Coset classes, Lagrange's theorem
Let be H group subgroup G... Left contiguous element a by subgroup H called a set of elements ahwhere h belongs H... Left contiguous class denotes aH... The right adjacent class of the element is introduced similarly a by subgroup Hwhich denote Ha.
Since there is always a neutral element in a subgroup, each element acontained in the coset aH (Ha).
Property 2.7. The elements a and b belong to the same left coset by subgroup H if and only if
Evidence... If, then b=ah, and therefore b belongs to the left coset aH... Conversely, let, then there are that, and.
Theorem 2.2. If left (right) cosets of elements a and b have a common element over the subgroup H, then they coincide.
Evidence... Let be . Then there will be that. An arbitrary element from the left coset aH contained in the left coset bH. Indeed, for, and, therefore,. The inclusion is proved similarly. This proves the theorem.
Corollary 2.1. The left cosets are either disjoint or coincide.
Evidence obvious.
Corollary 2.2. The left (right) coset is equal to H.
Evidence. Set the correspondence between the elements of the subgroup H and coset members aH according to the formula . The correspondence is one-to-one. This proves the statement.
Theorem 2.3 (Lagrange). The order of a finite group is divisible by the order of its subgroup.
Evidence... Let be G - order group n, and H - subgroup G order k.There is equality. Let's remove duplicate terms from the right side of the equality. As a result, there will be no overlapping cosets. Since the number of elements in the coset is equal, then where m the number of different cosets. This establishes the equality n=mk, as required.
The number of different cosets is called the subgroup index H in Group G.
A set of elements from a group G is called a generating set if G is obtained by the closure of this set with respect to a group operation.
A group generated by one element is called cyclic.
Corollary 2.3. Any group contains a cyclic subgroup.
Evidence. Let be a - group element G... The set is a cyclic subgroup.
The order of the cyclic subgroup generated by the element a, is called the order of the element.
Property 2.8. If element a has order nthen a n=e.
Evidence... Consider the sequence. Since the number of terms in the sequence is infinite, and for the powers of the element a there are a finite number of possibilities, the sequence will contain the same members. Let, where k<j and k the first repeating member. Then, and therefore, the term k-j +1 is repeated. Hence, j\u003d 1 (otherwise). Thus, the sequence consists of repeating sets of the view and in it k-1 different items. Hence, k=n+1. Since, then.
The order of any element is a divisor of the order of the group, therefore, a | G | =e for any element of the group.
Corollary 2.4. The order of the group is divisible without a remainder by the order of any element in the group.
Evidence obvious.
Theorem 2.4 (on cyclic groups)
I. For any natural n there is a cyclic group of order n.
II. Cyclic groups of the same order are isomorphic to each other.
III. A cyclic group of infinite order is isomorphic to the group of integers.
IV. Any subgroup of a cyclic group is cyclic.
V. For each divisor m numbers n (and only for them) in a cyclic group nth order there is a unique subgroup of order m.
Evidence... Many complex roots of degree n from 1 with respect to the operation of multiplication forms a cyclic group of order n... This proves the first statement.
Let the cyclic group Gorder n spawned by element aand the cyclic group H, of the same order, is generated by the element b... The match is one-to-one and preserves the operation. The second statement is proved
Cyclic group of infinite order generated by the element a, consists of elements. The match is one-to-one and preserves the operation. Thus, the third statement is proved.
Let be H - a subgroup of a cyclic group Ggenerated by the element a... The elements H are the degree a... Let's choose in H a... Let it be an element. Let us show that this element is a generator in the subgroup H... Let's take an arbitrary element from H... The work is contained in H for any r... Let's choose r equal to the quotient of division k on jthen k-rj is the remainder of the division k on j and therefore less j... Since in H there are no elements that are not degree zero a, less than jthen k-rj \u003d0, and. The fourth statement is proved.
Let the cyclic group Gorder n spawned by element a... The subgroup generated by the element has the order m... Consider the subgroup H order m... Let's choose in H element that is the smallest absolute value non-zero degree a... Let it be an element. Let us show that j \u003d n/m. Item belongs to H... Therefore, a nonzero number of the form rj-nv in absolute value not less j, which is only possible if n divided by j without a remainder. The subgroup generated has the order n/j=m, hence, j \u003d n/m... Since the generating element of a subgroup is uniquely determined by its order, the fifth statement is proved.
Finite groups
A group (semigroup) is called the ultimateif it consists of a finite number of elements. The number of elements of a finite group is called its orderly... Any subgroup of a finite group is finite. And if HÍ G - group subgroup G, then for any element andÎ G lots of On={x: x=h◦a, for any hÎ H) is called left adjacency class for G relatively H... It is clear that the number of elements in On equal to order H... (Similarly, we can formulate the definition a H - right adjacency relative to H).
It is important that for any subgroup H group G any two left (right) cosets by H either coincide or do not intersect, so any group can be represented as a union of disjoint left (right) cosets by H.
Indeed, if two classes H a and H bwhere a, bÎ Ghave a common element xthen exists tÎ H such that x = t◦a... And then the left class for x: H x={y: y=h◦x= h◦(t◦a) = (h◦t)◦a} Í H abut a= t ‑1 ◦x and H a={y: y=h◦a= h◦(t ‑1 ◦x) = (h◦t ‑1)◦x} Í H x... From here H x= H a... Similarly, one can show that H x= H b... And therefore H a= H b... If the classes H a and H b do not have common elements, then they do not intersect.
Such a partition of a group into left (right) cosets is called decomposition of the group into the subgroup Н.
Theorem 2.6.1. The order of a finite group is divisible by the order of any of its subgroups.
Evidence. Because GIs a finite group, then any of its subgroups H has finite order. Consider a decomposition of a group into a subgroup H... In each coset in this decomposition, the number of elements is the same and equal to the order H... Therefore, if n - group order G, and k - subgroup order Hthen n=m× kwhere m Is the number of adjacency classes by H in group decomposition G.
If for any element aÎ G Þ H a= a H (left and right cosets by subgroup H coincide), then H called normal divisor group G.
Statement: if a G Is a commutative group, then any of its subgroups H is a normal divisor G.
In view of the associativity of the action in a group (semigroup), we can speak of the "product" of three elements ( and◦b◦c) =(and◦b)◦c = and◦(b◦c). The concept of a complex product from n elements: and 1 ◦and 2 ◦…◦a n = ◦ a n = = ◦.
Composition nidentical elements of the group is called degree of element and denoted a n\u003d. This definition makes sense for any natural n... For any element of the group aÎ G denote and 0 =e - neutral element of the group G... And negative powers of an element a ‑ n defined as ( a ‑1) n or ( a n) ‑1, where a ‑1 - inverse element to and... Both definitions a ‑ n coincide, because a n◦(a ‑1) n = (and◦and◦ ¼◦ and)◦(a ‑1 ◦a ‑1 ◦ ¼◦ a ‑1) = and◦and◦¼◦( and◦a ‑1)◦a ‑1 ◦¼◦ a ‑1 =e n =e... In this way, ( a ‑1) n = (a n) ‑1 .
In the additive group, the analogue of the degree of the element a n will be n- multiple to it, usually denoted nawhich should not be taken as a work n on and, insofar as nÎℕ and maybe nÏ G... So na⇋, where nÎℕ, and 0 and=e⇋0, and (- n)a = ‑(na) = n(‑a) for any natural n, where (- a) - inverse to aÎ G.
It is easy to show that under the chosen notation for any integers m and n and for anyone aÎ G the known properties are fulfilled: and) with multiplicative notation a n ◦a m = a n + m and ( a n) m = a nm; b) with the additive notation na+ma = (n+m)a and n(ma)=(nm)a.
Consider a subset of the group Gcomposed of all powers of an arbitrary element gÎ G... Let's denote it A g... In this way, A g ={g 0 , g 1 , g ‑1 , g 2 , g ‑2, ¼). Obviously A g is a subgroup of the group Gsince for any elements x,atÎ A g follows that ( x◦at)Î A g, and for any element xÎ A g there will be x ‑1 Î A g, Besides, g 0 =eÎ A g.
Subgroup A g called cyclic subgroup group Ggenerated by the element g... This subgroup is always commutative, even if itself G not commutative. If the group G coincides with one of its cyclic subgroups, then it is called cyclic groupgenerated by the element g.
If all degrees of an element g different, then the group G called endless cyclic group, and element g - element infinite order.
If among the elements of the cyclic group there are equal, for example, g k=g m at k>mthen g k - m=e; and denoting k-m through n, we get g n=e, nÎℕ.
Least natural rate n such that g n=eis called order of element g, and the element itself g called element of finite order.
Such an element will always be found in a finite group, but it can also be in an infinite group.
Groups, all elements of which are of finite order, are called periodic.
Since any element of a finite group is of finite order, all finite groups are periodic. In addition, all cyclic subgroups of a finite group are periodic, since they are finite, and each element of finite order n generates a cyclic group of the same order nconsisting of elements ( g 0 , g 1 , g 2, ¼, g n -1 ). Indeed, if the number of elements were equal to some k<nthen g k=e=g n, which contradicts the choice nas the least degree such that g n=e; on the other hand, k>n also impossible, since in this case, there would be the same elements.
Statement: 1) all degrees g 0 , g 1 , g 2, ¼, g n ‑1 are different because if there were equal, for example, g i=g j (i>j), then g i - j=ebut ( i‑j)<nand by definition n - the smallest degree such that g n=e.
2) Any other degree g, positive or negative, is equal to one of the elements g 0 , g 1 , g 2, ¼, g n ‑1, because any integer k can be represented by the expression: k=nq+rwhere q,rÎℤ and £ 0 r<n, r - remainder and g k=g nq + r= g nq° g r= (g n) q° g r= e q° g r= g r.
1) Every group has a unique first-order element ( e) generating a first-order cyclic subgroup consisting of one element e.
2) Consider the permutation group S 3, consisting of the elements:,,,,,. Order S 3 \u003d 6. Item order and is equal to 2, because ... Item order b is also equal to 2, since ... Item order from is equal to 3, because and. Item order f is also equal to 3, because and. And finally order d is equal to 2, because ... Thus, the cyclic subgroups S 3 generated by elements e, a, b, d, c and f, respectively equal: ( e}, {e, a}, {e, b}, {e, d}, {e, c, f) and ( e, f, c), where the last two coincide. Note also that the order of each cyclic subgroup divides the order of the group without remainder. The following theorem is true.
Theorem 2.7.1. (Lagrange) The order of a finite group is divisible by the order of any of its elements (since the order of the element and the order of the cyclic subgroup generated by it coincide).
This also implies that any element of a finite group, when raised to a power of the order of the group, gives the unit of the group. (Because g m=g nk=e k=ewhere m - group order, n - item order g, k Is an integer).
Group S has 3 subgroup H={e, c, f) is a normal divisor, and 2nd order subgroups are not normal divisors. This can be easily verified by finding the left and right cosets by H for each element of the group. For example, for the element and left adjacency class On={f ◦ a, from◦ and, f◦ a} = {and, b, d) and the right coset a H={a ◦ f, and◦ c, and◦ f} = {and, d, b) match. Similarly for all other elements S 3 .
3) The set of all integers with addition forms an infinite cyclic group with the generating element 1 (or –1), since any integer is a multiple of 1.
4) Consider the set of roots n-Th degree of unit: E n\u003d. This set is a group with respect to the operation of multiplying roots. Indeed, the product of any two elements e k and e m of E nwhere k, m £ n-1 will also be an element E nsince \u003d \u003d, where r=(k + m) mod n and r £ n-1; multiplication associative, neutral element e=e 0 \u003d 1 and for any element e k there is a reverse and. This group is cyclic, its generating element is the primitive root. It is easy to see that all degrees are different:, further for k³ n the roots begin to repeat themselves. On the complex plane, the roots are located on a circle of unit radius and divide it by n equal arcs, as shown in Figure 11.
The last two examples exhaust essentially all cyclic groups. Since the following theorem is true.
Theorem 2.7.2. All infinite cyclic groups are isomorphic to each other. All finite cyclic groups of order n are isomorphic to each other.
Evidence. Let be ( G, ∘) is an infinite cyclic group with a generator g... Then there is a bijective mapping f: ℤ ® G such that for any integers k and m their images f(k) and f(m), equal respectively g k and g mare elements G... And wherein f(k+m)=f(k)∘f(m), insofar as g k + m=g k∘ g m.
Now let ( G, ∘) is a finite cyclic group of order nwith parent element g... Then each element g kÎ G the only way is to match an element e kÎ E n(0£ k<n), according to the rule f(g k)=e k... And at the same time for any g k and g mÎ G follows that f(g k∘ g m)= f(g k) ∘ f(g m), insofar as f(g k∘ g m)= f(g k + m)= f(g r), where r=(k+m) mod nand f(g r)=e r=e k× e m... It is clear that such a comparison is a bijective mapping.
Consider the multiplicative group of all integer powers of two (2Z,), where 2Z \u003d (2 n | p e Z). An analogue of this group in the additive language is the additive group of even integers (2Z, +), 2Z \u003d (2n | n e Z). Let us give a general definition of groups, particular examples of which are given groups.
Definition 1.8. Multiplicative group (G, ) (the additive group (G, +)) is called cyclical, if it consists of all integer powers (respectively, all integer multiples) of one element a e G, those. G \u003d (a n | n e Z) (respectively, G - (pa | n e Z)). Designation: (a), reads: cyclic group generated by element a.
Let's consider some examples.
- 1. An example of a multiplicative infinite cyclic group is the group of all integer powers of some fixed integer a f ± 1, it is denoted and Mr. In this way, a d - (a).
- 2. An example of a multiplicative finite cyclic group is the group C „of nth roots of unity. Recall that the nth roots of unity are found
according to the formula e k \u003d cos --- hisin ^ -, where k \u003d 0, 1, ..., p - 1. Next n n
therefore, Cn \u003d (ex) \u003d (ex \u003d 1, ex, ef \u003d e 2, ..., e "-1 \u003d?" _x). Recall that the complex numbers e k, k = 1, ..., p - 1, are depicted by the points of the unit circle that divide it by p equal parts.
- 3. A typical example of an additive infinite cyclic group is the additive group of integers Z, it is generated by the number 1, that is, Z \u003d (1). Geometrically, it is depicted as whole points of a number line. In essence, the multiplicative group is depicted in the same way 2 7 - = (2) in general a z \u003d (a), where integer a f ± 1 (see fig. 1.3). We will discuss this similarity of images in Section 1.6.
- 4. In an arbitrary multiplicative group, choose Gsome element and. Then all integer powers of this element form a cyclic subgroup (a) \u003d (a n n e Z) G.
- 5. Let us prove that the additive group of rational numbers Q itself is not cyclic, and any two of its elements lie in a cyclic subgroup.
A. Let us prove that the additive group Q is not cyclic. Suppose the opposite: let Q \u003d (-). There is an integer B,
not dividing t. Since - eQ \u003d (-) \u003d sn- | neZ\u003e, then there is
B t / (t J
there is an integer rc 0 such that - \u003d n 0 -. But then m \u003d n 0 kb,
from where t: b - came to a contradiction.
B. Let us prove that two arbitrary rational numbers -
from „ /1
and - belong to the cyclic subgroup (-), where t is the most d t /
lesser common multiple b and d. Indeed, let m-bu
, and ai 1 /1 from cv 1/1
and m \u003d av, u, v ∈ Z, then - \u003d - \u003d ai-e (-) and - \u003d - \u003d cv- e (-).
b b and t t / a dv t t /
Theorem 1.3. The order of a cyclic group is equal to the order of the generating element of this group, i.e. | (a) | \u003d | a |.
Evidence. 1. Let | a | \u003d "\u003e. Let us prove that all natural powers of the element and are different. Suppose the opposite: let a k \u003d a t and 0 k Then t - to is a natural number and a m ~ k \u003d e. But this contradicts the fact that | a \u003d °°. Thus, all natural degrees of an element and are different, whence the infinity of the group (a) follows. Therefore, | (a) | \u003d °° \u003d | a |.
2. Let | and | \u003d n. Let us prove that (a) \u003d (e - a 0, a, a 2, ..., a "-1). The definition of a cyclic group implies the inclusion (a 0, a, a 2, ..., o" 1-1) c (a). Let us prove the reverse inclusion. An arbitrary element of a cyclic group (and) has the form and t, Where those Z. Divide the schnapps with the remainder: m-nq + r, where 0 p. Since a n \u003d e, then and t = a n i + r \u003d a n h? a d \u003d a d f (a 0, a, a 2, ..., a "- 1). Hence (a) c (a 0, a, a 2, ..., Thus, (a) \u003d (a 0, a, a 2, ..., a" -1 ).
It remains to prove that all elements of the set (a 0, a, a 2,..., a ”-1) are different. Suppose the opposite: let 0 i p,but a "\u003d and). Then oh - e and 0 j - i - came to a contradiction with the condition | and | \u003d p. The theorem is proved.
Subgroups of cyclic groups
The following theorem describes the structure of subgroups of cyclic groups.
Theorem 1.4. A subgroup of a cyclic group is cyclic. If G \u003d (a) uH - nontrivial subgroup of the group G, moH \u003d (a P), where n - the smallest natural number such that a ne N.
Evidence. Let G \u003d (a) and H - group subgroup G.If the subgroup H single, then H \u003d (f) is a cyclic group. Let be H - a non-unit subgroup. Let us denote by p smallest natural number such that ape N, and prove that H \u003d (a p). Inclusion ( a n) from H obvious. Let us prove the reverse inclusion. Let be h e N. Insofar as G \u003d (a), then there is a whole indicator to, such that h \u003d a to. Divide to on p with the remainder: to = nq + r, where 0 n. If we assume that g f 0, then we get h \u003d a k \u003d a na p h a g, whence a r \u003d a ~ ppN e N. We came to a contradiction with the minimum indicator p. Therefore, r \u003d 0 and k - nq. From here h \u003d a k \u003d a p h e a "). Thus, H from ( and n), which means H \u003d (a P). The theorem is proved.
Cyclic group generators
What elements can generate a cyclic group? The following two theorems answer this question.
Theorem 1.5. Let a cyclic group G \u003d (a) of infinite order be given. Then (a) - (and to) if and only if k - ± 1.
Evidence. Let be G \u003d (a), | a | \u003d °° and (a) \u003d (a to). Then there is an integer p, so what a \u003d a kn. Hence a * "-1 \u003d e,and since | a \u003d then kp - 1 \u003d 0. But then kp \u003d 1 uk- ± 1. The converse is obvious.
Theorem 1.6. Let a cyclic group G \u003d (a) of order m be given.Then (a) \u003d (a k) if and only if GCD (/ s, t) = 1.
Evidence. (\u003d\u003e) Let (a) \u003d (a to), we prove that gcd (/ c, t) - 1. Let's denote NODCs, t) - d. Insofar as and e (a) - (a k), then a \u003d a kp with some general p. By the property of the orders of the elements, this implies that (1 - kp) : t, those. 1 - kp \u003d mt for some integer t. But then 1 \u003d (kp + mt) : d, whence d \u003d 1 and gcd (/ c, t)= 1.
(Let gcd (k, t) \u003d 1. Let us prove that (and) = (a to). Inclusion (a to) c (a) is obvious. Conversely, from the condition GCD #, t) \u003d 1 implies the existence of integers and and v such that ki + mv \u003d 1. Taking advantage of the fact that | and | - t, we get a \u003d a ku + mv \u003d a ku a mv \u003d a ku e (a k). Hence, (a) \u003d (a k). The theorem is proved.
Recall that euler function φ (t) is defined as the number of natural numbers not exceeding natural number t and mutually simple with t. Hence we get a corollary.
Consequence. Cyclic group (and) order t has φ (m) different generating elements.
To give geometric clarity to Theorem 1.5, we represent the cyclic group G \u003d (a) order t points of a circle A 0, A b ..., A t _ b dividing it into t equal parts. Element and to of this group corresponding to the point A to, will be generating if and only if, connecting successively the points A 0, A k, A 2k etc., we will come to point A]. We will find all such toat t \u003d 10 by a simple enumeration of cases (Fig. 1.5). As a result, we get k \u003d1,3, 7, 9. For a cyclic group (and) this means that (a) \u003d (a 3) \u003d (a 7) \u003d (a 9). Conversely: by finding to, coprime with a given number t, you can safely draw the corresponding "asterisk", firmly knowing that sooner or later you will get to every point, because (a) \u003d ( and to).
subgroup is called cyclic subgroup... Term exponentiation here means multiple application to a group operation element:The set obtained as a result of this process is denoted in the text as ... Note also that a 0 \u003d e.
Example 5.7
From the group G \u003d< Z 6 , +> four cyclic subgroups can be obtained. it H 1 \u003d<{0},+>, H 2 \u003d<{0, 2, 4}, +>, H 3 \u003d<{0, 3}, +> and H 4 \u003d G. Note that when the operation is addition, then a n means multiplication of n by a. Note also that in all these groups the operation is addition mod 6. Shown below is how we find the elements of these cyclic subgroups.
a. The cyclic subgroup generated from 0 is H 1 has only one element (neutral element).
b. The cyclic subgroup generated from 1 is H 4, which is G itself.
1 0 mod 6 \u003d 0 1 1 mod 6 \u003d 1 1 2 mod 6 \u003d (1 + 1) mod 6 \u003d 2 1 3 mod 6 \u003d (1 + 1 + 1) mod 6 \u003d 3 1 4 mod 6 \u003d (1 + 1 + 1 + 1) mod 6 \u003d 4 1 5 mod 6 \u003d (1 + 1 + 1 + 1 + 1) mod 6 \u003d 5 (stop, then the process repeats)
in. The cyclic subgroup generated from 2 is H 2, which has three elements: 0, 2, and 4.
2 0 mod 6 \u003d 0 2 1 mod 6 \u003d 2 2 2 mod 6 \u003d (2 + 2) mod 6 \u003d 4 (stop, then the process repeats)
d. The cyclic subgroup generated from 3 is H 3, which has two elements: 0 and 3.
e. Cyclic subgroup generated on the basis of 4, - H 2; this is not a new subgroup.
4 0 mod 6 \u003d 0 4 1 mod 6 \u003d 4 4 2 mod 6 \u003d (4 + 4) mod 6 \u003d 2 (stop, then the process repeats)
e. The cyclic subgroup generated from 5 is H 4, it is the group G itself.
5 0 mod 6 \u003d 0 5 1 mod 6 \u003d 5 5 2 mod 6 \u003d 4 5 3 mod 6 \u003d 3 5 4 mod 6 \u003d 2 5 5 mod 6 \u003d 1 (stop, then the process repeats)
Example 5.8
Three cyclic subgroups can be derived from a group. G has only four elements: 1, 3, 7, and 9. Cyclic subgroups - 
and. Below is how we find the elements of these subgroups.
a. The cyclic subgroup generated based on 1 is H 1. The subgroup has only one element, namely neutral.
b. The cyclic subgroup generated from 3 is H 3, which is the G group.
3 0 mod 10 \u003d 1 3 1 mod 10 \u003d 3 3 2 mod 10 \u003d 9 3 3 mod 10 \u003d 7 (stop, then the process repeats)
in. The cyclic subgroup generated from 7 is H 3, which is the G group.
7 0 mod 10 \u003d 1 7 1 mod 10 \u003d 7 7 2 mod 10 \u003d 9 7 3 mod 10 \u003d 3 (stop, then the process repeats)
d. The cyclic subgroup generated from 9 is H 2. The subgroup has only two members.
9 0 mod 10 \u003d 1 9 1 mod 10 \u003d 9 (stop, then the process repeats)
Cyclic groups
Cyclic group - a group that is its own cyclic subgroup. In Example 5.7, the group G has a cyclic subgroup H 5 \u003d G. This means that the group G is a cyclic group. In this case, the element that generates the cyclic subgroup can also generate the group itself. This element is hereinafter referred to as "generator". If g is a generator, the elements in the finite cyclic group can be written as
(e, g, g 2,… .., g n-1), where g n \u003d e.
Note that a cyclic group can have many generators.
Example 5.9
and. Group G \u003d
b. Group - a cyclic group with two generators, g \u003d 3 and g \u003d 7.
Lagrange's theorem
Lagrange's theorem shows the relationship between the order of a group and the order of its subgroup. Suppose G is a group and H is a subgroup of G. If the order of G and H is | G | and | H | , respectively, then according to this theorem | H | divides | G | ... In example 5.7 | G | \u003d 6. Subgroup order - | H1 | \u003d 1, | H2 | \u003d 3, | H3 | \u003d 2 and | H4 | \u003d 6. Obviously, all these orders are divisors of 6.
Lagrange's theorem has a very interesting application. When a group G and its order | G | , orders of potential subgroups can be easily determined if divisors can be found. For example, the order of the group G \u003d
Item order
Item order in the group ord (a) (order (a)) is the smallest integer n such that a n \u003d e. In other words: the order of an item is the order of the group it generates.
Example 5.10
a. In the group G \u003d
b. In the group G \u003d