# 9. FINITELY GENERATED ABELIAN GROUPS

Save this PDF as:

Size: px
Start display at page:

## Transcription

2 = + =. = The set of all 2 2 invertible real matrices, under matrix multiplication: This is a group, but not an abelian one the commutative law doesn t hold. The set of all positive real numbers under addition: This isn t an abelian group because there s no identity zero, the identity for addition, isn t in this set. The set of all integers, under multiplication: This isn t an abelian group because it doesn t contain inverses for all its elements ½ isn t an integer, for example. xample 4: Here s an abelian group where the set consists of numbers but where the operation is neither addition nor multiplication. Instead it s a combination of the two. (G, * ) where G is {x R x 1} and where * is defined by x * y = xy + x + y. Closure Law: Suppose x, y G. We have to show that xy + x + y G. Clearly it s a real number. ll we have to do is to check that it can t be 1. Suppose that xy + x + y = 1. Then xy + x + y + 1 = 0, so (x + 1)(y + 1) = 0. Being real numbers we can conclude that x + 1 = 0 or y + 1 = 0, that is, x = 1 or y = 1. But, being elements of G, neither is equal to 1. ssociative Law: Let x, y, z G. Then x * (y * z) = x(y * z) + x + (y * z) = x(yz + y + z) + x + (yz + y + z) = xyz + xy + xz + x + yz + y + z On the other hand (x * y) * z = (x * y)z + x * y + z = (xy + x + y)z + xy + x + y + z = xyz + xz + yz + xy + x + y + z. These are clearly equal. Commutative Law: This is the only one of the five axioms that s immediately obvious. Identity Law: x * 0 = x0 + x + 0 = x for all x G so 0 is the identity of this group. In other words e = 0 for this group. Inverse Law: It s not nearly so obvious what should be the inverse of x so let s suppose that it s y. Then we d have to have xy + x + y = 0, the identity. Thus y(x + 1) = x. x Now x + 1 is never zero if x G so we may write y = x + 1 x x x x 2 + x(x + 1) x Let s check: x * x x + 1 x + 1 x + x + 1 x ll of the examples given so far have been infinite. We ll now give some finite examples. The order of a finite abelian group is its size, that is, the number of elements that it contains. We denote the order of a finite abelian group G (or any finite group for that matter) by G. xample 5: {1, 1, i, i} is an abelian group of order 4 under multiplication. 144

3 xample 6: The following table defines a binary operation under which {a, b, c, d} is an abelian group. * a b c d a c d a b b d c b a c a b c d d b a d c Closure is obvious. The associative law is somewhat tedious to check, but this operation is indeed associative. Commutativity is obvious. The identity is clearly c. Finally, since c appears all the way down the diagonal, every element of this group is its own inverse so the inverse law holds. This is an abelian group of order 4, like the one in example 5. But they re quite different in structure. It s not just the different names for the elements. The group in example 6 satisfies the property that x 2 is the identity for all x. The group in example 5 doesn t. We ll formalise this later by saying that these two groups are not isomorphic to one another. xample 7: Let G = {0, 1, 2, 3, 4, 5, 6} and let x * y be the remainder on dividing x + y by 7. x + y if x + y < 7 So x * y = x + y 7 if x + y 7 (G, * ) is an abelian group and G = 7. This group is called the group of integers modulo 7 (or mod 7 for short). We denote it by 7 and we usually use the symbol + instead of *. You just have to be careful to know whether you re talking about ordinary arithmetic or mod 7 arithmetic. lthough = 8 in ordinary arithmetic it s 1 in mod 7 arithmetic. We can describe mod 7 arithmetic by a group table. This group may look strange, but remember that it s precisely the group you can use when calculating days of the week. If today is Monday and something is happening 5 days later and something else is happening 3 days after that you can simply add 5 and 3 modulo 7 and get 1. So that second event will be on a Tuesday. Never mind that = 8 in ordinary arithmetic. If we re only interested in days of the week 7 days are equivalent to no days! xample 8: The group table for Z 4 is: This has the same pattern as the group in example 5. If you replace 1 by 0, i by 1, 1 by 2 and i by 3 in that group, you ll get exactly the same operation as in this table. More generally, for any positive integer n, we define the group of integers modulo n to be {0, 1, 2,, n 1} under the operation of addition modulo n. We denote this group by Z n and Z n = n. 145

4 9.2. Isomorphic Groups We capture the idea of two groups having essentially the same structure by introducing the concept of isomorphisms and isomorphic groups. We define an isomorphism between groups (G, * ) and (H, ) to be a 1-1 and onto function f:g H such that f(x * y) = f(x) f(y) for all x, y G. In this case we d take f(1) = 0, f(i) = 1, f( 1) = 2 and f( i) = 3. Two groups are isomorphic if there s an isomorphism from one to the other. If (G, * ) is isomorphic to (H, ) we write (G, * ) (H, ). Usually we drop the notation for the operations and simply write G H. Isomorphism is an equivalence relation. This means that for all groups G, H, K: G G; if G H then H G; if G H and H K then G K. More informally we can say that two groups are isomorphic if the elements of one can be renamed as the elements of the other so that the operation is the same in both groups. xample 9: The following two groups, G, H are defined by means of group tables. 0BG BH a b c d e f a f d e b c a b d e f c a b c e f d a b c d b c a e f d e c a b f d e f a b c d e f The following is an isomorphism from G to H: x f(x) a c b f d e To check this we translate the group table for G by means of this function: f(g) a c b f d e a f e d a b c c e d f c a b b d f e b c a f a c b f d e d b a c d e f e c b a e f d Now we rearrange rows: f(g) a c b f d e a f e d a b c b d f e b c a c e d f c a b d b a c d e f e c b a e f d f a c b f d e 146

5 and finally we rearrange columns: f(g) a b c d e f a f d e b c a b d e f c a b c e f d a b c d b c a e f d e c a b f d e f a b c d e f This is now identical to the group table for H Direct Sums If (G, * ) and (H, ) are abelian groups their direct sum is the group (G H, #) whose elements are the ordered pairs (g, h) for g G and h H and where: (g 1, h 1 ) # (g 2, h 2 ) = (g 1 * g 2, h 1 h 2 ). Usually we use additive notation for both groups and their direct sum. This means that we write all operations as +. So in G H we have (g 1, h 1 ) + (g 2, h 2 ) = (g 1 + g 2, h 1 + h 2 ). Clearly if G, H are finite groups, with G = m and H = n then G H = mn. xample 10: Z 2 Z 2 is an abelian group of order 4. Its four elements are (0, 0), (0, 1), (1, 0) and (1, 1) and the direct sum operation can be described by the following group table. Z 2 Z 2 (0, 0) (0, 1) (1, 0) (1, 1) (0, 0) (0, 0) (0, 1) (1, 0) (1, 1) (0, 1) (0, 1) (0, 0) (1, 1) (1, 0) (1, 0) (1, 0) (1, 1) (0, 0) (0, 1) (1, 1) (1, 1) (1, 0) (0, 1) (0, 0) Compare this with the group table given in example 6. That group, and this one, are isomorphic. Notice that Z 2 Z 2 is not isomorphic to Z 4. xample 11: Z 2 Z 3 Z 6 Remember that, when adding ordered pairs in Z 2 Z 3, you must add modulo 2 in the first component but modulo 3 in the second. Z 2 Z 3 (0, 0) (0, 1) (0, 2) (1, 0) (1, 1) (1, 2) (0, 0) (0, 0) (0, 1) (0, 2) (1, 0) (1, 1) (1, 2) (0, 1) (0, 1) (0, 2) (0, 0) (1, 1) (1, 2) (1, 0) (0, 2) (0, 2) (0, 0) (0, 1) (1, 2) (1, 0) (1, 1) (1, 0) (1, 0) (1, 1) (1, 2) (0, 0) (0, 1) (0, 2) (1, 1) (1, 1) (1, 2) (1, 0) (0, 1) (0, 2) (0, 0) (1, 2) (1, 2) (1, 0) (1, 1) (0, 2) (0, 0) (0, 1) The following is an isomorphism from Z 2 Z 3 to Z 6.: 147

6 x (0, 0) (0, 1) (0, 2) (1, 0) (1, 1) (1, 2) f(x) , Subgroups From now on, except in some specific examples, we ll use + as the symbol for the binary operation in an abelian group and we ll denote the identity by the symbol 0. If g is an element of an abelian group its inverse under addition will be denoted by g. We ll also define g h to be g + ( h). This is the notation we use for groups of numbers under addition, but remember we re not assuming that the elements of our groups are numbers or that addition is the one we learnt many years ago. Suppose G is an abelian group and g G. If n is a positive integer we define ng to be the sum of n copies of g. We define 0g = 0 and if n = m where m > 0 we define ng = m( g). So, for example, 3g = g + g + g. If there exists a positive integer n such that ng = 0 then we say that g has finite order. The smallest positive n for which ng = 0 is called the order of g. xample 12: The order of 2 Z 6 is 3 because, mod 6, 2 0, 2 +2 = 4 0 but = 0. subset H of an abelian group G is called a subgroup of G (denoted by H G) if: (Closure under addition) h 1, h 2 H implies that h 1 + h 2 H; (Closure under identity) 0 H; (Closure under inverses) h H implies that h H. These properties ensure that H is itself an abelian group under addition. (We don t need to stipulate associativity, because if addition is associative for all elements of G it will certainly apply within H.) Suppose that G is an abelian group and H G. If g G we define the coset containing g to be {g + h h H} and we denote it by g + H. The element g is called the representative of the coset. xample 13: If G = Z 12 and H = {0, 3, 6, 9} then H G. The cosets of H are: H = 0 + H = {0, 3, 6, 9}, 1 + H = {1, 4, 7, 10} and 2 + H = {2, 5, 8, 11}. There are other possible representatives, but they lead to the same cosets. For example 5 + H = {5, 8, 11, 2}, remembering that = 2 mod 12, and this is the same as 2 + H. Theorem 1: If G is an abelian group and H G then distinct cosets of H are disjoint. (This means that if two cosets have one element in common they must be the same coset.) Proof: Suppose g a + H and g b + H. Then g = a + h 1 = b + h 2 for some h 1, h 2 H. Hence b = a + (h 1 h 2 ). So for any h H we have b + h = a + (h 1 h 2 + h) a + H. This shows that b + H a + H. Similarly a + H b + H and so the two cosets are equal. Theorem 2 (Lagrange s Theorem): If G is a finite abelian group and H G then H divides G. Proof: Suppose there are m distinct cosets of H. Since every element of G belongs to exactly one coset and since they all have the same size as H, we have G = m. H. The set {ng n Z} is called the cyclic subgroup generated by g and is denoted by g. This is a subgroup of G since mg + ng = (m + n)g, 0 = 0g and (ng) = ( n)g. If finite, the order of this cyclic subgroup is the same as the order of the element g. 148

7 Theorem 3 (LGRNG): If G is a finite abelian group then every element of G has finite order. Moreover the order of g must divide G. Proof: Let g G and suppose that G is finite. Since {g, 2g, 3g, } must be finite we must have mg = ng for some integers m, n with m > n. Then (m n)g = 0. The second part follows from Theorem Generators and Relations In abstract algebra, of which the theory of abelian groups forms a small but important part, the most important features are the underlying patterns, not what the elements actually are. One abstract way to define abelian groups is to provide a group table. We can use this to find the sum of any two elements without having to be bothered what the elements really are. But this is clearly no good for infinite groups. nother way to define abelian groups abstractly is to use generators and relations. Suppose x 1, x 2,, x n are n symbols, representing nothing in particular. We can consider formal linear combinations of these x i. These are abstract expressions of the form: a 1 x a n x n where the coefficients, a i, are integers. We don t ask what the x i are, or what addition means. If we have to add x 1 to x 2 the answer is simply x 1 + x 2. But we can add two of these formal linear combinations in the obvious way: (a 1 x a n x n ) + (b 1 x b n x n ) = (a 1 + b 1 )x (a n + b n )x n and the answer is yet another formal linear combination. So the set of all formal linear combinations of the x i is closed under addition. The associative law clearly holds. The identity is the formal linear combination 0x x n and the inverse of the formal linear combination a 1 x a n x n is the formal linear combination ( a 1 )x ( a n )x n. We denote the group of all formal linear combinations of x 1,, x n by [x 1,, x n ]. xample 14: [x, y] = {mx + ny m, n Z}. In this group the sum of 3x + 2y and 5x + ( 7)y is 8x + ( 5)y. Now a moment s thought will make it clear that [x, y] Z Z under the isomorphism f(mx + ny) = (m, n). Similarly for more generators and more copies of Z. So generators, on their own, give us nothing new. It s the relations that really make things interesting. zero. Suppose we have a finite set of these formal linear combinations and we put each of these to a 11 x a 1n x n = 0 a 21 x a 2n x n = 0. a m1 x a mn x n = 0 s a consequence of these relations we have to identify many of the formal linear equations in our group to one another. xample 15: Suppose we have two generators and one relation 3x + 5y = 0. Then we also have the relation 7x + 10y = 7x + 10y + (3x + 5y) = 10x + 15y. But we can t conclude from 10x + 15y = 0 that 2x + 3y = 0. ddition and subtraction are possible, but not division. This is because we re working entirely with integers and division by 5 means multiplication by 1/5, which isn t an integer. 149

8 = =. We can, from 3x + 5y = 0, deduce that 3x = 5y and so we can replace every 3x in a formal linear combination by this multiple of y. This means that the elements of our abelian group can be split into three disjoint subsets: {ny n Z}, {x + ny n Z} and {2x + ny n Z}. If the coefficient of x is anything else we can replace 3x by 5y as many times as we need to to bring the coefficient of x to being one of these three values. These subsets are disjoint because if, for example, x + ny = 2x + my for some m, n Z we d have x + (n m)y = 0 which is a relation that s not a consequence of the one given. We always assume that there are no further relations other than those that follow from the ones given. If we have such a set of relations R 1 = R 2 = = R m = 0 we denote the abelian group that s generated by x 1, x 2,, x n subject to these relations by [x 1, x n R 1 = R 2 = = R m = 0]. xample 16: Consider [x 7x = 0]. formal linear combination of one generator is simply a formal multiple. So the elements of this group have the form nx for n Z. But, because of the relation 7x = 0, the distinct elements are 0, x, 2x, 3x, 4x, 5x and 6x and, for example 4x + 5x = 9x = 2x + 7x = 2x + 0 = 2x. It s clear that this group is simply the group of integers mod 7 in disguise. In other words [x 7x = 0] Z 7. xample 17: [x, y 3x = 0, 5y = 0] Z 3 Z 5. [x, y 3x = 0, 5y = 0] = {0, y, 2y, 3y, 4y, x, x + y, x + 2y, x + 3y, x + 4y, 2x, 2x + y, 2x + 2y, 2x + 3y, 2x + 4y}. The function f(mx + ny) = (m, n) is an isomorphism onto Z 3 Z Relation Matrices Rather than manipulate relations it s easier to manipulate just their coefficients, as the rows of a matrix. If we have m relations in n variables the relation matrix is the m n matrix whose rows are the coefficients of the successive relations. xample 18: The set of relations: 3x + 4y 5z 7x + 9z can be coded as the 2 3 matrix We can use this notation to provide a convenient way of representing abelian groups that are given in terms of generators and relations. We simply write down the relation matrix and, to show that what we have in mind is the abelian group, and not just the matrix, we use square brackets instead of the usual parentheses. If is an m n integer matrix, with a ij being the entry in the i-j position, then [] denotes the abelian group [x 1, x n a 11 x a 1n x n = 0,., a m1 x a mn x n = 0]. We can use any variable names. The above group is clearly isomorphic to: [y 1, y n a 11 y a 1n y n = 0,. a m1 y a mn y n = 0]. Where we have three variables we generally use the symbols x, y and z. 3 7 xample 19: [x, y 3x + 7y = 2x + 4y = 0] xample 20: 3 7. Here we ve swapped the two rows, which means that we ve simply swapped the relations. 150

9 = = = that.. when. and and are are xample 21: 4 2. Here we ve swapped the two columns. Strictly speaking these groups aren t equal, because in one group 3x + 7y = 0 while in the other we have 3y + 7x = 0. But the only difference is a swapping of the variables. This gives us a group that s isomorphic to the one we began with [x, y 3x + 7y = 0, 2x + 4y = 0] [x, y 7x + 3y = 0, 4x + 2y = 0] = 4 2 The isomorphism here is f(ax + by) = bx + ay. So we can rearrange rows of a relation matrix without changing the corresponding abelian group and, if we rearrange the columns, we get a different group but one that s isomorphic to the original one. Since we re treating isomorphic groups as if they were the same (they have the same structural pattern) we haven t changed the group under either of these operations xample 22: 2 4. Here we ve added 10 times row 2 to row 1. This has the effect of replacing the equation 3x + 7y = 0 by 3x + 7y + 10(2x + 4y) = 23x + 47y = 0. The new equation is a consequence of the original two. Furthermore if we have the two equations: 23x + 47y = 0 2x + 4y = 0 we can deduce the equation 3x + 7y = 0 by subtracting 10 times the second equation from the first. 3x + 7y = 0 23x + 47y = 0 So the two sets of equations 2x + 4y = 0 2x + 4y = 0 equivalent (they have the same consequences) and so the abelian groups 2 4 identical xample 23: Here we ve added 10 times the second column to the first. This new group is different to the first, but it s isomorphic to it. Let s see why. 3 7 [x, y 3x + 7y = 0, 2x + 4y = 0]. Now let z = y 10x. This means that y = z + 10x, so any formal linear combination in x and y can be expressed in terms of x and z. For example 5x + 7y = 5x + 7(z + 10x) = 75x + 7z. Hence x and z are equally good generators as x and y. But the relations translate into different relations when we express them in terms of x and z. 3x + 7y = 0 The system of equations 2x + 4y = 0 expressed in terms of x and z become: 3x + 7(z + 10x) = 0 73x + 7z = 0 2x + 4(z + 10x)y = 0 is, 42x + 4z = Hence [x, y 3x + 7y = 0, 2x + 4y = 0] [x, z 73x + 7z = 0, 42x + 4z = 0] [x, y 73x + 7y = 0, 42x + 4y = 0] xample 24: Let s see if we can simplify 3 7 [x, y 3x + 7y = 0, 2x + 4y = 0] [x, y (3x + 7y) (2x + 4y) = 0, 2x + 4y = 0] 151

10 in is [x, y x + 3y = 0, 2x + 4y = 0] [x, y x + 3y = 0, 2(x + 2y) = 0]. Let z = x + 2y. Then [x, y x + 3y = 0, 2(x + 2y) = 0] [y, z (z 2y) + 3y = 0, 2z = 0] [y, z y + z = 0, 2z = 0]. Now the relation y + z = 0 allows us to write y = z. So any formal linear combination that can be expressed in terms of y and z can be expressed in terms of z alone. So we can drop y as a generator and remove the relation y + z = 0. Hence [y, z y + z = 0, 2z = 0] [z 2z = 0] Z 2. ll of these steps can be done purely in terms of the relation matrix by making suitable row and column operations. We ll now repeat it, just using matrices: [2] Z 2. R 1 R 2 C 2 2C 1 C 1 C 2 C 1 C xample 25: By similar methods we can show that 1 2 the abelian group of order 1, consisting just of the identity. We denote this group by the symbol 1. It s clear that this isn t 3 7 isomorphic to the group example 24. This illustrates the fact that we re not permitted to divide the second row by 2. So we represent an abelian group, given in terms of finitely many generators and relations, by a matrix and we use elementary row and column operations to simplify it. These operations are very much like the elementary row and column operations from linear algebra, except that we don t allow multiplication or division by a constant (though multiplication by 1 is permitted). lementary integer row operations: (1) R i R j : Swap rows i and j. (2) R i R i : Change the sign of row i. (3) R i kr j : Subtract k times row j from row i (where k is any integer). lementary integer column operations: (1) C i C j : Swap columns i and j. (2) C i C i : Change the sign of column i. (3) C i kc j : Subtract k times column j from column i (where k is any integer). Theorem 4: If a matrix B is obtained from a matrix by a sequence of elementary row and column operations then [] [B]. Proof: Let G = [x 1, x n R 1 = = R m = 0] where R i = a i1 x a in x n and let = (a ij ) be the matrix of coefficients. It s clear that the elementary row operations don t change the group. ll they do is to change one set of relations into an equivalent set of relations. The column operations are a little more difficult to see, since we need to change the generators. C i C j has the effect of swapping the generators x i and x j. C i C i has the effect of replacing x i by x i. 152

11 is. C i kc j is equivalent to replacing x j by x j + kx i. This one is a little more difficult to see. Suppose we define new generators y 1,, y n by: y i = x i if i j; y j = x j + kx i. It s clear that these new generators y 1, y n will generate the group, because we can express the old generators in terms of the new ones: x i = y i if i j; x j = y j kx i = y i ky i (remember that i j so x i = y i ). Now suppose that a 1 x a n x n = 0 is a relation between the old generators. We can write this as a 1 x (a i ka j )x i + + a j (x j + kx i ) + + a n x n = 0. [Here the omitted terms all have the form a r x r.] xpressing this in terms of the new generators it becomes: a 1 y (a i ka j )y i + a i x j + a n x n = 0. [Here the omitted terms all have the form a r y r.] xpressing all the relations in terms of these new generators the coefficient matrix has k times column j subtracted from column i Direct Sums of Cyclic Groups n abelian group G is cyclic if it can be generated by a single element, g. If g has order n then G Z n. If g has infinite order then G Z. In the special case where g has order 1, g is itself the identity and so G = {0}. We denote the group of order 1 by the symbol 0. So the complete list of cyclic groups is: 0, Z 2, Z 3, Z 4, Z 5, and Z. Moreover, no two of these cyclic groups are isomorphic since they all have different orders. So if we have two abelian groups and we find that they re isomorphic to different groups in this list then we know that they can t possibly be isomorphic to one another. matrix = (a ij ) is a diagonal matrix if a ij = 0 whenever i j. So any non-zero entries lie on the (top-left to bottom-right) diagonal. Normally we reserve the term diagonal matrix for a square matrix, but not here. xample 26: a diagonal matrix. So is If a relation matrix is diagonal the abelian group [] is a direct sum of cyclic groups xample 27: Z 2 Z 5 Z 8. If there are more rows than columns in a diagonal matrix we must have one or more rows of zeros at the bottom. We can simply remove these rows they simply represent the redundant relation 0 =

12 .. xample 28: Z 2 Z 5 Z 8. If there are more columns than rows in a diagonal matrix we must have one or more columns of zeros at the right-hand end. ach one represents a variable that enters into no relation and contributes a summand of Z to the direct sum of cyclic groups decomposition xample 29: Z 2 Z 5 Z 8 Z Z. If there s a 1 on the diagonal of a diagonal matrix it represents a variable that is equated to zero. We can clearly ignore it, removing the row and column in which it lies xample 30: Z 2 Z 8. finitely generated abelian group is one that has a finite set of generators. That is, where there s a finite set {x 1, x 2,, x n } such that every element of the group can be written as an integral linear combination a 1 x a n x n for some a 1,, a n Z. concrete example of a finitely generated abelian group is any direct sum of finitely many cyclic groups. The Fundamental Theorem of belian Groups states that, up to isomorphism, this is all there is. In other words every finitely generated abelian group is isomorphic to one of the form: Z n1 Z n2 Z nr Z Z where the number of copies of Z is finite (possibly zero). In particular every finite abelian group is isomorphic to one of the form Z n1 Z n2 Z nr. Theorem 5: very finitely generated abelian group is a direct sum of cyclic groups. Proof: Suppose first that G is given by generators and relations as [x 1,, x n R 1 = = R m = 0]. We can represent G by the m n matrix of integer coefficients = (a ij ). We ll show that any integer matrix can be converted to a diagonal matrix by a suitable sequence of elementary integer row and column operations. (1) If there s a non-zero entry in the matrix choose one with the smallest absolute value and move it to the top-left-hand corner by suitable row and column swaps. (If the matrix is completely zero, we ve finished.) xample 31: C 1 C 3 R 1 R 3 (2) If the top-left-hand corner is negative make it positive by changing the sign of the top row. Call this integer m xample 32: R 1 R Here m =

13 ... we etc. and where (3) Subtract suitable multiples of the first row from the remaining rows so that the entries in the first column, from row 2 down, are in the range {0, 1, 2,, m 1}. If any of these are positive we continue the process from step (1). The fact that the top-left corner entry is decreasing each time means that this process must eventually terminate xample 33: (4) Once the first column has the form R 2 + 3R 1, R 3 R 1 R 1 R 2 m subtract suitable multiples of the first column from 0 the remaining ones so that the entries in the first row, from the second column on, are in the range {0, 1, 2,, m 1). If any of these are positive we continue the process from step (1). gain, the fact that the top-left corner entry is decreasing each time means that this process must eventually terminate xample 34: C 2 7C 1, C 3 + 9C 1 (5) ventually we arrive at a matrix where the first column has the form m 0 the first row has 0 the form (m, 0,, 0). We now have a direct summand of Z m (ignore it if m = 1) and we can remove the first row and column to produce a smaller matrix. We then begin anew on this smaller matrix xample 35: Z xample 36: (6) If the matrix is square we finally get a 1 1 matrix of the form (m) and so our final summand is: Z m if m > 1 nothing if m = 1 Z if m = 0 xample 37: [5] Z 5, [1] 0, [0] Z. If the matrix has more columns than rows we ll eventually get an n 1 matrix m 0 n > 1. 0 This will have the form. This gives us just Z m, nothing or Z, depending on m as above. The 0 s are redundant. 155

14 (C xample 38: [7] Z 7. If the matrix has more rows than columns we ll eventually get a 1 n matrix where n > 1. This will have the form (m, 0, 0). This gives us Z m, nothing or Z, as above, plus n 1 copies of Z. xample 39: [2, 0, 0, 0] Z 2 Z Z Z. In the general case, suppose that G is generated by x 1, x 2,, x n. Let S be the set of all formal linear combinations that evaluate to 0 in G. This set will be but we could list them in some way and put the coefficients into a matrix with n columns and infinitely many rows. (If you ve heard of the concepts of countably infinite and uncountable the set S is countably infinite.) The above proof will still be valid for such an infinite matrix. xample 40: [x, y, z 4x y + 5z = 0, 14x + 7y + 7z = 0] C 2 ) xample 41: Z 2 Z 8 xample 42: Z R 1 R 4 R 4 4R 1 C 2 C 1, C 3 C 1 R 2 +R Z 2 Z 8 R 2 +R Z 2 Z 8 Z R 1 R 4 R 2 R 1, R 3 3R 1, R 4 R 1 R 1 R 2 R 2 2R 1, R 3 R 1, R 4 R Z 42 Z. 8 0 Z C 2 4C 1, C 3 C 1, C 4 4C 1 C 1 C 2 R 2 +5R 1, R 3 2R 1 R 1 R R 2 2R 1 C 2 +66C 1, C 3 +78C 1 R 1 R 2 C 1 C 2 R 1 R R 2 153R 1 C 2 +5C [890] Z

15 9.8. The Isomorphism Problem There remains the problem of deciding when two direct sums of cyclic groups are isomorphic. For a start we can rearrange the summands and the group, up to isomorphism, doesn t change. xample 43: Z 2 Z 4 Z 4 Z 2. But there are other ways in which different expressions of sums of cyclic groups can represent isomorphic groups. xample 44: Z 2 Z 3 Z 6. Let g = (1, 1). The cyclic subgroup generated by g must include 0g = (0, 0), 1g = (1, 1), 2g = (0, 2), 3g = (1, 0), 4g = (0, 1) and 5g = (1, 2). But this includes all 6 elements of Z 2 Z 3 and hence this group is generated by (1, 1). It is therefore cyclic and must be isomorphic to Z 6. Theorem 6: If GCD(m, n) = 1 then Z m Z n Z mn. Proof: Suppose GCD(m, n) = 1. Let g = (1, 1) Z m Z n. We shall show that g has order precisely mn. For suppose rg = (r, r) = (0, 0). This means that r must be a multiple of m (in order to give 0 mod m). Hence r = mk, for some k Z. But, by a similar argument, r must be a multiple of n. Since m and n are coprime it follows that n divides k. So r must be a multiple of mn. Hence the smallest value of r for which rg = 0 is clearly mn. Now the group Z m Z n has order mn and so it must coincide with the cyclic group generated by g, which also has order mn. Consequently Z m Z n Z mn. xample 45: Z 3 Z 8 Z 24. Putting this another way we ve shown that every finite cyclic group is isomorphic to a direct sum of cyclic groups of prime power order. xample 46: Z 1500 Z 3 Z 4 Z 125. Now how we decide whether two directs sums of cyclic groups are isomorphic if they have prime power orders (for the same prime)? The answer is that they are isomorphic only if the orders of the cyclic summands are the same, after possible rearrangement. Theorem 7: Suppose n 1, n 2, n r, m 1, m 2,, m s are all prime powers (not necessarily for the same prime) and u, v are non-negative integers. Suppose G = Z n1 Z n2 Z nr Z Z Z (u copies of Z) and H = Z m1 Z m2 Z ms Z Z Z (v copies of Z). Then G H if and only if u = v, r = s and n i = m π(i) for i = 1, 2,, r where π is some permutation of {1, 2,, r}. Proof: The if part of the proof is obvious. The only if comes from considering the number of elements of each order. If the two groups are isomorphic they must have the same number of elements of each order and we can recover from these numbers, the orders of the finite cyclic summands. We need slightly different techniques to show that u = v, but we omit all further details here. However in specific cases it s quite easy to do prove that the groups are not isomorphic when they differ in their direct sum decomposition. But remember, this only applies where the orders of the cyclic summands are powers of the same prime. 157

16 xample 47: Show that G = Z 2 Z 8 is not isomorphic to H = Z 2 Z 4. Solution: G = 16 while H = 8. xample 48: Show that G = Z 2 Z 8 is not isomorphic to H = Z 4 Z 4. Solution: G has elements of order 8 while H doesn t. For any integer m and any prime p the group Z p m has exactly p elements, g, where pg = 0, namely 0, m, 2m,, (p 1)m. In a direct sum we multiply the numbers of such elements so that, for example, Z p m Z p n has p 2 such elements. In general the total number of such elements g with pg = 0, in a direct sum of cyclic groups is p N where N is the number of finite summands whose order is divisible by p. xample 49: Show that G = Z 2 Z 4 Z 8 is not isomorphic to H = Z 8 Z 8. Solution: G has 8 elements, g, where 2g = 0 while H has only 4 such elements. xample 50: Show that G = Z 4 Z 4 Z 8 is not isomorphic to H = Z 2 Z 8 Z 8. Solution: G has = 64 elements, g, where 4g = 0 while H has only = 32 such elements. 158

17 by by by by as is is XRCISS FOR CHPTR 9 xercise 1: Write down the relation matrix for the group [a, b, c, d 6a + 7b = 2c + 11d, 12a 6c = 27d 18b, 3c + 9d = 3a + 6b] xercise 2: Write the group a direct sum of cyclic groups. xercise 3: Write the group in xercise 1 as a direct sum of cyclic groups xercise 4: Show that the group xercise 5: Show that the group cyclic. Find its order. cyclic. Find its order xercise 1: xercise 2: SOLUTIONS FOR CHPTR 9 Z 5 Z 25 Z Z. xercise 3: rearranging the columns and changing the sign of two of them This brings the component with smallest positive absolute value to the top left corner subtracting suitable multiples of row 1 from rows 2 and rearranging rows subtracting twice the 1 st row from the 2 nd, and then changing the sign of the 3 rd row. 159

18 by by by subtracting suitable multiples of the 1 st column from the others Z 3 [9, 3] Z 3 [3, 9] Z 3 [3, 0] Z 3 Z 3 Z xercise 4: subtracting the 1 st row from the 3 rd subtracting suitable multiples of the 1 st column from the others Z 5 Z 2 Z 7 Z 70 since 2, 5 and 7 are coprime Notice that we departed slightly from the standard algorithm. Instead of moving the 2, as the smallest positive component, to the top left hand corner, we used other operations because we could see that in this case it would simplify the arithmetic xercise 5:

19 [70] Z

20 162

### 4. FIRST STEPS IN THE THEORY 4.1. A

4. FIRST STEPS IN THE THEORY 4.1. A Catalogue of All Groups: The Impossible Dream The fundamental problem of group theory is to systematically explore the landscape and to chart what lies out there. We

### = 2 + 1 2 2 = 3 4, Now assume that P (k) is true for some fixed k 2. This means that

Instructions. Answer each of the questions on your own paper, and be sure to show your work so that partial credit can be adequately assessed. Credit will not be given for answers (even correct ones) without

### Introduction to Finite Fields (cont.)

Chapter 6 Introduction to Finite Fields (cont.) 6.1 Recall Theorem. Z m is a field m is a prime number. Theorem (Subfield Isomorphic to Z p ). Every finite field has the order of a power of a prime number

### Mathematics Course 111: Algebra I Part IV: Vector Spaces

Mathematics Course 111: Algebra I Part IV: Vector Spaces D. R. Wilkins Academic Year 1996-7 9 Vector Spaces A vector space over some field K is an algebraic structure consisting of a set V on which are

### SUBGROUPS OF CYCLIC GROUPS. 1. Introduction In a group G, we denote the (cyclic) group of powers of some g G by

SUBGROUPS OF CYCLIC GROUPS KEITH CONRAD 1. Introduction In a group G, we denote the (cyclic) group of powers of some g G by g = {g k : k Z}. If G = g, then G itself is cyclic, with g as a generator. Examples

### Elements of Abstract Group Theory

Chapter 2 Elements of Abstract Group Theory Mathematics is a game played according to certain simple rules with meaningless marks on paper. David Hilbert The importance of symmetry in physics, and for

### U.C. Berkeley CS276: Cryptography Handout 0.1 Luca Trevisan January, 2009. Notes on Algebra

U.C. Berkeley CS276: Cryptography Handout 0.1 Luca Trevisan January, 2009 Notes on Algebra These notes contain as little theory as possible, and most results are stated without proof. Any introductory

### 4. CLASSES OF RINGS 4.1. Classes of Rings class operator A-closed Example 1: product Example 2:

4. CLASSES OF RINGS 4.1. Classes of Rings Normally we associate, with any property, a set of objects that satisfy that property. But problems can arise when we allow sets to be elements of larger sets

### Introduction to Modern Algebra

Introduction to Modern Algebra David Joyce Clark University Version 0.0.6, 3 Oct 2008 1 1 Copyright (C) 2008. ii I dedicate this book to my friend and colleague Arthur Chou. Arthur encouraged me to write

### Cryptography and Network Security. Prof. D. Mukhopadhyay. Department of Computer Science and Engineering. Indian Institute of Technology, Kharagpur

Cryptography and Network Security Prof. D. Mukhopadhyay Department of Computer Science and Engineering Indian Institute of Technology, Kharagpur Module No. # 01 Lecture No. # 12 Block Cipher Standards

### Elementary Matrices and The LU Factorization

lementary Matrices and The LU Factorization Definition: ny matrix obtained by performing a single elementary row operation (RO) on the identity (unit) matrix is called an elementary matrix. There are three

### 11 Ideals. 11.1 Revisiting Z

11 Ideals The presentation here is somewhat different than the text. In particular, the sections do not match up. We have seen issues with the failure of unique factorization already, e.g., Z[ 5] = O Q(

### WHAT ARE MATHEMATICAL PROOFS AND WHY THEY ARE IMPORTANT?

WHAT ARE MATHEMATICAL PROOFS AND WHY THEY ARE IMPORTANT? introduction Many students seem to have trouble with the notion of a mathematical proof. People that come to a course like Math 216, who certainly

### The Dirichlet Unit Theorem

Chapter 6 The Dirichlet Unit Theorem As usual, we will be working in the ring B of algebraic integers of a number field L. Two factorizations of an element of B are regarded as essentially the same if

### GROUPS ACTING ON A SET

GROUPS ACTING ON A SET MATH 435 SPRING 2012 NOTES FROM FEBRUARY 27TH, 2012 1. Left group actions Definition 1.1. Suppose that G is a group and S is a set. A left (group) action of G on S is a rule for

### 9. POLYNOMIALS. Example 1: The expression a(x) = x 3 4x 2 + 7x 11 is a polynomial in x. The coefficients of a(x) are the numbers 1, 4, 7, 11.

9. POLYNOMIALS 9.1. Definition of a Polynomial A polynomial is an expression of the form: a(x) = a n x n + a n-1 x n-1 +... + a 1 x + a 0. The symbol x is called an indeterminate and simply plays the role

### Notes on Factoring. MA 206 Kurt Bryan

The General Approach Notes on Factoring MA 26 Kurt Bryan Suppose I hand you n, a 2 digit integer and tell you that n is composite, with smallest prime factor around 5 digits. Finding a nontrivial factor

### PUTNAM TRAINING POLYNOMIALS. Exercises 1. Find a polynomial with integral coefficients whose zeros include 2 + 5.

PUTNAM TRAINING POLYNOMIALS (Last updated: November 17, 2015) Remark. This is a list of exercises on polynomials. Miguel A. Lerma Exercises 1. Find a polynomial with integral coefficients whose zeros include

### Linear Algebra I. Ronald van Luijk, 2012

Linear Algebra I Ronald van Luijk, 2012 With many parts from Linear Algebra I by Michael Stoll, 2007 Contents 1. Vector spaces 3 1.1. Examples 3 1.2. Fields 4 1.3. The field of complex numbers. 6 1.4.

### Chapter 4, Arithmetic in F [x] Polynomial arithmetic and the division algorithm.

Chapter 4, Arithmetic in F [x] Polynomial arithmetic and the division algorithm. We begin by defining the ring of polynomials with coefficients in a ring R. After some preliminary results, we specialize

### r + s = i + j (q + t)n; 2 rs = ij (qj + ti)n + qtn.

Chapter 7 Introduction to finite fields This chapter provides an introduction to several kinds of abstract algebraic structures, particularly groups, fields, and polynomials. Our primary interest is in

### Solutions to TOPICS IN ALGEBRA I.N. HERSTEIN. Part II: Group Theory

Solutions to TOPICS IN ALGEBRA I.N. HERSTEIN Part II: Group Theory No rights reserved. Any part of this work can be reproduced or transmitted in any form or by any means. Version: 1.1 Release: Jan 2013

### FACTORING POLYNOMIALS IN THE RING OF FORMAL POWER SERIES OVER Z

FACTORING POLYNOMIALS IN THE RING OF FORMAL POWER SERIES OVER Z DANIEL BIRMAJER, JUAN B GIL, AND MICHAEL WEINER Abstract We consider polynomials with integer coefficients and discuss their factorization

### ABSTRACT ALGEBRA: A STUDY GUIDE FOR BEGINNERS

ABSTRACT ALGEBRA: A STUDY GUIDE FOR BEGINNERS John A. Beachy Northern Illinois University 2014 ii J.A.Beachy This is a supplement to Abstract Algebra, Third Edition by John A. Beachy and William D. Blair

### Group Theory. Contents

Group Theory Contents Chapter 1: Review... 2 Chapter 2: Permutation Groups and Group Actions... 3 Orbits and Transitivity... 6 Specific Actions The Right regular and coset actions... 8 The Conjugation

### Chapter 7. Permutation Groups

Chapter 7 Permutation Groups () We started the study of groups by considering planar isometries In the previous chapter, we learnt that finite groups of planar isometries can only be cyclic or dihedral

### Mathematical Induction

Mathematical Induction (Handout March 8, 01) The Principle of Mathematical Induction provides a means to prove infinitely many statements all at once The principle is logical rather than strictly mathematical,

### Some facts about polynomials modulo m (Full proof of the Fingerprinting Theorem)

Some facts about polynomials modulo m (Full proof of the Fingerprinting Theorem) In order to understand the details of the Fingerprinting Theorem on fingerprints of different texts from Chapter 19 of the

### Groups in Cryptography

Groups in Cryptography Çetin Kaya Koç http://cs.ucsb.edu/~koc/cs178 koc@cs.ucsb.edu Koç (http://cs.ucsb.edu/~koc) ucsb cs 178 intro to crypto winter 2013 1 / 13 Groups in Cryptography A set S and a binary

### If A is divided by B the result is 2/3. If B is divided by C the result is 4/7. What is the result if A is divided by C?

Problem 3 If A is divided by B the result is 2/3. If B is divided by C the result is 4/7. What is the result if A is divided by C? Suggested Questions to ask students about Problem 3 The key to this question

### 1 Sets and Set Notation.

LINEAR ALGEBRA MATH 27.6 SPRING 23 (COHEN) LECTURE NOTES Sets and Set Notation. Definition (Naive Definition of a Set). A set is any collection of objects, called the elements of that set. We will most

### Linear Codes. Chapter 3. 3.1 Basics

Chapter 3 Linear Codes In order to define codes that we can encode and decode efficiently, we add more structure to the codespace. We shall be mainly interested in linear codes. A linear code of length

### MATH 289 PROBLEM SET 4: NUMBER THEORY

MATH 289 PROBLEM SET 4: NUMBER THEORY 1. The greatest common divisor If d and n are integers, then we say that d divides n if and only if there exists an integer q such that n = qd. Notice that if d divides

### SOLUTIONS TO EXERCISES FOR. MATHEMATICS 205A Part 3. Spaces with special properties

SOLUTIONS TO EXERCISES FOR MATHEMATICS 205A Part 3 Fall 2008 III. Spaces with special properties III.1 : Compact spaces I Problems from Munkres, 26, pp. 170 172 3. Show that a finite union of compact subspaces

### 1 VECTOR SPACES AND SUBSPACES

1 VECTOR SPACES AND SUBSPACES What is a vector? Many are familiar with the concept of a vector as: Something which has magnitude and direction. an ordered pair or triple. a description for quantities such

### (0, 0) : order 1; (0, 1) : order 4; (0, 2) : order 2; (0, 3) : order 4; (1, 0) : order 2; (1, 1) : order 4; (1, 2) : order 2; (1, 3) : order 4.

11.01 List the elements of Z 2 Z 4. Find the order of each of the elements is this group cyclic? Solution: The elements of Z 2 Z 4 are: (0, 0) : order 1; (0, 1) : order 4; (0, 2) : order 2; (0, 3) : order

### Vector Spaces. Chapter 2. 2.1 R 2 through R n

Chapter 2 Vector Spaces One of my favorite dictionaries (the one from Oxford) defines a vector as A quantity having direction as well as magnitude, denoted by a line drawn from its original to its final

Factoring the trinomial ax 2 + bx + c when a = 1 A trinomial in the form x 2 + bx + c can be factored to equal (x + m)(x + n) when the product of m x n equals c and the sum of m + n equals b. (Note: the

### 7. Some irreducible polynomials

7. Some irreducible polynomials 7.1 Irreducibles over a finite field 7.2 Worked examples Linear factors x α of a polynomial P (x) with coefficients in a field k correspond precisely to roots α k [1] of

### Some Polynomial Theorems. John Kennedy Mathematics Department Santa Monica College 1900 Pico Blvd. Santa Monica, CA 90405 rkennedy@ix.netcom.

Some Polynomial Theorems by John Kennedy Mathematics Department Santa Monica College 1900 Pico Blvd. Santa Monica, CA 90405 rkennedy@ix.netcom.com This paper contains a collection of 31 theorems, lemmas,

### 2. Let H and K be subgroups of a group G. Show that H K G if and only if H K or K H.

Math 307 Abstract Algebra Sample final examination questions with solutions 1. Suppose that H is a proper subgroup of Z under addition and H contains 18, 30 and 40, Determine H. Solution. Since gcd(18,

### 3 1. Note that all cubes solve it; therefore, there are no more

Math 13 Problem set 5 Artin 11.4.7 Factor the following polynomials into irreducible factors in Q[x]: (a) x 3 3x (b) x 3 3x + (c) x 9 6x 6 + 9x 3 3 Solution: The first two polynomials are cubics, so if

GENERATING SETS KEITH CONRAD 1 Introduction In R n, every vector can be written as a unique linear combination of the standard basis e 1,, e n A notion weaker than a basis is a spanning set: a set of vectors

### Chapter 7: Products and quotients

Chapter 7: Products and quotients Matthew Macauley Department of Mathematical Sciences Clemson University http://www.math.clemson.edu/~macaule/ Math 42, Spring 24 M. Macauley (Clemson) Chapter 7: Products

### December 4, 2013 MATH 171 BASIC LINEAR ALGEBRA B. KITCHENS

December 4, 2013 MATH 171 BASIC LINEAR ALGEBRA B KITCHENS The equation 1 Lines in two-dimensional space (1) 2x y = 3 describes a line in two-dimensional space The coefficients of x and y in the equation

### RESULTANT AND DISCRIMINANT OF POLYNOMIALS

RESULTANT AND DISCRIMINANT OF POLYNOMIALS SVANTE JANSON Abstract. This is a collection of classical results about resultants and discriminants for polynomials, compiled mainly for my own use. All results

### Revised Version of Chapter 23. We learned long ago how to solve linear congruences. ax c (mod m)

Chapter 23 Squares Modulo p Revised Version of Chapter 23 We learned long ago how to solve linear congruences ax c (mod m) (see Chapter 8). It s now time to take the plunge and move on to quadratic equations.

### Cryptography and Network Security Department of Computer Science and Engineering Indian Institute of Technology Kharagpur

Cryptography and Network Security Department of Computer Science and Engineering Indian Institute of Technology Kharagpur Module No. # 01 Lecture No. # 05 Classic Cryptosystems (Refer Slide Time: 00:42)

### it is easy to see that α = a

21. Polynomial rings Let us now turn out attention to determining the prime elements of a polynomial ring, where the coefficient ring is a field. We already know that such a polynomial ring is a UF. Therefore

### COMMUTATIVITY DEGREES OF WREATH PRODUCTS OF FINITE ABELIAN GROUPS

COMMUTATIVITY DEGREES OF WREATH PRODUCTS OF FINITE ABELIAN GROUPS IGOR V. EROVENKO AND B. SURY ABSTRACT. We compute commutativity degrees of wreath products A B of finite abelian groups A and B. When B

### Mathematics for Computer Science/Software Engineering. Notes for the course MSM1F3 Dr. R. A. Wilson

Mathematics for Computer Science/Software Engineering Notes for the course MSM1F3 Dr. R. A. Wilson October 1996 Chapter 1 Logic Lecture no. 1. We introduce the concept of a proposition, which is a statement

### Algebra 2 PreAP. Name Period

Algebra 2 PreAP Name Period IMPORTANT INSTRUCTIONS FOR STUDENTS!!! We understand that students come to Algebra II with different strengths and needs. For this reason, students have options for completing

### Unique Factorization

Unique Factorization Waffle Mathcamp 2010 Throughout these notes, all rings will be assumed to be commutative. 1 Factorization in domains: definitions and examples In this class, we will study the phenomenon

### Boolean Algebra Part 1

Boolean Algebra Part 1 Page 1 Boolean Algebra Objectives Understand Basic Boolean Algebra Relate Boolean Algebra to Logic Networks Prove Laws using Truth Tables Understand and Use First Basic Theorems

### 3. INNER PRODUCT SPACES

. INNER PRODUCT SPACES.. Definition So far we have studied abstract vector spaces. These are a generalisation of the geometric spaces R and R. But these have more structure than just that of a vector space.

### Factoring Polynomials

Factoring Polynomials Sue Geller June 19, 2006 Factoring polynomials over the rational numbers, real numbers, and complex numbers has long been a standard topic of high school algebra. With the advent

### CHAPTER SIX IRREDUCIBILITY AND FACTORIZATION 1. BASIC DIVISIBILITY THEORY

January 10, 2010 CHAPTER SIX IRREDUCIBILITY AND FACTORIZATION 1. BASIC DIVISIBILITY THEORY The set of polynomials over a field F is a ring, whose structure shares with the ring of integers many characteristics.

### 9.2 Summation Notation

9. Summation Notation 66 9. Summation Notation In the previous section, we introduced sequences and now we shall present notation and theorems concerning the sum of terms of a sequence. We begin with a

### G = G 0 > G 1 > > G k = {e}

Proposition 49. 1. A group G is nilpotent if and only if G appears as an element of its upper central series. 2. If G is nilpotent, then the upper central series and the lower central series have the same

### minimal polyonomial Example

Minimal Polynomials Definition Let α be an element in GF(p e ). We call the monic polynomial of smallest degree which has coefficients in GF(p) and α as a root, the minimal polyonomial of α. Example: We

### Similar matrices and Jordan form

Similar matrices and Jordan form We ve nearly covered the entire heart of linear algebra once we ve finished singular value decompositions we ll have seen all the most central topics. A T A is positive

### 1.5. Factorisation. Introduction. Prerequisites. Learning Outcomes. Learning Style

Factorisation 1.5 Introduction In Block 4 we showed the way in which brackets were removed from algebraic expressions. Factorisation, which can be considered as the reverse of this process, is dealt with

### A Second Course in Mathematics Concepts for Elementary Teachers: Theory, Problems, and Solutions

A Second Course in Mathematics Concepts for Elementary Teachers: Theory, Problems, and Solutions Marcel B. Finan Arkansas Tech University c All Rights Reserved First Draft February 8, 2006 1 Contents 25

### Chapter 13: Basic ring theory

Chapter 3: Basic ring theory Matthew Macauley Department of Mathematical Sciences Clemson University http://www.math.clemson.edu/~macaule/ Math 42, Spring 24 M. Macauley (Clemson) Chapter 3: Basic ring

### Math/Stats 425 Introduction to Probability. 1. Uncertainty and the axioms of probability

Math/Stats 425 Introduction to Probability 1. Uncertainty and the axioms of probability Processes in the real world are random if outcomes cannot be predicted with certainty. Example: coin tossing, stock

### 1.3 Polynomials and Factoring

1.3 Polynomials and Factoring Polynomials Constant: a number, such as 5 or 27 Variable: a letter or symbol that represents a value. Term: a constant, variable, or the product or a constant and variable.

### Ideal Class Group and Units

Chapter 4 Ideal Class Group and Units We are now interested in understanding two aspects of ring of integers of number fields: how principal they are (that is, what is the proportion of principal ideals

### Algebra Unpacked Content For the new Common Core standards that will be effective in all North Carolina schools in the 2012-13 school year.

This document is designed to help North Carolina educators teach the Common Core (Standard Course of Study). NCDPI staff are continually updating and improving these tools to better serve teachers. Algebra

### The Ideal Class Group

Chapter 5 The Ideal Class Group We will use Minkowski theory, which belongs to the general area of geometry of numbers, to gain insight into the ideal class group of a number field. We have already mentioned

### You know from calculus that functions play a fundamental role in mathematics.

CHPTER 12 Functions You know from calculus that functions play a fundamental role in mathematics. You likely view a function as a kind of formula that describes a relationship between two (or more) quantities.

### Introduction to Algebraic Geometry. Bézout s Theorem and Inflection Points

Introduction to Algebraic Geometry Bézout s Theorem and Inflection Points 1. The resultant. Let K be a field. Then the polynomial ring K[x] is a unique factorisation domain (UFD). Another example of a

### 2.3. Finding polynomial functions. An Introduction:

2.3. Finding polynomial functions. An Introduction: As is usually the case when learning a new concept in mathematics, the new concept is the reverse of the previous one. Remember how you first learned

### COMMUTATIVITY DEGREES OF WREATH PRODUCTS OF FINITE ABELIAN GROUPS

Bull Austral Math Soc 77 (2008), 31 36 doi: 101017/S0004972708000038 COMMUTATIVITY DEGREES OF WREATH PRODUCTS OF FINITE ABELIAN GROUPS IGOR V EROVENKO and B SURY (Received 12 April 2007) Abstract We compute

### Math Review. for the Quantitative Reasoning Measure of the GRE revised General Test

Math Review for the Quantitative Reasoning Measure of the GRE revised General Test www.ets.org Overview This Math Review will familiarize you with the mathematical skills and concepts that are important

### ADDITIVE GROUPS OF RINGS WITH IDENTITY

ADDITIVE GROUPS OF RINGS WITH IDENTITY SIMION BREAZ AND GRIGORE CĂLUGĂREANU Abstract. A ring with identity exists on a torsion Abelian group exactly when the group is bounded. The additive groups of torsion-free

### Systems of Linear Equations

Chapter 1 Systems of Linear Equations 1.1 Intro. to systems of linear equations Homework: [Textbook, Ex. 13, 15, 41, 47, 49, 51, 65, 73; page 11-]. Main points in this section: 1. Definition of Linear

### INTRODUCTION TO ALGEBRAIC GEOMETRY, CLASS 24

INTRODUCTION TO ALGEBRAIC GEOMETRY, CLASS 24 RAVI VAKIL Contents 1. Degree of a line bundle / invertible sheaf 1 1.1. Last time 1 1.2. New material 2 2. The sheaf of differentials of a nonsingular curve

### ON GALOIS REALIZATIONS OF THE 2-COVERABLE SYMMETRIC AND ALTERNATING GROUPS

ON GALOIS REALIZATIONS OF THE 2-COVERABLE SYMMETRIC AND ALTERNATING GROUPS DANIEL RABAYEV AND JACK SONN Abstract. Let f(x) be a monic polynomial in Z[x] with no rational roots but with roots in Q p for

### Adding vectors We can do arithmetic with vectors. We ll start with vector addition and related operations. Suppose you have two vectors

1 Chapter 13. VECTORS IN THREE DIMENSIONAL SPACE Let s begin with some names and notation for things: R is the set (collection) of real numbers. We write x R to mean that x is a real number. A real number

### CAHSEE on Target UC Davis, School and University Partnerships

UC Davis, School and University Partnerships CAHSEE on Target Mathematics Curriculum Published by The University of California, Davis, School/University Partnerships Program 006 Director Sarah R. Martinez,

### Copy in your notebook: Add an example of each term with the symbols used in algebra 2 if there are any.

Algebra 2 - Chapter Prerequisites Vocabulary Copy in your notebook: Add an example of each term with the symbols used in algebra 2 if there are any. P1 p. 1 1. counting(natural) numbers - {1,2,3,4,...}

### Alex, I will take congruent numbers for one million dollars please

Alex, I will take congruent numbers for one million dollars please Jim L. Brown The Ohio State University Columbus, OH 4310 jimlb@math.ohio-state.edu One of the most alluring aspectives of number theory

### An Innocent Investigation

An Innocent Investigation D. Joyce, Clark University January 2006 The beginning. Have you ever wondered why every number is either even or odd? I don t mean to ask if you ever wondered whether every number

### DEGREES OF ORDERS ON TORSION-FREE ABELIAN GROUPS

DEGREES OF ORDERS ON TORSION-FREE ABELIAN GROUPS ASHER M. KACH, KAREN LANGE, AND REED SOLOMON Abstract. We construct two computable presentations of computable torsion-free abelian groups, one of isomorphism

### BEGINNING ALGEBRA ACKNOWLEDMENTS

BEGINNING ALGEBRA The Nursing Department of Labouré College requested the Department of Academic Planning and Support Services to help with mathematics preparatory materials for its Bachelor of Science

### Prime Numbers and Irreducible Polynomials

Prime Numbers and Irreducible Polynomials M. Ram Murty The similarity between prime numbers and irreducible polynomials has been a dominant theme in the development of number theory and algebraic geometry.

### PROBLEM SET 6: POLYNOMIALS

PROBLEM SET 6: POLYNOMIALS 1. introduction In this problem set we will consider polynomials with coefficients in K, where K is the real numbers R, the complex numbers C, the rational numbers Q or any other

### (Q, ), (R, ), (C, ), where the star means without 0, (Q +, ), (R +, ), where the plus-sign means just positive numbers, and (U, ),

2 Examples of Groups 21 Some infinite abelian groups It is easy to see that the following are infinite abelian groups: Z, +), Q, +), R, +), C, +), where R is the set of real numbers and C is the set of

### Continuous Groups, Lie Groups, and Lie Algebras

Chapter 7 Continuous Groups, Lie Groups, and Lie Algebras Zeno was concerned with three problems... These are the problem of the infinitesimal, the infinite, and continuity... Bertrand Russell The groups

### Module MA3411: Abstract Algebra Galois Theory Appendix Michaelmas Term 2013

Module MA3411: Abstract Algebra Galois Theory Appendix Michaelmas Term 2013 D. R. Wilkins Copyright c David R. Wilkins 1997 2013 Contents A Cyclotomic Polynomials 79 A.1 Minimum Polynomials of Roots of

### Switching Algebra and Logic Gates

Chapter 2 Switching Algebra and Logic Gates The word algebra in the title of this chapter should alert you that more mathematics is coming. No doubt, some of you are itching to get on with digital design

### Group Theory via Rubik s Cube

Group Theory via Rubik s Cube Tom Davis tomrdavis@earthlink.net http://www.geometer.org ROUGH DRAFT!!! December 6, 2006 Abstract A group is a mathematical object of great importance, but the usual study

### 9 MATRICES AND TRANSFORMATIONS

9 MATRICES AND TRANSFORMATIONS Chapter 9 Matrices and Transformations Objectives After studying this chapter you should be able to handle matrix (and vector) algebra with confidence, and understand the

### FOUNDATIONS OF ALGEBRAIC GEOMETRY CLASS 22

FOUNDATIONS OF ALGEBRAIC GEOMETRY CLASS 22 RAVI VAKIL CONTENTS 1. Discrete valuation rings: Dimension 1 Noetherian regular local rings 1 Last day, we discussed the Zariski tangent space, and saw that it

### AN ALGORITHM FOR DETERMINING WHETHER A GIVEN BINARY MATROID IS GRAPHIC

AN ALGORITHM FOR DETERMINING WHETHER A GIVEN BINARY MATROID IS GRAPHIC W. T. TUTTE. Introduction. In a recent series of papers [l-4] on graphs and matroids I used definitions equivalent to the following.

### GROUP ALGEBRAS. ANDREI YAFAEV

GROUP ALGEBRAS. ANDREI YAFAEV We will associate a certain algebra to a finite group and prove that it is semisimple. Then we will apply Wedderburn s theory to its study. Definition 0.1. Let G be a finite

### discuss how to describe points, lines and planes in 3 space.

Chapter 2 3 Space: lines and planes In this chapter we discuss how to describe points, lines and planes in 3 space. introduce the language of vectors. discuss various matters concerning the relative position