Fourier Analysis on Finite Abelian Groups and Applications Summer Course Sponsored by GSM Tefjol Pllaha July 22, 2018 Contents 1 Toolbox 1 2 Characters of a Finite Abelian Group6 3 The Space L2 G 10 4 The Discrete( Fourier) Transform 12 4.1 Fast Fourier Transform..................................... 15 5 Discrete Uncertainty Principle 16 6 Error-Correcting Codes: MacWilliams Theorem 21 7 Quadratic Reciprocity Law 23 8 Graphs over Finite Abelian Groups 27 8.1 Four Questions about Cayley Graphs............................. 28 8.2 Random Walks in Cayley Graphs............................... 29 8.3 Hamming graphs......................................... 31 9 Solutions to Selected Exercises 32 10 Written Assignment 36 The intent of these notes is to facilitate going through the first part of the book \Fourier Analysis on Finite Groups and Applications" by Audrey Terras. Please read with caution and be aware of typos as they are unedited. 1 Toolbox This section will be roughly based on chapter one of the book. We will review the necessary machinery from modular arithmetic, and then give some motivation by drawing connections with analysis. Since we will be dealing almost exclusively with finite abelian groups, unless otherwise stated, any group will be finite and abelian. Let Z be the ring of whole numbers. For any natural number n N, denote Zn Z nZ. That is, elements of Zn are equivalence classes a b Z n b a . Two representatives of the same equivalence class are called congruent (modulo n), and in this case we∈ will write a ∶b=mod~ n. Then Zn a 0 a n is again a ring with multiplication∶= { ∈ S andS( − addition)} given by ≡ = { S ≤ < } a b a b and a b a b: (1.1) We will refer to the ring Zn as the finite+ ∶ circle= +. Convince⋅ ∶= yourself⋅ that this is an appropriate name. Throughout we will pay special attention to the finite circle for two major reasons. The first one is that every cyclic group of order n is isomorphic with the additive group of Zn. The second reason is the following. Theorem 1.1 (Fundamental Theorem of Finite Abelian Groups). Every finite abelian group is the direct product of some finite circles. ∗ We will denote Zn the group of units of Zn, that is, ∗ Zn a Zn b Zn such that a b 1 : (1.2) ∗ Proposition 1.2. a Zn ∶gcd= { a;∈ n S ∃1: ∈ ⋅ = } ∗ Proof. By the very definition∈ ⇐⇒ of the( group) = of units, a Zn iff there exists b Zn such that a b 1. The latter happens precisely when n 1 ab . In other words, precisely when there exists k Z such that 1 ab kn. But this is just the B´ezout'sIdentity∈1 for a and n. Thus gcd∈ a; n 1. ⋅ = S( − ) ∈ Theorem= + 1.3. Zn is a field iff n is prime. ( ) = ∗ Proof. Since Zn is a field, every nonzero element has a multiplicative inverse. That is Zn Zn 0 . Thus for all 1 a n, by Proposition 1.2 we have gcd a; n 1. This implies that n is prime. Now note that all the implications in the proof of the forward direction are actually equivalences.= −{ } ≤ < ( ) = Remark 1.4. Let p be a prime. Then Zp is a field by Theorem 1.3. In particular this implies that 2 Zp is a domain . However this can also be seen directly as follows. Assume a b a b 0. Thus p ab. Since p is prime it follows that p a or p b. Thus a 0 or b 0. ⋅ = ⋅ = AS Sweet Little Trick 1.5. Many thingsS getS simplified= a lot in= a finite world. One of the reasons is Exercise 1.20. Let's see Exercise 1.20 in action. Let R be a finite commutative ring, which in addition is also a domain. Fix 0 a R. Define the map ma R R; x ax. Note first that ma is injective. Indeed ma x ma y iff a x y 0. Since R is a domain and a 0 we may conclude x y. But then ma is also surjective.≠ ∈ Thus, for 1 R, there∶ existsÐ→ b Rz→such that ab ma b 1. In other words, every nonzero( ) = element( ) is( a− unit) = and thus R is a field. ≠ =1 ∈ ∈ = ( ) = B´ezout'sIdentity: gcd a; b d r; s Z such that d ra sb. 2a b 0 a 0 or b 0. ( ) = ⇐⇒ ∃ ∈ = + ⋅ = Ô⇒ = = 1 Theorem 1.6 (Chinese Reminder Theorem). Let n; m N be two natural numbers such that gcd n; m 1. Then Zn Zm Znm ∈ 3 Proof.( See) = Dummit and× Foote,≅ along with the observation that gcd n; m 1 iff the ideals nZ and mZ are comaximal. ( ) = Definition 1.7. The map φ Z Z given by φ ∶n Ð→ a 1 a n 1 and gcd a; n 1 ∗ Zn (By Proposition 1.2) ( ) ∶= S{ S ≤ ≤ − ( ) = }S is called Euler's function. = S S A useful tool for computing Euler's function is the following. Theorem 1.8. Euler's function is multiplicative. That is φ nm φ n φ m for all n; m N such that gcd n; m 1. ( ) = ( ) ( ) ∈ Proof. One( can) = prove the statement using elementary counting arguments. However, this is an immediate consequence of the definition and Theorem 1.6. Example 1.9. By the very definition of φ we have φ n n 1 iff n is prime; compare this with Theorem 1.3. Convince yourself that for any prime power we have ( ) = − 1 φ pk pk pk−1 pk 1 : (1.3) p ( ) = − = − k1 kr ki kj Let n p1 pr be written in its prime decomposition. Then gcd pi ; pj 1 for all i j, and thus by Theorem 1.8 we have = ⋯ ( ) = ≠ r 1 φ n φ pk1 φ pkr n 1 : (1.4) 1 r p i=1 i ( ) = ( )⋯ ( ) = M − The latter is called Euler's product formula. Exercise 1.10. Show that Euler's function satisfies n φ d φ ; (1.5) d dSn dSn Q ( ) = Q and then use (1.5) to show that dSn φ d n: Example 1.11 (Public Key Cryptography)∑ ( ) = . In here we will describe the RSA4 cryptosystem. Think of your to be sent message as a number m. Assume p q are primes, and fix t such that gcd t; φ pq 1 (where φ is Euler's function). The encryption of the message m is mt mod pq . Typically neither p nor q divide m. The pair t; pq is known to≠ the entire world and that is why the( name( \public".)) = So how can we recover m from mt mod pq ? That is, we are looking( for s such) that ( ) mts m mod(pq : ) (1.6) 3 This follows easily from the B´ezoutIdentity. ≡ ( ) 4RSA is the acronym for Rivest Shamir Adleman. − − 2 By making use of Exercise 1.21 it is not difficult to see that it suffices to find s such that ts 1 mod φ pq ; (1.7) By making use of Exercise 1.21 once again,≡ it( suffices( that)) s satisfies s tφ(φ(pq))−1 mod φ pq : (1.8) So in oder to compute s one must have≡ in hand (other( than( the)) publicly know t) φ pq p 1 q 1 which practically impossible due to the difficulty of prime factorization. Knowing the product of two large primes doesn't say anything about the primes, and thus, knowing pq is( harmless) = ( − (as)( − one) must know p and q for the decryption). Definition 1.12. The M¨obiusfunction is defined as 1; if n is the product of an even number of distinct primes, µ n 1; if n is the product of an odd number of distinct primes, ⎧ ⎪ 0; otherwise. ( ) ∶= ⎨ − ⎪ k1 kr ⎪ Let n p1 pr be⎩ written in its prime decomposition. It follows directly by the definition that µ n 0 iff there exists i such that ki 1. = ⋯ Theorem( ) = 1.13. > 1; if n=1; µ d 0; otherwise. dSn Q ( ) = k1 kr Proof. Let n p1 pr be written in its prime decomposition. Then r r = ⋯ µ d 1 i 1 1 r 0: i dSn i=0 Q ( ) = Q (− ) = ( − ) = By making use of Theorem 1.13 we can relate Euler's and M¨obiusfunctions as follows. First, convince yourself of the following identity: r 1 r 1 1 1 1 (1.9) p p p p i=1 i i=1 i 1≤i<j≤r i j M − = − Q + Q − ⋯ Now combine (1.4) and (1.9), and apply Theorem 1.13 to obtain φ n µ d : (1.10) n d ( ) dSn ( ) = Q Note that the right-hand-side of (1.10) gives the proportion of numbers smaller than n that are relatively prime with n, and has vast applications to number theory (especially with regards to the distributions of primes). Theorem 1.14 (M¨obiusInversion Formula). Let f and g be two functions defined for every natural number and assume that they satisfy f n dSn g d . Then g satisfies n g( n) = ∑ µ (d )f : (1.11) d dSn ( ) = Q ( ) 3 Proof. n n n µ d f µ f d µ g d′ d d d ′ dSn dSn dSn d Sd Q ( ) = Q gd′ ( ) = µQm Q ( ) ′ ′ d Sn mS(n~d ) = gQn (; ) Q ( ) since by Theorem 1.13 we have = ( ) 1; if d′ n; µ m 0; otherwise. m n d′ S( ~ ) = Q ( ) = Definition 1.15. Let f and g be defined for any natural number. The convolution f g is defined by n ∗ f g n f d g f a g b : (1.12) d dSn ab=n Exercise 1.16.