Ultraproducts of Finite Groups

Transcription

1 Ultraproducts of Finite Groups Ben Reid May 11, Background 1.1 Ultrafilters Let S be any set, and let P (S) denote the power set of S. We then call ψ P (S) a filter over S if the following conditions hold: 1. / ψ.. A, B ψ A B ψ. 3. (A B S A ψ) B ψ. Furthermore, we say that ψ is an ultrafilter if it has the following property: For each A S, either A or its complement A c is in ψ. There are two major types of ultrafilters, principal ultrafilters and non-principal ultrafilters. A principal ultrafilter is one that is generated by any single element a S, and has the form a = {A S a A}. However, these principal ultrafilters are not incredibly interesting, and will be used only as an example in this document. On the other hand, we have non-principal ultrafilters. We can say that an ultrafilter ψ is non-prinicpal if it contains all cofinite subsets of S, that is all subsets whose complement is finite. From this, we can infer that (A ψ) =. If instead, we had x (A ψ), then ψ = x. In this document, we will be considering ultrafilters over the set of natural numbers, N. Except where noted otherwise, we will use ψ to denote a nonprincipal ultrafilter over N. 1. Ultraproducts We can think of Ultraproducts as similar in structure to an infinite Cartesian product. Let {A i i N} be a countably infinite collection of objects (sets, groups, rings, fields, etc.), though all must be the same type of object. We denote the ultraproduct of this collection of objects as U(A i ) = (A 1 A A 3... If x U(A i ), we can write x in component form, where x = (x 1, x, x 3,... ) and each x i A i. 1

2 The operation(s) inherent to the structures in our collection carry over intuitively to the ultraproduct in component form. For example, in the case of addition in a collection of groups, if we have x, y U(A i ), then x + y = (x 1, x, x 3,... ) + (y 1, y, y 3,... ) = (x 1 + y 1, x + y, x 3 + y 3,... We now let ψ be an arbitrary ultrafilter over N. We can now define an equivalence relationship among elements of U(A i We say that for x, y U(A i ), x y {i N x i = y i } ψ. That is, x and y are equivalent iff the set of indices on which their components are equal is a member of ψ. From this, we see why principal ultrafilters are not interesting to study. If ψ is a principal ultrafilter, it is generated by some n N. We then have that the set {n} ψ. Elements of U(A i ) can then be grouped into equivalence classes based on their n th element. Thus, the number of distinct equivalence classes of U(A i ) is equal to the order of A n. This set of equivalence classes, and thus U(A i ) itself, can then be said to be isomorphic to A n. From here on out, we will be dealing exclusively with non-principal ultrafilters. Some Simple Examples.1 (Z Z Z... ) Let Z represent the group of integers modulo under modular addition. We now consider the ultraproduct U(Z ) = (Z Z Z... ) constructed with respect to some non-principal ultrafilter ψ. Let x U(Z ) = (x 1, x, x 3,... ) where each x i Z. Thus, we can think of x as a string of zeros and ones. Let I x = {i N x i = 0}. We then have that (I x ) c = {i N x i = 1}. Since ψ is an ultrafilter, we know that we must either have I x ψ or (I x ) c ψ. If the first is true, then we can say that x (0, 0, 0,... Otherwise, we know x (1, 1, 1,... Thus, all elements of U(Z ) are equivalent to either (0, 0, 0,... ) or (1, 1, 1,... Now, let a, b U(Z We construct I a and I b as above. We know that for any i N, we have (a + b) i = 0 a i = b i. Thus, I a+b = {i N a i = b i } = (I a I b ) ((I a ) c (I b ) c If both a, b (0, 0, 0,... ), then I a, I b ψ. Since ψ is an ultrafilter, we then have that I a I b I a+b N is in ψ. Thus, we have that I a+b ψ. Similarly, if both a, b (1, 1, 1,... ), then (I a ) c, (I b ) c ψ, and then (I a ) c (I b ) c ψ. It then follows that I a+b ψ. Thus, if a b, then a + b (0, 0, 0,... On the other hand, we have that for any i N, we have (a + b) i = 1 a i b i. Thus, (I a+b ) c = {i N a i b i } = (I a (I b ) c ) ((I a ) c I b Using a similar construction as above, we can show that if a b, then either (I a (I b ) c ) ψ or ((I a ) c I b ) ψ. In either case, we have (I a+b ) c ψ, and thus a + b (1, 1, 1,... From examining in detail the properties of addition in U(Z ), we can see that there exists an isomorphism θ : U(Z ) Z defined by { 0 : x (0, 0, 0,... ) θ(x) = 1 : x (1, 1, 1,... )

3 . (Z n Z n Z n... ) for some n N Let n N, and Z n be the group of integers modulo n under modular addition. We now consider the ultraproduct U(Z n ) = (Z n Z n Z n... ) constructed with respect to some non-principal ultrafilter ψ. Now, let x U(Z n ) = (x 1, x, x 3,... ), where each x i Z n. For each y {0,..., n }, define I x,y = {i N x i = y}, and subsequently (I x,y ) c = {i N x i y}. For each pair of I x,y and (I x,y ) c, one of the two must be in ψ. Note that if both I x,a and I x,b ψ, then I x,a I x,b = {i N (x = a) (x = b)} = (for a b) ψ. Thus at most one of the I x,y ψ. If none of the I x,y ψ, then (I x,0 ) c (I x,1 ) c (I x,n ) c = {i N (x i 0) (x i 1) (x i n )} = {i N x i = n 1} = I x,n 1 ψ. Thus, only one of the sets {I x,0,..., I x,n 1 } will be in ψ. If I x,y ψ, we can say that x (y, y, y,... We have now established that U(Z n ) has n distinct equivalence classes. Using arguments similar to the example of U(Z ) above, we can see that we can say U(Z n ) = Z n. 3 Ultraproducts of groups of prime order 3.1 Introduction and Motivation Let Z p be the group of integers modulo p under modular addition for some odd prime p. We then consider the ultraproduct U(Z p ) = (Z 3 Z 5 Z 7... ) constructed with respect to some non-principal ultrafilter ψ. Upon initial consideration of this ultraproduct, we wished to determine whether or not U(Z p ) = C. 3. Initial Findings Some digging through existing material led to the following example from The Classical Fields by Salzmann: Example. An ultraproduct F = U(F p ) of all Galois fields of prime order has the following properties: 1. char F = 0. each element in F is a sum of two squares 3. the element 1 is a square in F if, and only if, the set {i N p i 1 mod 4} belongs to ψ 4. F has cardinality ℵ = ℵ0 5. the algebraic closure F of F is isomorphic to C 6. F is a proper subfield of U(F p [1, Example 1.8] Based on these results, we see that U(Z p ) = C U(Z p ) = (U(Z p )). 3

4 3.3 Algebraic closure of U(Z p ) We now turn to the question of whether U(Z p ) can ever be algebraically closed. Consider Z pi, where p i is the i th odd prime. Let x Z pi. Then x has an additive inverse: x = (p i x) mod p i. Note that 0 = 0. Then x x = ( x) = x = x x. Thus, the set Q = {x x Z pi } has pi+1 elements. Since p i 3, we have pi+1 < p, so there exists a Z pi such that a y for any y Z pi. Denote this element a i. Thus we can construct an element a U(Z p ), where a = (a 1, a, a 3,... From this construction, we can see that the equation x a = 0 has no solutions in U(Z p Thus, we can say that regardless of our choice of ultrafilter ψ, U(Z p ) can never be algebraically closed, and thus never isomorphic to C. 3.4 U(Z p) We now examine the ultraproduct of the algebraic closures of the Z p s, denoted as U(Z p We now investigate the possibility that this ultraproduct is isomorphic to C. We have that each Z p = n 1 (Z p n Thus Z p is countably infinite, and we can say that the cardinality of Z p is ℵ 0 for all p P. Thus, we have p card(z p) = ℵ 0 ℵ0, and we can conclude that the cardinality of U(Z p) = ℵ0. [1, Theorem 1.9] As stated above, we have that Z p = n 1 (Z p n Each Z pn has characteristic p, so we can say that the characteristic of Z p is p. Thus, we again have that the characteristic of U(Z p) = 0. From these two facts, and the proof of part 5 of the above example in Salzmann, we know that (U(Z p)), the algebraic closure of the ultraproduct is isomorphic to C. Thus we must show (U(Z p)) = U(Z p), that is, that our original ultraproduct is algebraically closed. Let f (U(Z p))[x], that is a polynomial with coefficients in U(Z p We can write f(x) = c n x n + + c 1 x + c 0, where c i U(Z p) for i {0,..., n}. We can then write each c i in component form as c i = (c i,1, c i,, c i,3,... ) where each c i,j Z p j with p j equal to the j th odd prime. Similarly, we can write x U(Z p) as x = (x 1, x, x 3,... We then have an infinite system of polynomials: c n,1 x n c 1,1 x 1 + c 0,1 = 0 c n, x n + + c 1, x + c 0, = 0 c n,3 x n c 1,3 x 3 + c 0,3 = 0... However, by definition, we know by our definition that each Z p i is algebraically closed, so there exists x i such that c n,i x n i + + c 1,i x i + c 0,i = 0. Thus, we can construct x = (x 1, x, x 3,... ) such that f(x) = c n x n + + c 1 x + c 0. Thus, we have shown that for any f (U(Z p))[x] has a solution in U(Z p), and we can say that it is algebraically closed. Thus, (U(Z p)) = U(Z p), and since (U(Z p)) = C, we have that U(Z p) = C. 4

5 4 Ultraproducts of groups of order p 1 for p prime 4.1 Introduction and Motivation Let Z p 1 be the group of integers modulo p 1 under modular addition for some odd prime p. We then consider the ultraproduct U(Z p 1 ) = (Z Z 4 Z 6... ) constructed with respect to some non-principal ultrafilter ψ. In the following section, we will be trying to construct a non-principal ultrafilter ψ such that U(Z p 1 ) has no elements of finite order (other than 4. Elements of order We can show that U(Z p 1 ) must have an element of order. For each i N, let p i be the i th odd prime. Then for each Z pi 1, we can find an element of order, namely pi 1. Now, construct x U(Z p 1) where x = ( p1 1, p 1, p3 1,... Then, x = x + x = (p 1 1, p 1, p 3 1,... ) = (0, 0, 0,... Thus, x =, and no matter our choice of ψ, we see that U(Z p 1 ) will have an element of order. 4.3 Elements of order p Let x be some odd prime. We want to choose some nonprincipal ultrafilter ψ, such that U(Z p 1 ) has no element of order x. We know that Z pi 1 has an element of order x if and only if p i 1 mod x. Define I x = {i N p i 1 mod x}. By Dirichlet s Theorem on Arithmetic Progressions, we know that there are infinitely many primes of the form xn + 1 (where n N), since x and 1 are relatively prime. Thus, I x is infinite. Now, consider (I x ) c = {i N p i 1 mod x}. For each y N, 1 < y < x, we know that x and y are relatively prime. Thus, by Dirichlet s Theorem, there are infinitely many primes of the form xn + y. From this, we can see that (I x ) c must also be infinite. Since ψ is an ultrafilter, and both I x and (I x ) c are infinite, one of the two sets must be in ψ. Thus, we can say that if I x ψ, then U(Z p 1 ) will have an element of order x. Otherwise, (I x ) c ψ, and U(Z p 1 ) will not have any elements of order x. Since x was arbitrary, we can apply this argument to any odd prime, creating a custom ultrafilter ψ to eliminate elements of order x from U(Z p Removing all elements of prime order In the section above, we saw that we could choose ψ such that we could eliminate elements of any given odd prime order. We now want to create some nonprincipal ultrafilter ψ such that U(Z p 1 ) contains no elements of any odd prime order. Our first step is to create an infinite sequence of prime numbers. Define p i P with i N to be the ith odd prime number. Now, define q n P to be the second smallest odd prime number such that q n > q n 1 and q n 5

6 mod p 1 p p n. We know that p 1 p p n is relatively prime to, so we can apply Dirichlet s Theorem on Arithmetic Progressions to show that infinitely many prime numbers exist that are equivalent to mod p 1 p p n and can pick the second smallest such prime. Now, define the set Q = {i N p i = q n for some n N}, that is, the indices of the q n s in the set of prime numbers. Because we chose each q n > q n 1, we can say that Q is infinite. Because we skipped at least one prime number between q n 1 and q n, we know that the complement of Q, denoted Q c, is also infinite. Since both Q and Q c are infinite and ψ is an ultrafilter, we must have either Q ψ or Q c ψ. We choose to let Q ψ. Now, consider an arbitrary p i P. Since for any n i, we have q n mod p 1 p p i p n, we can say p i q n 1. Thus, Z qn 1 cannot have an element of order p i. Then, p i can only divide finitely many numbers of the form q n 1, and only finitely many of the groups Z qn 1 can have an element of order p i. Now, assume we have an element x U(Z p 1 ) such that x has odd prime order a P. We can write x in component form as x = (x 1, x, x 3,... ) where x i Z pi 1. Define the set S x = {i N such that x i = a}. We must then have that S x ψ. In order for Z pi 1 to have an element of order a, we must have p i 1 mod a for each i S x. Since ψ is an ultrafilter, we have that it is closed under finite intersections. And thus, since Q ψ and S x ψ, we must have that Q S x ψ. However, we have a p i 1 for all i S x, but from our argument above, a q n 1 for only finitely many q n. Thus we must have that Q S x must be finite. But since ψ is nonprincipal, it cannot contain any finite sets. Thus, we cannot have any elements x U(Z p 1 ) of order a. Since a was an arbitrary odd prime, we can claim that if Q ψ, then U(Z p 1 ) contains no elements of odd prime order. 4.5 Elements of infinite order First, let ψ be a nonprincipal ultrafilter. Define x U(Z p 1 ) = (1, 1, 1,... In each Z p 1, 1 has order p 1. Now, choose any set S ψ. For each i S, we can say x i, the i th component of x, has order p i 1. Then, x has order lcm({p i 1}) for i S. Since S ψ, we know S is infinite, and thus the sequence {p i 1} is an infinite, increasing sequence. Thus, lcm({p i 1}) =, and we can say that the order of x is infinite. Since our choice of S ψ was arbitrary, we can say the the ultraproduct U(Z p 1 ) constructed with ψ has elements of infinite order. And since our initial choice of ψ was arbitrary, we can say that we cannot construct an ultrafilter such that U(Z p 1 ) has no elements of infinite order. 4.6 Removing elements of other finite order Through the arguments in the previous sections, we have shown that U(Z p 1 ) will always have elements of order, since each group Z p 1 will have an element of order two. We have also seen that we can construct an ultrafilter ψ such that we can remove all elements of odd prime order, and thus all elements that are multiples of odd primes. After removing these, we are left with elements 6

7 of infinite order and those with an order that is a power of. We showed in section 5 that it is impossible to remove all elements of infinite order, as (1, 1, 1,... ) U(Z p 1 ) will always have infinite order. We now examine if we can remove the elements whose order is a power of. Define p i to be the i th odd prime as above. Now let S = {i N p i 1 mod 4}, that is, the primes of the form 4n + 1 for some n N. Since 4 and 1 are relatively prime, Dirichlet s Theorem on Arithmetic Progressions states that S is infinite. We also have though, that S c = {i N p i 3 mod 4} is infinite. Since we are only concerned with the odd primes, we know that S S c = N, and since ψ is an ultrafilter, it must contain either S or S c. We then choose ψ so that it contains S c. Now, consider p i for any i S c. Then, p i 1 mod 4. Thus, we can say that n p i 1 for any n 1. Thus, Z pi 1 cannot contain any element whose order is a power of, other than itself. By this argument, we have created an ultrafilter ψ such that U(Z p 1 ) only contains elements with order or with infinite order. 5 Ultraproducts of Semidirect Products 5.1 Semidirect Products We construct a group of order p(p 1), where p is a prime, in the following way: Let N = Z p and let A = Z p 1. We know then that A = Aut(N), and can thus find a homomorphism ϕ : A Aut(N Let denote the left action of A on N as determined by ϕ. Now, let G = {(n, a) n N, a A}. Then define multiplication in G as: (n 1, a 1 )(n, a ) = (n 1 a 1 n, a 1 a ) = (n 1 a 1 n a 1 1, a 1a We then have that G is a group of order p(p 1), and we can denote G as the semidirect product of N and A, written N A, more specifically Z p Z p 1. We also have that N = {(n, 1) n N} and A = {(1, a) a A}. With this isomorphic construction of N, we have that N G. 5. U(Z p Z p 1 ) Now, consider the ultraproduct over some nonprincipal ultrafilter ψ of these semidirect products: U(Z p Z p 1 ) where p is a prime. We first investigate whether U(Z p ) U(Z p Z p 1 As was done in the construction of G above, we can say that each Z p = {(z, 1) z Zp }. Thus, we can consider U(Z p ) as a subset of U(Z p Z p 1 We must first show that it is a subgroup. Let a, b U(Z p ), and write a = ((a 1, 1), (a, 1), (a 3, 1)... ), b = ((b 1, 1), (b, 1), (b 3, 1),... ) where a i, b i Z pi. Then, using the multiplication defined above, we have: ab = ((a 1, 1)(b 1, 1), (a, 1)(b, 1), (a 3, 1)(b 3, 1),... ) = ((a 1 1b 1 1 1, 1)(a 1b 1 1, 1), (a 3 1b 3 1 1, 1),... ) 7

8 = ((a 1 b 1, 1), (a b, 1), (a 3 b 3, 1),... ) Since each a i, b i Z pi, we know that a i b i Z pi and thus, ab U(Z p Thus, U(Z p ) is closed under multiplication. Since we know that each Z p Z p Z p 1, we can say that for every x Z p, there exists x 1 such that xx 1 = 1. Equivalently: (x, 1)(x 1, 1) = (x1x 1 1 1, 1) = (xx 1, 1) = (1, 1 Thus, for every a U(Z p ) as defined above, we can construct an a 1 U(Z p ) by taking a 1 = ((a 1 1, 1), (a 1, 1), (a 1, 1),... We then have: 3 aa 1 = ((a 1, 1)(a 1 1, 1), (a, 1)(a 1, 1), (a 3, 1)(a 1 3, 1),... ) = ((a 1 1a , 1), (a 1a ), (a 3 1a , 1),... ) = ((a 1 a 1 1, 1), (a a 1, 1), (a 3a 1 3, 1),... ) = ((1, 1), (1, 1), (1, 1),... ) Thus, every element in U(Z p ) has an inverse, and we can conclude that U(Z p ) U(Z p Z p 1 We will now show U(Z p ) U(Z p Z p 1 Let x U(Z p ), written, as above, as x = ((x 1, 1), (x, 1), (x 3, 1),... Now, let y U(Z p Z p 1 ), written as y = ((y 1, a 1 ), (y, a ), (y 3, a 3 ),... ) where y i Z pi and a i Z pi 1. We must show that yxy 1 U(Z p We have: yxy 1 = ((y 1, a 1 )(x 1, 1)(y 1, a 1 ) 1, (y, a )(x, 1)(y, a ) 1,... ) From earlier, we have that Z p Z p Z p 1. Thus, for each component of yxy 1 we can say that (y i, a i )(x i, 1)(y i, a i ) 1 Z pi. Then we have that yxy 1 = ((z 1, 1), (z, 1), (z 3, 1),... ) where each z i Z p. Thus, yxy 1 U(Z p Since x and y were arbitrarily chosen, this proves that U(Z p ) U(Z p Z p 1 6 Ultraproducts of Dihedral Groups of order p First, let D p be the dihedral group of order p, for p prime and p >, or the group of symmetries of the regular p-gon. Let H p be the normal cyclic subgroup of D p with order p and containing all of the rotations of the regular p-gon (including the identity rotation Let ψ be a nonprincipal ultrafilter. Consider the ultraproduct U(D p ) constructed over ψ. For each x U(D p ), write x = (x 1, x, x 3,... ) where x i D pi and let I x = {i N x i H pi } Now, define H U(D p ) by H = {x U(D p ) I x ψ} We want to show that s a subgroup of U(D p ) of index. First we shall show that s a subgroup of U(D p Certainly, s nonempty, as it must contain the identity of U(D p ) and the identity element of each D p is in the corresponding H p. Now, let x, y H. Thus I x and I y ψ. Thus, since ψ is an ultrafilter, we have I x I y ψ. Since each H p is a subgroup, they are all closed under composition, we know I x I y I xy. Thus, by the properties of ultrafilters, I xy ψ, and thus xy H. Now, let 8

9 x 1 = (x 1 1, x 1, x 1 3,... ) be the inverse of x in U(D p Again, since each H p is a subgroup, they are all closed under inverses, so I x 1 ψ and x 1 H. Thus s nonempty and closed under composition and inverses, so we can conclude that H U(D p ) Now, we must show that s of index. Let y U(D p ) such that y / H. Thus I y / ψ, but (I y ) c = {i N such that y i / H pi }, the complement of I y in N, must be in ψ since it is an ultrafilter. Now, let z U(D p We must show that we either have z H or yz H. If z H, we are done, so assume z / H. Thus, like above, we have I z / ψ and (I z ) c ψ. Since ψ is an ultrafilter, we have that (I y ) c (I z ) c ψ. Since for any D p we can say that if a, b D p but a, b / H p, then ab H p, we can then say that (I y ) c (I z ) c I yz. Thus, since ψ is an ultrafilter, we have I yz ψ, and yz H. Thus we have proved that either z H or yz H, and we can conclude that s a subgroup of U(D p ) of index. 7 General Results 7.1 The Center of an Ultraproduct of Groups Theorem. Let be the center of G i, and let U( ), U(G i ) denote the ultraproducts of and G i respectively with respect to some non-principal ultrafilter ψ. We then have an obvious mapping θ : U( ) U(G i ) by θ(x) = x for x U( It then follows that: (i) θ is a homomorphism. (ii) θ is one to one. (iii) θ(u( )) = Center(U(G i ) (iv) U(G i) U( ) = U( G i Proof. (i) Since x U( ) x U(G i ), we certainly have that θ(xy) = xy = θ(x)θ(y), and thus θ is a homomorphism. (ii) Let x, y U( ) such that θ(x) = θ(y It immediately follows that we must have x = y, and thus θ is one to one. (iii) Let x U( ) be represented by x = (x 1, x, x 3,... ) where x i. Thus, for every i N, we have that x i y = yx i for all y G i. Now, let y U(G i ), writing y = (y 1, y, y 3,... ), where y i G i. We can then say that x i y i = y i x i for all i N, and thus that xy = yx. Thus x commutes with all elements of U(G i ), and we can say that x is in the center of U(G i Thus, θ(u( )) Center(U(G i ) Now, let x be in the center of U(G i ) and written in component form as x = (x 1, x, x 3,... ) with x i G i. Then, for every y U(G i ), written again as 9

10 y = (y 1, y, y 3,... ) with y i G i, we define I y = {i N such that x i y i = y i x i }. Since x is in the center of U(G i ), we know that for each y, I y ψ. Suppose then, to the contrary, that there exists some set I N, defined as: I = {i N such that x i / } and I ψ. Thus, for each i I, we can find some z i G i such that x i z i z i x i. Then define z U(G i ) by z = (z 1, z, z 3,... ) where if i I, z i is as described above, otherwise z i is an arbitrary element of G i. By the definitions above, we have that I z = {i N such that x i z i = z i x i } = I c. And since I ψ and ψ is an ultrafilter, we must have that I z / ψ. However, this contradicts our earlier conclusion that I z ψ for all z U(G i Thus we can conclude that I =, and that each x i, and subsequently, x θ(u( ) Thus, Center(U(G i )) θ(u( ) We have proven the inclusion in both directions, so we can conclude that θ(u( )) = Center(U(G i ) And since θ is the identity mapping, we can say that θ(u( )) = U( ) = Center(U(G i ) (iv) From part (iii), we found that U( ) = Center(U(G i ) Thus we can say U( ) U(G i Thus, the quotient group: makes sense. U(G i) U() Consider the mapping η : U(G i ) U( Gi ) defined by η(x) = U( )x = (Z 1 x 1, Z x, Z 3 x 3,... ) for every x U(G i This mapping is certainly well defined. Now, let y U( Gi Write y = (y 1, y, y 3,... ) where y i Gi. Then we can write y = (Z 1 a 1, Z a, Z 3 a 3,... ) where a i G i. Now, construct a U(G i ) with a = (a 1, a, a 3,... Then η(a) = U( )a = (Z 1, Z, Z 3,... )a = (Z 1 a 1, Z a, Z 3 a 3,... ) = y. Thus η is onto. We now show that U( ) is the identity element of U( Gi Certainly U( ) U( Gi ) since Gi. Let y U( Gi ) as above, written in component form as y = (Z 1 a 1, Z a, Z 3 a 3,... ) where a i G i. We can also write U( ) = (Z 1 e 1, Z e, Z 3 e 3,... ) where e i is the identity element of G i. Then yu( ) = ((Z 1 a 1 )(Z 1 e 1 ), (Z a )(Z e ), (Z 3 a 3 )(Z 3 e 3 ),... ) = (Z 1 (a 1 e 1 ), Z (a e ), Z 3 (a 3 e 3 ),... ) = (Z 1 a 1, Z a, Z 3 a 3,... ) = y Thus U( ) is the identity element of U( Gi We let x, y U(G i ) and consider η(xy) = U( )(xy) = (U( )x)(u( )y) = η(x)η(y Thus, η is a homomorphism. We can then say that if a U(G i ), then a Ker η iff η(a) = U( )a = U( ) since U( ) is the identity of U( Gi Thus, we can say that a Ker η iff a U( ), and we can say Ker η = U( Thus, U( ) is the kernel of a homomorphism between U(G i ) and U( Gi Then, by the Fundamental Homomorphism Theorem, we have the desired conclusion: U(G i) U() = U( Gi 7. Ultraproducts of Normal Subgroups Theorem. Let G i for groups G i. Let U( ) and U(G i ) denote the ultraproducts of and G i respectively, with respect to some nonprincipal ultrafilter 10

11 ψ. Define the mapping θ : U( ) U(G i ) by θ(x) = x for all x U( It then follows that: (i) θ is a homomorphism. (ii) θ is one to one. (iii) U( ) U(G i (iv) U(G i) U( ) = U( G i Proof. (i) Since x U( ) x U(G i ), we certainly have that θ(xy) = xy = θ(x)θ(y), and thus θ is a homomorphism. (ii) Let x, y U( ) such that θ(x) = θ(y It immediately follows that we must have x = y, and thus θ is one to one. (iii) Let x be any arbitrary element of U( Now, let y U(G i We must show that yxy 1 U( We can write x, y in component form as: x = (x 1, x, x 3,... ), y = (y 1, y, y 3,... Then we have yxy 1 = (y 1, y, y 3,... )(x 1, x, x 3,... )(y 1 1, y 1, y 1 3,... ) = (y 1 x 1 y 1 1, y x y 1, y 3x 3 y 1 3,... ) For each y i x i y 1 i, we have that y i G i and x i. Since G i, we can say that y i x i y 1 i. Thus, we can say yxy 1 U( ), and conclude that U( ) U(G i U(G i) U() makes (iv) Since we have shown U( ) U(G i ) the quotient group: sense. Consider the mapping η : U(G i ) U( Gi ) defined by η(x) = U( )x = (H 1 x 1, H x, H 3 x 3,... ) for every x U(G i This mapping is certainly well defined. Now, let y U( Gi Write y = (y 1, y, y 3,... ) where y i Gi. Then we can write y = (H 1 a 1, H a, H 3 a 3,... ) where a i G i. Now, construct a U(G i ) with a = (a 1, a, a 3,... Then η(a) = U( )a = (H 1, H, H 3,... )a = (H 1 a 1, H a, H 3 a 3,... ) = y. Thus η is onto. We now show that U( ) is the identity element of U( Gi Certainly U( ) U( Gi ) since Gi. Let y U( Gi ) as above, written in component form as y = (H 1 a 1, H a, H 3 a 3,... ) where a i G i. We can also write U( ) = (H 1 e 1, H e, H 3 e 3,... ) where e i is the identity element of G i. Then yu( ) = ((H 1 a 1 )(H 1 e 1 ), (H a )(H e ), (H 3 a 3 )(H 3 e 3 ),... ) = (H 1 (a 1 e 1 ), H (a e ), H 3 (a 3 e 3 ),... ) = (H 1 a 1, H a, H 3 a 3,... ) = y 11

12 Thus U( ) is the identity element of U( Gi We can then let x, y U(G i ) and thus see that η(xy) = U( )(xy) = (U( )x)(u( )y) = η(x)η(y Thus, η is a homomorphism. We can then say that if a U(G i ), then a Ker η iff η(a) = U( )a = U( ) since U( ) is the identity of U( Gi Thus, we can say that a Ker η iff a U( ), and we can say Ker η = U( Thus, U( ) is the kernel of a homomorphism between U(G i ) and U( Gi Then, by the Fundamental Homomorphism Theorem, we have the desired conclusion: U(G i) U() = U( Gi References [1] H. Salzmann, T. Grundhöfer, H. Hähl, and R. Löwen. The classical fields, volume 11 of Encyclopedia of Mathematics and its Applications. Cambridge University Press, Cambridge, 007. Structural features of the real and rational numbers. 1