Decomposition of a group into cyclic subgroups. Cyclic subgroups

• 1. Group Z integers with the addition operation.
• 2. Group of all complex roots of degree n from one with the multiplication operation. Since the cyclic number is isomorphism

the group is cyclic and the element is generative.

We see that cyclic groups can be either finite or infinite.

3. Let be an arbitrary group and an arbitrary element. The set is a cyclic group with generator element g. It is called the cyclic subgroup generated by the element g, and its order is the order of the element g. According to Lagrange's theorem, the order of an element is a divisor of the order of the group. Display

operating according to the formula:

is obviously a homomorphism and its image coincides with. A mapping is surjective if and only if the group G- cyclic and g its constituent element. In this case we will call the standard homomorphism for the cyclic group G with selected generatrix g.

Applying the homomorphism theorem in this case, we obtain an important property of cyclic groups: every cyclic group is a homomorphic image of the group Z .

In any group G can be determined degrees element with integer indicators:

The property holds

This is obvious if . Let's consider the case when . Then

The remaining cases are treated similarly.

From (6) it follows that

Moreover, by definition. Thus, the powers of an element form a subgroup in the group G. It is called a cyclic subgroup generated by an element, and is denoted by .

Two fundamentally different cases are possible: either all degrees of an element are different, or not. In the first case, the subgroup is infinite. Let us consider the second case in more detail.

Let ,; Then. Smallest natural number T, for which, is called in this case in order element and is denoted by .

Sentence 1. If , That

Proof. 1) Divide m on P with remainder:

Then, by virtue of the definition of order

Due to the previous

Consequence. If mo subgroup contains n elements.

Proof. Really,

and all of the listed elements are different.

In the case where there is no such natural T, that (i.e. the first of the cases described above occurs), it is believed . Note that; the orders of all other elements of the group are greater than 1.

In the additive group we are not talking about powers of an element , and about him multiples, which are denoted by . In accordance with this, the order of the element of the additive group is G-- is the smallest natural number T(if such exist) for which

EXAMPLE 1. The characteristic of a field is the order of any non-zero element in its additive group.

EXAMPLE 2. It is obvious that in a finite group the order of any element is finite. Let us show how the orders of the elements of a group are calculated. The substitution is called cycle length and is denoted by if it cyclically rearranges

and leaves all other numbers in place. Obviously, the order of the cycle length is equal to R. The cycles are called independent, if among the numbers they actually rearrange there are no common ones; in this case . Every substitution can be uniquely decomposed into a product of independent cycles. For example,

which is clearly shown in the figure, where the substitution action is depicted by arrows. If the substitution is decomposed into a product of independent cycles of length , That

EXAMPLE 3. The order of a complex number c in a group is finite if and only if this number is a root of some power of unity, which, in turn, occurs if and only if a is commensurate with c, i.e. .

EXAMPLE 4. Let us find elements of finite order in the group of motions of the plane. Let be. For any point

cyclically rearranged by movement , so their center of gravity O relatively motionless. Therefore, - either a rotation by the view angle around the point O, or reflection relative to some straight line passing through O.

EXAMPLE 5. Let's find the order of the matrix

as an element of the group. We have

So. Of course, this example is specially selected: the probability that the order of a randomly chosen matrix will be finite is zero.

Proposal 2. If , That

Proof. Let

So. We have

Hence, .

Definition 1 . Group G called cyclical, if such an element exists , What . Any such element is called generating element groups G.

EXAMPLE 6. The additive group of integers is cyclic because it is generated by the element 1.

EXAMPLE 7. Additive group of modulo deductions n is cyclic because it is generated by the element .

EXAMPLE 8. The multiplicative group of complex nth roots of 1 is cyclic. Indeed, these roots are numbers

It's clear that . Therefore, the group is generated by the element.

It is easy to see that in an infinite cyclic group the only generating elements are and. Thus, in the group Z the only generating elements are 1 and -- 1.

Number of elements of the final group G called her in order and is denoted by. The order of a finite cyclic group is equal to the order of its generating element. Therefore, from Proposition 2 it follows

Sentence 3 . Cyclic group element of order n is generating if and only if

EXAMPLE 9. The generating elements of a group are called primitive roots n th power of 1. These are the roots of the species , Where. For example, primitive roots of the 12th degree from 1 are.

Cyclic groups are the simplest groups imaginable. (In particular, they are Abelian.) The following theorem gives their complete description.

Theorem 1. Every infinite cyclic group is isomorphic to a group. Every finite cyclic group of order n is isomorphic to a group.

Proof. If is an infinite cyclic group, then by formula (4) the mapping is an isomorphism.

Let be a finite cyclic group of order P. Consider the mapping

then the mapping is well defined and bijective. Property

follows from the same formula (1). Thus, it is an isomorphism.

The theorem has been proven.

To understand the structure of a group, knowledge of its subgroups plays an important role. All subgroups of the cyclic group can be easily described.

Theorem 2. 1) Every subgroup of a cyclic group is cyclic.

2)In a cyclic group of order n the order of any subgroup divides n and for any divisor q of the number n there is exactly one subgroup of order q.

Proof. 1) Let be a cyclic group and N-- its subgroup, different from (The identity subgroup is obviously cyclic.) Note that if for any, then and . Let T-- the smallest of the natural numbers for which . Let's prove that . Let . Let's divide To on T with remainder:

whence, by virtue of the definition of number T it follows that and, therefore, .

2) If , then the previous reasoning applied to (in this case ), shows that . Wherein

And N is the only subgroup of order q in Group G. Back if q-- any number divisor P And , then a subset N, defined by equality (9), is a subgroup of order q. The theorem has been proven.

Consequence . In a cyclic group of prime order, any non-trivial subgroup coincides with the entire group.

EXAMPLE 10. In a group, every subgroup has the form where.

EXAMPLE 11. In a group of nth roots of 1, any subgroup is a group of roots q- th degree of 1, where.

A group O is called cyclic if all its elements are powers of the same element. This element is called a generator of the cyclic group O. Any cyclic group is obviously Abelian.

A cyclic group is, for example, the group of integers by addition. We will denote this group by the symbol 2. Its generator is the number 1 (as well as the number - 1). A cyclic group is also a group consisting of only one element (one).

In an arbitrary group O, the powers of any element g form a cyclic subgroup with generator g. The order of this subgroup obviously coincides with the order of the element g. From here, by virtue of Lagrange's theorem (see page 32), it follows that the order of any element of the group divides the order of the group (note that all elements of a finite group are elements of finite order).

Therefore, for any element g of a finite group of order the equality holds

This simple remark is often helpful.

Indeed, if the group O is cyclic and its generator, then the order of the element is equal to . Conversely, if a group O has an element of order , then among the powers of this element there are different ones, and therefore these powers exhaust the entire group O.

We see, therefore, that a cyclic group can have several different generators (namely, any element of the order is a generator).

Task. Prove that any group of prime order is a cyclic group.

Task. Prove that a cyclic group of order has exactly generators, where is the number of positive numbers less than and relatively prime to .

Along with the order, any finite group can be attributed to a number - the least common multiple of the orders of all its elements.

Task. Prove that for any finite group O the number divides the order of the group.

Obviously, for a cyclic group the number coincides with the order. The opposite is, generally speaking, not true. Nevertheless, the following statement characterizing cyclic groups in the class of finite Abelian groups holds:

a finite Abelian group O for which the number is equal to its order is a cyclic group.

Indeed, let

The orders of all possible non-unit elements of a finite Abelian group O are of order , and let be their least common multiple.

Let's expand the number into the product of powers of different prime numbers:

Let Since a number is, by definition, the least common multiple of numbers (1), among these numbers there is at least one number that is divisible exactly by, i.e., has the form , where b is coprime with . Let this number be the order of the element g. Then the element has order (see Corollary 1) on page 29).

Thus, for anyone in the group O there is at least one element of order. Having chosen one such element for each, we consider their product. According to the statement proven on pages 29-30, the order of this product is equal to the product of orders, i.e., equal to the number. Since the last number by condition is equal to , it is thereby proven that there is an element of order n in the group O. Consequently, this group is a cyclic group.

Now let O be an arbitrary cyclic group with a generator and H be some of its subgroups. Since any element of the subgroup H is an element of the group O, it can be represented in the form , where d is some positive or negative integer (generally speaking, it is not uniquely defined). Let's consider the set of all positive numbers for which the element belongs to the subgroup H. Since this set is non-empty (why?), then it contains the smallest number. It turns out that any element h of the subgroup H is a power of the element. Indeed, by definition, there is a number d such that (the number d can be negative). Divide (with remainder) the number d by the number

Since , then, due to the minimality of the number, the remainder must be equal to zero. Thus, .

This proves that the element is a generator of the group H, i.e., that the group H is cyclic. So, any subgroup of a cyclic group is a cyclic group.

Task. Prove that the number is equal to the index of the subgroup H and, therefore, divides the order of the group O (if the group O is finite).

Note also that for any order divisor of a finite cyclic group Q in the group O there is one and only one subgroup H of order (namely, the subgroup with the generator

This implies that if a finite cyclic group is simple, then its order is prime (or unity).

Let us finally note that any quotient group (hence, any homomorphic image) of a cyclic group Q is a cyclic group.

To prove it, it is enough to note that the generator of the group is the coset containing the generator of the group O.

In particular, any quotient group of the group of integers Z is a cyclic group. Let us study these cyclic groups in more detail.

Since the group Z is Abelian, any of its subgroups H is a normal divisor. On the other hand, according to what was proven above, the subgroup H is a cyclic group. Since quotient groups by trivial subgroups are known to us, we can consider the subgroup H to be nontrivial. Let the number be a generator of the subgroup H. We can consider this number to be positive (why?) and, therefore, greater than one.

The subgroup N. obviously consists of all integers divisible by . Therefore, two numbers belong to the same coset in the subgroup H if and only if their difference is divisible by , i.e., when they are comparable in modulus (see Course, p. 277). Thus, cosets in the subgroup H are nothing more than classes of numbers comparable to each other in modulus.

In other words, the quotient group of a group Z by subgroup H is a group (by addition) of classes of numbers comparable to each other in modulus. We will denote this group by Its generator is the class containing the number 1.

It turns out that any cyclic group is isomorphic either to the group Z (if it is infinite) or to one of the groups (if its order is finite).

Indeed, let be a generator of group O. Let us define a mapping from group 2 to group O, setting

Definition 1.22. Let R- Prime number. Group G called p-group, if the order of each element of the group is equal to some power of a prime number R.

Definition 1.23. Silovsky r-subgroup finite group G a p-subgroup of a given group is called that which is not contained in a larger p-subgroup of a given group.

Theorem 1.25. A finite Abelian group is equal to the direct product of its Sylow p-subgroups.

Proof. Consider a finite Abelian group G of order n and let n = R" ! p 2 2 p* 1 k - number expansion P into the product of powers of different prime numbers. For 1, 2,..., To let us denote by I, the Sylow r subgroup and by I, the subgroup generated by all I; For; * i. It is easy to prove that I, n I, = (e). Therefore I = (N 1,H 2,...,N k) = N 1 xN 2 x...xN k. Suppose that there is an element g e G, such that g g Y. By Corollary 2 from Lagrange’s theorem |G| : |g|. It follows that

|g| = pf"pjf 2 Pk k > g D e Pi - a i D for any i = 1, 2, To. By the corollary of Theorem 1.23, there are elements g 1; g2, ..., gk e G, such that = x x... x (g k) and | g,-1 = pf 1 for i = 1, 2, ..., /s. If we assume that g, g I, for some g, then we obtain a p-subgroup (gi, I,) F I, which contradicts the definition of a Sylow p-subgroup. Thus, for any i = 1, 2,..., /eg, e e Where am I from? g e N. Hence, H = G and the theorem is proven.

Theorem 1.26. A finite Abelian p-group is equal to the direct product of cyclic subgroups.

Proof. Let a finite Abelian p-group be given G. Let's select an element in it A of maximal order p", and let H be a maximal subgroup such that (a) n H = (e). Then (a, R) = (a) x R. Let us denote Gj = (a) x R.

Let's pretend that G Ф G y From all elements not belonging to G x , we choose an element g of minimum order рР. Assuming that gPg Gb then since |gp| = рР- 1, we come to a contradiction with the choice of element g. Consequently, gP e G x = (a) x I and there are an integer /c and an element h e I, such that gP = a fc /i. From here a k= gp/i -1 . If gcd(/c, p) = 1, then gcd(/c, p°9 = 1 and there exist integers u, v such that /c + p a v = 1. Then

Due to maximality | a = p a we have gP“ = e and e F aR“ _1 = = (gP"/i _u)P“ _1 =gP“h~ u P a ~ 1 =/i _u p““ 1 e I, which contradicts condition (a) p I = (e). Therefore, /s: r.

Let To= r/s x. Then aP fc i = a k =gPh~ 1 , where h = a~P k igP == (a _fc ig)P. Let us denote gj=a _/c ig. Then gf -heH. Assuming that gj =ar fc "geG] =(a)xN, then g е G x , which contradicts the choice of element g. Consequently, g x g G x, and therefore gj g I. Since I is a maximal subgroup with the condition (A) n I = (e), then (a) n (g x , I) ^ (e). Therefore, there are t, p e Z and element hj e I, such that e * a t= gf

Assuming that p:p,top=pp 1 at some n,eZ and e g a m = gf/ij = gf ni /ii e I, which contradicts condition (a) n I = = (e). Therefore, GCD(n,p) = 1 Hgf =a m /if 1 . If |g x | =pY, then GCD(n, p’O = 1 and there exist u x , v x g Z, such that gsh x -t-pYv x = 1. Hence g, =gf u i + P Yv i = gf Ul gf Yvi = gf Ul =(a m /i 1 - 1) u i Again we came to a contradiction. Thus, it remains to accept that G - (a)x R. Now in the subgroup R we similarly select by a direct factor the cyclic subgroup of the maximum in N order, etc., until we obtain a decomposition of the group G into a direct product of cyclic subgroups. The theorem has been proven.

Theorem 1.27. A finite Abelian group is equal to the direct product of cyclic p-subgroups.

The proof follows from Theorems 1.25 and 1.26.

To conclude the chapter on groups, we note that a group can be considered as a set with one binary operation, which is associative, and for any elements A And Kommersant equations are uniquely solvable ax = b uua-b. This view of the group leads to two generalizations. On the one hand, one can focus on studying the meaning of the associativity of an operation, and this leads to the concept of a semigroup as a set with one associative operation (see work). On the other hand, one can ignore the associativity requirement, and this leads to the concept of a quasigroup as a set with one binary operation, with respect to which the named equations are uniquely solvable. A quasigroup with identity is called a loop (see work). The theory of semigroups and the theory of quasigroups have turned into two independently developing substantive theories. We do not mention them in the main text for reasons of “maximum possible minimum” volume.

Finite groups

A group (semigroup) is called ultimate, if it consists of a finite number of elements. The number of elements of a finite group is called its in order. Any subgroup of a finite group is finite. And if NÍ G– subgroup of the group G, then for any element AÎ G a bunch of On={X: x=h◦a, for any hÎ H) is called left coset For G relatively N. It is clear that the number of elements in On equal to the order N. (The definition can be formulated similarly a N– right coset with respect to N).

The important thing is that for any subgroup N groups G any two left (right) cosets according to N either coincide or do not intersect, therefore any group can be represented as a union of disjoint left (right) cosets by N.

Indeed, if two classes N a And Hb, Where a, bÎ G, have a common element X, then there is tÎ H such that x = t◦a. And then the left class is for X: N x={y: y=h◦x= h◦(t◦a) = (h◦t)◦a} Í H a, But a=t ‑1 ◦x And N a={y: y=h◦a= h◦(t ‑1 ◦x) = (h◦t ‑1)◦x} Í H x. From here N x=N a. Similarly, it can be shown that N x=N b. And therefore N a=N b. If the classes N a And Hb 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 H.

Theorem 2.6.1. The order of a finite group is divided by the order of any of its subgroups.

Proof. Because G is a finite group, then so is any of its subgroups N has finite order. Consider the decomposition of a group into a subgroup N. In each coset in this decomposition the number of elements is the same and equal to the order N. Therefore, if n– group order G, A k– subgroup order N, That n=m× k, Where m– number of cosets according to N in the group decomposition G.

If for any element aÎ G Þ N a=a N(left and right cosets by subgroup N coincide), then N called normal divisor groups G.

Statement: If G is a commutative group, then any subgroup of it N is a normal divisor G.

Due to the associative nature of the action in a group (semigroup), we can talk about the “product” of three elements ( A◦b◦c) =(A◦b)◦c = A◦(b◦c). Similarly, the concept of a complex product of n elements: A 1 ◦A 2 ◦…◦a n = ◦ a n = = ◦.

Work n identical elements of a group is called element degree and is designated a n=. This definition makes sense for any natural n. For any group element aÎ G denote A 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 A. Both definitions a ‑ n coincide, because a n◦(a ‑1)n = (A◦A◦ ¼◦ A)◦(a ‑1 ◦a‑1◦ ¼◦ a ‑1) = A◦A◦¼◦( A◦a ‑1)◦a‑1 ◦¼◦ a ‑1 =e n =e. Thus, ( a ‑1)n = (a n) ‑1 .

In an additive group, the analogue of the degree of an element is a n will n its multiple, usually denoted na, which should not be taken as a work n on A, because the nÎℕ and perhaps nÏ G. That. na⇋, where nОℕ, and 0 A=e⇋0, and (‑ n)a = ‑(na) = n(‑a) for any natural n, Where (- a) – inverse to aÎ G.

It is easy to show that with the chosen notation for any integers m And n and for anyone aÎ G known properties are satisfied: A) in multiplicative notation a n ◦a m = a n + m And ( a n)m = a nm; b) in additive notation na+ma = (n+m)a And n(ma)=(nm)a.

Consider a subset of the group G, composed of all powers of an arbitrary element gÎ G. Let's denote it A g. Thus, A g ={g 0 , g 1 , g ‑1 , g 2 , g‑2,¼). Obviously, A g is a subgroup of the group G, because 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 О A g, Besides, g 0 =eÎ A g.

Subgroup A g called cyclic subgroup groups G, generated by the element g. This subgroup is always commutative, even if it itself G not commutative. If the group G coincides with one of its cyclic subgroups, then it is called cyclic group, generated by the element g.

If all powers of an element g are different, then the group G called endless cyclic group, and the element g– element infinite order.

If among the elements of a cyclic group there are equal, for example, g k=g m at k>m, That g k‑m=e; and, designating k-m through n, we get g n=e, nÎℕ.

Lowest natural indicator n such that g n=e, 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 whose elements all have finite order are called periodic.

Since any element of a finite group has finite order, all finite groups are periodic. Moreover, all cyclic subgroups of a finite group are periodic, since they are finite, and every element of finite order n generates a cyclic group of the same order n, consisting of elements ( g 0 , g 1 , g 2 ¼, g n-1 ). Indeed, if the number of elements were equal to some k<n, Then g k=e=g n, which contradicts the choice n, as the least degree such that g n=e; on the other side, k>n also impossible, because in this case there would be identical 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), That g i - j=e, But ( i‑j)<n, and by definition n – the smallest degree is such that g n=e.

2) Any other degree g, positive or negative, 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+r, Where 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 element of first order ( e), generating a cyclic subgroup of the first order consisting of one element e.

2) Consider the group of substitutions S 3, consisting of the elements: , , , , , . Order S 3 =6. Element order A is equal to 2, because . Element order b is also equal to 2, because . Element order With is equal to 3, because And . Element order f is also equal to 3, because And . And finally, order d is equal to 2, because . Thus, 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 divided 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).

It also follows 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=e, Where m– group order, n– element order g, k– integer).

There is 3 subgroup in group S N={e, c, f) is a normal divisor, but 2nd order subgroups are not normal divisors. This can be easily verified by finding the left and right cosets by N for each group element. For example, for an element A left coset On={e ◦ a, With◦ A, f◦ a} = {A, b, d) and right coset a N={a ◦ e, A◦ c, A◦ f} = {A, d, b) match up. Likewise for all other elements S 3 .

3) The set of all integers with addition forms an infinite cyclic group with a generating element 1 (or –1), because any integer is a multiple of 1.

4) Consider a set of roots n‑th power of unity: E n=. 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 from E n, Where k, m £ n-1 will also be an element E n, since = = , where r=(k+m) mod n And r £ n-1; multiplication associative, neutral element e=e 0 =1 and for any element e k there is reverse and . This group is cyclic, its generating element is a primitive root. It is easy to see that all powers are distinct: , 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 into n equal arcs, as shown in Figure 11.

The last two examples essentially exhaust 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.

Proof. Let ( G, ∘) is an infinite cyclic group with a generating element 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 m, are elements G. And wherein f(k+m)=f(k)∘f(m), because the g k + m=g k∘ g m.

Let now ( G, ∘) is a finite cyclic group of order n with a generating element g. Then each element g kÎ G the only way to match an element is 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), because the f(g k∘ g m)=f(g k + m)=f(g r), Where r=(k+m) mod n, And f(g r)=e r=e k× e m. It is clear that such a mapping is a bijective mapping.