NOTES for MATH 300.1 – FALL 2010 UNIVERSITY of MASSACHUSETTS These Notes Are a Complement to the Textbook by Gilbert and Vanst

Home , Equivalence class, Kernel (set theory)

NOTES FOR MATH 300.1 – FALL 2010 UNIVERSITY OF MASSACHUSETTS

EDUARDO CATTANI

These notes are a complement to the textbook by Gilbert and Vanstone and to the notes by Farshid Hajir. They are in preliminary form and intended only for use by students in Math. 300.

1. Relations We begin by defining the notion of a relation in a set X. Definition 1.1. A relation in a set X is a subset R X X. ⊆ × If a pair (x1,x2) R we say that x1 is related to x2. The following examples∈ will be used throughout our discussion or relations. Example 1.2. Let X = a, b, c, d and R = (a, b), (b, a), (c, d), (d, a) . Then, a is related to b, b is related{ to a,} etc. { } Example 1.3. Let f : X X be a map. Then the graph of f is a subset of X X: → × Graph(f) := (x ,x ) X X : x = f(x ) { 1 2 ∈ × 2 1 } and consequently defines a relation in X: x1 is related to x2 if and only if x = f(x ). In this manner, every map f : X X defines a relation in X. 2 1 → Example 1.4. Let X = (Z); i.e. X is the set whose elements are the subsets P of Z. Let R := (A, B) X X : A B . { ∈ × ∼ } Then, A is related to B if and only if A is equivalent to B; that is, if and only if there exists a bijection f : A B. → Example 1.5. Again, let X = (Z), but now we define P R := (A, B) X X : A B . { ∈ × ⊆ } Then, A is related to B if and only if A is contained to B; that is, every element of A is also an element of B. Example 1.6. Let X = Z and let R be the relation R := (m, n) X X : m n . { ∈ × ≤ } In other words, m is related to n if and only if m is less than or equal to n. 1 2 EDUARDO CATTANI

Example 1.7. Same example as above but now let X = R and let R be the relation R := (x ,x ) X X : x x . { 1 2 ∈ × 1 ≤ 2} In other words, x1 is related to x2 if and only if x1 is less than or equal to x2. Example 1.8. We may modify Example 1.6 by taking X = Z and letting R be the relation R := (m, n) X X : m < n . { ∈ × } In other words, m is related to n if and only if m is strictly less than n.

Example 1.9. Let X = R and let 2 2 R := (x1,x2) R R : x1 + x2 =1 . { ∈ × } For example, for any α R, we have that cos α is related to sin α. ∈ Example 1.10. Let f : X Y be a map and deﬁne → R := (x ,x ) X X : f(x )=f(x ) . { 1 2 ∈ × 1 2 } Note that if f is injective then (x ,x ) R if and only if x = x . 1 2 ∈ 1 2 Example 1.11. Let X = Z>0 and consider the relation

R := (m, n) Z>0 Z>0 : m divides n . { ∈ × } Note that in this case (1,n) R for all n Z>0. ∈ ∈

Definition 1.12. A relation R X X is said to be reflexive if the following property holds: ⊆ × x X, (x, x) R; ∀ ∈ ∈ that is, for all x X, x is related to x. ∈ Definition 1.13. A relation R X X is said to be symmetric if the following property holds: ⊆ × x X, x X, ((x ,x ) R (x ,x ) R); ∀ 1 ∈ ∀ 2 ∈ 1 2 ∈ ⇒ 2 1 ∈ that is, if x1 is related to x2 then x2 is related to x1. We will also be interested in the case when the symmetry property holds in only the “stupid” case, when x1 = x2: Definition 1.14. A relation R X X is said to be antisymmetric if the following property holds: ⊆ × x X, x X, ((x ,x ) R and (x ,x ) R x = x ); ∀ 1 ∈ ∀ 2 ∈ 1 2 ∈ 2 1 ∈ ⇒ 1 2 that is, if x1 is related to x2 and x2 is related to x1, then x1 = x2. MATH 300 3

Definition 1.15. A relation R X X is said to be transitive if the following property holds: ⊆ × x X, x X, x X, ((x ,x ) R and (x ,x ) R (x ,x ) R); ∀ 1 ∈ ∀ 2 ∈ ∀ 3 ∈ 1 2 ∈ 2 3 ∈ ⇒ 1 3 ∈ that is, if x1 is related to x2 and x2 is related to x3, then x1 is related to x3. Let us now check which, if any, of the above properties are satisfied in each of our examples: i) The relation defined by Example 1.2 doesn’t satisfy any of the above properties. Indeed, it is not reflexive since, for example, (a, a) R. It is not symmetric because even though (c, d) R, the pair (d,)∈ c) R. It is also not antisymmetric since (a, b) and (b,∈ a) are elements of R)∈but a = b. Finally, it is not transitive because even though (c, d) and (d, a) are) in R, the pair (c, a) R. )∈ ii) Consider now the relation defined by Example 1.3: In this case, the properties of R will depend on the map f. For example, R is reflexive if and only if (x, x) R for all x X, that is, if x = f(x) for all x X which means that∈ f is the identity∈ map in which case, the other properties∈ hold for trivial reasons. (Why?). (For which maps f is the relation symmmetric?) iii) The relation defined by Example 1.4 is reflexive (A A for all A X), symmetric (if A B then B A) and transitive (if∼A B and B∈ C, then A C) but∼ is not antisymmetric∼ since ∼ ∼ ∼ A B and B A A = B. ∼ ∼ )⇒ iv) The relation defined by Example 1.5 is reflexive (A A for all A X), antisymmetric (if A B and B A, then A = B⊆) and transitive∈ (if A B and B C, then⊆ A C)⊆ but is not symmetric since ⊆ ⊆ ⊆ A B B A. ⊆ )⇒ ⊆ v) It is easy to check that the relations defined in Examples 1.6 and 1.7 are also reflexive, antisymmetric, and transitive but not symmetric. vi) An interesting issue arises when discussing Example 1.8. Clearly R is not reflexive or symmetric, but it is transitive. On the other hand, R is antisymmetric since it is always false that x1

vii) The relation defined by Example 1.9 is not reflexive ((1, 1) R), but it 2 2 2 2 )∈ is symmetric: if x1 + x2 = 1 then clearly x2 + x1 = 1. However it is not antisymmetric ((1, 0) and (0, 1) are in R but 1 = 0. Finally, we can see that it is not transitive either since (1, 0) and (0) , 1) are in R but (1, 1) is not. viii) Next we consider the relation in Example 1.10. The relation is reflexive, symmetric, and transitive but is not in general antisymmetric. ix) Finally, we consider the relation in Example 1.11. It is reflexive since every positive integer divides itself, but is not symmetric (2 divides 4 but 4 does not divide 2). In fact, it is anti-symmetric, if m divides n and n divides m, then n = m (Prove it!). It is also transitive.

Definition 1.16. A relation is said to be an equivalence relation if it is reflexive, symmetric, and transitive. A relation is said to be an order relation if it is reflexive, antisymmetric, and transitive. Thus, the relations in Examples 1.4 and 1.10 are equivalence relations, while the relations in Examples 1.5, 1.6, 1.7, and 1.11 are order relations. The other relations in our examples are neither equivalence nor order relations.

Notation: If R X X is an equivalence relation and (x1,x2) R we will usually write x ⊆ x .× ∈ 1 ∼ 2 2. Order Relations We will now consider some additional properties that order relations may have∗. We begin with some notation: If R X X is an order relation and (x ,x ) R we will usually write x x⊆. If×x x but x = x we 1 2 ∈ 1 * 2 1 * 2 1 ) 2 will write x1 x2. Of course, if the order relation are the ones defined in Examples 1.5,≺ 1.6, and 1.7 we will continue to use the usual notations , . Whenever we have an order relation on a set R we will say simply that⊆ we≤ have defined an order in X. Example 2.1. Let X = R2. We define an order in X by: (x ,y ) (x ,y ) 1 1 * 2 2 if and only if x1

∗We will study equivalence relations in more detail later. MATH 300 5

Clearly, the orders in Examples 1.6 and 1.7 are total orders, while the orders in Examples 1.5 and 1.11 are not total orders. For example, 1 and 2 are elements in X but neither 1 2 nor 2 1 . Similarly,{ } we{ } may consider the integers 3 and 7.{ Clearly,}⊆{ } neither{ 3} divides⊆{ } 7 nor 7 divides 3. As an exercise, you should determine whether the lexicographic order is a total order. Definition 2.3. A set X with a total order is said to be well-ordered if every non empty subset A of X has a smallest element.* That is, if the following condition holds for every subset A X, A = : ⊆ ) ∅ m A, (x A m x). ∃ ∈ ∈ ⇒ * It is generally a relatively easy problem to determine that a given set with a total order is not well-ordered. For example, Z with the usual order is not ≤ well ordered since Z itself does not have a minimal element. Indeed, suppose m Z was a minimal element, then m 1 0 with the usual order is not well-ordered either since, again, R>0 has no minimal element. On the other hand, it seems intuitively clear that Z>0 = Z>0 with the usual order is a well-ordered set. Clearly, 1 is the minimum element of Z>0 and, given any non-empty subset A Z>0 we could check one-by-one, all the positive integers to see whether they⊆ were elements of A. Since A is non empty we should eventually find a smallest positive integer belonging to A and that would be the smallest element of A. Unfortunately that one-by-one process is not really well-defined and, in fact, defining it properly turns out to be equivalent to well-ordering. In fact, it is not possible to prove that Z>0 is well-ordered simply from the basic properties of natural numbers and it is one of the basic axioms (Peano axioms) that define the natural numbers.

Axiom (Well-ordering): Every non-empty subset A Z>0 has a smallest element. ⊆

We can use this axiom to prove a very nice result that goes back to Pithago- ras: Theorem 2.4. There is no rational number p/q such that (p/q)2 =2. In other words, the square root of 2 is not a rational number. Proof. We prove this result by contradiction. Suppose there exists p/q such that p/q = √2 . We may assume without loss of generality that p and q are positive integers (Why?). Then, the set

A := q Z>0 : q √2 Z>0 { ∈ · ∈ } 6 EDUARDO CATTANI is not empty. Hence, since Z>0 is well-ordered, A has a smallest element, which will be called m. Consider then the number k := m (√2 1). · − The number k is an integer since k = m √2 m and both m √2 and m are integers, and k>0 since m>0 and √2· 1 >− 0(Why?). We· also have that − k √2 Z>0. Indeed, · ∈ k √2=m (√2 1) √2=m (√2)2 m √2 = 2m m √2 · · − · · − · − · and both 2m and m √2 are integers. Hence, k A. But, we claim that k < m and this will contradict· the assumption that m∈was the smallest element of A. In fact, k = m (√2 1) and √2 1 < 1(Why?) and therefore k < m. · − − ! 3. Induction Principles The well-ordering axiom allows us to prove some crucial properties of the natural numbers:

Theorem 3.1 (Induction Principle). Let A Z>0 be a subset such that ⊆ i) 1 A ii) If∈n A then n +1 A. ∈ ∈ Then A = Z>0.

Proof. We prove the theorem by contradiction. Suppose A = Z>0 then the complement of A, B = Ac is a non-empty set. By the well-ordering) axiom B has a smallest element m B. That smallest element cannot be 1 since 1 A. ∈ ∈ Therefore m 1 Z>0 and since m is the smallest element of B we must have m 1 B, that− ∈ is m 1 A. But then our second assumption about A says that− )∈ − ∈ m =(m 1) + 1 A − ∈ which contradicts the assumption that m B. ∈ ! Remark: We have taken the Well-Ordering as an axiom and used it to prove the Principle of Induction. In fact, these two statements are equivalent and we could take the Principle of Induction as an axiom and deduce from it the fact that Z>0 is well-ordered. We can use very similar arguments to those in the proof of the Induction Principle to obtain two other induction methods which will be very useful in the sequel.

Theorem 3.2 (Generalized Induction Principle). Let A Z>0 and suppose that ⊆ i) s A ii) If∈n s and n A then n +1 A. ≥ ∈ ∈ MATH 300 7

Then n Z>0 : n s A. { ∈ ≥ }⊆ Proof. The proof is almost identical to that of the previous theorem. Except that now B = n Z>0 : n s and n A . { ∈ ≥ )∈ } Then, if the conclusion of the theorem doesn’t hold, B = and it has a minimal element m. Again, m s since m B but m = s. Therefore) ∅ m 1 s but once again: ≥ ∈ ) − ≥ m =(m 1) + 1 A − ∈ which contradicts the assumption that m B. ∈ ! We note that the Induction Principle is just the Generalized Induction Prin- ciple with s = 1.

Theorem 3.3 (Strong Induction Principle). Let A Z>0 be a subset such that ⊆ i) 1 A ii) If∈1, 2, ,n A then n +1 A. { ··· }⊆ ∈ Then A = Z>0.

Proof. We again argue by contradiction. Suppose A = Z>0 then the complement of A, B = Ac is a non-empty set. By the well-ordering) axiom B has a smallest element m B. That smallest element cannot be 1 since 1 A. Since m is the smallest element∈ of B, none of the elements 1, 2, ,m 1∈ B since they are smaller than m. Hence, 1, 2, ,m 1 A···and by− assumption∈ (ii), we have that { ··· − }⊆ m =(m 1) + 1 A − ∈ which contradicts the assumption that m B. ∈ ! The induction principles are one of the most useful tools to prove mathematical statements depending on a positive integer n. For example, the general strategy to apply the Induction Principle is the following: Suppose that we have a statement P (k) depending on k Z>0. We consider the set ∈ A = k Z>0 : P (k) is true . { ∈ } If we can then prove that: i) 1 A; i.e. P (1) is true, and ii) If∈n A then n +1 A; i.e. if the statement is true for n then it is true∈ for n + 1, ∈ then A = Z>0 and P (k) is true for all k Z>0. ∈ Important Remark: Note that you do not need to prove that P (n) is true. What you have to do is assume that P (n) is true and using that assumption 8 EDUARDO CATTANI prove that P (n+1) is true. The assumption that P (n) is true is usually called the inductive assumption or the inductive hypothesis. Let us illustrate this strategy with several examples Example 3.4. Let us use induction to prove that: k k(k + 1)(2k + 1) j2 = . 6 j=1 ! We begin by checking that the statement is true for k = 1. This is clear since in this case the left-hand side is equal to 1 and the right-hand side is 1 2 3/6 = 1. Suppose now that the statement is true for n. Using this inductive· · assumption we will now prove that the statement is true for n + 1: n+1 j2 = 1 + 22 + + n2 +(n + 1)2 ··· j=1 ! = (1 + 22 + + n2)+(n + 1)2 ··· n(n + 1)(2n + 1) = +(n + 1)2 6 n(n + 1)(2n + 1) + 6(n + 1)2 = 6 (n + 1)(2n2 + n) + 6(n + 1)2 = 6 (n + 1) ((2n2 + n) + 6(n + 1)) = 6 (n + 1) (2n2 +7n + 6) = 6 (n + 1) ((n + 2)(2n + 3)) = 6 (n + 1)(n + 2)(2(n + 1) + 1) = 6

Therefore, by the Induction Principle, the statement is true for all n Z>0. Where did we use the inductive hypothesis? Make sure you under-∈ stand every step in the calculations. Example 3.5. Note that as a corollary of the above result we have that for every k Z>0: ∈ 6 divides k(k + 1)(2k + 1). We can prove this result directly by induction. Clearly the statement is true for k = 1 since in this case k(k+1)(2k+1) = 6. Suppose then, that the statement is true for k = n and let us consider the case MATH 300 9 k = n + 1. We have (n + 1)(n + 2)(2(n + 1) + 1) = (n + 1)((n + 1) + 1))((2n + 1) + 2) = n(n + 1)(2n + 1) + n(2n + 1) + 2n(n + 1) + 2n +(n + 1)(2n + 1) + (2n + 1) + 2(n + 1) + 2 = n(n + 1)(2n + 1) + 6n2 + 12n +6. The first term in the last sum is divisible by 6 by inductive hypothesis and the remaining terms are clearly divisible by 6. Hence the statement is true for k = n + 1. Example 3.6. Let X be a finite set. Let us prove by induction that X (X) =2| |. |P | Let X = k. Then our statement P (k) will be | | P (k): If X = k then (X) =2k. | | |P | The statement is true if k = 1. In this case, X consists of a single element x and therefore, (X)= , x has 2 elements. Suppose now thatP the statement{∅ { }} is true for sets of cardinality n and suppose X is a set with n + 1 elements; i.e. X = x , ,x ,x . { 1 ··· n n+1} We then decompose the set (X) into two disjoint subsets: P = A (X):x A Q0 { ∈P n+1 )∈ } = A (X):x A Q1 { ∈P n+1 ∈ } Clearly, (X)= P Q0 ∪Q1 and 0 1 = .(Does 0 or 1?) ThereforeQ ∩Q ∅ ∅∈Q ∅∈Q (X) = + . |P | |Q0| |Q1| Now, we can compute 0 and 1 using the inductive hypothesis: The elements of are all the|Q subsets| of|Q the| set x , . . . , x ; that is, Q0 { 1 n} = ( x , . . . , x ) Q0 P { 1 n} n and, by inductive hypothesis, 0 =2 . On the other hand, we may|Q define| a bijection F : ( x , . . . , x ) P { 1 n} →Q1 by F (A)=A xn+1 , for any A ( x1, . . . , xn ). Why is F a bijection? Therefore, since∪{ ( x} , . . . , x ) ∈P {we must have} P { 1 n} ∼Q1 ( x , . . . , x ) = |P { 1 n} | |Q1| 10 EDUARDO CATTANI and, by inductive hypothesis =2n. |Q1| Consequently, (X)= =2n +2n =2 2n =2n+1. P Q0 ∪Q1 · Example 3.7. Using a very similar arguments we can count the set Sk of all permutations of 1, ,k . We claim that { ··· } S = k! := 1 2 3 k. | k| · · ··· Clearly, there exists only one bijection from the set 1 to itself and, therefore { } S1 = 1 = 1!. So our statement is true for k = 1. Suppose now that the statement| | is true for n and let us prove it for k = n + 1. Let us define a partition† of Sn+1 into n + 1 disjoints subsets Ti, i =1, 2, . . . , n + 1: T := σ S : σ(n + 1) = i . i { ∈ n } Clearly S = T T T , and the sets T are pairwise disjoint; that is, n+1 1 ∪ 2 ···∪ n+1 i i = j T T = Why?. ) ⇒ i ∩ j ∅ Therefore S = T + T + + T . | n+1| | 1| | 2| ··· | n+1| We now apply the inductive hypothesis to compute Ti . Notice that Tn+1 consists of those permutations that leave fixed the element| | n + 1, hence they only permute the first n elements. This means that Tn+1 Sn and therefore, by the inductive hypothesis, T = n!. On the other hand,∼ we can define a | n+1| bijection from Ti to Tn+1 by: F : T T ; F (σ)=τ σ, i i → n+1 i i,n+1 ◦ where τi,n+1 is the transposition that exchanges i and n + 1. (Why is Fi a bijection? Compute its inverse.) This means that for every i, we have T = T = n! and | i| | n+1| S = T + T + + T =(n + 1) n! = (n + 1)! | n+1| | 1| | 2| ··· | n+1| · We prove now that the “pigeon-hole” principle which we used in our discussion of cardinality follows from the Induction Principle. Theorem 3.8 (Pigeon-hole Principle). If f : 1, . . . , n 1, . . . , m is a map and n > m then f is not injective. In other{ words} if→ we{ try to place} n pigeons into m holes and n > m then some hole will contain at least two pigeons.

†A partition of a set X is a collection of pairwise disjoint, non-empty subsets whose union is X. MATH 300 11

Proof. We prove this by induction on m. That means that we consider the set A consisting of all positive integers m for which the pigeon-hole principle holds for maps from 1, . . . , n to 1, . . . , m , n > m. Clearly 1 A for if we have a map f : 1, . . . ,{ n 1} and{ n>1 then} 1, 2 1, . . . , n ∈and we must have f(1) = f{(2) = 1,} so→f {is} not injective. ∈{ } We assume now that the result holds for maps f : 1, . . . , n 1, . . . , m , n > m and prove that it holds for maps f : 1, . . .{ , n }1→, .{ . . , m +1}, n > m + 1. Suppose f is such a map. We consider{ four} cases:→{ } i) Suppose there exist i, j 1, . . . , n , i = j and f(i)=f(j)=m + 1. Then, f is not injective∈ and{ we are} done.) ii) Suppose there is no i 1, . . . , n such that f(i)=m + 1. Then we may regard the map f∈as{ a map} from 1, . . . , n to 1, . . . , m and, since n > m +1>m, it follows from the{ inductive} hypothesis{ } that f is not injective. iii) Suppose that f(n)=m + 1 and that there is no other element in 1, . . . , n with that property. Then, we may regard f as a map { } f : 1, . . . , n 1 1, . . . , m { − }→{ } and since n>m+ 1, n 1 >mand then it follows by the inductive hypothesis f may not be− injective. iv) Suppose there exists a unique i 1, . . . , n , i = n, such that f(i)= m + 1. Let τ S be the transposition∈{ that} ) exchanges i and n in in ∈ n 1, . . . , n . Then the map g = f τin satisﬁes g(n)=m + 1. Hence, we are{ in the} situation of case (iii) and◦ g is not injective. But this implies that f is not injective. (Why?). ! We will now give an example where we need to use the Generalized Induction Principle. Example 3.9. Suppose we want to compare 2k and k2. We compute the ﬁrst few values:

k 2k k2 1 2 2 2 4 4 3 8 9 4 16 16 5 32 25 6 64 36 7 128 49 8 256 64 12 EDUARDO CATTANI

From these values it seems reasonable to conjecture that 2k k2 if k 4. We prove this using the Generalized Induction Principle. Let ≥ ≥ k 2 A := k Z>0 :2 k . { ∈ ≥ } Then, the table shows that 4 A. Suppose now that n 4 and n A, that is 2n n2. We want to prove∈ that then 2n+1 (n + 1)2.≥ Now, ∈ ≥ ≥ 2n+1 =22n 2 n2. · ≥ · But, 2 n2 =(n + 1)2 + n2 2n 1 = (n + 1)2 +(n 1)2 2 (n + 1)2 · − − − − ≥ since n 4 so n 1 3 and (n 1)2 2 0. The Generalized Induction ≥ − ≥ − − ≥ Principle now implies that A = k Z>0 : k 4 . Make sure you understand each step in the proof{ and∈ that you≥ } can point out the place where the inductive hypothesis was used. We conclude this section with an example where we apply the Strong In- duction Principle. Recall that an integer p>1 is said to be prime if it is only divisible by 1 and by itself. Theorem 3.10. Every integer k>1 is divisible by a prime number. Proof. We will prove this result by Strong Induction. The result is true for k = 2 since 2 is prime and divides itself. We assume now that the result is true for all integers k such that 2 k n and we prove that it is true for n + 1. If n + 1 is prime then we are≤ done≤ since n + 1 is divisible by itself. Suppose then that n + 1 is not prime. Then there exists some integer k, diﬀerent from 1 and n + 1 that divides n + 1. In other words we can write n + 1 = k $ ;1

4. Equivalence Relations In this section we study in more detail equivalence relations on a set X. We recall that a relation R X X is called an equivalence relation if R is reflexive, symmetric, and transitive⊆ × (see Definitions 1.12, 1.13, and 1.15). Usually we will denote equivalence relations by . Examples 1.4 and 1.10 are key examples of equivalence∼ relations. In fact, we will show in this section that every equivalence relation on X is, in an abstract sense, given by a surjective map f : X Y to a suitable space Y . In order to make this more precise we introduce→ the important notion of equivalence classes. Definition 4.1. Let be an equivalence relation on X, then the equivalence class of x X is the set:∼ ∈ x" X : x" x X. { ∈ ∼ }⊆ There are many standard notations for the equivalence class of x. The most common are [x] and Cl(x). Proposition 4.2. Let be an equivalence relation on X. Then, for every x X, the set [x] = . ∼ ∈ ) ∅ Proof. This follows from the reflexive property of , since x [x] for all x X. ∼ ∈ ∈ ! Theorem 4.3. Let be an equivalence relation on X and x, x" X. Then the following three statements∼ are equivalent: ∈

i) x x". ∼ ii) [x]=[x"]. iii) [x] [x"] = . ∩ ) ∅ Proof. In order to prove that (i) (ii) (iii), it suﬃces to prove that ⇔ ⇔ (i) (ii) (iii) (i). ⇒ ⇒ ⇒ We prove then, these three implications. Suppose (i) holds, that is x x". Then, if z [x], we have z x, but since ∼ ∈ ∼ by assumption x x", it follows by transitivity of that z x" and, therefore, ∼ ∼ ∼ z [x"]. Hence [x] [x"]. Since is symmetric we also have x" x and we ∈ ⊆ ∼ ∼ can then repeat the argument to obtain that [x"] [x]. Hence, [x]=[x"] as asserted by (ii). ⊆ It is clear that (ii) (iii) since if [x] = [x"], then [x] [x"] = [x] = by Proposition 4.2. ⇒ ∩ ) ∅ Finally, let us assume that (iii) holds, i.e. [x] [x"] = . Let z [x] [x"]. ∩ ) ∅ ∈ ∩ Then, since z [x], we have that z x and, since z [x"] we have that z x". ∈ ∼ ∈ ∼ Therefore, by symmetry and transitivity of we have that x x" which is ∼ ∼ what we wanted to prove. ! 14 EDUARDO CATTANI

If we put together Proposition 4.2 and Theorem 4.3 we see that the equivalence classes of an equivalence relation deﬁne a partition of X. That is a collection of non-empty subsets which are pairwise disjoints and whose union is the total space X. We will denote by X/ the set of all equivalence classes of . That is, ∼ ∼ X/ = C X : C =[x] for some x X . ∼ { ⊂ ∈ } We call X/ the quotient of X by the equivalence relation . If C X/ and C =[x]∼ we refer to x as a representative of the equivalence∼ class C∈. Thus,∼ any element in C is a representative of C.

Example 4.4. Let X = R2 and deﬁne the equivalence relation 2 2 2 2 (x, y) (x",y") x + y =(x") +(y") . ∼ ⇔ Then, given (a, b) R2, the equivalence class ∈ [(a, b)] = (x, y) R2 : x2 + y2 = a2 + b2 . { ∈ } In other words, the equivalence class of the point (a, b) consists of all the points in the circle centered at the origin and passing through (a, b). Of course, if (a, b) = (0, 0) then [(0, 0)] = (0, 0).

Example 4.5. Let X = R2 and deﬁne the equivalence relation 2 2 2 2 (x, y) (x",y") x y =(x") (y") . ∼ ⇔ − − Then, given (a, b) R2, the equivalence class ∈ [(a, b)] = (x, y) R2 : x2 y2 = a2 b2 . { ∈ − − } To understand this equivalence class we must consider three diﬀerent cases a2 b2 > 0, a2 b2 < 0, and a2 b2 = 0. Suppose− a2 −b2 = k2 > 0, then− the equivalence class is given by − x2 y2 (x, y) R2 : =1 { ∈ k2 − k2 } which is a hyperbola with axis the x-axis and vertices at ( k, 0). If a2 b2 = k2 < 0, then the equivalence class is given± by − − y2 x2 (x, y) R2 : =1 { ∈ k2 − k2 } which is a hyperbola with axis the y-axis and vertices at (0, k). On the other hand, if a2 b2 = 0, then the equivalence class± is given by − (x, y) R2 : x2 y2 =0 { ∈ − } which is the union of the two lines y = x and y = x. − MATH 300 15

Example 4.6. Let U = u1, . . . , un be a finite set and X = (U). Consider the equivalence relation in{ X: } P A B if and only if there is a bijection from A to B. Since∼ A and B are finite sets, we know that A B if and only if A = B . Therefore, the equivalence class of a set A U∼consists of all the| | subsets| | B U with the same number of elements as ⊆A. ⊆ [A]= B U : A = B . { ⊆ | | | |} For example, if U = 1, 2, 3 , then there are 4 equivalence classes in X = (U). One is the equivalence{ class} of and [ ]= . The other equivalence classesP consist of the sets with 1, 2, and∅ 3 elements,∅ ∅ respectively. Namely: 1 , 2 , 3 ; 1, 2 , 1, 3 , 2, 3 ; 1, 2, 3 . {{ } { } { }} {{ } { } { }} {{ }} These three examples are particular instances of a general situation which we describe in the following Theorem: Theorem 4.7. Let be an equivalence relation on X and suppose there exists a surjective map f :∼X Y such that → x x" f(x)=f(x"). ∼ ⇔ Then, there is a bijection Φ between the set of equivalence classes of and Y defined by: ∼ Φ(C)=f(x), where C is an equivalence class and x is any point in C. Moreover, for every y Y , the set ∈ ←−f (y) := x X : f(x)=y { ∈ } is an equivalence class of and the inverse map of Φ is defined by: ∼ Ψ(y)=←−f (y). Proof. We begin by showing that Φis well-defined; that is, that it does not depend on the choice of representative x in the equivalence class C. But this is clear since if x, x" C, then x x" but this implies that f(x)=f(x"). So Φis well defined. ∈ ∼ Next we show that for each y Y , ←−f (y) is an equivalence class in X. By assumption, f is surjective so there∈ exists a X such that f(a)=y. We will ∈ prove that ←−f (y) = [a]. Indeed,

←−f (y)= x X : f(x)=y = x X : f(x)=f(a) { ∈ } { ∈ } = x X : x a =[a]. { ∈ ∼ } Therefore Ψdeﬁnes a map Ψ: Y X/ . → ∼ 16 EDUARDO CATTANI

To show that Φis a bijection it suffices to show that, as asserted,Ψ is its inverse. Suppose C X/ and let a C, i.e C =[a]. Then ∈ ∼ ∈ Ψ(Φ(C)) = Ψ(f(a)) = ←−f (f(a)) = x X : f(x)=f(a) { ∈ } = x X : x a =[a]=C. { ∈ ∼ } Therefore Ψ Φ: X/ X/ is the identity map. Conversely,◦ let y ∼→Y and let∼ us compute Φ(Ψ(y)). Let a X be such that ∈ ∈ f(a)=y. Then Ψ(y)=←−f (y) = [a] and Φ(Ψ(y)) = Φ([a]) = f(a)=y. Then Φ Ψ= id and Φand Ψare inverse maps. ◦ Y ! Definition 4.8. Let be an equivalence relation in X and let X/ be the quotient space. The surjective∼ map ∼ π : X X/ → ∼ defined by π(x) = [x] is called the natural projection from X to the quotient space X/ . ∼ While the notion of the quotient space is a very abstract notion, it allows us to study all equivalence relations as the equivalence relation associated with a surjective map f : X Y , namely given any equivalence relation on X, we take Y = X/ and→ f = π. We may, on the other hand, have∼ a natural realization of the∼ equivalence relation as the equivalence relation of a surjective map f : X Y , in that case, we∼ may let Y replace the quotient space X/ , since Theorem→ 4.7 tells us that the two spaces are bijectively equivalent,∼ and let f replace the natural projection π. In this case we think of Y as a concrete realization of the quotient space X/ . We can obtain such concrete realizations in Examples∼ 4.4, 4.5 and 4.6. In the first case we have 2 2 2 f : R R 0 ; f(x, y)=x + y . → ≥ 2 Then we can say that R / ∼= R 0. To each r R 0 we associate the equivalence class ∼ ≥ ∈ ≥ (x, y) R2 : x2 + y2 = r { ∈ } that is the circle of radius √r centered at the origin. Similarly, in the second example, we consider the surjective map f : R2 R ; f(x, y)=x2 y2. → − In Example 4.6 we may consider the surjective map card : ( u , . . . , u ) 0, 1, . . . , n P { 1 n} →{ } defined by card( ) = 0 and card(A)= A if A = . Hence, the quotient space (U)/ is bijectively∅ equivalent to 0|, 1|, . . . , n) .∅ P ∼ { } MATH 300 17

Remark: It is important to note that a quotient space X/ may have diﬀerent concrete realizations. For example, consider again Example∼ 4.4. Then, we may consider the surjective map

2 x2+y2 g : R R 1 ; g(x, y)=e . → ≥ Clearly (x, y) (x",y") g(x, y)=g(x",y"). Therefore X/ = R 1. On the ∼ ⇔ ∼ ∼ ≥ other hand, it follows from Theorem 4.7 that there is a bijection between R 0 ≥ and R 1. This is given explicitly by: ≥ t h: R 0 R 1 ; h(t)=e . ≥ → ≥ So, all concrete realizations of a quotient space are bijectively equivalent.

Another situation when we can obtain a concrete realization of the quotient space is when we have a natural way to choose a particular representative in each equivalence class. In this case we can choose as Y the subset of these special representatives. We illustrate this in the following two examples, the second of which will be studied in more detail later on. Example 4.9. Consider again Example 4.5. We can then choose a distinguished representative in each equivalence class [(a, b)]: if a2 b2 = c2 > 0 we choose as representative the point ( c , 0); if a2 b2 = c2 <−0 we choose as representative the point (0, c ), and| if| a2 b2 =− c2 =− 0 we choose as representative the origin (0, 0). The| | concrete realization− − Y in this case consists of the origin and the two positive axes.

Example 4.10. Let X = Z, and let q Z>1. Define ∈ m n m n = q k, for some k Z. ∼ ⇔ − · ∈ We want to choose a distinguished representative in each equivalence class [m]. We note first of all, that for each equivalence class C X/ , ∈ ∼ C Z 0 = . ∩ ≥ ) ∅ Indeed, let m C, if m Z 0 we are done and if m<0 then (q 1) m = m + q m C∈ and, clearly∈ ≥ (q 1) m > 0. We then choose as− our·| special| ·| |∈ − ·| | representative for C the smallest element r in C Z 0 (the existence of r is ∩ ≥ guaranteed by the well-ordering of Z 0 which follows from the well-ordering of ≥ Z>0). We note that r must always be between 0 and q 1. Indeed, r 0 by definition but, on the other hand, if r q then r q −r and r q 0≥ which ≥ − ∼ − ≥ contradicts the assumption that r was the smallest element in C Z 0. This means that the set Y = 0, 1, 2, 3,q 1 is a concrete realization∩ ≥ of the quotient space X/ and f : X { Y is the− map} that assigns to an equivalence class C its special∼ representative→r Y . ∈ 18 EDUARDO CATTANI

Remark: The argument above allows us to prove the Division Theorem, namely:

Theorem 4.11 (Division Theorem). Let p Z and q Z>1. Then we can write p in a unique way as ∈ ∈ p = k q + r ; with 0 r q 1. · ≤ ≤ − Proof. It follows from Example 4.10 that we can write p = k q + r ; with k, r Z, 0 r q 1. · ∈ ≤ ≤ − We need to show that this expression is unique. Suppose we have p = k" q +r" · with k",r" Z, 0 r" q 1. Since r p and r" p, it follows that r r", ∈ ≤ ≤ − ∼ ∼ ∼ but since they are both between 0 and q 1 this is possible if and only if r = r". − But then, k q = k" q and since q = 0 this implies k = k". · · ) ! We will return to the Division Theorem later in the course.

5. Countable and Denumerable Sets The next two sections may be considered as an in-depth study of the equivalence relation given by the existence of a bijection. Suppose we consider again Example 4.6 but now U = Z>0 = Z>0. If A is a finite subset of Z>0 then [A] is again the collection of finite subsets of Z>0 of the same cardinality as A. On the other hand, we must also consider the equivalence class of infinite subsets of Z>0 such as, for example, [Z>0] itself. In other words, we ask ourselves: what are the subsets of Z>0 which are bijectively equivalent to Z>0? We have already seen, for example, that the map f(m)=2m defines a bijection between Z>0 and the even positive integers. Similarly g(m) = 2m 1 − is a bijection from Z>0 to the odd positive integers. Therefore, those three sets are in the same equivalence class. More generally we define Definition 5.1. A set A is said to be denumerable if and only if there is a bijection f : Z>0 A. A set is said to be countable if it is either denumerable or finite. A set is→ said to be uncountable if it is not countable.

So, we may now say that the equivalence class of Z>0 in (Z>0) consists of P all denumerable subsets of Z>0. But, of course, giving these sets a name is not a great sign of progress! The following theorem, on the other hand, represents a real advance:

Theorem 5.2. Let A Z>0 be a non-empty subset. Then A is countable. ⊆ Proof. Suppose A is not a ﬁnite subset, then we need to deﬁne a bijection

f : Z>0 A. → MATH 300 19

We define f recursively using the Principle of Strong Induction. First we define f(1) as the smallest element in A (such element exists by the well-ordering of Z>0). Suppose now that we have defined f(1), . . . , f(n). Then we define f(n + 1) as the smallest element in B = A f(1), . . . , f(n) . n \{ } Note that Bn = since, otherwise, A would be finite. We need to) show∅ that f is a bijection. First we show that f is 1 : 1. Suppose i = j and f(i)=f(j). We may assume without loss of generality that i < j, but) then by construction f(j) A f(1), . . . , f(i), . . . , f(j 1) . ∈ \{ − } which contradicts the fact that f(i)=f(j). Finally, let us prove that f is surjective. Let r A. Then C = A 1, . . . , r is a finite set. Let c = C , there are exactly c ∈1 elements in A smaller∩{ than} | | − r, so r = f(c). !

This remarkable Theorem says that all inﬁnite subsets of Z>0 are bijectively equivalent. In other words, the quotient space (Z>0)/ , where is bijective equivalence may be concretely realized as P ∼ ∼

(Z>0)/ = Z 0 0 , P ∼ ∼ ≥ ∪{ℵ } where for each k Z>0 we get the equivalence class of finite subsets of Z>0 of cardinality k, the∈ empty set corresponds to 0, and corresponds to the ℵ0 equivalence class of infinite subsets of Z>0. Remark: is a letter in the hebrew alphabet called aleph used to denote the cardinalityℵ (i.e. the equivalence class with respect to bijective equivalence) of infinite sets. There are cardinals denoted , , etc. ℵ1 ℵ2 Corollary 5.3. A subset of a countable set is countable. Proof. Let A be a countable set and B A. If A is finite then B is finite. ⊆ Suppose then that A is denumerable, then there exists a bijection f : A Z>0. → Let g : B Z>0 be the restriction of f to B. Then g is a 1 : 1 map and, by → Theorem 5.2, is a countable set. Hence, by transitivity, B is countable. !

We can use Theorem 5.2 and its Corollary to prove properties of countable sets that generalize the fact that the union and cartesian product of ﬁnite sets are ﬁnite: Theorem 5.4. Let A be a denumerable set and B a countable set. i) A B is denumerable. ii) A ∪ B is denumerable. × 20 EDUARDO CATTANI

Proof. We begin with (i): Since A is denumerable there exists a bijection f : A Z>0. Since B is countable there exists an injective map g : B N (Why?→) → Let h: A B Z>0 by ∪ → 2f(x), if x A h(x)= ∈ 2g(x)+1, if x B A. " ∈ \ We claim that h is injective. Indeed, suppose h(x)=h(y). Then either h(x)=h(y) is even or odd. If even, we have that both x, y A and h(x)= 2f(x)=2f(y)=h(y). Then f(x)=f(y) and since f is injective∈ it follows that x = y. The odd case is completely similar. Since the image of h contains all the even natural numbers (Why?) it is an infinite set and therefore denumerable. Since A B is bijectively equivalent to a denumerable set, it is denumerable. We will∪ now prove (ii). We use again a trick as in the proof of (i). Let f : A Z>0 and g : B Z>0 be bijections and define → → f(a) g(b) h: A B Z>0 ; h(a, b)=2 3 . × → · The fact that h is 1 : 1 follows from the unique factorization of integers. But since we have not proved that theorem yet, let us sketch a proof without using that result: Suppose h(a, b)=h(a",b"). Then either we may consider four possible cases depending on whether f(a) f(a") or f(a) >f(a") and the similar cases for ≤ g. Suppose for example that f(a) f(a") and g(b) g(b"). Then ≤ ≤ f(a ) f(a) g(b ) g(b) 1 = 2 " − 3 " − · but the right-hand side is strictly greater than 1 unless f(a)=f(a") and g(b)=g(b"), but since f and g are bijections this implies a = a" and b = b". On the other hand, if f(a) f(a") but g(b) >g(b"). Then ≤ g(b) g(b ) f(a ) f(a) 3 − " =2 " − but that’s impossible unless f(a)=f(a") and g(b)=g(b") since, otherwise one of the terms would be even and the other odd. The other cases are completely similar. Again we consider the image Im(h) Z>0. This is clearly not a finite set and, therefore, by Theorem 5.2, it must⊆ be denumerable. Hence, by transitivity, A B is denumerable. × ! Statement (i) in Theorem 5.4 implies that the union of two denumerable sets is denumerable. An inductive argument (do it!) then shows that a finite union of denumerable sets is denumerable. But in fact, more is true: MATH 300 21

Theorem 5.5. A denumerable union of denumerable sets is denumerable. In other words, suppose that for each j Z>0 we have a denumerable set Aj, then ∈ ∞ Aj := x : x Aj for some j N { ∈ ∈ } j=1 # is denumerable.

Proof. Since each Aj is denumerable there exists a bijection fj : Aj N. On → the other hand, given x ∞ Aj, we deﬁne i(x) N as the smallest index j ∈ j=1 ∈ such that x A .(How do we know i(x) exists?). We now deﬁne ∈ j $ ∞ F : Aj N N → × j=1 # by F (x)=(i(x),fi(x)(x)). We claim that F is 1 : 1. Indeed, suppose F (x)=F (y), then

(i(x),fi(x)(x)) = (i(y),fi(y)(y)). But then i(x)=i(y) := j and both x, y A . But we also have f (x)=f (y), ∈ j j j but since fj is a bijection this implies x = y. Hence, F deﬁnes a bijection from j∞=1 Aj to the image of F which is an inﬁnite (Why?) subset of the countable set N N, therefore it is denumerable. $ × !

The Rational Numbers We are all familiar with the intuitive definition of rational numbers as those which can be written as a quotient p/q, where p, q Z and q = 0. However, this definition is very flawed because we are using∈ the idea) of quotient of integers which is not defined. If, on the other hand we simply think of the expression p/q as a formal expression then we need to deal with the ambiguity that different expressions p/q may represent the same rational number and we need to identify all those different expressions. A rigorous definition of the rational numbers may be accomplished with an equivalence relation.

Deﬁnition 5.6. Let X = Z Z∗, where Z∗ = Z 0 , and let be the equivalence relation in X given× by: \{ } ∼

(p, q) (p",q") p q" = p" q. ∼ ⇔ · · Then the rational numbers Q are the quotient space X/ . ∼ Remark: Note that the definition of the equivalence relation only involves multiplication of integers which is well defined. Morally, our equivalence relation is identifying different expressions of the same “rational number” as a quotient of integers. 22 EDUARDO CATTANI

We can deﬁne a map ϕ: Z Q = X/ by: → ∼ ϕ(m) := [(m, 1)]. Lemma 5.7. The map ϕ is injective.

Proof. Suppose ϕ(m)=ϕ(m"). Then [(m, 1)] = [(m", 1)] but, by Theorem 4.7 this is only possible if (m, 1) (m", 1) and therefore m 1=m" 1; i.e., ∼ · · m = m". ! Lemma 5.8. Every equivalence class in Q = X/ contains an element (p, q) ∼ with q Z>0. ∈ Proof. Let X be an equivalence class. Then, there exists (a, b) Z Z∗ such that C=⊆ [(a, b)]. If b>0 then we are done. Suppose b<0 (remember∈ × that b = 0).C Then we have that ) (a, b) ( a, b) ∼ − − since a ( b) = ( a) b. Since b>0 we are done. · − − · − ! We can now use Lemma 5.8 to choose a distinguished representative in each equivalence class X/ : Given an equivalenceC∈ class∼ , let C D := q Z>0 : p Z, (p, q) . { ∈ ∃ ∈ ∈C} By Lemma 5.8, D = , and therefore by the well-ordering of N there exists a ) ∅ smallest element q0 D and a unique (Why?) element (p0,q0) . Remark: We will∈ see later on that this distinguished representative∈C is the unique one such that q is positive, and p and q have no common factors. Note also that if m Z, the distinguished representative of ϕ(m) is (m, 1) since 1 D and therefore∈ is the smallest element. ∈ Proposition 5.9. The rational numbers Q are denumerable. Proof. Consider the map f : Q Z N that maps an equivalence class → × C∈ Q to its distinguished representative (p0,q0). This map is clearly injective. The image of f is a subset of the denumerable set Z N and therefore, by Corollary 5.3 is countable. But, since the image of f contains× all elements of the form (m, 1) : m Z (Why?), it is an inﬁnite set and therefore it is { ∈ } denumerable. !

6. Infinite Sets In this section we study infinite sets and some of their properties. We recall that an infinite set X is simply a non-empty set that is not finite; that is, there is no bijection from X to 1, . . . , n for any n Z>0. { } ∈ Theorem 6.1. Every infinite set X contains a denumerable subset. MATH 300 23

Proof. It suffices to define an injective map f : Z>0 X. Then, the image of f will be a denumerable subset of X. We define→f recursively using the Principle of Strong Induction: Since X is infinite it is non-empty; let a X and define f(1) = a. ∈ Suppose now that we have defined f(1), . . . , f(n). Since A is infinite, the set B = A f(1), . . . , f(n) = , so there exists b B, define f(n + 1) = b. Note that this\{ is the same idea}) that∅ we used in the∈ proof of Theorem 5.2, so we can use the same argument we used there to prove that f is 1 : 1. ! Theorem 6.2. A set X is infinite if and only if there exists a subset Y X, Y = X and a bijection f : X Y . ⊂ ) → Proof. If X is finite, then for every Y X, Y = X, we have Y < X and therefore there is no bijection f : X Y⊂. ) | | | | → Suppose now that X is infinite and let A = a1, . . . , an,... be a denumerable subset of X. Let Y = X a and define{f : X Y by:} \{ 1} → a , if x = a A f(x)= n+1 n ∈ x, if x A. " )∈ It is left as an exercise to check that f is a bijection. !

Remark: When the argument above is applied to the set Z>0 we obtain the so called Hilbert’s Hotel Paradox: Suppose H is a hotel whose set of rooms is denumerable. Thus, the rooms may be listed as: r ,r , . . . , r ,... { 1 2 n } A traveller named Hilbert arrives late at night and is told by the concierge that there are no more rooms available. Hilbert, being a famous mathematician from G¨ottingen,tells the concierge that there is a very simple solution: move the guest from room r1 to room r2, the guest from room r2 to room r3 and, inductively, the guest from room rn to room rn+1. Then Hilbert goes to sleep in room r1 and everybody is happy...

There is one glaring omission in our discussion so far of inﬁnite sets, namely we have not yet seen any example of an uncountable set; i.e an inﬁnite set which is not bijectively equivalent to Z>0.

Theorem 6.3. The set (Z>0) is not countable. P This theorem is a consequence of the following remarkable result due to the german mathematician Georg Cantor (1845-1918) who was the father of modern set theory (Why is it a consequence?): Theorem 6.4. Let X be a set. Then, there is no surjective map f : X (X). →P 24 EDUARDO CATTANI

Proof. We prove the theorem by contradiction. Suppose such a map exists, then let A = x X : x f(x) . { ∈ )∈ } Note that f(x) (X); that is, f(x) is a subset of X and it makes sense to ask whether x ∈fP(x) or x f(x). Now, if f is surjective then A is in the image of f; i.e.∈ there is some)∈a X such that f(a)=A. We will reach a contradiction∈ by asking whether a A? If a A then, by definition of A, a f(a)=A, which∈ is a contradiction. But if a A∈ = f(a) then, by definition of)∈A, a A. Again a contradiction!! )∈ ∈ ! Theorem 6.5. The set of real numbers R is not countable. Proof. The set of real numbers R is bijectively equivalent to the set of real numbers in the interval (0, 1) (Prove it! This is Problem 8 in HW 8.). So, it suffices to prove that the set of real numbers in (0, 1) is not countable. We will give two proofs of this fact. The first one is the famous Cantor’s Diagonal Argument and it goes as follows. Suppose the interval (0, 1) was countable. Then there would be a bijection f : N (0, 1). We can write each number f(n) (0, 1) as a decimal expression: → ∈

1 1 1 f(1) = 0.a1a2 . . . ak ... 2 2 2 f(2) = 0.a1a2 . . . ak ...... ! ! ! f($) = 0.a1a2 . . . ak ......

i where the aj are digits from 0 to 9. We will now construct a real number in (0, 1) that cannot be in the image of f. Consider a number

x =0.b1b2 . . . br ...... i with the property that bi = ai. Then this number x = f($) for all $ N and therefore it cannot be in the) image of f. ) ∈ The second proof consists in showing that the interval (0, 1) is bijectively equivalent to the set (Z>0) and then applying Theorem 6.3. In fact, we have already seen in classP that, for any X, the set (X) is bijectively equivalent to the set of maps P f : X 0, 1 , { →{ }} where the bijection assigns to a subset A X the characteristic function ⊆ χ : X 0, 1 . A →{ } MATH 300 25

Now, a map f : Z>0 0, 1 is a sequence an of 0’s and 1’s. Given such a sequence we define →{ } { } ∞ a x = n . 2n n=1 ! (Why does this series converge?) This is the decimal binary expansion of a number x (0, 1) and defines a ∈ bijection between (Z>0) and (0, 1). P ! We may summarize the discussion in the following result: Theorem 6.6. The following sets are bijectively equivalent and, hence, have the same cardinality. i) (N). P ii) The set of maps f : N 0, 1 . →{ } iii) Sequences an : n N with an =0, 1 for all n N. iv) The interval{ (0, 1)∈. } ∈ v) The real line R. A natural question to ask now is whether we can generalize Theorem 5.2 to subsets of real numbers. In other words: Question: Is every uncountable subset of the real numbers bijectively equivalent to R? The positive answer to this question is known as the Continuum Hypothesis. In 1940, Kurt Gödel(1906-1978), an Austrian logician and mathematician, proved that the Continuum Hypothesis is consistent with the “standard axioms” of Set Theory. Later, in 1963, Paul Cohen (Stanford University) proved that it is independent of those axioms; i.e. it is impossible to either prove or disprove the Continuum Hypothesis from the axioms. Therefore there are two perfectly valid set theories, one where the continuum hypothesis holds and one where it fails! We note that there are sets which are neither countable, nor equivalent to R. In fact, by Theorem 6.4, it follows that (R) is not bijectively equivalent P to R. Of course, ( (R)) is not bijectively equivalent to (R), and so forth. P P P We conclude this section by defining a relation in any given collection of sets X (U) in some universe U. ⊆P Definition 6.7. Let A and B be sets in X, then say that A B if there is an injective map f : A B. * → It is easy to check that the relation is reflexive (the identity map is an injective map A A) and transitive (the* composition of injective maps is injective). → 26 EDUARDO CATTANI

What is more interesting is that we can prove a variation of anti-symmetry.‡ We should point out that the following theorem is the first truly difficult theorem in the course. Theorem 6.8 (Cantor-Schröder-BernsteinTheorem). Suppose A, B X and there are injective maps f : A B and g : B A, then there is∈ a bijection → → h: A B. → Proof. § We will prove the theorem in two steps. The first step will be to prove a particular case of the theorem. By a particular case we mean that we make additional assumptions to simplify the argument. In the second step we will use the first case to prove the statement without the additional assumptions. This is a very common technique when proving mathematical statements. Case 1: Let us assume that B is a subset of A and therefore the map g is simply the inclusion map. The statement in this particular case is then:

Let B A and suppose there exists an injective map f : A B, then • there exists⊆ a bijective map h: A B. → →

We will define a collection Cj,j Z 0 of subsets of A inductively: { ∈ ≥ } We define C0 = A B. • Let C = f(C ). \ • 1 0 Assuming that Cn, n Z>0 has been defined we define Cn+1 = f(Cn). • ∈ By the Principle of Induction the subsets Cj are defined for all j Z>0 and, ∈ since we have already defined C0, for all n Z 0. We now define ∈ ≥ ∞ C = Cj. j=0 # Note that: i) Since C C, A C A C = B. 0 ⊆ \ ⊆ \ 0 ii) For all j Z>0 we have that C0 Cj = . Indeed, for all j Z>0, C Im(f∈) B but C B = by∩ definition.∅ ∈ j ⊆ ⊆ 0 ∩ ∅ We now define the map h that we are looking for: f(x), if x C h(x)= ∈ x, if x A C. " ∈ \

‡Strictly speaking, Theorem 6.8 says that deﬁnes a partial order in (U)/ , where is bijective equivalence. * P ∼ ∼ §The following proof is adapted from the one appearing in Professor Hajir’s Lecture Notes which, in turn, is a variation of a proof you may ﬁnd in planetmath.org MATH 300 27

We need to check that h is a bijection from A to B. Note, ﬁrst of all, that if x C then h(x)=f(x) B while, if x A C then h(x)=x B since, by (i),∈ A C B. Hence h(∈x) B for all x∈ A\. ∈ \ ⊆ ∈ ∈ Next we prove that h is injective. Suppose x, x" A are such that h(x)= ∈ h(x"). We must consider three cases:

If x and x" are both in C then h(x)=h(x") f(x)=f(x") but this • ⇒ implies that x = x" since f is injective. If x and x" are both in A C then h(x)=h(x") x = x" by definition • of h. \ ⇒ Suppose x C but x" A C. Then h(x)=f(x) and h(x")=x". • ∈ ∈ \ Therefore if h(x)=h(x") we have x" = f(x). Now, since x C we must ∈ have x Ck for some k Z 0 but this implies that x" Ck+1 C which is∈ a contradiction. ∈ ≥ ∈ ⊆ Finally, we prove that h is surjective. Let y B. Then, there are two possibilities: y C or y C. In the latter case,∈ y = h(y) and therefore y Im(h). Suppose∈ now that)∈ y C, then y C for some j 0. But it is ∈ ∈ ∈ j ≥ not possible for y to be in C0 since C0 B = and y B. So we must have that y C for some j 1. By definition,∩ if j ∅ 1, ∈ ∈ j ≥ ≥ Cj = f(Cj 1) − and therefore y = f(x) for some x Cj 1. But since x C, f(x)=h(x) and we have y = h(x). So h is surjective∈ and− the proof of Case∈ 1 is complete. Case 2: This is the general case where we no longer assume that B A. Let A and B be arbitrary sets and suppose we have injective maps f : A ⊆ B and g : B A and we want to show that there is a bijective map h: A →B. We will reduce→ this case to the previous case. → Let B˜ = Im(g) A. The map g now defines a bijection: ⊆ g˜: B B.˜ → We will apply Case 1 for B˜ A. Since f : A B is injective, we can define an injective map (recall that⊆ composition of injective→ maps is injective!): f˜ :=g ˜ f : A B˜ ◦ → By Case 1, there exists a bijection h˜ : A B˜. But then the composition → 1 h :=g ˜− h˜ : A B ◦ → is a bijection (composition of bijections is a bijection!). ! 28 EDUARDO CATTANI

7. Groups We begin with a series of definitions: Definition 7.1. Let G be a set. A (binary) operation or a composition law in G is a map: ∗ : G G G. ∗ × → Instead of writing (a, b) to indicate the result of applying the map to the pair (a, b) G G∗, we will usually write a b. ∗ ∈ × ∗ Definition 7.2. An operation on G is called associative if and only if ∗ (7.1) (a b) c = a (b c) for all a, b, c G. ∗ ∗ ∗ ∗ ∈ Definition 7.3. An operation on G is called commutative if and only if ∗ (7.2) a b = b a for all a, b G. ∗ ∗ ∈ Definition 7.4. Given a set G with a binary operation , an element e G is called an identity element for if and only if: ∗ ∈ ∗ (7.3) a e = e a = a for all a G. ∗ ∗ ∈ Even though, in principle, a set G with a binary operation could have different identity elements, it is easy to prove that, in fact, if∗ an identity element exists then it must be unique: Proposition 7.5. Let G be a set with a binary operation . Then if an identity element exists it is unique. ∗

Proof. Suppose e1 and e2 are elements in G satisfying (7.3). Then we have e = e e = e , 2 1 ∗ 2 1 where in the ﬁrst equality we use the fact that e1 is an identity and in the second equality that e2 is an identity. ! Deﬁnition 7.6. Given a set G with a binary operation and an identity element e, we say that b G is an inverse of a G if and only∗ if ∈ ∈ (7.4) a b = b a = e ∗ ∗ Once again, it is easy to prove that, for associative operations, if an inverse exists then it must be unique: Proposition 7.7. Let G be a set with a binary operation and suppose there exists an identity e. Then if an inverse of an element a exists,∗ it is unique. MATH 300 29

Proof. Suppose b and c are inverses of a. Then we have: b = b e (since e is the identity) ∗ = b (a c) (since c is an inverse of a) ∗ ∗ =(b a) c (by associativity) ∗ ∗ = e c (since b is an inverse of a) ∗ = c (since e is the identity)

! Deﬁnition 7.8. A group is a set G with an operation such that: ∗ i) is associative. ii)∗ There exists an identity element e G. iii) Every element a G has an inverse.∈ ∈ If, in addition, is commutative then we say that G is a commutative group or an abelian group.∗ Example 7.9. The integers Z with = + is an abelian group. We know that addition is associative and commutative∗ and 0 is the identity element. Moreover, given any a Z, a is the inverse of a. ∈ − Exactly the same arguments show that the rational numbers Q or the real numbers R with the operation of addition are abelian groups whose identity element is 0. Example 7.10. Consider now the same set Z but with the operation = product. We know that the product of integers is associative and commutative.∗ Moreover, the element 1 is the identity element since 1 m = m 1=m for · · all m Z. However, it is not true that every element has an inverse. In fact, ∈ the only elements with an inverse are 1 and 1. So, Z with the multiplication operation is not a group. − Consider next the set Q with the multiplication operation. Again, the product of rational numbers is associative, commutative and 1 is the identity element. What about existence of inverse? Every element except 0 has an inverse: if a = p/q and p, q = 0 then b = q/p is the inverse of a. But the element 0 does not have an inverse:) We cannot ﬁnd a rational number b such that 0 b = 1! Since 0 is the only element without an inverse and the product of non-zero· numbers is not zero we can restrict the product operation to the set

Q∗ := a Q : a =0 { ∈ ) } to get an abelian group. Similarly, we can deﬁne a multiplicative group (R∗, ), · where R∗ = R 0 . \{ } Example 7.11. Note that there is no a priori restriction on what an operation is, just that it is a map that to each ordered pair of elements of G assigns 30 EDUARDO CATTANI another element of G. For example G could be the set G = p, s, r and the operation be: { } p p = p ; s s = s ; r r = r ; ∗ ∗ ∗ p s = s p = s ; p r = r p = p ; r s = s r = r. ∗ ∗ ∗ ∗ ∗ ∗ (This is the operation derived from the paper/scissors/rock game.) Note that in this case the operation is commutative by deﬁnition. However, we see by inspection that there is no identity element. Is it associative? We have:

(p s) r = s r = r, ∗ ∗ ∗ but p (s r)=p r = p. ∗ ∗ ∗ So, is not associative. ∗ The following is one of the key examples and the one from which the term composition law is derived.

Example 7.12. Let X be an arbitrary non-empty set and let

(X) := f : X X : f is bijective , B { → } and set f g = f g, the composition of maps. This makes sense since the composition∗ of bijections◦ is a bijection. We have already shown that the composition of maps is associative and that the identity map idX satisﬁes that

f id = id f = f. ◦ X X ◦

Therefore, idX is the identity for ( (X), ). Moreover, we have also proved thatB every∗ bijection f (X) has an inverse 1 ∈B map f − (X) satisfying: ∈B 1 1 f f − = f − f = id . ◦ ◦ X Thus, the inverse in the sense of maps is also the inverse for . Therefore ( (X), ) is a group. Note that in general this group is not commutative.∗ For exampleB ∗ if X = 1, . . . , n { } then (X) is the set of permutations of 1, . . . , n and we have already seen that theB composition of permutations is not{ commutative.} Recall that in this case we denote by S the group of permutations of 1, . . . , n . n { } MATH 300 31

8. The integers mod m. Modular Arithmetic. In this section we will study in detail operations defined in a space of equivalence classes. This is the most important example of a group that we will study in this course. We begin by recalling that given an integer m>1, we have defined an equivalence relation on Z: a b m (a b). ∼ ⇔ | − Since we will be studying this particular equivalence relation in detail we introduce a specific notation to replace the generic notation . We will say that: ∼ a b (mod m) m (a b) ≡ ⇔ | − Let Rm : Z 0, 1, . . . , m 1 be the map that assigns to each a Z the →{ − } ∈ remainder of dividing a by m; i.e. Rm(a) is the unique integer between 0 and m 1 such that there exists k Z with − ∈ a = k m + R (a). · m (What theorem guarantees the existence and uniqueness of this de- composition?) Then, since m (a b) if and only if R (a)=R (b) we have that | − m m a b (mod m) R (a)=R (b). ≡ ⇔ m m This means that the map R defines the equivalence relation a b (mod m) m ≡ and, consequently, the quotient space of this equivalence relation Z/ is bijectively equivalent to 0, 1, . . . , m 1 . To keep track of the integer m∼used { − } to define the equivalent relation we will denote Z/ by Zm. ∼ We now define an operation in the set Z/ of equivalence classes in the ⊕ ∼ following way: Let C1 and C2 be equivalence classes, pick a1 C1 and a2 C2 then define: ∈ ∈ (8.1) C C := [a + a ] 1 ⊕ 2 1 2 Before we can accept this as a valid definition we need to check that the result of the operation does not depend on our pick of representatives a1 and a2 in the equivalence classes C1 and C2. Suppose we pick different elements, say b C and b C , then 1 ∈ 1 2 ∈ 2 a b (mod m) m (a b ), and 1 ≡ 1 ⇒ | 1 − 1 a b (mod m) m (a b ) 2 ≡ 2 ⇒ | 2 − 2 But then m divides (a b )+(a b ) = (a + a ) (b + b ). Therefore 1 − 1 2 − 2 1 2 − 1 2 a + a b + b (mod m) 1 2 ≡ 1 2 and [a1 + a2] = [b1 + b2]. 32 EDUARDO CATTANI

For example, let m = 4, then there are 4 equivalence classes in Z4 which we can list as [0], [1], [2], [3] and we have the following table for ⊕

[0] [1] [2] [3] [0]⊕ [0] [1] [2] [3] [1] [1] [2] [3] [0] [2] [2] [3] [0] [1] [3] [3] [0] [1] [2]

Remark: Since the set Z/ = Zm is bijectively equivalent to the set 0, 1, . . . , m 1 we may think of as an∼ operation on the set 0, 1, . . . , m 1 .{ With this − point} of view, the table⊕ above describes an operation{ in the− set} 0, 1, 2, 3 . Once one is used to the notion of quotients it is common to forget{ the square} brackets and to replace the symbol by the standard +. But, for the remaining of these notes we will keep the clumsier⊕ notation so we can be sure of where we are working.

We now check the conditions for (Zm, ) to be a group: ⊕ Associativity: ([a]+[b]) + [c]=[a + b]+[c] = [a + b + c] = [a]+[b + c]=[a] + ([b]+[c]). Identity: The class [0] is the identity element since: [a] + [0] = [a + 0] = [a] = [0 + a] = [0] + [a].

Existence of Inverse: For each [a] Zm the element [ a] is the inverse of [a] since: ∈ − [a]+[ a] = [a a] = [0] = [ a + a] = [ a]+[a]. − − − − Hence, (Zm, ) is a group. In fact, it is equally easy to check that (Zm, ) is an abelian group.⊕ We leave the verification to the reader. ⊕ Remark: We make an important notational comment. When operating in Zm the expressions: a + b c (mod m) ≡ and [a] [b]=[c] ⊕ are completely equivalent. Notice that the first expression conveys more in- formation since it makes explicit what m is. The second expression is simpler once m is fixed. For example the statements 5 + 11 2 (mod 7) ≡ is equivalent to saying: In the group Z7 [5] [11] = [2]. ⊕ MATH 300 33

But note that for the second expression to make sense we need to specify the group where the operation takes place.

We can similarly define a multiplication in the set Z/ = Zm as follows: Let C and C be equivalence classes, pick ⊗a C and a ∼ C then define: 1 2 1 ∈ 1 2 ∈ 2 (8.2) C C := [a a ] 1 ⊕ 2 1 · 2 Once again we need to check that this definition does not depend on the representatives a1 and a2 that we picked. Suppose we pick different elements, say b1 C1 and b2 C2, then since a b (mod m) we have from the Division Theorem∈ that: ∈ 1 ≡ 1 a = k m + r ; b = $ m + r ;0 r < m. 1 1 1 1 1 1 ≤ 1 (Why are the remainders the same?) Similarly, a = k m + r ; b = $ m + r ;0 r < m. 2 2 2 2 2 2 ≤ 2 We then have: a a = k k m + k r m + k r m + r r r r (mod m), 1 · 2 1 2 1 2 2 1 1 2 ≡ 1 · 2 b b = $ $ m + $ r m + $ r m + r r r r (mod m). 1 · 2 1 2 1 2 2 1 1 2 ≡ 1 · 2 Which means that [a a ] = [b b ]. So, the modular product is well defined. 1 · 2 1 · 2 ⊗ Example 8.1. Let us write the table for the modular product in Z4.

[0] [1] [2] [3] [0]⊗ [0] [0] [0] [0] [1] [0] [1] [2] [3] [2] [0] [2] [0] [2] [3] [0] [3] [2] [1]

Just as we did in the case of we can easily check that is associative, commutative, and that [1] is the⊕ identity element. However, we⊗ cannot expect to have an inverse since the element [0] will never have an inverse. We already encountered this problem in the example of the product operation in Q and we solved it by considering the set Q∗ = Q 0 . Can we do the same thing \{ } here? The table for Z4 tells us that the answer is NO since the element [2] has no inverse either!! It is then natural to ask:

Question: When does an element [a] Zm have an inverse in Zm; i.e. when ∈ can we ﬁnd b Z so that [a] [b] = [1]? Fortunately,∈ the answer to⊗ this question is very easy:

Theorem 8.2. Let [a] Zm then [a] has an inverse in (Zm, ) if and only if gcd(a, m)=1. ∈ ⊗ 34 EDUARDO CATTANI

Proof. Suppose that [a] has an inverse [b], then [a] [b] = [1]; i.e. ⊗ a b 1 (mod m) · ≡ and this means that there exists k Z such that ∈ a b = k m +1. · · But then 1=a b k m · − · and it follows that gcd(a, m) = 1. The converse is identical: if gcd(a, m) = 1 then there exist integers x, y such that 1=a x + m y · · But this implies that a x 1 (mod m) · ≡ or, equivalently, that [a] [x] = [1] and [x] is the inverse of [a] in (Zm, ). ⊗ ⊗ ! We can illustrate Theorem 8.2 in the case of Z4. The elements [1] and [3] have inverses since gcd(1, 4) = gcd(3, 4) = 1. But the elements [0] and [2] do not since gcd(0, 4) = 4 and gcd(2, 4) = 2. We have the following important Corollary to Theorem 8.2:

Corollary 8.3. If p is prime then every non-zero element in (Zp, ) has an ⊗ inverse. Therefore if we denote by Zp∗ = Zp [0] , then (Zp∗, ) is an abelian group. \{ } ⊗

Proof. Let [a] Zp and suppose that [a] = [0]. Then, p does not divide a and since p is prime∈ we must have gcd(a, p) =) 1. Then by the Theorem, [a] has an inverse in Zp. !

Example 8.4. Consider the multiplication table for Z5:

[0] [1] [2] [3] [4] [0]⊗ [0] [0] [0] [0] [0] [1] [0] [1] [2] [3] [4] [2] [0] [2] [4] [1] [3] [3] [0] [3] [1] [4] [2] [4] [0] [4] [3] [2] [1]

1 1 1 We see that in this case [2]− = [3], [3]− = [2], and [4]− = [4]. MATH 300 35

The fact that for a prime number p the operations in Zp, and , satisfy ⊕ ⊗ the same properties as the addition and product of rational or real numbers¶ means that we can operate with them just as we do with rationals or reals. Let’s illustrate this point with a few examples. Example 8.5. Solve the congruence equation 2x +3 4 (mod 5). ≡ We can view this expression as an equation in Z5: ([2] x) [3] = [4]. ⊗ ⊕ We then have

[2] x = [4] ( [3]) = [1] (here denotes the additive inverse in (Z5, ).) ⊗ ⊕ − − ⊕ Therefore 1 1 x = [2]− ([2] x) = [2]− [1] = [3] ⊗ ⊗ ⊗ 1 since [2]− = [3] in Z5 (see Example 8.4). We can verify our result: 2 3 + 3 = 9 4 (mod 5). · ≡ Example 8.6. Now let us try something harder. Solve the congruence equation 15x + 11 7 (mod 31). ≡ We can view this expression as an equation in Z31: ([15] x) [11] = [7]. ⊗ ⊕ We then have

[15] x = [7] ( [11]) = [7] [20] = [27] (here denotes the inverse in (Z5, ).) ⊗ ⊕ − ⊕ − ⊕ Therefore 1 1 x = [15]− ([15] x) = [15]− [27]. ⊗ ⊗ ⊗ 1 We now need to compute [15]− in Z31. The proof of Theorem 8.2 tells us how to proceed: Since gcd(15, 31) = 1 we can write 1 as an integral linear combination of 15 and 31. In this case this is very easy 1 = 31 + ( 2) 15. − ·

¶In addition to the properties we have already discussed, the modular product is distribu- tive with respect to modular addition, that is: [a] ([b] [c]) = ([a] [b]) ([a] [c]) ⊗ ⊕ ⊗ ⊕ ⊗ as is easily verified from the definitions. It is also easy to check that [a] [0] = [0] for all ⊗ [a] Zp. All of these properties together define the notion of a field. The reals R and the ∈ rationals Q are fields as is Zp for p prime. One big difference between them is that Zp is finite, Q is denumerable, and R is uncountable. 36 EDUARDO CATTANI

This means that ( 2) 5 1 (mod 31), in other words that [ 2] [15] = [1] 1 − · ≡ − ⊗ in Z31. So [15]− =[ 2] and x =[ 2] [27] = [ 54] = [8]. Again, it is worthwhile to verify our− result: − ⊗ − 15 8 + 11 = 131 = 4 31 + 7 7 (mod 31). · · ≡ Example 8.7. In this example we will use modular arithmetic to ﬁnd the test for divisibility by 11. Let m Z>0 be a positive integer and write m in its decimal expansion: ∈

m = akak 1 a1a0, − ··· where a are digits between 0 and 9 and a = 0. In other words, j k ) k k 1 m = ak 10 + ak 1 10 − + + a1 10 + a0. · − · ··· · Theorem 8.8. m is divisible by 11 if and only if the alternating sum k a a + a + +( 1)ka = ( 1)ja 0 − 1 2 ··· − k − j j=0 ! is divisible by 11.

Proof. We will work in Z11. If m is divisible by 11 then [m] = [0] in Z11 and therefore we have k k 1 [ak] [10 ] [ak 1] [10 − ] [a1] [10] [a0] = [0]. ⊗ ⊕ − ⊗ ⊕···⊕ ⊗ ⊕ But [10] = [ 1] in Z11, therefore the above expression may be rewritten as − k k 1 [ak] [( 1) ] [ak 1] [( 1) − ] [a1] [ 1] [a0] = [0], ⊗ − ⊕ − ⊗ − ⊕···⊕ ⊗ − ⊕ or, given the deﬁnitions of and as: ⊗ ⊕ k k 1 ( 1) ak +( 1) − ak 1 + a1 + a0] = [0] − · − · − ···− which means that a a + a + +( 1)ka 0 − 1 2 ··· − k is divisible by 11. ! So, for example, 385 is divisible by 11 since the alternating sum of the coeﬃcients is 5 8 + 3 = 0 and 0 is divisible by 11. But 749 is not divisible by 11 since, in this− case, the alternating sum is: 9 4 + 7 = 12 which is not divisible by 11. −

9. Subgroups - Cyclic Groups Deﬁnition 9.1. Let (G, ) be a group. A non-empty subset H G is called a subgroup if and only if ∗ ⊆

i) If h1,h2 H then h1 h2 H. ∈ ∗ ∈ 1 ii) For every h H, the inverse h− H. ∈ ∈ MATH 300 37

Remark: If H G is a subgroup of (G, ) then (H, ) is a group as well. We only need to check⊆ that H has an identity∗ element but∗ this is implied by i) and ii) in Deﬁnition 9.1. Indeed, pick any element h H (we required H to 1 ∈1 be non-empty), then by ii), h− H and by i), e = h− h H. But then e is also the identity element in H.∈ ∗ ∈

Example 9.2. The integers Z are a subgroup of (Q, +). Indeed, if m, n Z ∈ then m + n Z and m Z. On the other hand, Z>0 is not a subgroup of ∈ − ∈ Z. While it is true that for m, n Z>0, m + n Z>0, it is not true that for ∈ ∈ m Z>0, m Z>0. Therefore the second condition in Definition 9.1 fails. ∈ − ∈ Example 9.3. For any n Z>0, ∈ nZ := k Z : n k { ∈ | } is a subgroup of (Z, +). Indeed, if n divides k1 and k2 then n divides k1 +k2 and if n divides k then n divides k. Therefore both conditions in Definition 9.1 are satisfied. −

Example 9.4. Consider the group (Z4, ). It follows from the operation table ⊕ for this group that the subset H = [0], [2] is a subgroup. Indeed, in Z4: { } [0] [0] = [2] [2] = [0] ; [0] [2] = [2] [0] = [2] ⊕ ⊕ ⊕ ⊕ and [0] = [0] and [2] = [2], where as always, denotes the inverse in − − − (Z4, ). ⊕ Given a group (G, ) and an element g G we may define the n-th power of g recursively: ∗ ∈ g1 = g; • Assuming we have defined gn then we define gn+1 = gn g. • n 0 ∗ This defines g for every n Z>0. We also define: g = e and for n Z>0: ∈ ∈ n 1 n g− =(g− ) . 1 Note that g− is the inverse of g with respect to . The following Lemma will be useful in proving∗ a very important Theorem. Lemma 9.5. For every n Z, ∈ n 1 n g− =(g− ) . Proof. Note that the statement follows from the definition if n>0 while it is obvious if n = 0. We need to consider then the case n<0. But then 1 n 1 1 n n (g− ) = ((g− )− )− = g− . ! Theorem 9.6. Let (G, ) be a group and let g G. Then for all $, k Z: ∗ ∈ ∈ g! gk = g!+k. ∗ 38 EDUARDO CATTANI

Proof. This proof is a good example of an argument which appears to be obvious but that requires quite a bit of work in order to prove it with the ingredients at our disposal. The main difficulty is that we have different definitions for positive and negative powers. We will fix $ Z and prove the assertion of the Lemma for all k Z. We begin with∈ the easiest case: k = 0, then ∈

g! g0 = g! e = g! = g!+0. ∗ ∗

Next, we prove the result for k Z>0 by induction. We begin with the base case k = 1. We need to distinguish∈ three cases depending on whether $> 0, $ = 0, or $< 0. i) If $> 0, then g! g1 = g! g = g!+1 by the recursive deﬁnition of the powers of an element.∗ ∗ ii) If $ = 0, then g0 = e and g0 g1 = e g1 = g1 = g0+1. iii) Suppose now that $< 0, and∗ write∗$ = s with s>0. Then g! = 1 s 1 s 1 1 − (g− ) =(g− ) − (g− ) by the previous cases (remember that s 0). Then ∗ ≥

! 1 1 s 1 1 g g = ((g− ) − (g− )) g ∗ 1 s 1 ∗ 1 ∗ =(g− ) − ((g− )) g) 1 s 1 ∗ ∗1 s 1 =(g− ) − e =(g− ) − (s 1) ∗ = g− − = g!+1

! n !+n Suppose now that the g g = g for all all $ Z and some n Z>0. We want to prove that ∗ ∈ ∈ g! gn+1 = g!+n+1. ∗ Now, by deﬁnition gn+1 = gn g, therefore ∗ g! gn+1 = g! (gn g) ∗ ∗ ∗ =(g! gn) g (associativity) ∗ ∗ = g!+n g (inductive hypothesis) ∗ = g!+n+1 (base case)

Unfortunately, our work is not yet done because we have only proved the assertion of the Lemma for $ Z and k Z 0. We still need to consider the case k<0. However we can∈ use a trick∈ to≥ reduce it to the case we have k 1 r already proved! If k<0, then k = r with r>0 and g =(g− ) by − MATH 300 39

! 1 ! deﬁnition. Moreover, by Lemma 9.5, we have g =(g− )− . Then ! k 1 ! 1 r g g =(g− )− (g− ) ∗ 1 !+∗r =(g− )− (by the previous case since r>0) ! r = g − (by Lemma 9.5) = g!+k (since k = r) − !

Remark: One can prove by induction that if a, b Z, then (ga)b = gab. The proof is left as an exercise. ∈ The following result is essentially a Corollary of Theorem 9.6 but we state it as a Theorem because of its importance. Theorem 9.7. Let (G, ) be a group and let g G. Then, the subset ∗ ∈ H := gk : k Z { ∈ } is a subgroup of G. Moreover, as a group (H, ) is commutative. ∗ Proof. We need to check the two conditions in the definition of a group. Sup- k1 k2 pose h1,h2 H then there exist k1,k2 Z such that h1 = g and h2 = g . But then, Theorem∈ 9.6 says that ∈ h h = gk1 gk2 = gk1+k2 H. 1 ∗ 2 ∗ ∈ So, the first condition is satisfied. Suppose now that h H. Then h = gk for some k Z and, again from Theorem 9.6 we have: ∈ ∈ k k k+k 0 g− g = g− = g = e, ∗ k k and similarly g g− = e. Therefore ∗ 1 k h− = g− H. ∈ Finally, it is easy to see that, as a group, H is abelian. If h1,h2 H we k1 k2 ∈ have that h1 = g and h2 = g , for some k1,k2 Z. But then, Theorem 9.6 says that ∈ h h = gk1+k2 = h h . 1 ∗ 2 2 ∗ 1 !

Remarks: The subgroup H in Theorem 9.7 is called the subgroup generated by g and g is called a generator of H. We will often denote by g the subgroup generated by g. If the operation of the group is addition then< we= will usually write k g instead of gk. · 40 EDUARDO CATTANI

Example 9.8. If G = Z and n Z then the subgroup generated by n: ∈ n = k n : k Z < = { · ∈ } coincides with the subgroup nZ Z deﬁned in Example 9.3. ⊆

Example 9.9. Consider the multiplicative group (Z5∗, ) whose group table is given in Example 8.4. Note that every element g except⊗ [1] has the property that g = Z5∗. For example: < = [2]0 = [1]; [2]1 = [2]; [2]2 = [4]; [2]3 = [3]. Deﬁnition 9.10. A group G is said to be cyclic if there exists an element g G such that ∈ G = g . < = Example 9.11. The integers (Z, +) are a cyclic group since Z = 1 = 1 . < = <− = In other words 1 is generator of Z. Similarly, nZ is a cyclic subgroup with generator n (or n). − By Example 9.9, the group (Z5∗, ) is cyclic. We may take any element diﬀerent from 1 as a generator. ⊗ On the other hand, (Q∗, ) is not a cyclic group. Suppose r = p/q is a generator and assume, without· loss of generality, that gcd(p, q) = 1. Then if x = rn r , and x = a/b, gcd(a, b) = 1 then either p/b of q/b. Therefore it is ∈< = not possible to have Q∗ = r for any r Q∗. < = ∈ Theorem 9.12. If G is a cyclic group then G is countable. Proof. Let G be a cyclic group and suppose g is a generator; i.e. G = g . < = Let ϕ: Z G be the map → ϕ(n)=gn. Since g is a generator of G we know that ϕ is a surjective map. If ϕ is also injective then ϕ is a bijection and G is denumerable. Suppose then that ϕ is not injective. Then there exist m, n Z, m = n, such that ϕ(m)=ϕ(n). We may assume without loss of generality∈ that m) > n. Then m n m n m n g = g g g− = g − = e. ⇒ ∗ k Let now A = k Z>0 : g = e . We have that A = (Why?). Let p be the smallest element{ ∈ in A. We now} claim: ) ∅ Claim: ϕ(m)=ϕ(n) if and only if m n (mod p) ≡ MATH 300 41

Let us prove the Claim. Suppose m n (mod p), then m = kp + n but then ≡ ϕ(m)=gm = gkp+n =(gp)k gn = ek gn = gn = ϕ(n). ∗ ∗ In particular, if r is the remainder of division of m by p, we have that ϕ(m)= ϕ(r). Suppose then that ϕ(m)=ϕ(n) and let m r (mod p), m ≡ 1 ≡ r2 (mod p), with 0 r1,r2 p 1. Then ϕ(r1)=ϕ(r2) and if, say r1 0 or to be the smallest element in ) ∈ k A := k Z>0 : g = e { ∈ } if A is not empty.

Corollary 9.13. Let G be a group and suppose that G = m. Suppose g G is an element of order m. Then G = g ; i.e. G is a| cyclic| group and g ∈is a generator. < = 2 m 1 Proof. Consider the elements e, g, g , . . . , g − . We claim that all these ele- a b b a ments are diﬀerent. Suppose g = g with 0 a

10. Homomorphisms of Groups The natural maps between groups are those that preserve the operations. Such maps are called homomorphisms.

Deﬁnition 10.1. Let (G1, 1) and (G2, 2) be groups and let ϕ: G1 G2 be a map. We say that ϕ is a homomorphism∗ ∗ if and only if →

ϕ(g g")=ϕ(g) ϕ(g"), ∗1 ∗2 for all g, g" G . ∈ 1 Example 10.2. Let G1 = G2 =(Z, +), then for every n Z, the map ∈ ϕ(x)=nx is a homomorphism. This follows since

ϕ(x + x")=n(x + x")=nx + nx" = ϕ(x)+ϕ(x").

Example 10.3. Let G1 =(R, +) and G2 =(R∗, ), then the map ϕ: G1 G2 · → ϕ(x)=ex is a homomorphism. To prove this we ﬁrst note that ex is never zero so that ϕ is a map with values in R∗. Next we check that it is a homomorphism: x+x x x ϕ(x + x")=e " = e e " = ϕ(x) ϕ(x"). · · Theorem 10.4. Let (G1, 1) and (G2, 2) be groups and let ϕ: G1 G2 be a homomorphism. Then ∗ ∗ →

i) ϕ maps the identity element in G1 to the identitity element in G2. ii) For every g G1, ∈ 1 1 ϕ(g− )=ϕ(g)− .

Proof. Let e1 be the identity in G1, then we have e1 1 e1 = e1. This means that ∗ ϕ(e1)=ϕ(e1 1 e1)=ϕ(e1) 2 ϕ(e1) ∗ 1 ∗ since ϕ is an isomorphism. Now, let ϕ(e1)− be the inverse of ϕ(e1) in G2. Then 1 e2 = ϕ(e1)− 2 ϕ(e1) 1 ∗ = ϕ(e1)− 2 (ϕ(e1) 2 ϕ(e1)) 1∗ ∗ =(ϕ(e )− ϕ(e )) ϕ(e ) 1 ∗2 1 ∗2 1 = e ϕ(e ) 2 ∗2 1 = ϕ(e1) (Make sure you can justify each of the steps!) Now we can use i) to prove the second assertion. Let g G then ∈ 1 1 1 ϕ(g− ) ϕ(g)=ϕ(g− g)=ϕ(e )=e . ∗2 ∗1 1 2 44 EDUARDO CATTANI

1 and similarly, ϕ(g) ϕ(g− )=e . Therefore ∗2 2 1 1 ϕ(g− )=ϕ(g)− 1 since ϕ(g− ) satisﬁes the requirements for an element to be the inverse and we know from Proposition 7.7 that if the inverse exists, then it is unique. ! Note that in Example 10.2, ϕ(0) = 0 and that in Example 10.3 we have 0 x x 1 ϕ(0) = e = 1. Also, in the second example we have that e− =(e )− .

Definition 10.5. Let (G1, 1) and (G2, 2) be groups and let ϕ: G1 G2 be a homomorphism. Then the∗ subset ∗ → ker(ϕ) := g G : ϕ(g)=e = ϕ (e ) { ∈ 1 2} ←− 2 is called the kernel of ϕ. The subset of G2 Im(ϕ) := h G : h = ϕ(g) for some g G { ∈ 2 ∈ 1} is called the image of ϕ. Theorem 10.6. Let (G , ) and (G , ) be groups and let ϕ: G G be a 1 ∗1 2 ∗2 1 → 2 homomorphism. Then ker(ϕ) is a subgroup of G1 and Im(ϕ) is a subgroup of G2. Proof. We need to verify that the subset ker(ϕ) G satisfies the conditions ⊆ 1 of Definition 9.1. Suppose g, g" ker(ϕ), then by definition of ker(ϕ) we have ∈ ϕ(g)=ϕ(g")=e2. But then ϕ(g g")=ϕ(g) ϕ(g")=e e = e , ∗1 ∗2 2 ∗2 2 2 and therefore g g" ker(ϕ). Similarly we have ∗1 ∈ 1 1 1 ϕ(g− )=ϕ(g)− = e2− = e2, 1 which means that g− ker(ϕ). ∈ It remains to show that Im(ϕ) is a subgroup of G . Suppose h, h" Im(ϕ), 2 ∈ then there exist g, g" G such that h = ϕ(g) and h" = ϕ(g"), but then ∈ 1 h h" = ϕ(g) ϕ(g")=ϕ(g g") Im(ϕ). ∗2 ∗2 ∗1 ∈ Also, 1 1 1 h− = ϕ(g)− = ϕ(g− ) Im(ϕ). ∈ Therefore Im(ϕ) is a subgroup of G2. ! It is easy to check that in Examples 10.2 and 10.3, the kernel is the trivial subgroup 0 . In Example 10.2, the subgroup Im(ϕ) is the subgroup nZ of { } Z discussed in Examples 9.3 and 9.8. On the other hand, we know that the image of the exponential map is the set of positive real numbers; therefore, in Example 10.3, the the subgroup Im(ϕ) is the subgroup (R>0, ) of (R∗, ). · · MATH 300 45

Example 10.7. We can deﬁne a map

π :(Z, +) (Zm, ) → ⊕ by π(a)=[a]. This map is a homomorphism since we deﬁned as ⊕ [a] [b] = [a + b], ⊕ in other words, π(a) π(b)=π(a + b). This map is surjective and therefore ⊕ Im(π)=Zm but it is not injective, in fact: ker(π)= a Z : π(a) = [0] = a Z : a 0 (mod m) = mZ. { ∈ } { ∈ ≡ } Example 10.8. Let (G, ) be a cyclic group and g G a generator of G. Let ∗ ∈ ϕ: Z G → be the map ϕ(m)=gm introduced in the proof of Theorem 9.12. Then ϕ is a homomorphism. Indeed, this is a consequence of Theorem 9.6 since ϕ(m + m )=gm1+m2 = gm1 gm2 = ϕ(m ) ϕ(m ). 1 2 ∗ 1 ∗ 2 In the proof of Theorem 9.12 we proved that if ϕ is not injective then ϕ(m)= ϕ(n) if and only if m n (mod p), where p was the smallest positive integer ≡ such that gp = e. This means that ker(ϕ)=pZ. The following theorem shows that in the case of homomorphisms it is much easier to check whether the map is injective since we only need to check that there cannot be two diﬀerent elements being mapped to the identity.

Theorem 10.9. Let (G1, 1) and (G2, 2) be groups and let ϕ: G1 G2 be a homomorphism. Then ϕ is∗ injective if and∗ only if: → ker(ϕ)= e . { 1} Proof. If ϕ is injective then, for all h G , every set ϕ (h) consists of, at ∈ 2 ←− most, one element. In the case of the identity element e2 G2 we know that e ϕ (e ) and therefore we must have ∈ 1 ∈ ←− 2 e = ϕ (e ) = ker(ϕ). { 1} ←− 2 Conversely, assume ker(ϕ)=e1 and let us prove that ϕ is injective. For that we need to check that if ϕ(g)=ϕ(g"), g, g" G then g = g". We have ∈ 1 1 1 1 ϕ(g− g")=ϕ(g− ) ϕ(g")=ϕ(g)− ϕ(g")=e . ∗1 ∗2 ∗2 2 1 1 Therefore, g− g" ker(ϕ) which means that g− g" = e and g" = g. ∗1 ∈ ∗1 1 ! Deﬁnition 10.10. Let (G1, 1) and (G2, 2) be groups and let ϕ: G1 G2 be a homomorphism. Then we∗ say that ϕ is∗ an isomorphism if ϕ is a bijection.→ We write (G , ) = (G , ) or simply G = G if there exists an isomorphism 1 ∗1 ∼ 2 ∗2 1 ∼ 2 ϕ: G G . 1 → 2 46 EDUARDO CATTANI

We know that if ϕ: G1 G2 is a bijection then ϕ has an inverse map 1 → ϕ− : G2 G1. The following theorem says that if ϕ is a homomorphism then 1→ so is ϕ− .

Theorem 10.11. Let (G1, 1) and (G2, 2) be groups and let ϕ: G1 G2 be ∗ 1 ∗ → an isomorphism. Let ψ = ϕ− : G2 G1 be the inverse map of ϕ. Then ψ is a homomorphism. →

Proof. Let h, h" G . We need to show that ∈ 2 ψ(h h")=ψ(h) ψ(h"). ∗2 ∗1 Since ψ is the inverse map of ϕ, ψ(h)=g, where g is the unique element in G1 such that ϕ(g)=h. Similarly, ψ(h")=g", where g" is the unique element in G1 such that ϕ(g")=h". But then

ϕ(g g")=ϕ(g) ϕ(g")=h h" ∗1 ∗2 ∗2 and, since ϕ is injective, g g" is the unique element in G such that ϕ(g g")= ∗1 1 ∗1 h h" which means that ∗2 ψ(h h")=g g" = ψ(h) ψ(h"). ∗2 ∗1 ∗1 ! In the case of Example 10.3 we know that the inverse map of the exponential:

ϕ = exp: (R, +) (R>0, ) → · is the natural logarithm. So, Theorem 10.11 gives us the fact that

ψ = ln: (R>0, ) (R, +) · → is a homomorphism; i.e. that

ln(y y") = ln(y) + ln(y"). · 11. Cosets - Normal Subgroups - Quotient Groups Given a group G and a subgroup H G we can define an equivalence relation whose associated partition has a⊆ very “symmetric” form. We begin by describing the partitions: Definition 11.1. Let (G, ) be a group and H G a subgroup. Given an element a G we denote by∗ a H the set ⊆ ∈ ∗ (11.1) a H := a h : h H . ∗ { ∗ ∈ } The set a H is called the left-coset of a. Similarly we define the right-coset of a by: ∗ (11.2) H a := h a : h H . ∗ { ∗ ∈ } MATH 300 47

Clearly, if G is a commutative group, there is no distinction between left and right cosets and a H = H a for all a G. We note that if e is the∗ identity∗ element in∈G then e H = H e = H. Also, a a H since a subgroup always contains the identity∗ element.∗ In particular, left∈ (resp.∗ right) cosets are never empty. Theorem 11.2. Let (G, ) be a group, H G a subgroup, and a, b G. The following conditions are equivalent:∗ ⊆ ∈ i) a H b H = . ∗1 ∩ ∗ ) ∅ ii) a− b H. 1 ∗ ∈ iii) b− a H. iv) a H∗ =∈b H. ∗ ∗ Proof. We will prove that i) ii) iii) iv). Suppose, first of all, that i) holds. Then there exists c ⇒G such⇒ that c⇒ a H b H. This means that there exist elements h ,h ∈ H such that: ∈ ∗ ∩ ∗ 1 2 ∈ c = a h ; c = b h . ∗ 1 ∗ 2 But, this implies that a h b h and therefore ∗ 1 ∗ 2 1 1 a− b = h h− H, ∗ 1 ∗ 2 ∈ as asserted by ii). 1 Suppose now that ii) holds. Then a− b H; but then ∗ ∈ 1 1 1 b− a =(a− b)− H ∗ ∗ ∈ since H is a subgroup. Clearly, the same argument with the roles of a and b reversed shows that ii) iii). ⇔ 1 Next we show that iii) iv). We suppose that b− a H and, equivalently 1 ⇒ ∗ ∈ that a− b H, and show that a H = b H. Let c a H, then c = a h for some ∗h ∈H, but then ∗ ∗ ∈ ∗ ∗ ∈ 1 1 c = b b− a h = b ((b− a) h)=b h", ∗ ∗ ∗ ∗ ∗ ∗ ∗ 1 1 where h" =(b− a) h H since b− a H. Hence, c b H which means that ∗ ∗ ∈ ∗ ∈ ∈ ∗ a H b H. ∗ ⊆ ∗ The reverse containment follows in exactly the same way. Finally, we note that since a H = , if a H = b H then ∗ ) ∅ ∗ ∗ a H b H = a H = . ∗ ∩ ∗ ∗ ) ∅ ! Theorem 11.3. The left (resp. right) cosets define a partition of G. That is, they are non-empty subsets, pairwise disjoints, and their union is the total 48 EDUARDO CATTANI space G. The equivalence relation defined by the partition into left-cosets is given by: 1 (11.3) a b a− b H. ∼ ⇔ ∗ ∈ Analogously, the equivalence relation defined by the partition into right-cosets is given by: 1 (11.4) a b a b− H. ∼ ⇔ ∗ ∈ Proof. We prove the theorem in the case of left-cosets and leave the proof for right-cosets to the reader. The fact that the left-cosets are a partition follows from Theorem 11.2 since the only way in which two left-cosets may intersect is if they agree. We have already pointed out that left-cosets are non empty and clearly they union is all of G since given any a G we have a a H. The equivalence relation∈ defined by this∈ partition∗ is given by a b if and only if a and b belong to the same set in the partition; that is ∼ a b a H = b H. ∼ ⇔ ∗ ∗ but then Theorem 11.2 implies that 1 a b a− b H. ∼ ⇔ ∗ ∈ !

Remark: Theorem 11.3 means that for the equivalence relation (11.3) the equivalence class of a G is [a]=a H. We will denote by G/H the space of equivalence classes G/∈ deﬁned by∗ the left-cosets. Similarly, we will denote by H/G the space of equivalence∼ classes deﬁned by the right-cosets.

Example 11.4. Let G =(Z, +) and H = nZ, where n Z>0. Since Z is commutative, there is no distinction between left and right∈ cosets. Following tradition we work with left cosets. Note that the equivalence relation (11.3) is given by: a b b a nZ ∼ ⇔ − ∈ that is a b b a = k n for some k Z. ∼ ⇔ − · ∈ So, is the equivalence relation studied in Section 8. We have: ∼ 0+mZ = [0] = 0, m, 2m, . . . = mZ; { ± ± } 1+mZ = [1] = 1, 1+ m, 1+ 2m, . . . , { ± ± } and so on. As noted before, the space of cosets (or equivalence classes) is bijectively equivalent to the set Zm = 0, 1, . . . , m 1 . { − } MATH 300 49

In Section 8 we saw how to define a group operation which we denoted by in Zm. We will next show that if we make an additional assumption on the subgroup⊕ H, we can always define a group operation in the space G/H. Definition 11.5. Let (G, ) be a group. A subgroup H G is called a normal subgroup if and only if for∗ every a G, ⊆ ∈ 1 1 (11.5) a− H a = a− h a : h H = H. ∗ ∗ { ∗ ∗ ∈ } Example 11.6. It follows from the definition that if G is commutative then 1 every subgroup is normal since, in that case, a− h a = h for all a G. Thus, to find an example of a subgroup which is not∗ ∗ normal we must begin∈ with a non-commutative group G. Let G = S4 be be group of permutations of 1, 2, 3, 4 . We know that for any transpostion τ ,1 i

Theorem 11.8. Let (G, ) be a group and H a normal subgroup. Let G/H be the space of left-cosets. For∗ any a G let [a] denote the left-coset a H. Then the operation ∈ ∗ [a] [b] := [a b] ? ∗ turns G/H into a group. Moreover, the map π : G G/H → defined by π(a)=[a] is a homomorphism and ker(π)=H. Proof. Before we check that (G/H, ) satisfies the requirements of a grooup we need to verify that is well defined,? that is that it does not depend on our ? choice of element in each equivalence class (left-coset). So, suppose [a"]=[a] and [b"]=[b], we need to check that

[a b]=[a" b"]. ∗ ∗ 1 1 Now, we know from Theorem 11.2 that [a] = [b] if and only if a− b and b− a ∗ ∗ are elements of H. Similarly for a",b". But now: 1 1 1 (a b)− (a" b")=b− a− a" b" ∗ ∗ ∗ 1 ∗ 1∗ ∗ 1 = b− (a− a") (b" b− ) b 1 ∗ ∗ ∗ ∗ ∗ b− H b = H. ∈ ∗ ∗ Therefore [a b] = [a" b"]. It is now easy∗ to check∗ that is associative, that [e] is the identity and that 1 ? the inverse of [a] is [a− ] and therefore, (G/H, ) is a group. That the map π is an isomorphism follows from the definition of ? and since the identity element of G/H is [e], the kernel of π is [e]=e H = ?H. ∗ ! We conclude with a very important theorem which yields, as a consequence, a description of all cyclic groups: Theorem 11.9. Let (G , ) and (G , ) be groups and let ϕ: G G be a 1 ∗1 2 ∗2 1 → 2 homomorphism. Then there is an isomorphism from the group G1/ ker(ϕ) to the group Im(ϕ). Proof. We define a map ϕ˜: G / ker(ϕ) Im(ϕ) 1 → byϕ ˜([a]) = ϕ(a). We must begin by showing thatϕ ˜ is well-defined, i.e. its value does not depend on the representative we choose for an equivalence class in G1/ ker(ϕ). Thus, we have to show that if [a]=[b] in G1/ ker(f) then ϕ(a)=ϕ(b). 1 But, [a]=[b] if and only if a− 1 b ker(ϕ) or, equivalently, if and only if 1 ∗ ∈ ϕ(a− b)=e . But since ϕ is a homomorphism, we have ∗1 2 1 1 1 ϕ(a− b)=ϕ(a− ) ϕ(b)=ϕ(a)− ϕ(b)=e , ∗1 ∗2 ∗2 2 MATH 300 51 and therefore ϕ(a)=ϕ(b). Next, we show thatϕ ˜ is a homomorphism. Let us denote by 1 the product in the quotient space G / ker(ϕ), then we know that [a] [b]=[? a b]. Then 1 ?1 ∗1 ϕ˜([a] [b]) =ϕ ˜([a b]) ?1 ∗1 = ϕ(a b) ∗1 = ϕ(a) ϕ(b) ∗2 =˜ϕ([a]) ϕ˜([b]). ∗2 Hence,ϕ ˜ is a homomorphism. It is clear, by definition, thatϕ ˜ is surjective. So, it remains to show that it is injective. But, sinceϕ ˜ is a homomorphism it suffices to show that ifϕ ˜([a]) = e2 then [a] = [e1]. But, ifϕ ˜([a]) = e2 this means that ϕ(a)=e , i.e. that a ker(ϕ) but this implies that [a] = [e ]. 2 ∈ 1 ! Theorem 11.10. Let (G, ) be a cyclic group. Then either ∗ i) G is infinite and (G, ) is isomorphic to (Z, +), or ∗ ii) G is finite and (G, ) is isomorphic to (Zp, ), where p = G . ∗ ⊕ | | Proof. We will use the ideas introduced in the proof of Theorem 9.12. Let g G be a generator of G and consider again the map ∈ ϕ: Z G → defined by ϕ(m)=gm. We already saw in Example 10.8 that ϕ is a homomorphism and since g is a generator of the cyclic group G, ϕ is surjective. If ϕ is injective then ker(ϕ)= e and Theorem 11.9 says that Im(ϕ)=G is { } isomorphic to Z/ ker(ϕ)=Z. On the other hand, if ϕ is not injective then we showed in Example 10.8 that ker(f)=pZ for some p Z>0. Then, Im(ϕ)=G is isomorphic to ∈ Z/pZ = Zp. !