. By definition, for any subgroup H of G we have that a(P(H))^H. Now assume that H is closed. Then again by Theorem 1-18, H is the intersection of a family ^ of open subgroups of G. Since a and p are clearly order reversing,
P(H)=P( f]U)3 \JP(U)
and
a(p(H))Qa( [jP(U))c f]a(P(U))= f]U = H .
UeW VzW Ue.W
The point is that each of the open subgroups U has finite index, and thus in each case a(P(U))= Uby the finite theory.
Step 4. Finally, suppose that a(L)=Gal(K/L)=H, where L is some intermediate field. Let cr lie in G. Then from the diagram
cr
K-- K
F
36     1. Topological Groups
1.4. Pro-p-Groups 37
we deduce that Gal(K/o(L))= crHcT1. Thus according to parts (iH"i) above, we have that o(L)=L for all ere G if and only if aHa~i=H for all creG. This is to say that L is normal (and hence Galois) over F if and only if// is normal in G. □
Remark. We leave it to the reader to determine the effect of a°0 on an arbitrary subgroup of Gal(A7F). (See Exercise 15 below.)
1.4 Pro-/?-Groups
Our aim here is to introduce for profinite groups an analogue of the /?-Sylow subgroups that play such a crucial role in finite group theory. To begin, we must first generalize the notion of order.
Orders of Profinite Groups
Definition. A supernatural number is a formal product
Up"" p
where p runs over the set of rational primes and each n^eNufoo} .
Clearly the set of supernatural numbers is a commutative monoid with respect to the obvious product. If a is a supernatural number, we set vp{a) equal to the exponent of p occurring in a. We say that a divides b, and as usual write a|b, iivp(a)<vp(b) for all primes p. Note that if a\b, there exists a supernatural number c such that ac=b.
Given supernatural numbers a and b, we may define both their least common multiple and greatest common divisor by the formulas
lcm(a,6) = n/UP(Vp(aXVpW) and ffdia,b) = Y[p"«y*alv>m .
One extends these notions to arbitrary (even) infinite families of supernatural numbers in the obvious way.
Now let G be a profinite group. As previously, let denote the set of all open, normal subgroups of G. Recall that each quotient group GIN, for NejV, is finite.
Definition. Let Hbe a closed subgroup of G. Then we define [G:H], the index of H in G, by the formula
[G:H]= 1cm [GIN:HNIN\ .
In particular, [G:{e}], the index of the trivial subgroup, is called the order of G and denoted |G|.
Using the standard isomorphism between HNIN and H/Hr\N, we may recast the definition above as
[G:H]= 1cm [GIN.HIHr\N\. See also Exercise 16 below.
1-21 Proposition. Let G be a profinite group with closed subgroups H and K such thatHczK. Then [G:K] = [G:H][H:K].
Proof. Note that since H is closed, it is also profinite, and so the assertion is well defined. Now let N be any open normal subgroup of G. Then
[G/N:K/Kr\N] = [G/N:H/Hr>N] [H/Hr>N:K/Kn,N]. (1.3)
The 1cm (over NeJ^) of either side of the equation is, of course, [G:H]. Consider the factors on the right: if we replace N by any smaller subgroup JV, both indices are inflated (cf. Exercise 17). Hence, taking intersections, any pair of prime powers occurring in [G/N:H/Hr^N] and [HIHr^N:KIKr\N\, respectively, may be assumed to occur simultaneously. The upshot is that we can compute the 1cm of the product by separately computing the lem's of each factor. The first yields [G:H]; it remains only to show that the second yields [H:K\.
Let M be any open, normal subgroup of H. Then M=Hr\ U, where U is open in G. But by Lemma 1-17, U contains an open, normal subgroup N of G, and one argues as above that
[HIM:KIKr\M] \ [H/Hn>N:K/Kr>N].
Thus [H:K\ may be computed as the 1cm over subgroups of Hot the form Hr\N, where N is open and normal in G. Hence the second factor on the right of Eq. 1.3 indeed yields [H:K], as required. □
Remark. The proof shows that we may compute a profinite index as the 1cm over any cofinal family ^#c^f" of open normal subgroups of the ambient group; that is, if for every N&yV there exists an Me J? such that M^N, then
Icm [GIN-.HNIN] = 1cm [G/M.HM/M] .
We/ Mejf
38     1. Topological Groups
1.4. Pro-p-Groups 39
Examples
(1) Consider the p-adic integers
Z = lim(Z/p"Z) .
w *-
Let Hn denote the kernel of the projection map from Zp to Z/p"Z. Since this projection is surjective, we have Zp/Hn=Z/p"Z, and it follows that pm divides \Zp\. Conversely, every finite quotient of Zp has order a power of/?, and therefore \Zp\=px.
(2) Next consider
Z = lim(Z/wZ) .
Arguing as above, every factor group ZlnZ occurs as a quotient of Z, whence every positive integer is a divisor of its order. Thus
izi=rb" ■
p prime
Pro-p-Groups
Let p be a rational prime. Recall that a group is called a p-group if the order of every element is finite and a power of p. In the case that G is finite, this is equivalent to the statement that the order of G is a power of p.
Definition. A projective limit of finite p-groups is called a pro-p-group.
Of course, Zp is a pro-p-group; so is Hp, the projective limit of the Heisen-berg groups H(Zlp"Z). (See Exercise 18 below.)
1-22 Proposition. A profinite group G is a pro-p-group if and only if its order is a power of p (possibly infinite).
Proof. <=) We have already seen in the proof of Theorem 1-14 that G is the projective limit of its finite quotient groups GIN. If the order of G is a power of p, then each of these quotients must be a p-group, as required.
=>) Suppose that G is the projective limit of the projective system P, of p-groups. Then by definition of the topology of G, cofinal among the open normal subgroups of G are subgroups of the form
M = (l\Qi)r,G
where Q^P^ for all but finitely many indices, and 2,={e,} for the exceptions. Now given an arbitrary xeG and specifying any finite subset of its coordinates, there is clearly a finite exponent of the form q=pr such that x9 is trivial at each of the specified coordinates. Hence GIM is a p-group, and it follows by the remark following Proposition 1-21 that the order of G is a power of p. □
Definition. Let G be a profinite group. A maximal pro-p-subgroup of G is called a pro-p-Sylow subgroup of G (or more simply, a p-Sylow subgroup of G).
Note that the trivial subgroup may well be a pro-p-subgroup of G for some primes p. The following theorem shows among other things that this is the case if and only if p does not divide the order of G.
1-23 Theorem. Let G be a profinite group and let p be a rational prime. Then the following assertions hold:
(i) p-Sylow subgroups ofG exist.
(ii) Any pair of conjugate p-Sylow subgroups of G are conjugate.
(iii) IfP is a p-Sylow subgroup of G, then [G.P] is prime to p.
(iv) Each p-Sylow subgroup of G is nontrivial if and only ifp divides the order of G.
Proof. As usual, let jV denote the set of open normal subgroups of G and recall the explicit isomorphism
q>:G ->lim GIN x^(xN)NeJr .
Note in particular that if x,yeG and xN=yN for every open normal subgroup N, then x-y. A similar statement holds for arbitrary subsets of G.
(i) For each NeJ^, let £P(N) denote the set of p-Sylow subgroups of the finite group GIN. Then clearly £P(N) is finite and, moreover, nonempty. (If GIN has order prime to p, then the trivial subgroup is a p-Sylow subgroup.) Assume that M,N<EJ^-mt\\ JVcM Then there exists a surjective homomorphism of finite groups <pMN:G/N->G/M. Since this map sends a p-Sylow subgroup of GIN to a p-Sylow subgroup of GIM (refer again to Exercise 17), we obtain an induced map <pMN:3*!(N)-+3a(M). Thus we obtain a projective system (£P(N),(pMM) of finite nonempty sets, and the projective limit of this system is likewise nonempty by Proposition 1-11. This means that there exists a projective system of
40     1. Topological Groups
1.4. Pro-p-Groups 41
/p-Sylow subgroups (PN,<PMN), where for each Ne^f, we have PNg:GIN. Let P be the projective limit of the PN, which we can clearly identify with a subgroup of the projective limit of the GIN and hence with a subgroup of G via <p. Then P is a pro-/?-group by construction, and we shall now show that it is maximal. Let Q be any pro-/?-subgroup containing P. Then for every open normal subgroup N, QN/N^PN/N=PN. But Q is a pro-/?-group, so by the previous proposition, QNIN is a p-group and therefore equal to the p-Sylow subgroup PN. Thus for every open normal subgroup A7, QNIN=PNIN, and therefore Q and P have the same image under <p and accordingly are equal. Hence P is indeed maximal, as claimed.
(ii) Let P and Q be />-Sylow subgroups of G. For every NesK, we make the following definitions:
PN = PNIN QN = QNIN
YN = {yNeG/N;yNPNyN-i = QN}.
Note that each YN is finite and, by the Sylow theorems for finite groups, nonempty. Moreover, the subsets YN clearly constitute a projective system. Let Y denote the (nonempty) projective limit of the YN, which we again identify with a subset of G via <p, and let^ lie in Y. Then by construction, yPy~] and Q have equal projection in GIN for all open, normal N and are therefore equal. Hence P and Q are indeed conjugate.
(iii) Let P be a /7-Sylow subgroup of G. Then by definition
[G:P]=lcm[G/N:PN/N] .
But by Exercise 19, for each N, the subquotient PNIN is a p-Sylow subgroup of GIN, and so by finite group theory each index [GIN. PNIN] is prime to p. Hence [G.P] is likewise prime to p.
(iv) This follows at once from parts (i) and (iii). □
1-24 corollary. Let G be a commutative profinite group. Then the following assertions hold:
(i) For every prime p, G admits a unique pro-p-Sylow subgroup.
(ii) Let p and q be distinct primes and let P and Q be the corresponding Sylow subgroups. Then Pr\Q is trivial.
(iii) G is isomorphic to the direct product of its Sylow subgroups.
Proof, (i) In light of the commutativity of G, this follows at once from parts (i) and (ii) of the theorem above.
(ii) The order of Pr\Q must divide powers of both p and q, whence this intersection must be trivial.
(iii) Let A7 be an open normal subgroup of G. Then for each pro-/>-Sylow subgroup P we have a canonical projection from P onto PAW, the unique p-Sylow subgroup of GIN. Note that this projection is trivial for all but the finitely many primes p that divide the order of GIN. By the theory of finite commutative groups, we have
Y\PNIN = GIN
where the product is taken over all of the Sylow subgroups of G. We may lift this isomorphism to G as follows:
G = lim GIN
= ]imT]PN/N
= Y\limPN/N
= 11 lim P/Pr^N <-
= I\r
All products are over the set of Sylow subgroups of G; all projective limits are over the family of open, normal subgroups of G. The final line of the calculation is justified by the cofinality of subgroups of the form PnN among the open subgroups of P, which may be deduced from Lemma 1-17. □
Example. Recall that the abelian profinite group
Z = limZ/nZ
has order 11/?°°, where the product is taken over all primes. Given a prime p, let P be the unique corresponding /?-Sylow subgroup of Z. Let Pn denote the unique p-Sylow subgroup of Z/«Z. Then
P = limP„ = limZ/pVpin)Z = limZ//7mZ = Zp .
n n m
Thus according to the corollary, Z = FIZ^,.
42    1. Topological Groups
Exercises
1. Let G be a topological group. Show that the topology on G is completely determined by a system of open neighborhoods of the identity e.
2. Let G=Z and impose the following topology: Ugz G is open if either Og U or G-U is finite. Show that G is not a topological group with respect to this topology. [Hint: If so, the mapping a\-*a + l would be a homeo-morphism; show that it is not.]
3. This exercise shows that we may impose a nondiscrete topology on Z such that Z is nonetheless a topological group with respect to addition. Let S1 denote the multiplicative group of complex numbers of absolute value 1. Recall that an element of Hom(Z,51) is called a character of Z. We denote such a character Let
s?=Y\sl
X
where the product is taken over all characters. Then & is a compact topological group. Now consider the homomorphism
j : Z -> &
(a) Show that j is injective; that is, show that for any nonzero neZ there exists a character % such that %(n)*l.
(b) Let G=j(Z). Then G is a group algebraically isomorphic to Z and a topological group with respect to the subspace topology induced by 9. Show that G is not discrete with respect to this topology and conclude that Z itself admits a nondiscrete topological group structure with respect to addition. [Hint: Suppose thaty'(l) is open. Then there exists an open subset U of 9 such that £/nG=y(l); moreover, we may assume that all but finitely many projections of U onto its various coordinates yield all of S1. Noting that generates the infinite group G, one may now derive a contradiction]
4. Give an example of a topological group with a closed subgroup that is not open.
5. Let X be a topological space-and let C(X) denote the space of connected components ofX (This constitutes a partition of X). As usual, we impose
Exercises 43
the quotient topology on C(X)—the strongest topology such that the canonical projection p:X->C(X) is continuous. Show that C(X) is totally disconnected with respect to this topology. [Hint: We say that a subset Y of a topological space is saturated if whenever yeY, the entire connected component of v lies in Y. Let F be a connected component of C(X) that contains more than one point. Show that p~\F) is a saturated, closed, disconnected set. Write p'\F) as the disjoint union of two saturated, closed subsets of X, and apply p to this decomposition to show that F is in fact disconnected—a contradiction.]
6. Let G = GLn(R). Show that G° is the set of nxn matrices with positive determinant.
7. Let H be a subgroup of the topological group G. Show that its closure H is normal (respectively, abelian) if H is.
8. Let /: G -* G' be a surjective continuous homomorphism of topological groups. Show that / factors uniquely through G/Ker(/); that is, there exists a unique continuous homomorphism / such that the following diagram commutes:
G -- G'
G/Ker(/)
Show that / is moreover injective. Under what conditions is / a topological isomorphism onto its image?
9. Let /:X-> Y be a continuous bijective mapping of topological spaces and assume that X is compact and Y is Hausdorff. Show that / is moreover a homeomorphism. [Hint: It suffices to show that/is open. What can one say about the image of U° under/where U is any open subset of XI]
10. Let / be an index set with preordering defined by equality and let (GppJ be a projective system of sets defined with respect to /. What is the projective limit in this case?
11. Give an example of a projective system of finite nonempty sets over a pre-ordered, but not directed, set of indices such that the projective limit is nevertheless itself empty.
12. Let G be an arbitrary group. Show that in general G is not isomorphic to the projective limit of the quotient groups GIN, as N varies over all of the
44     1. Topological Groups
subgroups of G of finite index. Hence not every abstract group acquires a profinite structure by this device. [Hint: Take G=Z.]
13. Let (G{, <Py) and (Ht, <ptJ) be two projective systems of sets. (Note that we use the same map designators <ptj for both systems.) Suppose that we have a family of maps {^.'G(.-»//,} that is compatible with these systems in the sense that pjiC-Qop^fyt all pairs of indices i<j. Show that there exists a unique map £:G->H on their respective projective limits such that d°PrPi°C^0T au '. where pi denotes the appropriate projection map. Observe that this construction works equally well in the categories of groups, topological spaces, and topological groups. [Hint: In light of the universal property of projective limits, consider the family of composed maps
14. Let K/Fbc a Galois extension with Galois group G.
(a) Let L be an intermediate field that is finite over F. For any given creG, define NL(o)c:G to be the set of reG such that a and r agree on L. The subsets NL(o) constitute a subbase for a topology on G. Show (i) that this topology remains unchanged if we restrict the subbase to normal intermediate fields that are finite over F and (ii) that this topology is identical to the profinite topology on G.
(b) Now let L be an arbitrary intermediate field, and let H denote the Galois group of K over L. Use the characterization of the profinite topology given in part (a) to show that the topology induced on H by G is identical to the profinite topology defined directly on H as Gal(A7I).
15. Let K/Fbe a Galois extension (not necessarily finite) and let Hbe any subgroup of G=Gal(A7F) (not necessarily closed). Let a and B be defined as in Theorem 1-20. Show that a(8(HJ) = H,the closure ofH.
16. Let G be a profinite group and let //be a closed subgroup. Show that
[G:H]= 1cm [G:HN]
where-/f is the set of all open, normal subgroups of G. Show further that if M is any open subgroup of G containing H, then there exists an open normal subgroup N of G such that M^NH. Conclude from this and the previous equation that moreover,
Exercises 45
17. Let cp: G->G' be a surjective homomorphism of groups with kernel L. Let //be a subgroup of G of finite index and let H be the image of H under <p. Show that [G:H]=[G': HI ■ [HL :H].
18. For any commutative ring ,4 with unity, define the Heisenberg group H(A) o\erA by
H(A) =
(l   a c\ 0   1 b 0  0 1
a,b,c eA
[G:H]= 1cm [G:M]
M open
(a) Show that H(A) is a group under multiplication in the matrix ring M^iA) and that this construction is, moreover, functorial in A.
To continue, for n> 1, H(Z/p"Z) is clearly a group of order p3", and hence ap-group. If m\n, then by functoriality, we have that the canonical projection Zlp"Z^>Z/pmZ induces a homomorphism <pmn from H(Zlp"Z) to H(ZlpmZ).
(b) Show that (H(Z/p"Z),<pmn) is a projective system of groups.
Let Hp denote the projective limit of the H(Zlp"Z); by definition, this is a pro-p-group. ,
(c) Show that H(Zp) = Hp. [Hint: Consider the map
nn.H(Zp)^H(Zlp"Z)
induced by projection from Zp onto Z/p"Z. Show that this is a continuous surjective homomorphism and that moreover, the family {nn} is compatible with the system of homomorphisms {<pmn}- Finally, show that the map n obtained from the nn by the universal property of the direct limit is the desired isomorphism.]
19. Let G be a profinite group and p a rational prime. For each open, normal subgroup N in G, let HNbea p-subgroup of GIN (not necessarily a p-Sylow subgroup). Show that there exists a pro-/?-Sylow subgroup P of G such that PNIN^HN for all N. Conclude (i) that every pro-p-subgroup of G is contained in a pro-p-Sylow subgroup of G; and (ii) that if P is a pro-p-Sylow subgroup of G, then PNIN is a p-Sylow subgroup of GIN for each open, normal subgroup of G. [Hint: Generalize the argument from the proof of part (i) of Theorem 1-23.]