Fall 2006 Final

Final Examination Math 98/198:LATEX for Math/Science

November 30, 2006

Who should be taking this final? People taking the DeCal for two units, or people who will not be able to pass with their current homework score are required to take this final examination to pass this DeCal. If you are taking this DeCal for one unit, and you have finished the homework requirement for this class, you do not have to take this final examination!

Directions You must code everything following this page in LATEX . LATEX as many pages as you can, but understand that you need to code at least 90% of the content (with minimal errors) on the following pages to pass. This examination is due on December 8, 2006 at 10pm.

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0.1

Preliminaries

Recall that a homomorphism δ : G → G0 creates a natural factor group, G/Ker(δ). This is handy because there’s a theorem that says that each factor group G/H leads to a canonical homomorphism having H as its kernel: Theorem 0.1. Let H be a normal subgroup of G. Then δ : G → G/H given by δ(x) = xH is a homomorphism with kernel H. Proof. For a, b ∈ G, we have δ(ab) = δ(ab)H = (xH)(yH) = δ(a)δ(b), which by definition is a homomorphism. We see that aH = H iff a ∈ H, so at once we conclude that the kernel of δ is H. 

0.2

The Isomorphism Theorems

Using the results above, we can now say something about the factorization of a homomorphism into the product of a homomorphism and an isomorphism: Theorem 0.2. (First Isomorphism Theorem) Let δ : G → G0 be a homomorphism with kernel K, and let γK : G → G/K be the canonical homomorphism. Then, there exists a unique isomorphism α : G/K → δ[G] such that δ(x) = α(γK (x)) for each x ∈ G. 1 Note that the First Isomorphism Theorem is also called the Fundamental Homomorphism Theorem. I call this next one the Diamond Isomorphism theorem, because if you draw the diagram for this, it looks like...well, a diamond: Theorem 0.3. (Second Isomorphism Theorem) Let H be a subgroup of G and N be a normal subgroup. Then, (HN )/N ∼ = H/(H ∩ N ). Proof. Let δ : HN → H/(H ∩ N ) be defined as δ(hn) = h(H ∩ N ). For an isomorphism, we need to check that δ is well defined, a homomorphism, onto, and that ker(δ) = 1. The first three results can be computed directly. For the kernel, we see that ker(δ) = {hn|δ(hn) = 1(H ∩ N )} = {hn|h ∈ (H ∩ N )} = N. Since 1 ∈ (H ∩ N ), N ⊆ ker(δ), and ker(δ) ⊆ N .  Theorem 0.4. (Third Isomorphism Theorem) Let H and K be normal subgroups of a group G with K ≤ H. Then, G/H ∼ = (G/K)/(H/K). 1I

have omitted the proof for this, partly because I don’t have one off-hand. Sorry.

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Proof. Define δ : G/H → G/K by gH 7→ gK. At once we can say that it is well defined, since H ≤ K, and as a result is onto. The kernel is {gH|gK = K} = {gH|g ∈ K} = K/H. Then, use the First Isomorphism Theorem to obtain the desired isomorphism.  Theorem 0.5. (Fouth Isomorphism Theorem) Let G be a group. There exists a one to one correspondence between subgroups H of G containing N, and subgroups H = H/N of G/N . All subgroups of G/N are then of the form H/N , where H ≥ N is a subgroup of G. For all H, K ≤ G with N ≤ H, N ≤ K, we have: 1. H ≤ K ⇔ H ≤ K 2. H ≤ K ⇒ [K : H] = [K : H] 3. < H, K > =< H, K > and 4. H ∩ K = H ∩ K Proof. The preimage of a subgroup G/N is a subgroup of G, which contains N. So, everything else is computation.  The fourth isomorphism theorem is also called the Lattice Group Isomorphism Theorem, since the lattice for G/N appears in the lattice for G as a section above N. I found it easier to check by looking at normal subgroups of the 8th Dihedral group (D8 ).

0.3

Preliminaries

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It’s not so hard to understand finitely generated abelian groups (hereby abbreviated FGAG’s); there is a handy theorem we use to understand them: Theorem 0.1. (Fundamental Theorem of Finitely Generated Abelian Groups) Every finitely generated abelian group is isomorphic to some direct product of cyclic groups of the form Z(p1 )r1 × Z(p2 )r2 × . . . × Z(pn )rn × Z × Z × . . .

(1)

where pi are primes, and ri are positive integers. Note that the direct product above is unique up to reordering. However, this theorem is only good for FGAG’s. If we wanted any information on finite nonabelian groups, it would be a little harder. Thankfully (or not), the Sylow theorems make it a little easier to understand finite nonabelian groups. 2 Familiarity

with algebra up to group actions on sets is assumed.

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0.4

The Importance of Lagrange’s Theorem

Recall that Lagrange’s theorem for finite groups state that the order of subgroup of some group G must divide |G|. We also know that abelian groups contain subgroups of every factor of |G|, meaning that the converse of Lagrange’s theorem holds as well, or that if a the order of a subset of G divides |G|, then the subset is also a subgroup of G. For nonabelian groups, this is not true in all cases, meaning that sometimes, a nonabelian group H might not have any subgroup of some order d dividing |H|. This is where the Sylow theorems come in. Given that if d is a power of a prime, and a divisor of G, then G will have a subgroup of order d. The theorems also expand on the number of subgroups, and any relations they may hold together.

0.5

The Sylow Theorems

Recall that conjugation of h by g is the automorphism A : G −→ G, where A(h) = ghg − 1 for all h in G, where G is a group. To that end, let G be a group, and C be the collection of all possible subgroups of G. C is now a G-set through the group action of G on C, defined by the conjugate subgroup gSg −1 , where S is in C, g is in G, and S ≤ G. We define the normaliser GS of S in G as GS = N [H] = {g ∈ G|gSg −1 = S}

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and is the largest subgroup of G having S as a normal subgroup. Well so what? Normalisation isn’t even mentioned in any of the theorems! This will be used in proof shortly later, so for now, just keep it in mind. Recall that a p-group is a group such that every element has an order of a power of some prime p, and that a p-subgroup is a subgroup such that the subgroup itself satisfy the properties of a p-group. To that end, we will now lay the groundwork for the three Sylow theorems, starting with the definition of a Sylow p-subgroup: Definition. A Sylow p-subgroup of a given group G is the maximal psubgroup of G. That is, it has order pk , or the highest power of some prime p. Theorem 0.2. Let G be a group with order pn and let A be a finite G-set. Then, |A| ≡ |AG | (mod p). Proof. Using properties of G-sets, we know that |Gxi | divides |G|. Also, p r X divides |Gxi | for a + 1 ≤ i ≤ r, where |A| = |AG | + |Axi |. So, |A| − |AG | i=a+1

is divisible by p, so |A| ≡ |AG | (mod p). 

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Theorem 0.3. (Cauchy’s Theorem) If |G| is divisible by some prime n, then G must contain a subgroup with order n. Proof. Let S be the set of n-tuples (n is prime) {(a1 , a2 , . . . , an ) ∈ Gn |Πgi = e for all i}, and let Zn act on S by the mapping: k · (a1 , a2 , . . . , an ) −→ (a1 + k, a2 + k, . . . , an + k), where each additional i + k in the subscript is taken mod n. Since |Zn | = n, the class equation |A| ≡ |A0 | (mod p), where A0 = {x ∈ A|hx = x for all h ∈ Zn } works. Since |A| = |G|n−1 , and n divides |G|, n also must divide |A|. This implies that |A0 | ≡ 0 (mod p). A0 consists of all the elements of the form (a, a, a, a, . . . , a), where an = e. So A0 is non-empty, and thus we can say that there exists an element in G of order n.  We may now state the Sylow theorems: Theorem 0.4. (First Sylow) Let |G| = pn r, where r is not divisible by p. Then for each 0 ≤ r ≤ n − 1, every subgroup H of G of order pi is contained in a subgroup of order pi+1 , and is normal. Also, every p-subgroup of G is contained in a Sylow p-subgroup of G. Proof. By Cauchy’s theorem, we know that G contains a subgroup of order p. So, by induction, we can show that if a subgroup of order pi for i < n exists, then it implies that that a subgroup of order pi+1 exists. Since i < n for all i, p|(G : H)3 . We can then say that p|(N [H] : H]).4 Since H is normal subgroup of N [H], p divides |N [H]/H|. Using Cauchy’s theorem again, this factor group N [H]/H contains some subgroup A of order p. We define the canonical homomorphism δ : N [H] −→ N [H]/H, and thus can say that δ −1 [A] = {x ∈ N [H]|δ(x) ∈ A} is a subgroup of N [H], and a subgroup of G. Thus, this subgroup contains a subgroup of order pi+1 , and is known to you and I as H. Repeating the work above, it is interesting to note that H < δ −1 [K] ≤ N [H], where the order of δ −1 is pi+1 . Since H is normal in N [H], at once we can say that it is also normal in δ −1 .  Theorem 0.5. (Second Sylow) Let G1 and G2 be Sylow p-subgroups of the finite group G. G1 and G2 are thus conjugate subgroups of G. Proof. Fix G1 to act on the left cosets of G2 , using Theorem 0.2. Let C be the collection of left cosets of G1 , and let G2 act on C by the action y(xG1 ) = (yx)G1 , for y ∈ G2 . By this convention, C is a G2 -set. Theorem 0.2 says that |CG2 | ≡ |C| (mod p), and that |C| = (G : G1 ) is not divisible by p, so the order of CG2 is 3 Remember, 4 The

(G : H) is the index of H on G. proof for this is omitted.

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nonzero. Let xG1 ∈ CG2 . So, xyG1 = xG1 for all y ∈ G2 , so x−1 yxG1 = G1 for all y ∈ G2 . Since the orders of G1 and G2 are the same, G1 = x−1 G2 x, so G1 and G2 are conjugate subgroups.  Theorem 0.6. (Third Sylow) Given a finite group G and a prime p such that p divides |G|, the number of Sylow p-subgroups contained in G is congruent to 1 (mod p), and is a divisor of |G|. Proof. Let A be one Sylow p-subgroup of G. Let H be the set of all Sylow psubgroups and let A act on H through conjugation. So, we have xBx−1 , given B ∈ HA . We can say then that |HA | ≡ |H| (mod p), by Theorem 0.2. Using the definition of conjugation, xBx−1 = B for all x ∈ H. Thus, P ≤ N [B], and B ≤ N [B]. Since A and B are conjugate in N [T ], they are both normal. We have just proven that A = B, which implies that HA = {A}. Since we stated earlier that |HA | ≡ |H| (mod p), we see that the number of Sylow p-subgroups is congruent to 1 modulo p. To prove that this number divides |G|, conjugate H with G. Since all Sylow p-subgroups are conjugate by definition, there exists one and only one orbit in H. So, |H| = (G : GP ) for P ∈ H. Since (G : GP ) divides |G|, so does the number of Sylow p-subgroups. 

Problem:

Find all integer solutions to the equation 1 1 1 + = . x y z

Solution:

We will show that all solutions are of the form: x = k · a · (a + b) y = k · b · (a + b) z = k · a · b,

where k, a, and b are arbitrary non-zero integers, a + b 6= 0. We can rewrite the given equation as z=

xy , x+y

so it suffices to find all pairs of integers x, y such that x + y | xy. Lemma: Let r and s be relatively prime positive integers. Then r ± s and rs are relatively prime.

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Proof: Suppose not; then there exists an integer k > 1 such that k | r ± s and k | rs. We have k | r ± s ⇒ k | r2 ± rs ⇒ k | r2 . Similarly, k | s2 . Let p > 1 be any prime factor of k, so that p | r2 and p | s2 . Now simply note that for any positive integer m, p - m ⇒ p - m2 . Hence, we must have p | r and p | s, which contradicts our assumption that r and s are relatively prime. So r ± s and rs must be relatively prime. // Let x and y be integers such that x+y | xy. We clearly cannot have x+y = 0. If |x + y| = 1, then we have the solution sets x=c y = − (c + 1) z = c · (c + 1) and x=c+1 y = −c z = −c · (c + 1) , where c is an integer, c 6= 0, 1. We now assume |x + y| > 1. Let n = gcd (|x| , |y|). Suppose that |x| and |y| are relatively prime. Then by the lemma, |x + y| and |xy| are relatively prime, a contradiction, since x + y | xy and |x + y| > 1. Hence, n > 1. Choose integers a, b such that x = na and y = nb. This gives: (na + nb) | (na) · (nb) ⇔ n · (a + b) | n2 · a · b ⇔ a + b | n · a · b. Since n = gcd (|x| , |y|), |a| and |b| must be relatively prime. So by the lemma, the above holds if and only if a + b | n. Put n = k · (a + b). This gives the solution set: x = k · a · (a + b) y = k · b · (a + b) z = k · a · b. Finally, note that the previous two solution sets are contained within this one (for the first set, take k = −1, a = c, b = − (c + 1); for the second set, take k = 1, a = c + 1, b = −c). Hence, this is the entire family of solutions, as desired. 

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0.6

Motivation and History

One of the most useful properties of Rn is invariance under a linear transformation. That is, if a ∈ Rn and f is any Lebesgue–integrable function, then Z Z f (x) dx = f (y + a) dy. Rn

Rn

Similarly, if we consider the multiplicative group of positive real numbers, R× +, and let k be a positive real number and f a Lebesgue–integrable function, then Z Z dx dy f (x) = . f (ky) × × x y R+ R+

The notion of Haar measure is a generalization of the above two examples. It turns out that in any locally compact group G, there exists a measure µ such that Z Z f (x) dµ(x) = f (gx) dµ(x) G

G

for any integrable function f and any g ∈ G.

At some time in the early twentieth century, people started to wonder if there was an invariant measure on all topological groups. The first two people to make significant progress on this problem were Alfr´ed Haar and John von Neumann in 1933. Haar in 1933 proved that there exists an invariant measure on any separable compact group. Using Haar’s result, von Neumann proved the special case of David Hilbert’s Fifth Problem for compact locally Euclidean groups. The following year, von Neumann proved the uniqueness of invariant measures.

Ultimately, neither Haar nor von Neumann proved the existence of invariant measures on all locally compact groups. The first one to come up with a full proof was Andr´e Weil. This proof, however, was criticized for using the Axiom of Choice in the form of Tychonoff’s Theorem. Later, Henri Cartan proved the existence of invariant measures on locally compact groups without the Axiom of Choice. Since then, several other people have also proved this theorem.

0.7

Definitions

Definition 1 A topological group G is a group as well as a topological space with the property that the mapping (g1 , g2 ) 7→ g1−1 g2 is continuous for all g1 , g2 ∈ G. The multiplicative group of positive reals, for example, is a topological group since (g1 , g2 ) 7→ gg12 is continuous due to continuity of multiplication and nonzero division of real numbers. 8

Definition 2 A topological space X is said to be locally compact if for all x ∈ X, there is a compact set containing a neighborhood of x.

Definition 3 Let X be a topological space, and let A ⊂ X. Then A is σ– ∞ bounded if it is possible S∞ to find a sequence of compact sets {Kn }n=1 with the property that A ⊂ n=1 Kn . Definition 4 A left Haar measure µ on a topological group G is a Radon measure which is invariant under left translation, i.e. µ(gB) = µ(B) for all g ∈ G. A right Haar measure µ on a topological group G is a Radon measure which is invariant under right translation, i.e. µ(Bg) = µ(B) for all g ∈ G.

Definition 5 A content λ is a set function that acts on the set of compact sets C that is finite, nonnegative, additive, subadditive, and monotone. A content induces an inner content and an outer measure. The inner content λ∗ is defined by λ∗ (A) = sup{λ(K) | K ∈ C, K ⊂ A}. Let O denote the set of open sets. The outer measure µe is defined by µe (A) = inf{λ∗ (O) | O ∈ O, A ⊂ O}.

Definition 6 If µe is an outer measure, then a set A is said to be µe –measurable if for all sets B, µe (B) = µe (A ∩ B) + µe (Ac ∩ B).

0.8

Existence and Uniqueness

Theorem 1 On any locally compact group G, there exists a nonzero left Haar measure µ, and this Haar measure is unique up to a positive multiplicative constant of proportionality.

Proof The proof of this theorem relies on four lemmas.

Lemma 1 Let λ be a content, and let λ∗ and µe be the inner content and outer measure, respectively, induced by λ. Then for all O ∈ O and for all K ∈ C, λ∗ (O) = µe (O) and µe (int(K)) ≤ λ(K) ≤ µe (K).

Proof For any O ∈ O, it is clear that µe (O) ≤ λ∗ (O) since we can pick O as an open superset of O in the definition of µe . Now if O0 ∈ O with O ⊂ O0 , then

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λ∗ (O) ≤ λ∗ (O0 ). Hence λ∗ (O) ≤ inf0 λ∗ (O0 ) = µe (O). O

Therefore λ∗ (O) = µe (O).

Now if K ∈ C and O ∈ O with K ⊂ O, λ(K) ≤ λ∗ (O). Thus λ(K) ≤ inf λ∗ (O) = µe (K). O

If K 0 ∈ C with K 0 ⊂ int(K), then λ(K 0 ) ≤ λ(K), so µe (int(K)) = λ∗ (int(K)) = sup λ(K 0 ) ≤ λ(K). K0



Lemma 2 Let λ be a content, and let µe be the outer measure induced by λ. Then a σ–bounded set A is measurable with respect to µe if and only if for all O ∈ O, µe (A ∩ O) + µe (Ac ∩ O) ≤ µe (O).

Proof Let λ∗ be the inner content induced by λ, let B be a σ–bounded set, and let O ∈ O satisfying B ⊂ O. Since λ∗ (O) = µe (O) ≥ µ(A ∩ O) + µe (Ac ∩ O) ≥ µe (A ∩ B) + µe (Ac ∩ B), µe (B) = inf λ∗ (O) ≥ µe (A ∩ B) + µe (Ac ∩ B). O

The other direction and the converse follow from the definition of subadditivity and µe –measurability. 

Lemma 3 Let µe be the outer measure induced by a content λ. Then the measure µ that satisfies µ(A) = µe (A) for all Borel sets A is a regular Borel measure. µ is called the induced measure of λ.

Proof It suffices to show that each K ∈ C is µe –measurable. By Lemma 2, this would follow from showing that µe (O) ≥ µe (O ∩ K) + µe (O ∩ K c ) for all O ∈ O. e ∈ C be a subset of O ∩ K 0c . Clearly Let K 0 ∈ C be a subset of O ∩ K c , and let K c 0c e = ∅ and K 0 ∪ K e ⊂ O, O ∩ K ∈ O and O ∩ K ∈ O. Because K 0 ∩ K e = λ(K 0 ) + λ(K). e µe (O) = λ∗ (O) ≥ λ(K 0 ∪ K) Thus e = λ(K 0 ) + λ∗ (O ∩ K 0c ) µe (O) ≥ λ(K 0 ) + sup λ(K) e K

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= λ(K 0 ) + µe (O ∩ K 0c ) ≥ λ(K 0 ) + µe (O ∩ K). Therefore, µe (O) ≥ µe (O ∩ K) + sup λ(K 0 ) = µe (O ∩ K) + λ∗ (O ∩ K c ) K0

= λ(K 0 ) = µe (O ∩ K) + µe (O ∩ K c ). Now it is necessary to show that µ(K) is finite. To do so, take L ∈ C with K ⊂ int(L). Then µ(K) = µe (K) ≤ µe (int(L)) ≤ λ(L) < ∞. Finally, regularity follows from µ(K) = µe (K) = inf {λ∗ (O) | K ⊂ O, O ∈ O} = inf {µe (O) | K ⊂ O, O ∈ O} O

O

= inf {µ(O) | K ⊂ O, O ∈ O}. O



Lemma 4 Let Ω be a measurable space and let h : Ω → Ω be a homeomorphism. Let λ and κ be contents on Ω such that for all K ∈ C, λ(h(K)) = κ(K). Suppose that µ and ν are the induced measures of λ and κ, respectively. Then µ(h(A)) = ν(A) for any Borel measurable set A ∈ Ω.

Proof Let λ∗ and κ∗ be the inner contents induced by λ and κ, respectively, and let µe and νe be their respective outer measures. If O ∈ O, then {κ(K) | K ⊂ O, K ∈ C} = {λ(h(K)) | K ⊂ O, K ∈ C} = {λ(A) | A = h(K), K ⊂ O, K ∈ C} = {λ(A) | h−1 (A) ⊂ O, h−1 (A) ∈ C} = {λ(A) | A ⊂ h(O), A ∈ C}. Thus κ∗ (O) = λ∗ (h(O)). Let B a σ–bounded set. Then {κ∗ (O) | B ⊂ O, O ∈ O} = {λ∗ (h(O)) | B ⊂ O, O ∈ O} = {λ∗ (C) | C = h(O), B ⊂ O, O ∈ O} = {λ∗ (C) | h−1 (C) | h−1 (C) ⊂ B, h−1 (C) ∈ O} = {λ∗ (C) | C ⊂ h(B), C ∈ O}. Thus νe (B) = µe (h(B)). By the result of Lemma 3, if A is any Borel set, then µ(h(A)) = ν(A). 

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Because of Lemma 4, one must simply find a content λ on G which is invariant under left translation to demonstrate existence. By Lemma 1, the induced measure of λ will be nonzero.

Let A ⊂ G be a bounded set, and let B ⊂ G be a set with nonempty interior. Then let A : B denote the lowest positive integer Sn n such that there exists a set {gj }nj=1 ⊂ G with the property that A ⊂ j=1 gj B. Now let A ∈ C be a set with nonempty interior. Let N denote the set of all neighborhoods of the identity element of G. Fix O ∈ N . Now define λO (K) =

K:O A:O

for K ∈ C. Clearly λO (K) satisfies 0 ≤ λO (K) ≤ K : A. λO (K) clearly satisfies all the properties of a content other than additivity.

Q For each K ∈ C, consider the interval IK = [0, K : A], and let Ξ = IK . By Tychonoff’s Theorem, Ξ is compact. Ξ consists of points that are the direct products of functions φ acting on C with the property that 0 ≤ φ(K) ≤ K : A. λO ∈ Ξ for all O ∈ N .

Now define Λ(O) = {λO0 | O0 ⊂ O, O0 ∈ N } given O ∈ N . If {Oj }nj=1 ⊂ N , then Λ

n \

! Oj

⊂

j=1

n \

Λ(Oj ).

j=1


Tn Clearly Λ j=1 Oj is nonempty. Since Ξ is compact, there is some point in the intersection of the closures of all the Λs \ λ ∈ {Λ(O) | O ∈ N }. O

It is now necessary to prove that λ is in fact a content. For any K ∈ C, λ(K) is finite and nonnegative since 0 ≤ λ(K) ≤ K : A < ∞. To prove monotonicity and subadditivity, let ξK (φ) = φ(K). Then ξK is a continuous function. Thus if K1 and K2 are compact sets, then Θ = {φ | φ(K1 ) ≤ φ(K2 )} ⊂ Ξ is closed. Then let K1 ⊂ K2 and O ∈ N . Then λO ∈ Θ, and hence Λ(O) ⊂ Θ. Since Θ is closed, λ ∈ Λ(O) ⊂ Θ, which implies that λ is monotone and subadditive. 12

To prove additivity, first note the restricted additivity of λO . Let gO be a left translation of O, and fix K1 , K2 ∈ C so that K1 O−1 ∩K2 O−1 = 0. If K1 ∩gO 6= 0, then g ∈ K1 O−1 , and if K2 ∩ gO 6= ∅, then g ∈ K2 O−1 . Thus there are no left translations of O that do not intersect either K1 or K2 , and so λO has additivity given that K1 O−1 ∩ K2 O−1 = ∅. Let K1 , K2 ∈ C with K1 ∩ K2 = ∅. Then there is some O ∈ N satisfying K1 O−1 ∩ K2 O−1 = ∅. If O0 ∈ N and O0 ⊂ O, then K1 O0−1 ∩ K2 O0−1 = ∅ as well. Thus λO0 (K1 ∪ K2 ) = λO0 (K1 ) + λO0 (K2 ). Then if O0 ⊂ O, λO0 ∈ Θ0 = {φ | φ(K1 ∪ K2 ) = φ(K1 ) + φ(K2 )}. Thus λ is additive. This establishes the existence of a Haar measure on any locally compact group.

To establish uniqueness, let µ be a left Haar measures, and consider a nonnegative continuous R function f on a locally compact R group G that is not identically zero. Since G f dµ > 0, we may assume that G f dµ = 1. Let us write Z Ψ(g) = f (xg −1 ) dµ(x), G

where g ∈ G. Then Ψ : G → R+ is a continuous function and also a homomorphism. Now select a continuous function h on G and consider the convolution Z Z (f ∗ h)(g) = f (x)h(x−1 g) dµ(x) = f (gx)h(x−1 ) dµ(x). G

G

R

By the definition of Ψ and G f dµ = 1, Z Z h(x) dµ(x) = h(x−1 )Ψ(x−1 ) dµ(x). G

G

A right translation of h gives Z Z h(xg −1 ) dµ(x) = h(x−1 g −1 )Ψ(x−1 ) dµ(x) G

G

Z

h((gx)−1 )Ψ((gx)−1 ) dµ(x)

= Ψ(g) G

Z = Ψ(g)

h(x−1 )Ψ(x−1 ) dµ(x).

G

Thus

R Ψ(g) =

GR

h(xg −1 ) dµ(x) . h(x) dµ(x) G

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Now let υ and φ be two continuous functions on G, and let Ψ be defined as above. Also, let ν be another left Haar measure. Then Z Z Z Z υ(x) dµ(x) φ(y) dν(y) = υ(x) dµ(x)φ(y) dν(y) G

G

G

G

Z Z =

υ(xy) dµ(x)Ψ(y)φ(y) dν(y) G

G

Z Z =

υ(xy)φ(y)Ψ(y) dν(y) dµ(x) G

G

Z Z

υ(y)φ(x−1 y)Ψ(x−1 y) dν(y) dµ(x)

= G

G

Z Z = G

Z Z

G

φ(x−1 )Ψ(x−1 ) dµ(x)υ(y) dν(y)

= G

φ((y −1 x)−1 )Ψ((y −1 x)−1 ) dµ(x)υ(y) dν(y)

G

Z =

Z φ(x) dµ(x)

G

υ(y) dν(y). G

R R R R Thus G υ dµ G φ dν = G φ dµ G υ dν. Now letting υ be a positive continuous function and setting R υ dν c = RG υ dµ G R R gives G φ dν = c G φ dµ. 

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