CHARACTERS OF FINITE ABELIAN GROUPS (MATH 321, SPRING 2007)

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1 CHARACTERS OF FINITE ABELIAN GROUPS (MATH 321, SPRING 2007) 1. Introduction The theme we will study is an analogue for finite abelian groups of Fourier analysis on R. Fourier series on the real line are expansions in terms of sines and cosines: f(x) = a n cos(nx) + b n sin(nx). n 0 n 1 Since e inx = cos(nx) + i sin(nx) and e inx = cos(nx) i sin(nx), Fourier series can also be written in terms of complex exponentials: f(x) = n Z c n e inx, where the summation runs over all integers. The convenient algebraic property of e inx, which is not shared by sines and cosines, is that it is a group homomorphism from R to the unit circle S 1 = z C : z = 1}: e in(x+x ) = e inx e inx. We now replace the real line R with a finite abelian group. Here is the analogue of the functions e inx. Definition 1.1. A character of a finite abelian group G is a group homomorphism χ: G S 1. We will usually write abstract groups multiplicatively, with identity 1. Example 1.2. The trivial character of G is the homomorphism ε defined by ε(g) = 1 for all g G. Example 1.3. Let G be cyclic of order 4 with generator γ. Since γ 4 = 1, a character χ of G has χ(γ) 4 = 1, so χ takes only four possible values at γ, namely 1, 1, i, or i. Once χ(γ) is known, the value of χ elsewhere is determined by multiplicativity: χ(γ j ) = χ(γ) j. So we get four characters, whose values can be organized into a character table. See Table 1. When G has size n and g G, for any character χ of G we have χ(g) n = χ(g n ) = χ(1) = 1, so the values of χ lie among the nth roots of unity in S 1. More precisely, the order of χ(g) divides the order of g (which divides #G). Characters on finite abelian groups were first studied in number theory, since number theory is a source of many interesting finite abelian groups. For instance, Dirichlet used characters on the group (Z/mZ) to prove that when (a, m) = 1 there are infinitely many primes p a mod m. For a 1

2 2 CHARACTERS OF FINITE ABELIAN GROUPS (MATH 321, SPRING 2007) 1 γ γ 2 γ 3 ε χ χ 2 1 i 1 i χ 3 1 i 1 i Table 1 finite field F, characters on F are useful tools to count or estimate the number of solutions to equations over F, and people like Gauss and Jacobi were interested in this problem. The quadratic reciprocity law of elementary number theory is concerned with a deep property of a particular character, the Legendre symbol. To provide a context against which our development of characters over finite abelian groups can be compared, Section 2 discusses classical Fourier analysis on the real line. In Section 3 we will run through some properties of characters of finite abelian groups and introduce their dual groups. Section 4 uses characters of a finite abelian group to develop a finite analogue of Fourier series. The simplest aspect of this development is applied in Section 5 to count solutions to an equation over finite fields. In Section 6 we use characters to prove a structure theorem for finite abelian groups. In Section 7 we look at (finite) Pontryagin duality. Characters are used in Section 8 to factor the group determinant of any finite abelian group. Our notation is completely standard, but we make two remarks about it. For a complex-valued function f(x), the complex-conjugate function is usually denoted f(x) instead of as f(x) to stress that conjugation creates a new function. (We sometimes use the overline notation also to mean the reduction g into a quotient group.) For n 1, we write µ n for the group of nth roots of unity in the unit circle S 1. It is a cyclic group of size n. Exercises. 1. Make a character table for Z/2Z Z/2Z. 2. Let G be a finite nonabelian simple group. (Examples include A n for n 5.) Show the only group homomorphism χ: G S 1 is the trivial map. 2. Classical Fourier analysis This section on Fourier analysis on R serves as motivation for our later treatment of finite abelian groups, where there will be no delicate convergence issues (just finite sums!), so we take a soft approach and sidestep the analytic technicalities that a serious treatment of Fourier analysis on R would demand. Fourier analysis for periodic functions on R is based on the functions e inx for n Z. Any reasonably nice function f : R C which has period 2π

3 CHARACTERS OF FINITE ABELIAN GROUPS (MATH 321, SPRING 2007) 3 can be expanded into a Fourier series f(x) = n Z c n e inx, where the sum runs over Z and the nth Fourier coefficient c n can be recovered as an integral: (2.1) c n = 1 2π 2π 0 f(x)e inx dx. This formula for c n can be explained by replacing f(x) in (2.1) by its Fourier series and integrating termwise (for reasonably nice functions this termwise integration is analytically justifiable), using the formula 1 2π e imx e inx 1, if m = n, dx = 2π 0 0, if m n. An important link between a function f(x) and its Fourier coefficients c n is given by Parseval s formula n Z c n 2 = 1 2π 2π 0 f(x) 2 dx. Rather than working with functions f : R C having period 2π, formulas look cleaner using functions f : R C having period 1. The basic exponentials become e 2πinx and the Fourier series and coefficients for f are (2.2) f(x) = n Z c n e 2πinx, c n = Parseval s formula becomes (2.3) c n 2 = n Z f(x) 2 dx. f(x)e 2πinx dx. Note c n in (2.2) is not the same as c n in (2.1). We turn now from Fourier series to Fourier integrals. These are applied to functions f : R C which are not periodic but which instead decay to 0 at ±. The Fourier transform of such an f is the function f : R C defined by the integral formula f(x) = f(y)e 2πixy dy. R The analogue of the expansion (2.2) of a periodic function into a Fourier series is the Fourier inversion formula, which expresses f in terms of its Fourier transform: f(x) = f(y)e 2πixy dy. R

4 4 CHARACTERS OF FINITE ABELIAN GROUPS (MATH 321, SPRING 2007) Define a (Hermitian) inner product of two functions f 1 and f 2 on R by the integral f 1, f 2 = f 1 (x)f 2 (x) dx C, R Plancherel s theorem compares the inner product of two functions and the inner product of their Fourier transforms: (2.4) f 1, f 2 = f 1, f 2. In particular, when f 1 = f 2 = f this becomes Parseval s formula f(x) 2 dx = f(x) 2 dx, R which is an analogue of (2.3). The convolution of two functions f 1 and f 2 from R to C is a new function from R to C defined by (f 1 f 2 )(x) = f 1 (y)f 2 (x y) dy R and the Fourier transform turns this convolution into pointwise multiplication: f 1 f 2 (x) = f 1 (x) f 2 (x). There are several conventions for the Fourier transform, the inner product, and the convolution. These conventions differ on how 2π appears (if at all) in the defining formulas and in the properties (Fourier inversion, Plancherel theorem, Fourier transform of a convolution). Tables 2 and 3 collect a number of different 2π-conventions found in analysis books. The first two columns of Table 2 are definitions and the last two columns are theorems (Fourier inversion and Plancherel s theorem). Table 3 describes the effect of the Fourier transform on convolutions using different 2π-conventions. f(x) f 1, f 2 f(x) f 1, f 2 1 2π R f(y)e ixy dy R f 1 1(x)f 2 (x) dx 2π R f(y)e ixy dy f 1, f 2 1 2π R f(y)e ixy 1 dy 2π R f 1 1(x)f 2 (x) dx 2π R f(y)e ixy dy f 1, f 2 R f(y)e 2πixy dy R f 1(x)f 2 (x) dx f(y)e R 2πixy dy f 1, f 2 Table 2 R f(x) (f 1 f 2 )(x) f 1 f 2 (x) 1 2π R f(y)e ixy dy R f 1(y)f 2 (x y) dy 2π f1 (x) f 2 (x) 1 2π R f(y)e ixy 1 dy 2π R f 1(t)f 2 (x t) dt f1 (x) f 2 (x) R f(y)e 2πixy dy R f 1(y)f 2 (x y) dy f1 (x) f 2 (x) Table 3

5 CHARACTERS OF FINITE ABELIAN GROUPS (MATH 321, SPRING 2007) 5 Functions which equal their Fourier transform are called self-dual. For instance, using complex analysis one can show (2.5) R e π(y+ix)2 dy = 1 for all x R. Expanding the square in the exponent, this becomes e πy2 e 2πixy dy = e πx2, R which says e πx2 is self-dual when we define the Fourier transform as f(x) = R f(y)e 2πixy dy. When the Fourier transform is defined using f(x) = 1 2π R f(y)e ixy dy, the function e (1/2)x2 is self-dual. More generally, a change of variables in (2.5) shows the Fourier transform of a Gaussian exponential e cx2 is the scalar multiple of another Gaussian exponential: (2.6) R e cy2 e 2πixy dy = π c e π2 x 2 /c. This formula shows that a highly peaked Gaussian (large c) has a Fourier transform which is a spread out Gaussian (small π 2 /c) and vice versa. More generally, there is a sense in which a function and its Fourier transform can t both be highly localized; this is a mathematical incarnation of Heisenberg s uncertainty principle. A link between Fourier series and Fourier integrals is the Poisson summation formula: for a nice function f : R C that decays rapidly enough at ±, (2.7) n Z f(n) = n Z f(n), where f(x) = R f(y)e 2πixy dy. To (nonrigorously) prove the Poisson summation formula, we will use Fourier series: Periodize f(x) as F (x) = n Z f(x + n).

6 6 CHARACTERS OF FINITE ABELIAN GROUPS (MATH 321, SPRING 2007) Since F (x + 1) = F (x), write F as a Fourier series: F (x) = n Z c ne 2πinx. The formula for c n is c n = = = m Z F (x)e 2πinx dx ( ) f(x + m) e 2πinx dx m Z 1 0 m+1 f(x + m)e 2πinx dx = f(x)e 2πinx dx m Z m = f(x)e 2πinx dx R = f(n). Therefore the expansion of F (x) into a Fourier series is equivalent to (2.8) f(n)e 2πinx, n Z f(x + n) = n Z which becomes the Poisson summation formula (2.7) by setting x = 0. If we replace a sum over Z with a sum over any one-dimensional lattice L = az in R, where a > 0, the Poisson summation formula becomes f(λ) = 1 f(µ), a λ L µ L where L = (1/a)Z is the dual lattice : For example, Z = Z. Exercises. L = x : e 2πixλ = 1 for all λ L}. 1. Without dwelling on analytic subtleties, check from Fourier inversion that f(x) = f( x) (if the Fourier transform is defined suitably). 2. If f is a real-valued even function, show its Fourier transform is also real-valued and even (assuming the Fourier transform of f is meaningful). 3. For a function f : R C and c R, let g(x) = f(x + c). Define the Fourier transform of a function h by ĥ(x) = R h(y)e 2πixy dy. If f has a Fourier transform, show g has Fourier transform ĝ(x) = e 2πicx f(x).

7 CHARACTERS OF FINITE ABELIAN GROUPS (MATH 321, SPRING 2007) 7 4. The Poisson summation formula over R was obtained by setting x = 0 in (2.8). Conversely, show that (2.8) for the function f follows from the Poisson summation formula for the function g(t) = f(t+x). 5. Assuming the Fourier inversion formula holds for a definition of the Fourier transform as in Table 2, check that for all α and β in R that if we set (Ff)(x) = α f(y)e iβxy dy R for all x then f(x) = β (Ff)(y)e iβxy dy. 2πα R (If β = 2πα 2 then these two equations are symmetric in the roles of f and Ff except for a sign in the exponential term.) Considering Ff to be the Fourier transform of f, show e (1/2)βx2 is self-dual in this sense. 3. Finite characters We leave the real line and turn to the setting of finite abelian groups G. Our interest shifts from the functions e inx to characters: homomorphisms from G to S 1. The construction of characters on these groups begins with the case of cyclic groups. Theorem 3.1. Let G be a finite cyclic group of size n with a chosen generator γ. There are exactly n characters of G, each determined by sending γ to the different nth roots of unity in C. Proof. We mimic Example 1.3, where G is cyclic of size 4. Since γ generates G, a character is determined by its value on γ and that value must be an nth root of unity (not necessarily of exact order n, e.g., ε(γ) = 1), so there are at most n characters. We now write down n characters. Let ζ be any nth root of unity in C. Set χ(γ j ) = ζ j for j Z. This formula is well-defined (if γ j = γ k for two different integer exponents j and k, we have j k mod n so ζ j = ζ k ), and χ is a homomorphism. Of course χ depends on ζ. As ζ changes, we get different characters (their values at γ are changing), so in total we have n characters. To handle characters on non-cyclic groups, the following lemma is critical. Lemma 3.2. Let G be a finite abelian group and H G be a subgroup. Any character of H can be extended to a character of G in [G : H] ways. Proof. We will induct on the index [G : H] and we may suppose H G. Pick a G with a H, so H H, a G. Let χ: H S 1 be a character of H. We will extend χ to a character χ of H, a and count the number of possible χ.

8 8 CHARACTERS OF FINITE ABELIAN GROUPS (MATH 321, SPRING 2007) What is a viable choice for χ(a)? Since a H, χ(a) is not initially defined. But χ(a k ) is defined if a k H: χ(a k ) = χ(a k ). Pick k 1 minimal with a k H. That is, k is the order of a in G/H, so k = [ H, a : H]. If χ is going to be a character then χ(a) has to be an k-th root of χ(a k ). Define χ(a) S 1 to be a solution to z k = χ(a k ): χ(a) k = χ(a k ). Every number in S 1 has k different k-th roots in S 1, so there are k choices for χ(a). Once we have chosen χ(a), define χ on H, a by χ(ha i ) = χ(h) χ(a) i. Is χ well-defined? Perhaps H and a overlap nontrivially, so the expression of an element of H, a in the form ha i is not unique. Suppose ha i = h a i. Then a i i H, so i i mod k since k is the order of a in G/H. Write i = i + kq, so h = h a i i = h a kq. The terms h, h, and a k are in H, so χ(h ) χ(a) i = χ(h ) χ(a) i χ(a) kq = χ(h ) χ(a) i χ(a k ) q since χ(a) k = χ(a k ) = χ(h a kq ) χ(a) i = χ(h) χ(a) i. Therefore χ: H, a S 1 is a well-defined function and it is easily checked to be a homomorphism. It restricts to χ on H. The number of choices of χ extending χ is the number of choices for χ(a), which is k = [ H, a : H]. Since [G : H, a ] < [G : H], by induction on the index there are [G : H, a ] extensions of each χ to a character of G, so the number of extensions of χ to a character on G is [G : H, a ][ H, a : H] = [G : H]. Theorem 3.3. If g 1 in a finite abelian group G then χ(g) 1 for some character χ of G. The number of characters of G is #G. Proof. The cyclic group g is nontrivial, say of size n, so n > 1. The group µ n of n-th roots of unity in S 1 is also cyclic, so there is an isomorphism g = µ n. This isomorphism can be viewed as a character of g. By Lemma 3.2 it extends to a character of G and does not send g to 1. To show G has #G characters, apply Lemma 3.2 with H the trivial subgroup. Corollary 3.4. If g 1 g 2 in G then there is a character of G which takes different values at g 1 and g 2. Proof. Apply Theorem 3.3 to g = g 1 g 1 2. Corollary 3.4 shows the characters of G separate the elements of G: different elements of the group admit a character taking different values on them.

9 CHARACTERS OF FINITE ABELIAN GROUPS (MATH 321, SPRING 2007) 9 Corollary 3.5. If H G is a subgroup and g G with g H then there is a character of G which is trivial on H and not equal to 1 at g. Proof. We work in the finite abelian group G/H, where g 1. By Theorem 3.3 there is a character of G/H which is not 1 at g. Lift this function to G, where it is a character trivial on H and not equal to 1 at g. It is easy to find functions on G which separate elements without using characters. For g G, define δ g : G 0, 1} by 1, if x = g, (3.1) δ g (x) = 0, if x g. These functions separate elements of the group, but characters do this too and have better algebraic properties: they are group homomorphisms. Our definition of a character makes sense on finite nonabelian groups, but there will not be enough such characters for Theorem 3.3 to hold. If G is a nonabelian group, any homomorphism χ: G S 1 must equal 1 on the commutator subgroup [G, G], which is a nontrivial subgroup. Any g 1 in [G, G] has χ(g) = 1 for all χ. Definition 3.6. For a character χ on a finite abelian group G, the conjugate character is the function χ: G S 1 given by χ(g) := χ(g). Since any complex number z with z = 1 has z = 1/z, χ(g) = χ(g) 1. Definition 3.7. The dual group of a finite abelian group G is the set of homomorphisms G S 1 with the group law of pointwise multiplication of functions: (χψ)(g) = χ(g)ψ(g). The dual group of G is denoted Ĝ. The trivial character of G is the identity in Ĝ and the inverse of a character is its conjugate character. Note Ĝ is abelian since multiplication in C is commutative. Theorem 3.3 says in part that (3.2) #G = #Ĝ. In Theorem 6.8 we ll obtain a stronger result: the groups G and Ĝ are isomorphic. For now we check this on cyclic groups. Theorem 3.8. If G is cyclic then G = Ĝ as groups. Proof. We will show Ĝ is cyclic. Then since G and Ĝ have the same size they are isomorphic. Let n = #G and γ be a generator of G. Set χ: G S 1 by χ(γ j ) = e 2πij/n for all j. For any other character ψ Ĝ, we have ψ(γ) = e2πik/n for some integer k, so ψ(γ) = χ(γ) k. Then ψ(γ j ) = ψ(γ) j = χ(γ) jk = χ(γ j ) k, which shows ψ = χ k. Therefore χ generates Ĝ.

10 10 CHARACTERS OF FINITE ABELIAN GROUPS (MATH 321, SPRING 2007) The following simple calculation will play a crucial role in the next section. Theorem 3.9. Let G be a finite abelian group and H be a subgroup. For χ Ĝ, #H, if χ is trivial on H, χ(h) = 0, otherwise. In particular, h H χ(g) = g G #G, if χ = ε, 0, if χ ε. Proof. Let S = h H χ(h). If χ is trivial on H then S = #H. If χ is not trivial on H, choose h 0 H such that χ(h 0 ) 1. Then χ(h 0 )S = h H χ(hh 0) = h H χ(h) = S, so S = 0. Taking H = G, Theorem 3.9 says the sum of a nontrivial character over a group vanishes. In terms of the character table of G this means the sum of the elements in each row of a character table is zero except for the top row (corresponding to the trivial character). See Table 1. Corollary Let G be a finite abelian group and K be a subgroup of Ĝ. For g G, #K, if χ(g) = 1 for all χ K, χ(g) = 0, otherwise. In particular, χ K χ(g) = χ b G #G, if g = 1, 0, if g 1. Proof. The map evaluate at g gives a function K S 1, where χ χ(g). This is a character of K, so the sum χ K χ(g) can be viewed as the sum of a character over the group K. By Theorem 3.9 this sum vanishes unless the character is trivial on the group, which means χ(g) = 1 for all χ K. Corollary 3.10 with K = Ĝ tells us that the sum of the elements in each column of the character table is zero except for the first column (corresponding to the identity element of G). See Table 1. Characters reverse the roles of mth roots of the identity in a finite abelian group and mth powers in its dual group, as in the next two theorems. Theorem Let G be a finite abelian group and m Z. For g G, g m = 1 if and only if χ(g) = 1 for every χ Ĝ which is an mth power in Ĝ. Proof. If g m = 1 and χ = ψ m for some ψ Ĝ then χ(g) = ψ m (g) = ψ(g) m = ψ(g m ) = ψ(1) = 1. Conversely, suppose χ(g) = 1 whenever χ = ψ m for some ψ Ĝ. Then for all ψ Ĝ we have 1 = ψm (g) = ψ(g m ), so g m = 1 by Theorem 3.3.

11 CHARACTERS OF FINITE ABELIAN GROUPS (MATH 321, SPRING 2007) 11 Theorem Let G be a finite abelian group and m Z. For g G, g is an mth power in G if and only if χ(g) = 1 for every χ Ĝ satisfying χ m = ε. Proof. If g = x m for some x G then for every χ Ĝ such that χm = ε we have χ(g) = χ(x) m = χ m (x) = 1. Conversely, assume χ(g) = 1 for all χ such that χ m = ε. Then such χ are identically 1 on the subgroup G m of mth powers in G, and conversely every character of G which is trivial on this subgroup has mth power ε. Therefore g is in G m by Corollary 3.5. We have used in this section two important features of S 1 as the target group for characters: for any m 1 the mth power map on S 1 is m-to-1 (proof of Lemma 3.2) and for each n 1 there is a cyclic subgroup of order n in S 1 (proof of Theorem 3.3). Another property is that the values of characters on a finite abelian group are roots of unity. The subgroup of roots of unity in S 1 is isomorphic to Q/Z by r e 2πir, where µ n corresponds to (1/n)Z/Z. From this point of view, the characters of a finite abelian group G could be defined as the homomorphisms G Q/Z (additive target group). Table 4 illustrates this when G is cyclic of order 4, where the values are in (1/4)Z/Z. 1 γ γ 2 γ 3 ε χ 1 0 1/2 0 1/2 χ 2 0 1/4 1/2 3/4 χ 3 0 3/4 1/2 1/4 Table 4 For classical Fourier analysis the target group Q/Z is inadequate since e ix has values other than roots of unity as x runs over R. Exercises. 1. For abelian groups G 1 and G 2, show the set Hom(G 1, G 2 ) of group homomorphisms from G 1 to G 2 is an abelian group under pointwise multiplication of functions. Noting that Hom(G, S 1 ) = Hom(G, C ) when G is finite abelian, for any field F we could consider the group Hom(G, F ) as an analogue of the dual group. How could the treatment of such characters encounter a problem if F C? (Try F = R.) 2. Let s find the characters of the additive group (Z/mZ) r. (a) For k Z/mZ, let χ k : Z/mZ S 1 by χ k (j) = e 2πijk/m,

12 12 CHARACTERS OF FINITE ABELIAN GROUPS (MATH 321, SPRING 2007) so χ k (1) = e 2πik/m. Show χ 0, χ 1,..., χ m 1 are all the characters of Z/mZ and χ k χ l = χ k+l. (b) Let r 1. For r-tuples a, b in (Z/mZ) r, let a b = a 1 b a r b r Z/mZ be the usual dot product. For k (Z/mZ) r, let χ k (j) = e 2πi(j k)/m. Show the functions χ k are all the characters of (Z/mZ) r and χ k χ l = χ k+l. 3. For finite nonabelian G, show the characters of G (that is, homomorphisms G S 1 ) separate elements modulo [G, G]: if χ(g 1 ) = χ(g 2 ) for all χ then g 1 = g 2 in G/[G, G]. 4. Show the following are equivalent: χ(g) = ±1 for all g, χ(g) = χ(g) for all g, and χ 2 = ε. 4. Finite Fourier series We will introduce the analogue of Fourier series on finite abelian groups. Engineering applications of these ideas include signal processing (the fast Fourier transform [2, Chap. 9]) and error-correcting codes [2, Chap. 11]. Let G be a finite abelian group. Set L(G) = f : G C}, the C-valued functions on G. Every f L(G) can be expressed as a linear combination of the delta-functions δ g from (3.1): (4.1) f = g G f(g)δ g. Indeed, evaluate both sides at each x G and we get the same value. If the sum on the right side of (4.1) is zero, then f(g) = 0 for all g, so f = 0. Thus the functions δ g are a basis of L(G), so dim L(G) = #G. The next theorem is the first step leading to an expression for each δ g as a linear combination of characters of G. Theorem 4.1. Let G be a finite abelian group. Then #G, if χ = ε, #G, if g = 1, χ(g) = χ(g) = 0, if χ ε, 0, if g 1. g G χ b G Proof. This is a combination of Theorem 3.9 and Corollary Corollary 4.2. For characters χ 1 and χ 2, #G, if χ 1 = χ 2, χ 1 (g)χ 2 (g) = 0, if χ 1 χ 2. For g 1, g 2 G, g G χ(g 1 )χ(g 2 ) = χ b G #G, if g 1 = g 2, 0, if g 1 g 2.

13 CHARACTERS OF FINITE ABELIAN GROUPS (MATH 321, SPRING 2007) 13 Proof. In the first equation of Theorem 4.1 let χ = χ 1 χ 2. In the second equation of Theorem 4.1 let g = g 1 g 1 2. The equations in Corollary 4.2 say the character table of G has orthogonal rows and orthogonal columns when we measure orthogonality on row and column vectors over C using the usual Hermitian inner product on row and column vectors. Example 4.3. Let G = (Z/mZ). For a (Z/mZ) and p a prime not dividing m, 1 1, if p a mod m, χ(a)χ(p) = ϕ(m) 0, if p a mod m, χ mod m where the sum runs over the characters of (Z/mZ). (Since p is prime, p not dividing m forces p to be in (Z/mZ).) This identity was used by Dirichlet in his proof that there are infinitely many primes p a mod m. By the second equation in Corollary 4.2 we can express the delta-functions in terms of characters: δ g (x) = 1 χ(g)χ(x) #G χ G b for every x G. Substituting this formula for δ g into (4.1) gives f(x) = f(g) 1 χ(g)χ(x) #G g G = χ b G g G χ b G 1 #G f(g)χ(g)χ(x) (4.2) = χ b G c χ χ(x), where (4.3) c χ = 1 #G f(g)χ(g). g G The expansion (4.2) is the Fourier series for f. Equation (4.3) is similar to the formula for the coefficient c n of e inx in (2.1): an integral over [0, 2π] divided by 2π is replaced by a sum over G divided by #G and f(x)e inx is replaced by f(g)χ(g). The number e inx is the conjugate of e inx, which is also the relation between χ(g) and χ(g). Equation (4.2) shows Ĝ is a spanning set for L(G). Since #Ĝ = #G = dim L(G), Ĝ is a basis for L(G).

14 14 CHARACTERS OF FINITE ABELIAN GROUPS (MATH 321, SPRING 2007) Definition 4.4. Let G be a finite abelian group. Fourier transform is the function f L(Ĝ) given by f(χ) = f(g)χ(g). g G If f L(G) then its By (4.2) and (4.3), (4.4) f(x) = 1 #G χ b G f(χ)χ(x). Equation (4.4) is called the Fourier inversion formula since it tells us how to recover f from its Fourier transform. Remark 4.5. Classically the Fourier transform of a function R C is another function R C. The finite Fourier transform, however, is defined on the dual group instead of on the original group. This is also happening classically, if one looks carefully. For x R let χ x (y) = e ixy. Then χ x : R S 1 is a character and f(x) could be viewed as f(χ x ) = R f(y)χ x(y) dy, so f is a function of characters rather than of numbers. Example 4.6. Let f = δ g. Then f(χ) = χ(g) = χ(g 1 ). Notice f vanishes at all but one element of G while f is nonzero on all of Ĝ. Example 4.7. Let f = ψ be a character of G. Then f(χ) = g ψ(g)χ(g) = (#G)δ ψ (χ), so f = (#G)δ ψ. Here f is nonzero on all of G and f is nonzero at only one element of Ĝ. The Fourier transform on R interchanges highly spread and highly peaked Gaussians. Examples 4.6 and 4.7 suggest a similar phenomenon in the finite case. Here is a general result in that direction (a finite version of Heisenberg uncertainty). Theorem 4.8. Let f : G C be a function on a finite abelian group G which is not identically zero. Then (4.5) #(supp f) #(supp f) #G, where supp denotes the support of a function (the set of points where the function is nonzero). Proof. We expand f into a Fourier series and make estimates. Since f(x) = 1 #G f(χ)χ(x), χ G b we have (4.6) f(x) χ b G 1 #G f(χ) #(supp f) #G max f(χ). χ G b

15 CHARACTERS OF FINITE ABELIAN GROUPS (MATH 321, SPRING 2007) 15 By the definition of f(χ), (4.7) f(χ) g G f(g). Let m = max g G f(g), so m > 0 since f is not identically zero. Then (4.7) implies f(χ) m#(supp f), and feeding this into (4.6) yields f(x) m#(supp f)#(supp f). #G Maximizing over all x G implies m m#(supp f)#(supp f)/#g. Divide both sides by m and the desired inequality drops out. In Examples 4.6 and 4.7, inequality (4.5) is an equality, so Theorem 4.8 is sharp. Since L(G) is spanned by both the characters of G and the delta-functions, any linear identity in L(G) can be verified by checking it on characters or on delta-functions. Let s look at an example. Define a Hermitian inner product on L(G) by the rule (4.8) f 1, f 2 G = 1 #G f 1 (x)f 2 (x) C. x G We will prove Plancherel s theorem for G: f 1, f 2 G = 1 #G f 1, f 2 bg for all f 1 and f 2 in L(G). (Compare to (2.4).) To check this identity, which is linear in both f 1 and f 2, it suffices to check it when f 1 and f 2 are characters. By Corollary 4.2, for characters χ 1 and χ 2 of G we have 1, if χ 1 = χ 2, χ 1, χ 2 G = 0, if χ 1 χ 2. Since χ = (#G)δ χ (Example 4.7), 1 #G χ 1, χ 2 bg = #G δ χ1, δ χ2 bg = χ b G δ χ1 (χ)δ χ2 (χ) = 1, if χ 1 = χ 2, 0, if χ 1 χ 2. This verifies Plancherel s theorem for G. The special case where f 1 = f 2 = f is a single function from G to C gives us Parseval s formula for G: (4.9) f(g) 2 = 1 #G f(χ) 2. g G χ b G

16 16 CHARACTERS OF FINITE ABELIAN GROUPS (MATH 321, SPRING 2007) Finite abelian analogues of Poisson summation and the relation between Fourier transforms and convolutions are in Exercises 4.6 and Let s look at some examples of Fourier transforms for functions on a cyclic group. By writing a cyclic group in the form Z/mZ, we can make the isomorphism with the dual group explicit: every character of Z/mZ arises in the form χ k : j e 2πijk/m for a unique k Z/mZ (Exercise 3.2). The character χ 1 : j e 2πij/m is a generator of Ẑ/mZ. From this point of view the Fourier transform of a function f : Z/mZ C is a second function f : Z/mZ C also defined on Z/mZ: f(j) = f(k)e 2πijk/m. k Z/mZ Notice that j appears on the right side in the context of e 2πijk/m = χ j (k), showing the Fourier transform is still really a function of characters. But we can think of it as a function of integers modulo m since characters of Z/mZ are parametrized by Z/mZ. Example 4.9. Let f : Z/8Z C be a function with period 2 having values 1 and 2. See Table 5. The Fourier transform of f vanishes except at 0 and 4, which are the multiples of the frequency of f (how often the period repeats). n f(n) f(n) Table 5 Example Now let f : Z/8Z C have the periodic values 5, 3, 1, and 1. Both f and its Fourier transform are listed in Table 6. Now f has frequency 2 (its period repeats twice) and the Fourier transform vanishes except at 0, 2, 4, and 6, which are multiples of the frequency. n f(n) f(n) i i 0 Table 6 Example Consider a function f : Z/45Z C with the four successive repeating values 1, 8, 19, 17. It is not a periodic function on Z/45Z since 4 does not divide 45, but the sequence 1, 8, 19, 17 repeats nearly 11 times. A calculation of the Fourier transform of f shows there are sharp peaks in f(n) at n = 0, 11, 22, 23, and 34. (For example, f(11) is between 3 and 4 times larger than the adjacent values f(10) and f(12).) The peaks in f(n) occur approximately at multiples of the approximate frequency.

17 CHARACTERS OF FINITE ABELIAN GROUPS (MATH 321, SPRING 2007) 17 The Fourier transform of a periodic function on Z/mZ knows the frequency of the original function by the positions where the Fourier transform takes on its nonzero values, and for nearly periodic functions the approximate frequency is reflected in where the Fourier transform takes on its largest values. This is used in Shor s quantum algorithm for integer factorization [3, Chap. 17]. Exercises. 1. Let f : Z/8Z C take the four values a, b, c, and d twice in this order. Compute f(n) explicitly and determine some values for a, b, c, and d such that f(n) is nonzero for n = 0, 2, and 6, but f(4) = For a subgroup H of a finite abelian group G, let H = χ Ĝ : χ = 1 on H}. These are the characters of G that are trivial on H. For example, G = ε} and 1} = Ĝ. Note H Ĝ and H depends on H and G. Show H is a subgroup of Ĝ, it is isomorphic to Ĝ/H, and Ĝ/(H ) = Ĥ. In particular, #H = [G : H]. 3. For any subgroup H of a finite abelian group G, let δ H be the function which is 1 on H and 0 off of H. Show the Fourier transform of δ H is δ H = (#H)δ H. How do the supports of δ H and its Fourier transform compare with the inequality (4.5)? 4. Let H be a subgroup of a finite abelian group G. (a) Suppose f : G C is constant on H-cosets (it is H-periodic). For χ Ĝ with χ H, show f(χ) = 0. Thus the Fourier transform of an H-periodic function on G is supported on H. (b) If f : Z/mZ C has period d where d m, show f : Z/mZ C is supported on the multiples of m/d. (See Examples 4.9 and 4.10.) 5. Find the analogue of Exercise 2.3 for functions on a finite abelian group. 6. Let G be a finite abelian group and H be a subgroup. For any function f : G C, Poisson summation on G says 1 #H h H f(h) = 1 #G χ H f(χ), where H is as in Exercise 4.2. Prove this formula in two ways: copy the classical proof sketched in Section 2 (start with the function F (x) = h H f(xh), which is H-periodic so it defines a function on G/H) to obtain 1 #H h H f(xh) = 1 f(χ)χ(x) #G χ H

18 18 CHARACTERS OF FINITE ABELIAN GROUPS (MATH 321, SPRING 2007) for any x G and then set x = 1. Or, by linearity in f of both sides of the desired identity, verify Poisson summation directly on the delta-functions of G. (Corollary 3.5 and Example 4.6 will be useful.) 7. Let p be prime and χ be a nontrivial character of the multiplicative group (Z/pZ). Extend χ to a function on Z/pZ by setting χ(0) = 0. This extended function is not a character of the additive group Z/pZ, but the formula χ(xy) = χ(x)χ(y) holds even if one of x or y is 0. The Fourier transform χ: Z/pZ C is χ(x) = y Z/pZ χ(y)e 2πixy/p = y (Z/pZ) χ(y)e 2πixy/p. (Don t confuse the Fourier transform for a function on Z/pZ with the Fourier transform for a function on (Z/pZ).) (a) Show χ(x) = χ( 1)G(χ)χ(x), where G(χ) = y χ(y)e2πiy/p. The sum G(χ) is called a Gauss sum. (b) Show G(χ) = χ( 1)G(χ). (c) Although G(χ) is a sum of p 1 roots of unity, its magnitude is much smaller than the naive upper bound p 1: G(χ) = p. Show this by computing the Fourier expansion for χ as a function on Z/pZ to find G(χ)G(χ) = χ( 1)p and then use (b). (d) Derive the formula G(χ) = p in a second way by computing both sides of Parseval s formula (4.9) for the function χ: Z/pZ C where χ(0) = For a finite abelian group G and g G, let T g : L(G) L(G) by (T g f)(x) = f(gx). (a) Show the T g s are commuting linear transformations. (b) If f is a simultaneous eigenvector of all the T g s, show f(1) 0 (if f(1) = 0 conclude f is identically zero, but the zero vector is not an eigenvector) and then after rescaling f so f(1) = 1 deduce that f is a character of G. Thus the simultaneous eigenvectors of the T g s, suitably normalized, are the characters of G. (c) Show the T g s are simultaneously diagonalizable. Deduce from this and (b) that Ĝ is a basis of L(G), so #Ĝ = dim L(G) = #G. (This gives a different proof that G and Ĝ have the same size.) 9. Let T g : L(G) L(G) be as in the previous exercise. (a) Show T g f = χ(g) f for any f L(G). (b) Show for any f 1 and f 2 in L(G) that T g f 1, T g f 2 G = f 1, f 2 G. 10. Let G be a finite abelian group. For f 1 and f 2 in L(G), define their convolution f 1 f 2 : G C by (f 1 f 2 )(x) = y G f 1 (y)f 2 (xy 1 ).

19 CHARACTERS OF FINITE ABELIAN GROUPS (MATH 321, SPRING 2007) 19 (a) Show δ g δ h = δ gh, so L(G) under convolution is a commutative C-algebra isomorphic to the group ring C[G]. (b) Show f 1 f 2 (χ) = f 1 (χ) f 2 (χ), so the Fourier transform turns convolution into pointwise multiplication. (c) Show δ g f = T g 1(f) and χ f = f(χ)χ in two ways: by a direct calculation or by computing the Fourier transform of both sides and using (b). (d) For each χ Ĝ, the function h χ : L(G) C given by h χ (f) = f(χ) is a C-algebra homomorphism by (b). Does every C-algebra homomorphism from L(G) to C arise in this way? 11. Formulate the Fourier inversion formula and Plancherel s theorem for functions on a finite abelian group G using the rescaled definitions f(χ) = 1 f(g)χ(g), f 1, f 2 G = 1 f 1 (g)f #G #G 2 (g). g G g G Does this modified Fourier transform send convolution to multiplication if we use the definition (f 1 f 2 )(g) = 1 f 1 (h)f 2 (gh 1 )? #G h G 5. Counting over finite fields Let F be a finite field of size q with odd characteristic (e.g., q = p is an odd prime and F = Z/pZ). In the field F, every element is a sum of two squares. The simplest proof of this goes as follows. For each c F we want to show x 2 + y 2 = c has a solution (x, y) in F. Consider the sets (5.1) x 2 : x F}, c y 2 : y F}. These sets both have the same size: the number of squares in F. There are (q 1)/2 nonzero squares in F (squaring is 2-to-1 on nonzero elements), so including 0 = 0 2 shows the total number of squares in F is (q 1)/2 + 1 = (q + 1)/2. The two sets in (5.1) therefore each contain (q + 1)/2 elements. If the two sets were disjoint then their union would have size q + 1, which exceeds the size of F. Hence the two sets overlap, and that means x 2 = c y 2 has a solution in F, so c = x 2 + y 2. The above argument tells us we can solve x 2 +y 2 = c in F, but it does not tell us how many solutions there are (as a function of c). We will now use a character on F to count the number of solutions, and will find a surprising result: the count is actually independent of c as long as c 0. Write x = or x to indicate that an element x is or is not a square in F. Definition 5.1. Let η : F ±1} be given by the rule 1, if x =, η(x) = 1, if x.

20 20 CHARACTERS OF FINITE ABELIAN GROUPS (MATH 321, SPRING 2007) Extend η to all of F by η(0) = 0. The function η is a character on F : η(xy) = η(x)η(y) for all x and y in F. This is because the nonzero squares have size (q 1)/2, and hence index 2 in F. Indeed, if G is any abelian group and H is a subgroup with index 2, the map G ±1} which is 1 on H and 1 on its complement is a group homomorphism, as the reader can check (only interesting point: the product of two elements not in H is in H). With the definition η(0) = 0, the formula η(xy) = η(x)η(y) holds for all x and y in F, including 0. Lemma 5.2. For any t F, #x F : x 2 = t} = 1 + η(t). Proof. When t = 0, both sides of the counting formula are 1. When t 0 and t is a perfect square, both sides are 2. When t is not a square, both sides are 0. Theorem 5.3. For c F, let N c = #(x, y) : x 2 + y 2 = c}. Then N 0 = 2q 1, if 1 =, 1, if 1, and for c 0 N c = q 1, if 1 =, q + 1, if 1. In particular, N c is independent of c when c 0. Proof. If we can solve x 2 + y 2 = 0 with one (and hence both) of x and y nonzero then 1 is a square in F. Therefore if 1 is not a square in F the only solution to x 2 + y 2 = 0 is (0, 0), hence N 0 = 1. If 1 is a square in F, say 1 = r 2, then solutions are described by (x, ±rx) : x F}, which has size N 0 = 1 + 2(q 1) = 2q 1. For c 0, we will show N c = N 1. Then we will figure out what this common value is. Getting the independence of c will be the hard part, where characters get used. When c 0, N c = x = x #y : y 2 = c x 2 } (1 + η(c x 2 )) by Lemma 5.2 = q + x η(c x 2 ).

21 CHARACTERS OF FINITE ABELIAN GROUPS (MATH 321, SPRING 2007) 21 For each t F, the term η(t) occurs in the sum as often as we can solve t = c x 2, so N c = q + t = q + t = q + t #x : c x 2 = t}η(t) #x : x 2 = c t}η(t) (1 + η(c t))η(t) by Lemma 5.2 = q + t η(t) + t η(c t)η(t). Since η(0) = 0, t η(t) = t 0 η(t), and this sum is 0 by Theorem 3.9 because η is a nontrivial character (there are nonsquares in F ). Thus N c = q + t η(c t)η(t). Because c 0, we can do a change of variables from t to ct: N c = q+ t η(c ct)η(ct) = q+ t η(c)η(1 t)η(c)η(t) = q+ t η(1 t)η(t) since η(c) 2 = 1. This last formula for N c does not involve c, so N c is independent of c when c 0. Now we compute this common N c. Since c F N c = #(x, y)} = q 2, we get N 0 + (q 1)N 1 = q 2. Taking cases for the two formulas for N 0 gives the two formulas for N 1 (and thus N c for all nonzero c). The same method applies to ax 2 + by 2 = c where a, b F : the number of solutions (x, y) in F is independent of c when c 0. Specifically, this number is q 1 when ab = and q + 1 when ab. For further uses of characters to count solutions to equations over finite fields, see [1, Chap. 8]. This application involves characters of cyclic groups. Characters of non-cyclic finite abelian groups are also important in number theory; they arise in the definition of Dirichlet L-functions [1, Chap. 16]. 6. Structure of finite abelian groups We will now use characters to prove a structure theorem: every finite abelian group is a direct product of cyclic groups. The main technical result in this approach is Theorem 6.5. Lemma 6.1. For positive integers d and m with d m, the natural reduction map (Z/mZ) (Z/dZ) is onto: when (a, d) = 1, there is b such that b a mod d and (b, m) = 1. Proof. Let d be the product of the prime powers in m whose primes divide d, so m = dn with ( d, n) = 1. (For example, if m = 90 and d = 6 then

22 22 CHARACTERS OF FINITE ABELIAN GROUPS (MATH 321, SPRING 2007) d = 18 and n = 5.) Then d d. By the Chinese remainder theorem solve b a mod d, b 1 mod n. Then b a mod d and b is relatively prime to m since it is relatively prime to d (a factor of d) and to n. Lemma 6.2. Let G be a finite abelian group. If x G has order m and y G has order n then there is a character χ: G S 1 such that χ(x) has order m and χ(y) has order n. Proof. The subgroup x is cyclic of order m, so there is an isomorphism χ: x = µ m. In particular, χ(x) has order m. Following the proof of Lemma 3.2, we can extend χ to a character on x, y (and then all the way up to G) by sending y to any solution z S 1 of the equation z k = χ(y k ), where k 1 is minimal such that y k x. We will show z can be picked to have order n in S 1. Since y n = 1 x, k divides n. Then y k has order n/k, so χ(y k ) has order n/k because χ is one-to-one on x. Write χ(y k ) = e 2πil/(n/k) = e 2πilk/n, where (l, n/k) = 1. By Lemma 6.1, there is an l l mod n/k such that (l, n) = 1. Since l k lk mod n, χ(y k ) = e 2πil k/n. Set z = e 2πil /n, which has order n. Since z k = χ(y k ), we can set χ(y) = z. Lemma 6.2 does not extend to more than two arbitrary elements in G. For instance, if G = µ 2 µ 2 then no character on G sends all three non-identity elements in G to 1. (Why?) Theorem 6.3. Let G be a finite abelian group. The order of any element in G divides the maximal order of the elements of G. Proof. We will use characters to prove this theorem, which says nothing about characters in its statement. (There are character-free proofs.) Let n be the maximal order of the elements of G, say n is the order of g. Pick any h G, with order (say) m. We will use characters to construct an element of G with order [m, n]. Then since n is the maximal order in G and [m, n] n we must have [m, n] = n, so m n, which is what we wanted. By Lemma 6.2, there is a character χ on G such that χ(g) has order n and χ(h) has order m. Write χ(g) = e 2πia/n and χ(h) = e 2πib/m, where (a, n) = 1 and (b, m) = 1. The roots of unity e 2πi/n and e 2πi/m are in χ(g). For instance, letting aa 1 mod n, χ(g a ) = χ(g) a = e 2πiaa /n = e 2πi/n. The argument for e 2πi/m is similar. Write mu + nv = (m, n) for some integers u and v, so the equation mn = [m, n](m, n) can be rewritten as Thus 1 (m, n) = [m, n] mn = mu + nv mn = u 1 n + v 1 m. e 2πi/[m,n] = (e 2πi/n ) u (e 2πi/m ) v χ(g),

23 CHARACTERS OF FINITE ABELIAN GROUPS (MATH 321, SPRING 2007) 23 say e 2πi/[m,n] = χ(t). The order of t is divisible by the order of χ(t), so t has order divisible by [m, n]. Thus, raising t to a suitable power we obtain an element of G with order [m, n]. For example, if the maximal order of the elements of G is 20 then the order of any element of G is a factor of 20. This kind of result is false for nonabelian groups, e.g., the orders of the elements of S 3 are 1, 2, and 3. Corollary 6.4. Let G be a finite abelian group and let n be the maximum order of any element of G. Then every character of G has values in µ n. Proof. By Theorem 6.3, every g G satisfies g n = 1, so χ(g) n = 1 for any χ Ĝ. The following theorem shows that a maximal cyclic subgroup of a finite abelian group can always be split off as a direct factor. The theorem is just about groups, not characters, but characters get used in an essential way in the proof. Theorem 6.5. Let G be a finite abelian group and let g G have maximal order in G. Then there is a subgroup H G such that G = H g. Proof. Let n be the order of g. The subgroup g is cyclic of size n, so it is isomorphic to µ n. Pick an isomorphism of g with µ n. This isomorphism is an example of a character of g. Extend this to a character of G (Lemma 3.2), so we get a character χ: G S 1. The image of χ is µ n (Corollary 6.4) since χ(g) was chosen to generate µ n. Set H = ker χ. Then H g = 1} since χ is one-to-one on g by construction. For any x G, χ(x) is in µ n = χ( g ) so χ(x) = χ(g k ) for some k. Therefore x = hg k, so h H. This proves G = H g. Theorem 6.6. Every nontrivial finite abelian group G is isomorphic to a product of cyclic groups. More precisely, we can write where n 1 n 2 n k and n 1 > 1. G = Z/n 1 Z Z/n 2 Z Z/n k Z, Proof. Induct on #G. The result is clear when #G is prime, e.g., when #G = 2. Let n be the maximal order of the elements of G, so G = H Z/nZ by Theorem 6.5. Since #H < #G, by induction (6.1) H = Z/n 1 Z Z/n 2 Z Z/n r Z where n 1 n 2 n r. From (6.1) n r is the order of an element of H (in fact it is the maximal order of an element of H), so n r n by Theorem 6.3. Thus we can tack Z/nZ onto the end of (6.1) and we re done. Theorem 6.6 not only expresses G as a direct product of cyclic groups, but does so with the extra feature that successive cyclic factors have size divisible by the previous cyclic factor. When written this way, the n i s are uniquely

24 24 CHARACTERS OF FINITE ABELIAN GROUPS (MATH 321, SPRING 2007) determined by G, but we will not need this more precise information and don t prove it here. As an application of Theorem 6.6 we compare the group structure of G and Ĝ. Lemma 6.7. If A and B are finite abelian groups, there is an isomorphism  B =  B. Proof. Let χ be a character on A B. Identify the subgroups A 1} and 1} B of A B with A and B in the obvious way. Let χ A and χ B be the restrictions of χ to A and B respectively, i.e., χ A (a) = χ(a, 1) and χ B (b) = χ(1, b). Then χ A and χ B are characters on A and B and χ(a, b) = χ((a, 1)(1, b)) = χ(a, 1)χ(1, b) = χ A (a)χ B (b). So we get a map (6.2)  B  B by sending χ to (χ A, χ B ). It is left to the reader to check (6.2) is a group homomorphism. Its kernel is trivial since if χ A and χ B are trivial characters then χ(a, b) = χ A (a)χ B (b) = 1, so χ is trivial. Both sides of (6.2) have the same size by (3.2), so (6.2) is an isomorphism. Theorem 6.8. If G is a finite abelian group then G is isomorphic to Ĝ. Proof. The case when G is cyclic was Theorem 3.8. easily to several factors in a direct product: (H 1 H r ) b = Ĥ 1 Ĥr. Lemma 6.7 extends In particular, applying this to a cyclic decomposition of G from Theorem 6.6, each H j is isomorphic to Ĥj so we obtain G = Ĝ. Corollary 6.9. If G is a finite abelian group and H is a subgroup then there is a subgroup of G which is isomorphic to G/H. Proof. Every character of G/H lifts in a natural way to a character of G which is trivial on H and this is an embedding Ĝ/H Ĝ. Since the dual group is isomorphic to the group, this embedding can be converted (noncanonically) into an embedding of G/H into G. This corollary is not true for all finite groups, for instance Q 8 /±1} = Z/2Z Z/2Z but there is no subgroup of Q 8 which is isomorphic to this quotient group (there is only one element of order 2 in Q 8 ). Exercises. 1. What is the structure (as a direct product of cyclic groups) of the finite abelian groups whose nontrivial characters all have order 2? 2. Mimic the proof of Theorem 6.5 to decompose (Z/20Z) (of size 8) and (Z/45Z) (of size 24) into a direct product of cyclic groups as in Theorem 6.6.

25 CHARACTERS OF FINITE ABELIAN GROUPS (MATH 321, SPRING 2007) Show by an explicit counterexample that the following is false: if two subgroups H and K of a finite abelian group G are isomorphic then there is an automorphism of G which restricts to an isomorphism from H to K. 7. Pontryagin duality While G and Ĝ are isomorphic groups, there is no natural isomorphism between them. For instance, when G is cyclic the isomorphism we wrote down between G and Ĝ in the proof of Theorem 3.8 depends on a choice of generator, and there is not a unique generator for G (if #G > 2). Another indication that the isomorphism between G and Ĝ is not natural is that it breaks down for infinite G, e.g., the characters of Z (homomorphisms Z S 1 ) are parametrized by points on the unit circle as functions χ z, where χ z (n) = z n, so Ẑ = S 1 if dual groups for infinite abelian groups are defined in the right way. In particular, Ẑ = Z. We can map any finite abelian group G to its double-dual group Ĝ (the dual group of Ĝ) by sending each g G to the function evaluate at g, which associates to each χ Ĝ the number χ(g). Viewing evaluation at g as a character on Ĝ was met already in the proof of Corollary Theorem 7.1. Let G be a finite abelian group. The homomorphism G Ĝ associating to g G the function evaluate at g is an isomorphism. Proof. Since a finite abelian group and its dual group have the same size, a group and its double-dual group have the same size, so it suffices to show this homomorphism is injective. If g G is in the kernel then every element of Ĝ is 1 at g, so g = 1 by Theorem 3.3. Theorem 7.1 is called Pontryagin duality. This label actually applies to a more general result about characters on locally compact abelian groups. Finite abelian groups are a special case, where difficult analytic techniques can be replaced by counting arguments. The isomorphism between G and its double-dual group given by Pontryagin duality lets us think about any finite abelian group as a dual group (namely the dual group of Ĝ). For instance we can now say that the two parts of Corollary 4.2, as statements about all finite abelian groups G, are logically equivalent to each other since we can view G as a dual group by viewing its elements as the different characters of Ĝ. The isomorphism in Pontryagin duality is natural, i.e., it does not depend on any ad hoc choices along the way (unlike the isomorphism between a finite abelian group and its dual group). As an application of Pontryagin duality, recall that in Theorem 3.11 we saw in any finite abelian group G that g m = 1 χ(g) = 1 for every χ Ĝ such that χ = ψm for some ψ Ĝ.

26 26 CHARACTERS OF FINITE ABELIAN GROUPS (MATH 321, SPRING 2007) By Pontryagin duality this implies (in fact, it is equivalent to) the result χ m = ε χ(g) = 1 for every g G such that g = x m for some x G. Similarly, the dual version of Theorem 3.12 says that a character χ of G is an mth power of another character if and only if χ is trivial on every g G such that g m = 1. The set Hom(G 1, G 2 ) of all homomorphisms from the abelian group G 1 to the abelian group G 2 forms an abelian group under pointwise multiplication (Exercise 3.1). Pontryagin duality gives the following relation between two such Hom groups. Theorem 7.2. Let G 1 and G 2 be finite abelian groups. For any homomorphism f : G 1 G 2, set f : Ĝ2 Ĝ1 by f (χ) = χ f. Then f is a group homomorphism and the map sending f to f gives a group isomorphism Hom(G 1, G 2 ) = Hom(Ĝ2, Ĝ1). Proof. If f : G 1 G 2 is a homomorphism and χ Ĝ2, then for g and g in G 1 we have χ(f(gg )) = χ(f(g)f(g )) = χ(f(g))χ(f(g )), so f (χ) := χ f lies in Ĝ1. Thus we get the map from Hom(G 1, G 2 ) to Hom(Ĝ2, Ĝ1) as advertised. The reader can check (ff ) = f (f ) in Hom(Ĝ2, Ĝ1), so the map between Hom-groups is a homomorphism. The same construction gives a group homomorphism Hom(Ĝ2, Ĝ1) Hom(Ĝ 1,Ĝ 2 ). By Pontryagin duality this can be viewed as a homomorphism Hom(Ĝ2, Ĝ1) Hom(G 1, G 2 ) and the composite map Hom(G 1, G 2 ) Hom(Ĝ2, Ĝ1) Hom(G 1, G 2 ) turns out (after unwinding definitions, which is left to the reader) to be the identity function. Therefore our original map Hom(G 1, G 2 ) Hom(Ĝ2, Ĝ1) is a group isomorphism. More applications of Pontryagin duality are in the exercises. Exercises. 1. Let G be finite abelian and H G be a subgroup. (a) Viewing H = (H ) in G using Pontryagin duality, show H = H. (Hint: The inclusion in one direction is easy. Count sizes for the other inclusion.) (b) Show for each m dividing #G that #H G : #H = m} = #H G : [G : H] = m} by associating H to H and using a (fixed) isomorphism of G with Ĝ. (c) Show, for a prime p, that the number of subspaces of (Z/pZ) n with dimension d equals the number of subspaces with dimension n d.