Cyclic groups. Element order

Let g be an arbitrary element of the group G. Then, taking , we obtain a minimal subgroup
, generated by one element
.

Definition. Minimum subgroup
, generated by one element g of the group G, is called cyclic subgroup Group G.

Definition. If the entire group G is generated by one element, i.e.
, then it is called cyclic group.

Let element of the multiplicative group G, then the minimal subgroup generated by this element consists of elements of the form

Consider the powers of the element , i.e. elements

.

There are two possibilities:

1. All powers of the element g are different, i.e.

, then in this case we say that the element g has infinite order.

2. There are coincidences of degrees, i.e. , But
.

In this case, the element g has finite order.

Indeed, let, for example,
And
, Then,
, i.e. there are positive degrees
element
, equal to the unit element.

Let d be the smallest positive exponent of the element , for which
. Then they say that the element
has a finite order equal to d.

Conclusion. In any group G of finite order (
) all elements will be of finite order.

Let g be an element of a multiplicative group G, then the multiplicative subgroup
consists of all the different powers of the element g. Therefore, the number of elements in the subgroup
matches the order of the element i.e.

number of elements in group
equal to the order of the element ,

.

On the other hand, the following statement holds.

Statement. Order any element
equal to the order of the minimal subgroup generated by this element
.

Proof. 1.If – element of finite order , That

2. If is an element of infinite order, then there is nothing to prove.

If element has order , then, by definition, all elements

various and any degree matches one of these elements.

Indeed, let the exponent
, i.e. is an arbitrary integer and let
. Then the number can be represented in the form
, Where
,
. Then, using the properties of the degree of the element g, we obtain

.

In particular, if .

Example. Let
is an additive Abelian group of integers. The group G coincides with the minimal subgroup generated by one of the elements 1 or –1:

,

hence,
is an infinite cyclic group.

Cyclic groups of finite order

As an example of a cyclic group of finite order, consider group of rotations of a regular n-gon relative to its center
.

Group elements

are the rotations of the n-gon counterclockwise by the angles

Group elements
are

,

and from geometric considerations it is clear that

.

Group
contains n elements, i.e.
, and the generating element of the group
is , i.e.

.

Let
, then (see Fig. 1)

Rice. 1 – Group – rotations of regular triangle ABC relative to center O.

Algebraic operation  in group – sequential rotation counterclockwise, at an angle that is a multiple of , i.e.

Reverse element
– clockwise rotation at an angle 1, i.e.

.

Table Kuhwhether

The analysis of finite groups is most clearly carried out using the Cayley table, which is a generalization of the well-known “multiplication table”.

Let a group G contain n elements.

In this case, the Cayley table is square matrix having n rows and n columns.

Each row and each column corresponds to one and only one element of the group.

Element Cayley table, standing at the intersection of the i-th row and the j-th column, is equal to the result of the “multiplication” operation of the i-th element with the j-th element of the group.

Example. Let the group G contain three elements (g 1,g 2,g 3). The operation in the group is “multiplication”. In this case, the Cayley table looks like:

Comment. Each row and each column of the Cayley table contains all the elements of the group and only them. The Cayley table contains complete information about the group. What can be said about the properties of this group?

1. The unit element of this group is g 1.

2. Abelian group because the table is symmetrical about the main diagonal.

3.For each element of the group there are inverses -

for g 1 the inverse is the element g 1, for g 2 the element g 3.

Let's build for groups Keli's table.

To find the inverse of an element, for example, , required in the line corresponding to the element find columnj containing element . Element corresponding to the given column and is the inverse of the element , because
.

If the Keley table is symmetrical about the main diagonal, then this means that

– i.e. the operation in the group under consideration is commutative. For the example under consideration, the Keley table is symmetrical about the main diagonal, which means that the operation in commutative, i.e.
,

and the group – Abelian.

We can consider the complete group of symmetry transformations of a regular n-gon , adding to the rotation operation additional operations of spatial rotation around the axes of symmetry.

For a triangle
, and the group contains six elements

Where
these are rotations (see Fig. 2) around the height, the median, the bisector have the form:

;

,

,
.

Rice. 2.- Group – symmetry transformations of regular triangle ABC.

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 observation 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, 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

Cosets, Lagrange's theorem

Let H subgroup of the group G. Left adjacent element class a by subgroup H called a set of elements ah, Where h belongs H. The left coset is denoted by aH. The right adjacent class of the element is introduced similarly a by subgroup H, which denotes Ha.

Since there is always a neutral element in a subgroup, then each element a contained in adjacent class aH (Ha).

Property 2.7. Elements a And b belong to the same left coset by subgroup H then and only when

Proof. 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) adjacent classes of elements a And b have a common element in the subgroup H, then they coincide.

Proof. Let . 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 in a similar way. Thus the theorem is proven.

Corollary 2.1. The left cosets either do not intersect or coincide.

Proof obviously.

Corollary 2.2. The left (right) coset is equivalent to H.

Proof. Let us establish correspondence between the elements of the subgroup H and elements of the related class aH according to the formula . The correspondence is one-to-one. Thus the statement is proven.

Theorem 2.3 (Lagrange). The order of a finite group is divided by the order of its subgroup.

Proof. Let G– order group n, A H- subgroup G order k.Equality takes place. Let's remove duplicate terms from the right side of the equality. As a result, disjoint cosets will remain. Since the number of elements in the coset is equal to , then where m number of different related classes. This establishes equality n=mk, which is what was required.

The number of distinct cosets is called the subgroup index H in Group G.

A set of elements from a group G is called generating if G is obtained by the closure of this set with respect to the group operation.

A group generated by one element is called cyclic.

Corollary 2.3. Every group contains a cyclic subgroup.

Proof. Let a–group element G. The set is a cyclic subgroup.

Order of the cyclic subgroup generated by an element a, is called the order of the element.

Property 2.8. If element a has order n, That a n=e.

Proof. Consider the sequence. Since the number of terms in the sequence is infinite, and for powers of an element a There is a finite number of possibilities, then the sequence will contain identical terms. Let where k<j And k first repeating term. Then , and therefore the member k-j+ 1 is repeated. Hence, j=1 (otherwise ). Thus, the sequence consists of repeating sets of the form and in it k- 1 different elements. Hence, k=n+1. Since, then.

The order of any element is a divisor of the group order, hence a | G | =e for any element of the group.

Corollary 2.4. The order of the group is divided without remainder by the order of any element of the group.

Proof obviously.

Theorem 2.4 (about cyclic groups)

I. For any natural n there is a cyclic group of order n.

II. Cyclic groups of the same orders 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 the cyclic group n-th order there is a unique subgroup of order m.

Proof. Set of complex roots of degree n from 1 with respect to the operation of multiplication forms a cyclic group of order n. Thus the first statement is proven.

Let the cyclic group G order n generated by the element a, and the cyclic group H, of the same order, generated by the element b. The correspondence is one-to-one and preserves the operation. The second statement has been proven

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 proven.

Let H– subgroup of a cyclic group G, generated by the element a. Elements H are the degree a. Let's choose in H a. Let this be the element. Let us show that this element is generating in the subgroup H. Let's take an arbitrary element from H. The work is contained in H at any r. Let's choose r equal to the quotient of division k on j, Then k-rj there is a remainder after division k on j and therefore less j. Since in H there are no elements that are non-zero degree a, less than j, That k-rj= 0, and . The fourth statement has been proven.

Let the cyclic group G order n generated by the 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 in absolute value non-zero power a. Let this be the element. Let's show that j=n/m. Element belongs H. Therefore, a non-zero number of the form rj-nv in absolute value no less j, which is only possible if n divided by j without a trace. The subgroup generated by , has order n/j=m, hence, j=n/m. Since the generating element of a subgroup is determined uniquely by its order, the fifth statement is proven.

the subgroup is called cyclic subgroup. Term exponentiation here means repeatedly applying a group operation to an element:

The set resulting from this process is denoted in the text as . Note also that a 0 = e.

Example 5.7

From group G =< Z 6 , +>four cyclic subgroups can be obtained. This H 1 =<{0},+>, H 2 =<{0, 2, 4}, +>, H 3 =<{0, 3}, +> and H4 = G. Note that when the operation is addition, then a n means multiplying n by a. Note also that in all these groups the operation is addition modulo 6. Below is how we find the elements of these cyclic subgroups.

a. The cyclic subgroup generated from 0 is H1, has only one element (the neutral element).

b. The cyclic subgroup generated from 1 is H4, which is the group G itself.

1 0 mod 6 = 0 1 1 mod 6 = 1 1 2 mod 6 = (1 + 1) mod 6 = 2 1 3 mod 6 = (1 + 1 + 1) mod 6 = 3 1 4 mod 6 = (1 + 1 + 1 + 1) mod 6 = 4 1 5 mod 6 = (1 + 1 + 1 + 1 + 1) mod 6 = 5 (stop, then repeat the process)

V. The cyclic subgroup generated from 2 is H2, which has three elements: 0, 2, and 4.

2 0 mod 6 = 0 2 1 mod 6 = 2 2 2 mod 6 = (2 + 2) mod 6 = 4 (stop, then repeat the process)

d. The cyclic subgroup generated from 3 is H3, 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 = 0 4 1 mod 6 = 4 4 2 mod 6 = (4 + 4) mod 6 = 2 (stop, then repeat the process)

e. The cyclic subgroup generated on the basis of 5 is H 4, it is the group G itself.

5 0 mod 6 = 0 5 1 mod 6 = 5 5 2 mod 6 = 4 5 3 mod 6 = 3 5 4 mod 6 = 2 5 5 mod 6 = 1 (stop, then the process is repeated)

Example 5.8

Three cyclic subgroups can be obtained from the 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 from 1 is H1. The subgroup has only one element, namely neutral.

b. The cyclic subgroup generated from 3 is H3, which is the group G.

3 0 mod 10 = 1 3 1 mod 10 = 3 3 2 mod 10 = 9 3 3 mod 10 = 7 (stop, then repeat the process)

V. The cyclic subgroup generated from 7 is H3, which is a group G.

7 0 mod 10 = 1 7 1 mod 10 = 7 7 2 mod 10 = 9 7 3 mod 10 = 3 (stop, then repeat the process)

d. The cyclic subgroup generated from 9 is H2. The subgroup has only two elements.

9 0 mod 10 = 1 9 1 mod 10 = 9 (stop, then repeat the process)

Cyclic groups

Cyclic group is a group that is a proper cyclic subgroup. In Example 5.7, the group G has a cyclic subgroup H 5 = G. This means that the group G is a cyclic group. In this case, the element that generates the cyclic subgroup may also generate the group itself. This element is hereafter referred to as the “generator”. If g is a generator, the elements in a finite cyclic group can be written as

(e,g,g 2 ,….., g n-1) , where g n = e.

Note that a cyclic group can have many generators.

Example 5.9

A. Group G = is a cyclic group with two generators, g = 1 and g = 5.

b. The group is a cyclic group with two generators, g = 3 and g = 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|

Lagrange's theorem has a very interesting application. When given a group G and its order |G| , orders of potential subgroups can be easily determined if divisors can be found. For example, the group order G = - this is |17| . The divisors of 17 are 1 and 17. This means that this group can only have two subgroups - the neutral element and H 2 = G.

Element order

Element order in the group ord(a) (order(a)) is the smallest integer n such that a n = e. In other words: the order of an element is the order of the group it generates.

Example 5.10

a. In group G = , element orders: order ord(0) = 1, order ord(1) = 6, order ord(2) = 3, order ord(3) = 2, order ord(4) = 3, order ord(5) = 6.

b. In group G = , orders of elements: order ord (1) = 1 , order ord (3) = 4 , order ord (7) =4 , order (9) = 2 .

Let M be some subset of the group G. The set of all possible products of elements from M and their inverses is a subgroup. It is called the subgroup generated by the subset M and is denoted by hMi. In particular, M generates a group G if G = hMi. The following simple statement is helpful:

a subgroup H is generated by a subset M then and

If G = hMi and |M|< ∞, то G называется of course generated.

A subgroup generated by one element a G is called cyclic and is denoted by hai. If G = hai for some a G, then G is also called cyclic. Examples of cyclic groups:

1) group Z of integers relative to addition;

2) group Z(n) modulo deductions n relative to addition;

her elements are the sets of all integers that give the same remainder when divided by a given number n Z.

It turns out that these examples exhaust all cyclic groups:

Theorem 2.1 1) If G is an infinite cyclic group, then

G Z.

2) If G is a finite cyclic group of order n, then

G Z(n).

The order of an element a G is the smallest natural number n such that an = 1; if such a number does not exist, then the order of the element is considered to be infinity. The order of element a is denoted by |a|. Note that |hai| = |a|.

2.1. Calculate the orders of elements of groups S3, D4.

2.2. Let |G|< ∞, g G. Докажите, что |g| делит |G|.

2.3. Let g G, |g| = n. Prove that gm = e if and only if n divides m.

2.4. Let |G| = n. Prove that an = e for all a G.

2.5. Prove that a group of even order contains an element of order 2.

2.6. Let the group G have odd order. Prove that for every a G there is a b G such that a = b2.

2.7. Check that |x| = |yxy−1 |, |ab| = |ba|, |abc| = |bca| = |cab|.

2.8. Let a G, |a| = n and b = ak. Prove that |b| = n/GCD(n, k);

2.9. Let ab = ba. Prove that LCM(|a|, |b|) is divisible by |ab|. Give an example when LCM(|a|, |b|) 6= |ab|.

2.10. Let ab = ba, GCD(|a|, |b|) = 1. Prove that |ab| = |a||b|.

2.11. Let σ Sn be a cycle. Check that |σ| equal to the length σ.

2.12. Let σ Sn, σ = σ1. . . σm, where σ1, . . . , σm are independent cycles. Check that |σ| = LCM(|σ1 |, . . . , |σm |).

2.13. Are the groups cyclic: a) Sn ;

b) Dn;

c) µn := (z C | zn = 1)?

2.14. Prove that if |G| = p is a prime number, then G is cyclic.

2.15. Prove that a non-identity group G has no proper subgroups if and only if |G| = p, i.e. G is isomorphic to Z(p) (p is a prime number).

2.16. Prove that if |G| ≤ 5, then G is Abelian. Describe groups of order 4.

2.17. Let G be a cyclic group of order n with generator a. Let b = ak. Prove that G = hbi if and only if GCD(n, k) = 1, i.e. the number of generating elements in a cyclic group of order n is equal to ϕ(n), where ϕ is the Euler function:

	(k | k N, 1 ≤ k ≤ n, GCD(n, k) = 1) .

2.18.* Prove that

2.19. Let G be a cyclic group of order n, m|n. Prove that G contains exactly one subgroup of order m.

2.20. Find all generators of the groups: a) Z, b) Z(18).

2.21. Prove that an infinite group has an infinite number of subgroups.

2 .22 .* Let |G|< ∞. Докажите, что G циклична тогда и только тогда, когда |Gd | ≤ d для всех d N, где Gd = {x G | xd = e}.

2 .23 .* Let F be a field, G a finite subgroup of F . Prove that G is cyclic.

EXAMPLE 3

Homomorphisms. Normal subgroups. Factor groups

A group mapping f: G −→ H is called a homomorphism if f(ab) = f(a)f(b) for any a, b G (so the isomorphism

– a special case of homomorphism). Other types of homomorphism are often used:

monomorphism is an injective homomorphism, epimorphism is a surjective homomorphism, endomorphism is a homomorphism into itself, automorphism is an isomorphism into itself.

Subsets

Kerf = (a G | f(a) = 1) G

Imf = (b H | f(a) = b for some a G) H

are called the kernel and the image of the homomorphism f, respectively. Obviously, Kerf and Imf are subgroups.

Subgroup N< G называется нормальной (это обозначается N C G), если a−1 Na = N для всех a G; это эквивалентно тому, что Na = aN. Группа называется простой , если она не содержит собственных нормальных подгрупп.

The kernel of a homomorphism is a normal subgroup. The converse is also true: every normal subgroup is the kernel of some homomorphism. To show this, let us introduce on the set

16 Section 3. Homomorphisms, factor groups

G/N = (aN | a G) cosets by the normal subgroup N operation: aN · bN = abN. Then G/N turns into a group, which is called the quotient group by subgroup N. The mapping f: G −→ G/N is an epimorphism, and Kerf = N.

Every homomorphism f: G −→ H is a composition of an epimorphism G −→ G/Kerf, an isomorphism G/Kerf −→ Imf, and a monomorphism Imf −→ H.

3.1. Prove that these mappings are homomorphic

mother groups, and find their core and image. a) f: R → R, f(x) = ex;

b) f: R → C, f(x) = e2πix;

c) f: F → F (where F is the field), f(x) = ax, a F ; d) f: R → R, f(x) = sgnx;

e) f: R → R, f(x) = |x|; e) f: C → R, f(x) = |x|;

g) f: GL(n, F) → F (where F is the field), f(A) = det A;

h) f: GL(2, F) → G, where G is a group of linear fractional functions (see Problem 1.8), F is a field,

i) f: Sn → (1, −1), f(σ) = sgnσ.

3.2. Under what condition on a group G is the mapping f: G → G given by the formula

a) g 7→g2 b) g 7→g−1 ,

is it a homomorphism?

3.3. Let f: G → H be a homomorphism and let G. Prove that |f(a)| divides |a|.

3.4. Prove that the homomorphic image of a cyclic group is cyclic.

3.5. Prove that the image and inverse image of a subgroup under a homomorphism are subgroups.

3.6. We call the groups G1 and G2 anti-isomorphic if there is a bijection f: G1 → G2 such that f(ab) = f(b)f(a) for all a, b G1. Prove that antiisomorphic groups are isomorphic.

3.7.* Prove that there are no nontrivial homomorphisms Q → Z, Q → Q+.

3 .8 .* Let G be a group, g G. Prove that for the existence of f Hom(Z(m), G) such that f(1) = g, it is necessary and sufficient that gm = e.

3.9. Describe

a) Hom(Z(6), Z(18)), b) Hom(Z(18), Z(6)), c) Hom(Z(12), Z(15)), d) Hom(Z (m), Z(n)).

3.10. Check that

α, β R, α2 + β2 6= 0 .

3. 11. (Generalization of Cayley's theorem.) Prove that the assignment to an element a G of the permutation xH 7→axH on the set of cosets with respect to the subgroup H< G является гомоморфизмом G в группу S(G/H). Чему равно его ядро?

3. 12. Check that the set Aut G of all automorphisms of a group G forms a group with respect to composition.

3. 13. Check that the mapping f g : G → G, f g (x) = gxg −1 , where g G, is an automorphism of the group G (such automorphisms are called internal ). Check that the inner automorphisms form a subgroup Inn G< Aut G.

3.14. Find the automorphism group a) Z;

b) a non-cyclic group of order 4 (see Problem 2.16); c) S3;

18 Section 3. Homomorphisms, quotient groups

3.15. Is it true that: a) G C G, E C G;

b) SL(n, F) C GL(n, F);

c) scalar nonzero matrices form a normal subgroup in GL(n, F);

d) diagonal (upper triangular) matrices with non-zero diagonal elements form a normal subgroup in

e) An C Sn;

e) Inn G C Aut G?

3.16. Let = 2. Prove that H C G.

3.17. Let M, N C G. Prove that M ∩ N, MN C G.

3.18. Let N C G, H< G. Докажите, что N ∩ H C H.

3.19. Let N C G, H< G. Докажите, что NH = HN < G.

3.20. Let H< G. Докажите, что xHx−1 C G.

3.21. Let H< K < G. Докажите, что H C K тогда и только тогда, когда K NG (H).

3.22. Let M, N C G, M ∩ N = E. Prove that M and N commute elementwise.

3.23. Prove that:

a) The image of a normal subgroup under an epimorphism is normal; b) The complete inverse image of a normal subgroup (for any homo-

morphism) is normal.

3.24. Check that G/G E, G/E G.

3.25. Prove that Z/nZ is a cyclic group of order n.

3.26.* Prove that:

d) R /R (1, −1);

e) GL(n, F)/SL(n, F) F ;

E. A. Karolinsky, B. V. Novikov

where GL+ (n, R) := (A GL(n, R) | det A > 0).

3.27. Prove that Q/Z is a periodic group (i.e., the order of any of its elements is finite) which contains a unique subgroup of order n for every natural number n. Each such subgroup is cyclic.

3 .28 .* Prove that: a) C(G) C G,

b) Inn G G/C(G).

3 .29 .* Let N C G, H< G. Докажите, что NH/N H/H ∩ N.

3 .30 .* Prove that if M C N C G, M C G, then

(G/M)/(N/M) G/N.

3.31. Prove that if G/C(G) is cyclic, then G = C(G) (i.e. G/C(G) = E).

3.32. Let us call the commutator of elements x and y of group G the element := x−1 y−1 xy. A commutator subgroup of a group G is its subgroup G0 generated by all commutators. Prove that:

a) G0 C G;

b) The group G/G0 is Abelian;

c) G is Abelian if and only if G0 = E.

3.33. Let N C G. Prove that G/N is Abelian if and only if N G0 .

3.34. Let us define by induction G(0) = G, G(n) = (G(n−1) )0 . A group G is called solvable if G(n) = E for some n N. Check that:

a) subgroups and quotient groups of a solvable group are solvable;

b) if N C G is such that N and G/N are solvable, then G is solvable.

3.35. Prove that the group G is solvable if and only if there is a chain of subgroups

E = Gn C Gn−1 C . . . C G1 C G0 = G

20 Section 3. Homomorphisms, factor groups

such that all quotient groups Gk /Gk+1 are Abelian.

3.36. Check that a) are Abelian groups; b) groups S3 and S4;

c) the subgroup of all upper triangular matrices in GL(n, F) (where F is a field)

are solvable.

3.37. Let G(n) be a subgroup in G generated by the set (gn | g G). Prove that:

a) G(n) C G;

b) G/G(n) has period n (i.e., the identity xn = 1 is satisfied);

c) G has period n if and only if G(n) = E.

3.38. Let N C G. Prove that G/N has period n if and only if N G(n) .

3.39. Let G be the group (with respect to composition) of mappings

φ : R → R of the form x 7→ax + b (a 6= 0), H = (φ G | φ : x 7→x + b). Prove that H C G. What is G/H equal to?

3.40. Let us define the operation on the set G = Z × Z:

(a, b)(c, d) = (a + (−1)b c, b + d)

Prove that G is a group and H = h(1, 0)i C G.