GROUP ACTIONS AND FERMAT S LITTLE THEOREM

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1 GROUP ACTIONS AND FERMAT S LITTLE THEOREM Frank Blume John Brown University Siloam Springs, AR fblume@bu.edu phone: Abstract We use group actions to establish divisibilty properties that generalize Fermat s Little Theorem and thereby suggest an alternative route for presenting this material in undergraduate classrooms. 1 Introduction The standard formulation of Fermat s little theorem asserts that p n p n for all prime numbers p P and all n N. A proof of this statement is easily given using the theorem of Lagrange: if p n, then, trivially, p n p n; if not, then gcd(n, p = 1 and, by implication, n + pz (Z/pZ. Since n + pz =# n + pz p 1=#(Z/pZ, it follows that n p 1 + pz =1+pZ, and therefore p n p 1 1 n p n, as desired. An obvious generalization of this elementary result can be obtained by considering the multiplicative group (Z/p k Z for some k N. Arguing as above and observing that #(Z/p k Z = p k p k 1 = ϕ(p k (where we denote, as usual, by ϕ Euler s ϕ-function, i.e., ϕ(m :=#{k {1,...,m} gcd(m, k = 1}, we may infer that n pk p k 1 + p k Z =1+p k Z whenever gcd(n, p = 1. Furthermore, using again the case distinction p n versus gcd(n, p = 1, it is easy to see that ( ( n,k N p P gcd(n, p =1 p k n pk p k 1 1 n,k N p P p k n pk n pk 1. Thus we arrive at the following conclusion: 1

2 1.1 Theorem. For all n, k N and all p P it is the case that p k n p k n pk 1, and in particular, if p e 1 1 p e i i is the unique prime factorization of an integer m N {1}, then m (n pe n pe 1. for all n N. The purpose of the discussion that is to follow is to demonstrate how this result can be proven using a group-action construction and to further explore possible generalizations. 2 Group Actions and Divisibility Consider a wheel along which p colored discs are placed at equal angular intervals (see Figure 1. If p is a prime number (as in Figure 1 where p = 7, then the color pattern is nonperiodic, and every rotation by an angle 2πk/p with k {0, 1,...,p 1} rotate the colors Figure 1: rotating colors on a wheel. produces a different arrangement of colors in relation to the p fixed digits from 0 to p 1 at the wheel s center. The only exceptions to this rule are color patterns that are constant in the sense that all discs are filled with the same color. Put differently, we may therefore say that the operation of p-fold rotation partitions the set of all nonconstant color patterns into p-element subsets. Thus p divides n p n because n p n evidently equals the number of nonconstant patterns. In order to formalize and generalize the proof of Fermat s theorem thus obtained, we introduce the following definition: 2.1 Definition. For a given i N we denote by F (i the set of all functions σ : {1,...,i} {0, 1}, and for σ F (i we set δ(σ :=( 1 i σ(. 2

3 Furthermore, if p e 1 1 p e i i N {1}, then is, as above, the prime factorization of an integer m p(σ := p σ(. Finally, if A 1,...,A i are subsets of a given set S, then we define { A if σ( =1, B (σ := S if σ( = 0, for all {1,...,i} and set A(σ := i B (σ. 2.2 Theorem. If A 1,...,A i are finite subsets of a given set S, then i # A = δ(σ#a(σ. Remark. Theorem 2.2 generalizes the well known fact that #(A 1 A 2 =#A 1 +#A 2 #(A 1 A 2 and its proof is an easy exercise in mathematical induction. Furthermore, as a direct consequence we obtain the following familiar representation of Euler s function: ϕ(m =ϕ (p e 1 1 pe i i = ( p e p e 1. For if we set A := {k {1,...,m} p k} for all {1,...,i}, then ϕ(m =m # = m i A = m + δ(σ p(σ = m (1 1p = δ(σ#a(σ =m + ( p e p e 1. δ(σm p(σ (1 2.3 Definition. Given a nonempty set S and a group G, we set H(G, S :={f f is a map from G to S}. Furthermore, for a given x G we define r x : G G y yx. 3

4 2.4 Theorem. The map : G H(G, S H(G, S, (x, f x f := f r x is a group action. Proof. Since the map is clearly well defined, we only need to verify that x (y f = (xy f and e f = f. To do so we observe that (e f(z =(f r e (z =f(ze =f(z for all z G, and similarly (x (y f(z =((y f r x (z =(y f(zx =(f r y (zx = f(zxy =(f r xy (z =((xy f(z. 2.5 Theorem. For all f H(G, S the stabilizer of f in G is G f = {x G f y x f(y for all y G}. Equivalently, we may say that x G f if and only if the map is well defined. f x : G/ x S y x f(y Proof. Let x G f. Then f(yx =f(y for all y G and by implication f(yx 1 = f(yx 1 x = f(y. Iteration yields f(yx n =f(y for all n Z, or equivalently, f y x f(y as desired. Iff y x f(y for all y G, then in particular f(yx =f(y, or equivalently x f = f. Hence x G f. 2.6 Theorem. If G is a finite group and #S = n<, then #G ( n #G #{f H(G, S #G f > 1}. Proof. Since the orbit O(f of an element f H(G, S satisfies the equation #O(f = (G : G f =#G/#G f, it follows that #G f > 1 if and only if #O(f < #G. Since the orbits form a partition of H(G, S, we may infer that #G divides the number of elements f H(G, S for which #O(f =#G. Since the latter number is evidently equal to the total number of elements in H(G, S minus the number of elements f for which #O(f < #G, and since #H(G, S = n #G, we may conclude that #G ( n #G #{f H(G, S #G f > 1}, as desired. 2.7 Lemma. If G is a group of order m = p e 1 1 p e i i, then {f H(G, S #G f > 1} = {f H(G, S x G {e} y G f y x f(y} i { ( = f H(G, S x G x = p y G f y x f(y } 4

5 Proof. Since the first of these equations is a trivial rephrasing of Theorem 2.5, we only need to establish the second. To do so, it is sufficient to show that {f H(G, S #G f > 1} i { ( f H(G, S x G x = p y G f y x f(y } as the reverse inclusion is trivial. So let f H(G, S such that #G f > 1. Then there exists a z G {e} such that f y z f(y for all y G. Since z m and z > 1, there exists an x z such that x = p for some {1,...,i}. Furthermore, since x z, it follows that f y x f(y for all y G. 2.8 Lemma. For all elements x 1,...,x k G it is the case that k {f H(G, S x G f } = {f H(G, S y G f y x1,...,x k f(y}. Proof. Iff y x1,...,x k f(y for all y G, then, trivially, f y x f(y for all y G and all {1,...,k}, and therefore, x G f for all {1,...,k}, as desired. Let f be in H(G, S such that x G f for all {1,...,k}. Ifz y x 1,...,x k for some y G, then there exist integers m N, 1,..., m {1,...,k}, and n 1,...,n m Z such that z = yx n 1 1 x nm m. Thus z yx n 1 1 x n m 1 m 1 x m, and since x m G f, we may infer that ( f(z =f yx n 1 1 x n m 1 m 1. Proceeding in this manner iteratively, we arrive at the desired conclusion that f(z = f(y. 2.9 Theorem. If m = p e 1 1 pe i i, then m for all n N. δ(σn m/p(σ Proof. Let G be a cyclic group of order m (such as for instance Z/mZ and let S be a finite set of order n. Then, for every {1,...,i} there exists an element x G such that x = p. Since G is cyclic, the subgroup x is unique in the sense that x G x = p x = x. 5

6 Consequently, A := { ( f H(G, S x G x = p y G f y x f(y } = { f H(G, S y G f y x f(y } = {f H(G, S x G f }, and therefore Lemma 2.7 in conunction with Theorem 2.2 implies that i #{f H(G, S #G f > 1} =# A = δ(σ#a(σ = δ(σ# i B (σ, where { A if σ( =1, B (σ = H(G, S if σ( =0. Using Lemma 2.8, it follows that { } A(σ = f H(G, S y Gf y x f(y, σ(1 and since evidently we find that Hence #A(σ =#H 1,...,x σ(i i # x σ(1 1,...,x σ(i i = p(σ, ( x G/ x σ(1 1,...,x σ(i i,s = n #G/ σ(1 1,...,x σ(i i = n m/p(σ. #{f H(G, S #G f > 1} = and Theorem 2.6 therefore implies that m nm + δ(σn m/p(σ = δ(σn m/p(σ, δ(σn m/p(σ Corollary. For all n, k N and all p P it is the case that and in particular, if m = p e 1 1 pe i i, then m for all n N. p k n p k n pk 1, (n pe 6 n pe 1.

7 Proof. Using the statement of Theorem 2.9 with p k in place of m, we may infer that p k δ(σn m/p(σ = n pk n pk 1, σ F (1 and the statement of the corollary immediately follows. Remark. The most general form of the divisibilty statement derived from Lagrange s theorem when applied to Z/mZ (as explained in the introductory paragraphs asserts that m n ϕ(m 1 whenever gcd(n, m = 1 (this is known as Euler s theorem. Thus, using the representation of ϕ in (1, we have n,m N gcd(n, m =1 m n δ(σm/p(σ 1. This latter formulation reveals an interesting formal anology to the statement made in Theorem 2.9 as the terms and n δ(σm/p(σ = δ(σn m/p(σ are evidently structurally similar formally speaking. n δ(σm/p(σ 3 Open Questions for Undergraduate Research To conclude our discussion, we will now propose several avenues for further exploration that can perhaps be used as points of entry for undergraduate research proects. a According to the proof of Theorem 2.9, the number of functions f H(G, S for which #G f =1is δ(σn m/p(σ whenever G is a cyclic group of order m = p e 1 1 p e i i. If we consider instead of a cyclic group the k-fold product of Z/pZ with itself (for some p P, i.e., G = Z/pZ Z/pZ, }{{} k times then preliminary investigation reveals that #{f H(G, S #G f =1} n p2 n p p (n p n if k =2, = n p3 n p2 pa 2 (n p2 n p +p 3 (n p n if k =3, n p4 n p3 pa 3 (n p3 n p2 +p 3 a 3 (n p2 n p p 6 (n p n if k =4, 7

8 where a i := pi 1 p 1. Consequently, an interesting problem for further research might be to confirm and rigorously establish the pattern that these preliminary results appear to suggest. b In further generalization of Theorem 2.9 as well as of the problem proposed in a, we may ask what the number of functions f H(G, S with #G f = 1 turns out to be if G is an arbitrary finite abelian group, that is a group that is known to be a direct sum of finite cyclic groups. c Passing from abelian to nonabelian groups, it is natural to ask what the number of functions f H(G, S with #G f = 1 will be if G is a symmetric group, say, or a matrix group over Z/pZ. d How is the statement of Theorem 2.9 related to Euler s theorem? Is it weaker or stronger or equivalent? (Most likely, it isn t stronger, but which of the other two options is true may be a good question to ask. 8