Worksheets for MAT220, spring 2011

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1 Worksheets for MAT220, spring 2011 Morten Brun Matematisk Institutt Universitetet i Bergen mbr085@math.uib.no May 11, Partitions and equivalence relations Given a set S we let (S) be the set of all subsets of S. Exercise 0.1. Write a list with all the elements in ({1, 2, 3}). Definition 0.2. A partition of the set S is a subset T (S) consisting of non-empty subsets of S satisfying that every element in S is contained in exactly one of the subsets that T consists of. Figure 1: Illustration of a partition Exercise 0.3. Given sets X and Y, what do we mean when we say that f : X Y is a function? Definition 0.4. Given a partition T of the set S we can construct a function π: S T that takes x S to the subset of S in T that contains x. Exercise 0.5. Let T be the partition T = { {1}, {2, 3} } of S = {1, 2, 3} and let f : S {1, 2} be the function defined by f (1) = 1, f (2) = 1 and f (3) = 2 1

2 1 2 f 1 2 S = {1, 2, 3} T = {{1}, {2, 3}} 3 Write a list of the pairs (x, π(x)) for all x S. Does there exist a function f : T {1, 2} so that f (π(x)) = f (x) for all x S? Exercise 0.6. Write a list with all partitions of S = {1, 2, 3}. Exercise 0.7. Let T be a partition of the set S and let f : S A be a function satisfying the following: if x, y S are contained in a common element in T then f (x) = f (y). Explain why there exists a function f : T A so that f (π(x)) = f (x) for all x S. Exercise 0.8. Let f : S A be a function that is onto. Show that there exists a partition T of S consisting of the sets f 1 (a) = {x S f (x) = a} for a A. In Exercise 0.7 we have constructed a function f : T A. Explain why f is one-to-one and onto. Definition 0.9. A relation on the set S is a subset S S. We write x y instead of (x, y) Exercise Write a list with all relations on the set S = {1, 2}. Definition A relation on a set S is called an equivalence relation if the following three conditions are satisfied for all x, y, z S: 1. reflexivity: x x 2. symmetry: x y implies y x 3. transitivity: x y and y z implies x z. Exercise Write a list with all equivalence relations on the set S = {1, 2}. Example Let T be a partition of S and define S S by the formula = {(x, y) S S there exists an element A in T that contains both x and y.} 2

3 Exercise Explain why defined in Example 0.13 is an equivalence relation. Exercise Use Example 0.13 to find five different equivalence relations on the set S = {1, 2, 3}. Try to read the proof for the the theorem below. Be critical and tell which parts of the proof require extra consideration. Theorem Let S be a set. There exists a one-to-one and onto function from the set of equivalence relations on S to the set of partitions of S. More precisely there is a one-to-one and onto function that takes an equivalence relation to the partition T consisting of the sets for a S. a = {x S x a} Proof. There are three ting we must do in order to prove that we have defined a one-to-one and onto function. First we must make sure that we have defined a function at all. Next we must show that this function is one-to-one, and finally that it is onto. In our situation the most difficult part is to ensure that the above formula defines a function. Let us for the moment assume that we know this. Then x y x y gives that the function is one-to-one. (Why?) Conversely, given a partition T, in example 0.13 we have constructed a relation that maps onto T. It remains to show that the formula in the above theorem defines a function. The only thing which is not clear is why the formula defines a partition T consisting of the sets on the form a for a S. In order to see this we must show 1. every element in S is contained in a set on the form a 2. Given a, b S either a = b or a b =. Reflexivity for gives that a a for all a S, so it remains to show the second part. We do this by showing that if there exists an element x a b then a = b. Let us start by showing that a b: Let y a. Then y a and x a b gives x a and x b. Symmetry of gives a x. By transitivity of we have that y a and a x implies 3

4 y x, and the fact that y x and x b implies y b. We conclude that y b. We have shown a b. The inclusion b a is shown in the same way. Exercise Compare the above proof with the proof of Theorem 0.22 in Fraleigh s book. Definition Given a relation on a set S the set a = {x S x a} is called the equivalence class of the element a S. Remark A partition T on S is a subset of (S) satisfying a certain condition. We can consider T just as a set and forget that it is a subset of (S). This set is so important that it has a special notation: If is the equivalence relation that corresponds to T under the one-to-one correspondence in Theorem 0.16, then we mean this set when we write S/. We say that S/ is the set of equivalence classes for. Exercise 0.7 gives that if f : S A is a function so that for all x, y S we have that x y implies f (x) = f (y) then there exists a function f : S/ A satisfying f (a) = f (a) for all a S. 4

5 1 Introduction to groups To wet our apetite on groups we start by looking at symmetry groups of figures in the plane 2. To us a figure in 2 is just a subset of 2. Exercise 1.1. What are the symmetries of this figure? What do we mean when we say that a figure in the plane has a symmetry? Draw an example of a figure with mirrowing in the line x = y as symmetry. What does it mean when we say that a function f : 2 2 is a linear transformation? Is the composition of two linear transformations also linear? Let Q 2 be the square with the corners (1, 0), (0, 1), ( 1, 0) og (0, 1). Write a list of all linear transformations acting as symmetries of Q. (The black dot is Origo.) Q Write a list of the matrices corresponding to the above linear transformations. Explain why the following is true: 1. The matrices on the above list are all invertible and the inverse matrices are also on the list. 2. The composition og any two matrices on the above list is also on the list. More information about symmetry groups is given in the following interview with Ian Stewart who has written some very good populaized books about mathematics. 5

6 Definition 1.2. A linear symmetry of a figure T in 2 is a function f : 2 2 satisfying the following: 1. f takes the figure onto itself, that is, f (T) = T, 2. f is a linear transformation and 3. f is one-to-one and onto. Note that usally when we speak about symmetries we require angles and distances to be preserved, but for simplicity this requirement has been omitted here. Exercise 1.3. Give examples of figures with exactly one, two, three and four linear symmetries. Give an example of a figure with the set of matrices of the form 2 n n for n as symmetries. How many linear isometries does a regular n-gon with center in origo have? The figure below is a 6-gon. Complex numbers The complex numbers are the real plane = 2 with multiplicaition determined by the rule (a, b) (a, b ) = (aa bb, ab + a b). Usually we write 1 instead of (1, 0). Note that (1, 0) (a, b) = (a, b). Similarly we write i instead of (0, 1). With this convention we have (a, b) = (a, 0) 1 + (b, 0) i. In order to make this more readable we omit the sign for multiplication and we write a + bi instead of (a, b). Thus the rule for multiplication in is (a + bi)(a + b i) = aa bb + (ab + a b)i. 6

7 Exercise 1.4. Write (2 5i)(8 + 3i) and i 2 on the form a + bi for a, b. Exercise Write equations for sum and scalar multiplication in using the notation a + bi for (a, b) Given a, b, show the linear function f : with matrix a b b a corresponds to multiplication by a + bi. Exercise 1.6. Let S 1 = {a + bi a 2 + b 2 = 1}. Explain the following: 1. If (a + bi) S 1 and (a + b i) S 1, then (a + bi)(a + b i) S If (a + bi) S 1, then (a bi) S 1 and (a + bi)(a bi) = The associative identity holds, that is, ((a + bi)(a + b i))(a + b i) = (a + bi)((a + b i)(a + b i)). The properties of the above exersise together with the unit property 1(a + bi) = (a + bi) = (a + bi)1 are the defining properties of a group. Since = 2 the above two exercises show that we may consider multiplication by elements of S 1 as linear symmetries of S 1 2. Exercise 1.7. Every element of S 1 is of the form cos v + i sin v for some v. Use the trigonometric identities cos(u + v) = cos(u) cos(v) sin(u) sin(v) sin(u + v) = cos(u) sin(v) + sin(u) cos(v). to explain why these symmeties are rotations. 7

8 2 3 Binary operations Definition 2.1. A binary operation on a set S is a function µ: S S S. Example 2.2. µ(a, b) = ab for S equal to, or. µ(a, b) = a + b for S equal to or 2. µ(a, B) = AB for real 2 2-matrices A and B and with S equal to the set M 2 ( ) of real 2 2-matrices. Exercise 2.3. Mention other important binary operations. Definition 2.4. A subset H S is closed under µ if µ(h H) H. Exercise 2.5. Give an element-based version of the above definition. Exercise 2.6. Use symmetries of figures to give examples of subsets of M 2 ( ) that are closed under the binary operation µ with µ(a, B) equal to the matrix-product of two real 2 2-matrices A and B. Definition 2.7. Let S S be an equivalence relation on S and let µ: S S S be a binary operation on S. We say that is a congruence for µ if for all a, a, b, b in S we have that a a and b b implies µ(a, b) µ(a, b ). Example 2.8. Let S = {0, 1, 2}, let be the equivalence relation correponding to the partition T = {{0}, {1, 2}} and let µ: S S S be given by µ(a, b) = ab if ab 2 and µ(2, 2) = 0 as in the figure below. Then is not a congruence for µ T : 1 µ : Exercise 2.9. Explain why the formula µ(a, b) := µ(a, b) defines a binary relation µ: S/ S/ S/ if and only if is a congruence for µ in the sense of definition

9 The exercise below requires that you have read the passage on n on page 17 and 18 in Fraleigh s book. Exercise Let µ: be the binary operation with µ(a, b) = a + b. Given a positive integer n we define a relation n on by the formula n = {(a, a + nb) a, b }. Show that n is an equivalence relation. Show that given a, b we have that a n b if and only if a b is divisible by n. Show that n is a congruence for +, so we have a binary operation µ: / n / n / n with µ(a, b) = a + b. Show that the function ϕ : n / n given by the composite n / n of the inclusion of n = {0, 1,..., n 1} in and the function a a from to / n is one-to-one and onto. Show that ϕ above satisfies the relation ϕ(a + n b) = µ(ϕ(a), ϕ(b)). Remark It is common to write /n instead of / n. Example With respect to 3 all numbers at he same height in the following diagram are equivalent Example Let T be a set and let M be the set of functions g : T T. We can define a binary operation µ: M M M by the formula µ(h, g) = h g. Definition A binary operation µ: S S S is 1. commutative if µ(a, b) = µ(b, a) for all a, b S, 2. associative if µ(a, µ(b, c)) = µ(µ(a, b), c) for all a, b, c S. Exercise Let T be a figure in 2. Explain why the set of linear symmetries of T is closed in the set M of functions 2 2 under the compositionoperation µ from example

10 Explain why the composition-operation on the set of linear symmetries of T is associative. Find a figure T in 2 so that the composition-operation defined above on the set of linear symmetries of T not is commutative. Definition Let (S, µ) and (S, µ ) be sets with binary operations. A function ϕ : S S is a homomorphism if ϕ(µ(a, b)) = µ (ϕ(a), ϕ(b)) for all a, b S Example If H S is closed under a binary operation µ on S then the inclusion ϕ : H S is a homomorphism. Definition Let (S, µ) and (S, µ ) be sets with binary operations. 1. A homomorphism ϕ : S S is an isomorphism if the function ϕ is one-to-one and onto. 2. We say that (S, µ) and (S, µ ) are isomorphic if there exists an isomorphism ϕ : S S. Example In exercise 2.10 we have constructed an isomorphism ϕ : n /n. Exercise Prove the following two statements. If ϕ : S S is an isomorphism then also ϕ 1 : S S is an isomorphism. If (S, µ) is commutative and (S, µ ) is not commutative then is (S, µ) and (S, µ ) are not isomorphic. Definition Given a binary operation µ on S we say that an element e S is an identity element if µ(e, s) = µ(s, e) = s for all s S. Exercise Show that if e and e are identity elements for (S, µ) then e = e. Exercise Explain why the set of linear symmetries of every figure in 2 has an identity element. 10

11 4 Groups Definition 4.1. Let G be a set and let µ: G G G be a binary operation on G. We say that (G, µ) is a group if the axioms 1, 2 and 3 below are satisfied. 1 the operation µ is associative 2 there exists an identity element e G 3 for all a G there exists an element a G so that µ(a, a ) = µ(a, a) = e. Example 4.2. One of the most important groups at all is (, µ) with µ(a, b) = a + b. Let n be a positive integer. Since the function /n that takes a to a is onto it follows that /n with µ(a, b) = a + b is a group. (Why?) Exercise 4.3. Let (G, µ) be a group and let a be an element in G. Show that if a and a are elements in G so that µ(a, a ) = µ(a, a) = e and µ(a, a ) = µ(a, a) = e then a = a. Conclude that (ab) = b a. Example 4.4. In Exercise 1.1 we have seen that the set of linear symmetries of a figure T in 2 satisfies the axioms 1 3. Therefore it is natural so say the group of linear symmetries of T instead of the set of linear symmetries of T. Exercise 4.5. Let µ: be given by µ(a, b) = a + b. Find a subset S of that is closed under the operation µ so that (S, µ) is not a group. Find three different subsets S 1, S 2 and S 3 of that are closed under the operation µ so that (S i, µ) is a group for i = 1, 2, 3. Exercise 4.6. Below you find a theorem and a proof. Which parts of the proof need to be explained more carefully? Compare the proof with the proof of Theorem 4.15 Fraleigh s book. Theorem 4.7. Let (G, µ) be a group and let a, b and c be elements in G. If µ(a, b) = µ(a, c) then b = c. Proof. Let a be so that µ(a, a ) = µ(a, a) = e. Using the axioms 1 and 2 we see that µ(a, b) = µ(a, c) gives c = µ(e, c) = µ(µ(a, a), c) = µ(a, µ(a, c)) = µ(a, µ(a, b)) = µ(µ(a, a), b) = µ(e, b) 11

12 Exercise 4.8. Show that given elements a and b in a group (G, µ) there exists a unique element x in G so that µ(a, x) = b. Exercise 4.9. Write the group-table for the group of linear symmetries of the figure placed with center in 0. 12

13 5 Subgroups Definition 5.1. Let (G, µ) be a group. A group (H, µ ) is a subgroup of G if H is a subset of G and µ (a, b) = µ(a, b) for all a, b H. Formulated differently a subgroup of (G, µ) consists of a subset H of G so that the formula defines a group (H, µ ). µ (a, b) = µ(a, b) for a, b H Example 5.2. Let T be a figure in 2. The group of linear symmetries of T is a subgroup of the group of linear symmetries of the figure 2. Exercise 5.3. Give other examples of figures T 1 and T 2 in 2 so that the group of linear symmetries of T 1 is a subgroup of the group of linear symmetries of T 2. Example 5.4. The group (, +) is a subgroup of the group (, +) that again is a subgroup of (, +). However is n not a subgroup of. Exercise 5.5. Read the proof of the theorem below. further explanation? Which parts need Theorem 5.6. Let (G, µ) be a group and let H be a nonempty subset of G. The Formula µ (a, b) = µ(a, b) for a, b H defines a group (H, µ ) if and only if the following holds ( ) µ(a, b ) H for all a, b H. (Remember that b G is the element satisfying µ(b, b ) = µ(b, b) = e.) Proof. Assume first that (H, µ ) is a group. Then ( ) is satisfied because µ(a, b) = µ (a, b) for a, b H. Conversely, assume that ( ) is satisfied. Taking b = a in ( ) we get e = µ(a, a ) H so 2 is satisfied. Taking a = e in ( ) we get b = µ(e, b ) H for every b H. If b H then we have just seen that c = b H. Note that c = b. Thus given b H we have c H and c = b. Using this together with ( ) it follows that given a, b H the element µ(a, b) = µ(a, c ) is contained in H. Thus H is closed under the binary operation µ and so the axiom 1 for µ implies the axiom 1 for µ. Finally we have already checked the axiom 3 for µ. 13

14 Exercise 5.7. Let m and n be integers. Show that H = {nr + ms r, s } is a subgroup of. Exercise 5.8. Let ϕ : (G, µ) (G, µ ) be a homomorphism of groups and let e G be the identity element in G. Show that ϕ takes the identity element of G to the identity element of G. (Hint: take a close look at ϕ(µ(e, e)) with Theorem 4.7 in mind.) Show that ϕ(a) 1 = ϕ(a 1 ) for all a in G. Show that the set H = {a G ϕ(a) = e} is a subgroup of G. This subgroup is called the kernel of ϕ and we write Ker(ϕ) instead of H. Show that if Ker(ϕ) contains exactly one element then ϕ is one-toone. Show that the set H = {ϕ(a) a G} is a subgroup of G. This subgroup is called the image of ϕ and we write Im(ϕ) instead of H. Exercise 5.9. Let T : m n be a linear transformation from m to n. Suppose that T(x) = Ax for some matrix A. Explain how T can be considered as a homomorphism of groups. Show that Nul A is equal to Ker(T) and that Col A is equal to Im(T). Remark Let (G, µ) be a group. From now on we shall write ab instead of µ(a, b), and we shall write a 1 for the element in G such that a 1 a = aa 1 = e. Of course we make an exception if µ is defined by the formula µ(a, b) = a + b. Exercise Let G be a group and let a G. Explain how we can construct an element a n for every integer n so that a 0 = e, a 1 = a and a m a n = a m+n for all m, n. Exercise Let G be a group and let a G. Show that a := {a n n } is a subgroup of G. Show that if H G is a subgroup with a H then a is a subgroup of H. Conclude that a is the smallest subgroup of G that contains the element a. 14

15 Exercise Consider the group (, +) and let m. elements in the group m. Definition In the situation from exercise 5.12 we say: The Group a is the cyclic subgroup generated by a. Describe the The Group G is cyclic if there exists an element a G so that a = G. The element a G is a generator for G if a = G. Given groups (G 1, µ 1 ) and (G 2, µ 2 ) we can construct a group (G 1 G 2, µ 12 ) consisting of pairs (a 1, a 2 ) of elements with a i G i for i = 1, 2 and with µ 12 ((a 1, a 2 ), (b 1, b 2 )) = (µ 1 (a 1, b 1 ), µ 2 (a 2, b 2 )). Exercise Check that is (G 1 G 2, µ 12 ) a group. (Hint: see Theorem 11.2 in Fraleigh s book.) Exercise List the elements in the subgroup (1, 0) of the group /2 /2. We have seen that the property G is commutative is invariant under isomorphism. If G is finite we can also consider numbers depending on G that are invariant under isomorphism. Definition The order G of a finite group G is the number of elements in G. The order of a non-finite group is infinite. Definition The order of an element a G is the order of the group a. Exercise Show that if the order of a is finite, then it is equal the smallest positive integer n with the property that a n = e. Exercise Let ϕ : G G be an isomorphism of finite groups. Show the following G and G have the same order, G and G have the same number of subgroups, for every a G the elements a and ϕ(a) have the same order. Exercise Show that the groups /4 and /2 /2 both have order 4 and that they do not have the same number of subgroups. Show also that /4 contains an element of order 4 and that every element in /2 /2 has order 1 or 2. Conclude that these groups are not isomorphic. 15

16 Exercise Let G be a group and let a G. Show that the formula ϕ(n) = a n defines a homomorphism ϕ : a and that ϕ is onto. Exercise Let ϕ : G G be a homomorphism of groups so that the function ϕ is onto. Prove that if G is abelian then also G is abelian. Use exercise 5.22 to conclude that a is abelian for all a G. 16

17 6 Cyclic groups Exercise 6.1 (Division algorithm). How is division of integers performed by hand? Compute 168/7 by hand. What do we mean when we say that a division gives a remainder? Explain how it given two integers m and n with m 0 is possible to find integers r and q so that n = qm + r and 0 r < m. Exercise 6.2. Let m and n be non-negative integers with m > 0 and let Why is S non-empty? S = {a n < am}. Why does S have a smallest element? Let b be the smallest element of S, let q = b 1 and let r = n qm. Explain why n = qm + r with 0 r < m. Read the proof of the theorem below. Does anything need extra explanation? Theorem 6.3. Every subgroup of (, +) is on the form m for some integer m. Proof. Let H be a subgroup. If H = 0 then H = 0 so H is of the asserted form. Thus we may assume H 0. Choose m > 0 minimal so that m H. (Why does such an m exist?) We shall show that H = m, or more precisely that both m H and H m. Since m H the inclusion m H follows from Exercise Conversely, given an element n of H we shall show that n m. The division algorithm provides integers r and q so that n = qm + r and 0 r < m. We want to show that r = 0 so that n = qm = m + m + + m m. Since both m and n are in H we have that r = n qm is in H. If r 0 then r H and 0 < r < m which contradicts the minimality of m. Thus r = 0 and n m. This gives the inclusion H m. 17

18 Exercise 6.4. Let m and n be positive integers and let H be the subgroup H = {nr + ms r, s } of. By Theorem 6.3 the subgroup H is of the form H = d for some non-negative integer d. Show that d is the greatest common divisor of m and n. (This is explained on page 62 of Fraleigh s book.) Exercise 6.5. Let ϕ : G G be a homomorphism of groups. Show the following: If H is a subgroup of G, then is a subgroup of G. If H is a subgroup of G, then is a subgroup of G. ϕ(h) = {ϕ(a) a H} ϕ 1 (H ) = {a G ϕ(a) H } Exercise 6.6. Let ϕ : G G be a homomorphism of groups. Show that if H is a cyclic subgroup of G then ϕ(h) is a cyclic subgroup of G. Use Exercise 6.5, Exercise 5.22 and Theorem 6.3 with G = to conclude that if G = a is a cyclic group, then all subgroups of G are cyclic. Exercise 6.7. Let ϕ : G G be a homomorphism of groups, let H = Ker(ϕ) and let H be the relation H = {(a, ha) a G, h H} G G. Show that a H b if and only if ab 1 H. Show that a H b if and only if ϕ(a) = ϕ(b). Show that H is an equivalence relation. (This is related to Exercise 0.8.) Show that H is a congruence for the multiplication µ: G G G. Show that if ϕ is onto, then there is an isomorphism ϕ : G/ H G. 18

19 Exercise 6.8. Let ϕ : G be a group homomorphism which is onto. Show that G is a cyclic group. Use Exercise 6.7 to show that G is isomorphic to /n for some n 0. Use Exercise 5.22 to conclude that if G is a cyclic group then G is either isomorphic to (, +) or to ( /n, +) for some n > 0. Compare this exercise to Theorem 6.10 in Fraleigh s book. Exercise 6.9. Let m and n be integers and consider the element m = {m + rn r } in /n. By Exercise 5.22 the homomorphism ϕ : m defined by ϕ(k) = mk is onto, and Theorem 6.3 says that there exists an integer k such that Ker(ϕ) = k. Let k be the non-negative integer such that Ker(ϕ) = k. Explain why there is an isomorphism m = /k. (Hint: use Exercise 6.7) Why does there exist a least common multiple of m and n? Let lcm(m, n) be the least common multiple of m and n. Show that Ker(ϕ) = lcm(m, n)/m. Compare this exercise to Theorem 6.14 in the book of Fraleigh. Let gcd(m, n) be that greatest common divisor of m and n and let lcm(m, n) be the least common multiple of m and n. Is it true that gcd(m, n) lcm(m, n) = mn? Justify your answer. 19

20 7 Generating sets Exercise 7.1. Let G be the group of linear symmetries of the square with center in the origin of 2. Why is every element of G determined by its effect on the corners of the square? List the elements of G. Give them any names you like. What is the smallest subgroup of G containing a 90 degree clockwise rotation? What is the smallest subgroup of G containing both reflection in the x-axis and reflection in the y-axis? What is the smallest subgroup of G containing both reflection in the x-axis and in the line x = y? Given two elements a, b G how could we find the smallest subgroup of G containing a and b? Definition 7.2. Let {S i i I} be a collection of sets, that is, I is a set and for every element i I we are given a set S i. The union i I S i of the sets S i is the set S i = {x x S i for some i I}. i I Exercise 7.3. Describe i I S i in the following situations: I = {1, 2}, S 1 = {1, 2, 3}, S 2 = {3, 4, 5}. I =, S i =] i, i [ for i. I =, S i =]0, i ] for i. When I is the set I = {1, 2, 3,... } of positive integers we write i 1 S i instead of i I S i. Exercise 7.4. Let G be a group and let A G be a subset. Given a positive integer i we let S A i G be the subset of i-fold products of elements of A. S A i = {a 1 a i a 1,..., a i A} 16

21 Show that the subset S A = i 1 SA i in G. of G is closed under multiplication Show that if A = {a, a 1 } then S A = a. Show that if a 1 is in A for every a in A then S A is a subgroup of G. Let B = A A 1, where A 1 = {a 1 a A}. Show that if H is a subgroup of G containing A, then S B is contained in H, that is, A H implies S B H. Conclude that S B is the smallest subgroup of G containing A. Exercise 7.5. Show that if G is a finite group and A G is a subset of G then S A = S A A 1. Also explain why S A i = S A i+1 =... for some positive integer i. 17

22 8 Groups of Permutations Exercise 8.1. Let A be a set and let M be the set of functions g : A A. In Example 2.13 we defined a binary operation µ: M M M by the formula µ(h, g) = h g. In Exercise 2.15 we checked that µ is associative. Check again that the operation µ on M satisfies the axioms 1 and 2 in the definition of the concept of a group. Let S A M be the subset consisting of one-to-one and onto functions. Show that S A is closed under the binary operation µ. Conclude that S A satisfies the axioms 1 and 2. Explain why (S A, µ) is a group. Definition The group S A of one-to-one and onto functions of the form f : A A with multiplication given by composition of functions is the permutation group of A. 2. The elements of S A are called permutations of A. 3. We write S n instead of S {1,2,...,n} and call it the symmetric group on n letters. Exercise 8.3. Let us specify a function f : {1, 2, 3, 4, 5, 6} {1, 2, 3, 4, 5, 6} by writing f (1) f (2) f (3) f (4) f (5) f (6) Consider the functions f 1 = f 2 = f 3 = f 4 = Which of the functions specified above are permutations? For i = 1, 2, 3, 4, if it exists, write f 1 i on the same form as f i. Write f 1 f 3, f 3 f 1 and f 2 f 4 on the above form. 18

23 Does the above calculation show whether S 6 is commutative? Theorem 8.4 (Caley). Every group is isomorphic to a group of permutations. Remark 8.5. The above theorem says that given a group G there exists a set A and a subgroup G of S A such that G and G are isomorphic. Definition 8.6. Given a group G, a set A and a homomorphism ϕ : G S A we say that G acts on A. We also say that A is a G-set. The strategy we shall use to prove Caley s theorem is to let G act on itself. Exercise 8.7. Let G be a group. For x G we let λ x : G G be the function defined by the formula λ x (g) = x g. Show that λ x is one-to-one and onto. Show that there is a homomorphism ϕ : G S G with ϕ(x) = λ x. Show that ϕ is one-to-one. Conclude that G is isomorphic to Im(ϕ) and that Im(ϕ) is a subgroup of S G. Definition 8.8. We call ϕ : G S G defined in Exercise 8.7 the left regular representation of G. Remark 8.9. The phrases G acts on A, A is a G-set and A is a G- representation express the same thing. Exercise Let G be a group. For x G we let ρ x : G G be the function defined by the formula ρ x (g) = g x. Show that ρ x is one-to-one and onto. Show that there is a homomorphism ϕ : G S G with ϕ(x) = ρ x 1. Show that ϕ is one-to-one. Conclude that G is isomorphic to Im(ϕ) and that Im(ϕ) is a subgroup of S G. 19

24 9 Orbits, Cycles, and the Alternating Group Let G be a group acting on a set A, that is, G is a group, A is a set and there is a specified homomorphism ϕ : G S A. Let ϕ A A be the relation ϕ = {(a, ϕ(g)(a)) a A, g G}. Example 9.1. Let n be an integer, let G = A = and let ϕ : S be the homomorphism with ϕ(b)(a) = a + nb for a, b. Then ϕ = {(a, a + nb) a, b } is equal to the eqivalence relation n of Exercise Exercise 9.2. Show that a ϕ b if and only if b = ϕ(g)(a) for some g G and that ϕ is an equivalence relation. Definition 9.3. The equivalence calsses under ϕ of elements in A are called G-orbits. Given a A the G-orbit containing a is the denoted Ga. Exercise 9.4. Let G be a group acting on A. Show that Ga = {ϕ(g)(a) g G}. Exercise 9.5. Let H be a subgroup of a group G and let ϕ : H S G be the function with ϕ(h)(g) = hg. Show that ϕ is a homomorphism. Show that ϕ = {(g, hg) g G, h H}. Show that a ϕ b if and only if ab 1 H. Find for every g G a one-to-one and onto map H H g between the H-orbit of e and the H-orbit of g. Exercise 9.6. Let A = {1, 2, 3, 4, 5, 6}, let σ be the permutation σ =, let G = σ and let ϕ : G S A be the inclusion. Compute the G-orbit Ga for every a A. Explain the cyclic notation where for example (2, 5) =,

25 Check that σ = (2, 5)(3, 4) = (3, 4)(2, 5). Exercise 9.7. Show that if ϕ : G S A is the identity S A = SA then Ga = A for every a A. Show that if ϕ : G S G is the left regular representation, then G g = G for every g G. Let σ S A and let ϕ : G S A be the inclusion σ S A. Show that if Ga = {a} for every a A, then σ(a) = a for every a A. Definition 9.8. A permutation σ S A is a cycle if it has at most one orbit containing more than one element. Exercise 9.9. Explain why every non-identity cycle σ in S n is of the form σ = (a 1,..., a k ) for some pairwise distinct numbers a 1,..., a k {1,..., n}. Exercise Let σ S n be a permutation and let ϕ : σ S n be the inclusion. Use the following steps to see that σ can be written as a product of disjoint cycles. Show that for every σ -orbit X {1,..., n} there exists a permutation σ X S n of with σ(a) if a X σ X (a) = a if a / X. Show that if X and Y are σ -orbits then σ X σ Y = σ Y σ X. Show that if {X 1,..., X m } is the set of σ -orbits then σ = σ X1 σ X2 σ X m. Show that σ X is a cycle for every σ -orbit X. Definition Let σ S n have orbits X 1,..., X m of cardinality x 1, x 2,..., x m respectively. The parity of σ is the element in /2. pari(σ) = m (x i + 1) Theorem The function pari: S n /2 is a homomorphism. i=1 21

26 Definition The alternating group A n on n letters is the kernel of the homomorphism pari: S n /2. Exercise Use the following steps to prove Theorem Explain why (a 1, a 2, a 3,..., a m ) = (a 1, a m )(a 1, a m 1 ) (a 1, a 2 ). Show that every permutation is a product of transpositions. We have to show that pari(τσ) = pari(τ) + pari(σ) for all permutations τ and σ. Explain why it is enough to show this in the case where τ is a transposition. Explain why (i, j)(j, d 1,..., d k )(i, c 1,..., c h ) = (i, c 1,..., c h, j, d 1,..., d k ). Here i, j, c 1,..., c h, d 1,..., d k are pairwise distinct. (Compare to figure 9.16 in the book of Fraleigh.) What happens if h = 0? Use the above computations to finish the proof of Theorem Remark We shall see later that for n 4 there are very few homorphisms out of alternating group A n. More precisely it has the property that if ϕ : A n G is a homomorphism which is onto then either G = {e} or ϕ is an isomorhism. Corollary A permutation σ S n can bot be expressed both as a product of an even number of transpositions and as an odd number of transposotions. 22

27 10 Cosets and the Theorem of Lagrange Exercise Let H be a subgroup of a group G and let ψ: H S G be the function with ψ(h)(g) = gh 1. Show that ψ is a homomorphism. Show that R ψ = {(gh, g) g G, h H}. (By definition R ψ = {(g, ψ(h)(g) g G, h H}.) Recall that R ψ is an equivalence relation and show that the equivalence class of g G is the set gh = {gh h H}. Show that a ψ b if and only if a 1 b H. Find a one-to-one and onto map H gh for every g G. Remark The homomorphisms ϕ : H S G and ψ: H S G give two different acitons of H on G. In order to be able to distinguish the orbits for these two actions we write H g for the H-orbit of g G with respect to the action given by ϕ and gh for the H-orbit of g G with respect to the action given by ψ. Exercise Consider the subgroup of the symmetric group H = (1, 2) = {e, (1, 2)} S 3 = {e, (1, 2), (1, 3), (2, 3), (1, 2, 3), (1, 3, 2)}. Compute σh and Hσ for σ = (1, 2, 3). Definition Let H be a subgroup of G and let g G. we call the subset H g = {hg h H} of G the right coset of H containing g we call gh = {gh h H} the left coset of H containing g we write G/H for the set of left H-cosets gh in G. Exercise Let H be a subgroup of G. Show that the function t : G G with t(g) = g 1 is one-to-one and onto and that t(gh) = H g 1 for every g G. Show that t induces a function t : G/R ψ G/R ϕ which is one-toone and onto. 23

28 Exercise Let G be the group of linear symmetries of the square with center in the origin of 2. Let H be the cyclic subgroup of G generated by reflection in the line y = 0 and let ρ be anti-clockwise rotation by 90 degrees. Compute Hρ and ρh. Exercise Show that if H is a subgroup of a finite group G then the order of G is equal to the product of the order of H and the number of left cosets of H in G. (Hint: use the last part of Exercise 10.1.) The above result in particular gives the following: Theorem 10.8 (Lagrange). Let H be a subgroup of a finite group G. Then the order of H is a divisor of the order of G. Corollary Every group of prime order is cyclic. Proof. If a G and the order of G is p, then the order of a divides p, so either a = {e} or a = G. Thus a = G for every a e in G. Definition Let H be a subgroup of G. The index of H in G is the cardinality G/H of the the set of left H-cosets in G. Exercise Suppose that H g = gh for all g G. Show that the equivalently ϕ is a congruence for the product in G, that is, that a ϕ a and b ϕ b implies ab ϕ a b. Exercise Let S S be a congruence for µ: S S S in the sense of definition 2.7. Explain why the formula µ(a, b) := µ(a, b) defines a binary relation µ: S/ S/ S/. Definition We say that H is a normal subgroup of G if g H = H g for every g G. Exercise Use Exercise and Exercise to conclude that if H is a normal subgroup of G then there is a binary operation µ: G/H G/H G/H with µ(ah, bh) = abh. Recall that there exists a function π: G G/H with π(a) = ah. 24

29 Explain why π is onto. Use π to explain why (G/H, µ) is a group. Definition If H is a normal subgroup of G then the group G/H = (G/H, µ) constructed in Exercise is called the factor group. Example Let n be a non-negative integer, let G = and let H = n. The factor group /nz is equal to the cyclic group /n. 25

30 11 Finitely Generated Abelian Groups We have already discussed that given groups (G 1, µ 1 ) and (G 2, µ 2 ) we can construct a group (G 1 G 2, µ 12 ) consisting of pairs (a 1, a 2 ) of elements with a i G i for i = 1, 2 and with Exercise µ 12 ((a 1, a 2 ), (b 1, b 2 )) = (µ 1 (a 1, b 1 ), µ 2 (a 2, b 2 )). Prove that the projection pr 1 G 1 G 2 G 1 taking (a 1, a 2 ) to a 1 and the projection pr 2 : G 1 G 2 G 2 taking (a 1, a 2 ) to a 2 are homomorphisms of groups and that they are onto. Prove that the function G 1 G 1 G 2 taking a 1 to (a 1, e) is a homomorphism of groups and that it is one-to-one. Show that given homomorphisms ϕ i : G G i for i = 1, 2 there exists a unique homomorphism ϕ : G G 1 G 2 such that pr 1 (ϕ(g)) = ϕ 1 (g) and pr 2 (ϕ(g)) = ϕ 2 (g) for all g G. Let G 1, G 2 and G 3 be groups. Explain how the above construction specifies a group-structure on the set G 1 (G 2 G 3 ) consisting of elements of the form (a 1, (a 2, a 3 )) with a i G i for i = 1, 2, 3. This set is often denoted G 1 G 2 G 3 and its elements are often written on the form (a 1, a 2, a 3 ). Explain how we given groups G 1,..., G n can use the above construction to define a group-structure on G 1 G 2 G n = G 1 (G 2 G n ). Definition Let S S be an equivalence relation and let f : S A be a function. We say that f is compatible with if for all x, y S we have x y implies f (x) = f (y). Exercise Show that if f : S A is compatible with an equivalence relation S S, then the formula f (a) = f (a) defines a function f : S/ A. (Compare to Exercise 0.7.) 26

31 Exercise Let ϕ 2 : /2 be the function ϕ 2 (a) = a = a + 2 and let ϕ 3 : /3 be the function ϕ 3 (a) = a = a + 3. Show that ϕ 2 and ϕ 3 are compatible with the equivalence relation 6 of Exercise Conclude that there are functions ϕ 2 : /6 /2 and ϕ 3 : /6 /3 such that ϕ i (a + 6 ) = ϕ i (a) for i = 2, 3. Explain why ϕ 2 and ϕ 3 are homomorphisms. Let ϕ : /6 /2 /3 be the function with ϕ(x) = (ϕ 2 (x), ϕ 3 (x)). Compute ϕ(x) for all x /6. Conclude that ϕ is an isomorphism. Example In Exercise 5.21 we have seen that /2 /2 and /4 are non-isomorphic because no element of /2 /2 has order 4. Recall that we have introduced the symbol lcm(m, n) for the least common multiple of two numbers m and n. Note that if an element a in a group G satisfies a m = e, then m is a multiple of the order of a. In order to see this, let n be the order of a and use the division algorithm to get m = qn + r where 0 r < n. Then a m = a r = e, and so r = 0 by minimality of n. Read the proof of the theorem below. Does anything need further explanation? Theorem Let (a 1, a 2 ) G 1 G 2. If a 1 is of finite order r 1 in G 1 and a 2 is of finite order r 2 in G 2 then (a 1, a 2 ) is of order lcm(r 1, r 2 ) in G 1 G 2. Proof. Let H be the set of those integers m such that (a 1, a 2 ) m = (e, e). If m = ar 1 = br 2 is a common multiple of r 1 and r 2 then (a 1, a 2 ) m = (a ar 1 1, abr 2 2 ) = ((ar 1 1 )a, (a r 2 2 )b ) = (e, e) and so the set of common multiples of r 1 and r 2 is contained in H. Conversely, if (a 1, a 2 ) m = (e, e) then both a m 1 = e and am 2 = e. This implies that m is a multiple of both r 1 and r 2. We have shown that H is the set of common multiples of r 1 and r 2. The order of (a 1, a 2 ) is the least positive element in H, that is, it is the least common multiple of r 1 and r 2. 27

32 Exercise What is the order of the element (1, 2) in the group /4 /4? Exercise Let G 1, G 2 and G 3 be groups and let (a 1, (a 2, a 3 )) G 1 (G 2 G 3 ). Show that if the order of a i in G i is finite for i = 1, 2, 3 then the order of (a 1, (a 2, a 3 )) is the least common multiple of the orders of a 1, a 2 and a 3. Explain how the above extends by induction on the number of factors in the product. Exercise Let a and b be elements of finite order in a group G. Suppose that ab = ba and that a b = {e}. Show that the order of ab is equal to the least common multiple of the orders of a and b. (Hint: First use the assumptions to see that (ab) m = e implies that a m = b m and that this implies a m = b m = e, and then use the argument from the proof of Theorem 11.6.) Exercise Let σ S n be a permutation. Then we can find disjoint cycles c 1, c 2,... c r such that σ = c 1 c 2 c r Use the result of Exercise 11.9 to show that the order of σ is the least common multiple of the the lengths of the cycles c 1, c 2,, c r. The following is proved in Section 38 of Fraleigh s book. Theorem (Fundamental Theorem of Finitely Generated Abelian Groups). Let G be a finitely generated abelian group. There exist prime numbers p 1,..., p n with p 1 p 2... p n and positive integers r 1, r 2,... r n such p i = p i+1 implies r i r i+1 and an isomorphism ϕ : G = r /p 1 1 /pr 2 2 /pr n n. The prime numbers p i, the integers r i and the number of factors of are uniquely determined. Exercise Find all abelian groups, up to isomorphism, of order 72. (Hint: take a look at Example in Fraleigh s book.) 28

33 14 Factor Groups Recall the trigonometric identities cos(u + v) = cos(u) cos(v) sin(u) sin(v) sin(u + v) = cos(u) sin(v) + sin(u) cos(v). Exercise Let be the complex numbers. Use the above trigonometrical identities to prove that the function ϕ : with ϕ(u) = cos(u) + sin(u)i is a homomorphism from (, +) to (, µ). Consider H = 2π as a subgroup of G =. Show that ϕ is compatible with the equivalence relation R ψ on of Exercise 10.1, and conclude that there exists a function ϕ : / ψ. Explain why 2π is a normal subgroup of and why / ψ = /2π. Show that ϕ : /2π is a homomorphism and that it is one-toone. Recall that given a group (G, µ) and a normal subgroup N of G we have defined the factor group G/N consisting of the left N-cosets an = {ah h N} for a G. The multiplication in G/N is given by the formula (an)(bn) = abn. Also recall that G/N is the set of equivalence classes for the equivalence relation ψ = {(gh, g) g G, h N} G G. Exercise 14.2 (Compare to Exercise 10.14). Check (again) that ψ is a congruence for the binary operation µ: G G G. (Hint: N a = an implies that given h N there exists k N such that ha = ak.) Confirm that Exercise 2.7 provides a binary operation with µ(an, bn) = abn. Explain why (G/N, µ) is a group. µ: G/N G/N G/N 29

34 Let ϕ : G G be a homomorphism of groups with N Ker(ϕ). Show that ϕ is compatible with ψ. Explain how it follows from Exercise 0.7 that there exists a function ϕ : G/N G Show that ϕ is uniquely determined by requiring ϕ(an) = ϕ(a) for every a G. Explain why ϕ is a homomorphism. Conclude that the theorem below holds. Theorem 14.3 (Universal Property of Factor Groups). Let ϕ : G G be a homomorhism of groups and let N be a normal subgroup of G contained in the kernel of ϕ, that is, with N Ker(ϕ). There exists a unique homomorphism ϕ : G/N G with ϕ(an) = ϕ(a) for every a G. Exercise Use Theorem 14.3 to prove the theorem below. (In particular you need to show that Ker(ϕ) is a normal subgroup.) Theorem Let ϕ : G G be a homomorphism of groups and let N = Ker(ϕ). There exists a unique isomorphism ϕ : G/N Im(ϕ) with ϕ(an) = ϕ(a) for every a G. If π: G G/N is the homomorphism with π(a) = an, then ϕ(a) = ϕ(π(a)) for every a G. 30

35 15 Factor Groups and Simple Groups Exercise Let (G, µ), (G 1, µ 1 ) and (G 2, µ 2 ) be groups. Show that π: G G/{e} is an isomorphism. Show that π: G/G {e} is an isomorphism. Show that N = {(e, b) b G 2 } is a normal subgroup of G 1 G 2. Explain why the projection pr 1 : G 1 G 2 G 1 induces an isomorphism pr 1 : (G 1 G 2 )/N = G1. Exercise Let G be a group. Show that every subgroup of index two in G is normal. Exercise Show that a subgroup H of G is normal if and only if there exists a homomorhism of the form ϕ : G G such that Ker(ϕ) = H. Exercise Read the definition of the concept being a simple group in Fraleigh s book. Use Exercise 15.3 to show that the following conditions on a group G are equivalent. G is simple. If ϕ : G G is a homomorphism of groups which is onto then either G = {e} or ϕ is an isomorphism. 31

36 16 17 Counting with Actions Recall that an action of a group G on a set X is a homomorphism ϕ : G S X We say that X is a G-set. Given g G and x X we write g x instead of ϕ(g)(x). Example Let T be a figure in 2. The group of linear symmetries of T is a subgroup of the group G L( 2 ) of invertible linear transformations of 2. Moreover G L( 2 ) is a subgroup of S 2, so the group of linear symmetries of T acts on 2. Since a linear symmetry f : 2 2 of T restricts to a permutation f : T T the group of linear symmetries of T acts on T. Note that if f is a linear symmetry of T and x T then f x = f (x). Definition Let X be a G-set. 1. Given x X the G-orbit of x is G x = {g x g G}. 2. Given x X the isotropy group of x is the set G x = {g G g x = x}. 3. Given g G the set of g-fixed elements in X is the set X g = {x X g x = x}. In the rest of this section G and X are finite. Exercise Show that G x is a subgroup of G. Exercise Use the following steps to show that for every x X the number of elements in the G-orbit G x is equal to the index G/G x of the isotropy subgroup G x in G. Note that the formula f (g) = g x defines a function f : G Gx. Explain why f is onto. Show that given a, b G we have f (a) = f (b) if and only if a 1 b G x. 32

37 Conclude that given y G x there exist exactly G x elements in G mapping to y. Explain how the above gives that G = G x G x. Theorem 16.5 (Burnside s Formula). The number of G-orbits in X is 1 G Exercise Let S G X be the subset S = {(g, x) g G, x X and g x = x}. Use the following steps to prove Burnside s formula. g G X g. Explain why S = {g} X g = G x {x}. g G x X Explain why S = X g = G x. g G x X Use Exercise 16.4 to show that S = G x X Prove that for every y X we have x G y Use the above to explain that r = x X G-orbits in X. 1 G x 1 G x. 1 G x = 1. is equal to the number of Conclude that g G 1 X g = S = G G x = G r x X as claimed in Burnsied s Formula. Exercise The edges of the square are painted. Four different colors are available. Only one color is used on each edge and the same color may be used on different edges. We consider two paintings to be equivalent if there exists a linear symmetry of the figure transforming one painting into the other. Find the number of inequivalent paintings using the following steps: Explain why there are 4 4 = 256 ways of painting the edges in all. 33

38 Let G be the group of linear symmetries of and let X be the set of paintings of the square. Explain how G acts on X. Explain why X e = 256. Show that if ρ G is anti-clockwise rotation by 90 degrees, then X ρ = X ρ 3 = 4. Show that X ρ 2 = 4 2 = 16. Conclude that X g = 280 g G and that the number of G-orbits in X is = Why are there 70 inequivalent paintings of the square? How many inequivalent paintings have at most two edges of the same colour? Exercise Consider an octagon made of straight wires, all of the same length. Each edge is painted in one of two colors. How many distinguishable paintings can be made? (The octagon may be fliped.) 34

39 18 Rings and Fields Definition A ring is an abelian group (R, +) together with an associative binary operation µ: R R R satisfying the distributive laws (L) µ(a + b, c) = µ(a, c) + µ(b, c) (R) µ(a, b + c) = µ(a, b) + µ(a, c) We shall write ab instead of µ(a, b). Example Let (R, +) be an abelian group, and define µ(a, b) = 0 for all a, b R. Remark In Exercise 2.22 we have seen that there exists at most one identity element for the binary operation µ. Definition Let R be a ring with multiplication µ. If there exists an identity element for µ we write 1 this element and call it unity. Example The rings are rings with unity. Exercise There is a surjective homomorphism π: /n of abelian groups with π(a) = a + n. We would like to define a ring-structure on /n by defining π(a)π(b) = π(ab). Is this possible? Exercise Explain why matrix multiplication of n n-matrices is the product in a ring M n,n ( ) with unity. Exercise Let R be a ring. Explain why the following holds for all a, b R: 0b = a0 = 0 a( b) = ( a)b = (ab) ( a)( b) = ab Definition A homomorphism ϕ : R R of rings is a map that is a homomorphism for both the additive- and the multiplicative operations. That is, ϕ(a+ b) = ϕ(a) + ϕ(b) and ϕ(ab) = ϕ(a)ϕ(b) for all a, b R. 2. An isomorphism of rings is a homomorphism that is both one-to-one and onto. 35

40 3. A unitary homomorphism ϕ : R R of rings with unity is a homomorphism of rings satisfying that ϕ(1) = 1. Exercise Let G = (G, +) be an abelian group and let R = Hom(G, G) be the set of homomorphisms of the form ϕ : G G. Addition in R is defined by (ϕ + ψ)(g) = ϕ(g) + ψ(g). Explain why the binary operation defines a ring with unity. µ: R R R (ϕ, ψ) µ(ϕ, ψ) = ϕ ψ Exercise Use the following steps to give /n the structure of a ring with unity. Let R = Hom( /n, /n). Show that the function f : R /n with f (ϕ) = ϕ(1) is one-to-one and onto. Explain why f is an isomorphism of abelian groups. Explain how we can define a ring-structure on /n so that f becomes an isomorphism of rings. Let m be a positive integer. Explain why m is the sum m 1 = of m copies of 1. Show that if ϕ and ψ are elements in R with ϕ(1) = k and ψ(1) = m then f (ϕ ψ) = km. Conclude that the formula µ(k, m) = km defines an associative and bilinear binary pairing µ: /n /n /n. Explain why the map π: /n with π(m) = m is a homomorphism of rings. Exercise What is a commuative ring? And what is a field? 36

41 19 Integral domains In this section a ring R will be a commutative ring with unity, that is, R = (R, +,, 1) and ab = ba for all a, b R. Exercise Solve the equation 2a = 0 in the rings /2, /4 and /8. Definition Let a be a non-zero element in R. We say that a is a unit in R if there exists an element b in R with ab = ba = 1. We say that a is a zero divisor in R if there exists a non-zero element b in R with ab = ba = 0. Exercise List all zero-divisors in the ring /8. Exercise Let R be a ring with no zero divisors and let a, b R. Show that if ab = ac and a 0 then b = c. Exercise Let R =. Is (2, 1) a zero-divisor in R? What about (2, 0)? The following is Theorem 19.3 in Fraleigh s book. Does the proof in the book need further explanation? Theorem A non-zero element m in /n is a zero-divisor if and only if m and n are not relatively prime. Corollary If p is a prime, then /p has no zero-divisors. Definition An integral domain D is a commutative ring with unity 1 0 and containing no zero-divisors. Example The rings and /p for p a prime are integral domains. Exercise Show that every subring of a field is an integral domain. Theorem Every finite integral domain is a field. Proof. Let D be a finite integral domain and a be a non-zero element of D. By exercise 19.4 the function λ a : D D with λ a (b) = ab is one-to-one. Since D is finite λ a must also be onto (why?). In particular there exists an element b in D so that λ a (b) = ab = 1. Corollary If p is a prime then /p is a field. Definition The characteristic of a commutative ring R with unity 1 is the order of the additive subgroup 1 of the additive group (R, +). 37