Consider the multiplicative group of all integer powers of two (2Z, ), where 2Z= (2 n | P e Z). The additive language analogue of this group is the additive group of even integers (2Z, +), 2Z = (2n | p e Z). Let us give a general definition of groups, of which these groups are particular examples.
Definition 1.8. Multiplicative group (G,) (the additive group (G, +)) is called cyclic, if it consists of all integer powers (respectively, all integer multiples) of one element a e G, those. G=(a p | p e Z) (respectively, G - (pa | p e Z)). Designation: (a), reads: the cyclic group generated by the element a.
Consider 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. Thus, and d - (a).
- 2. An example of a multiplicative finite cyclic group is the group Cn of nth roots of unity. Recall that the nth roots of unity are
according to the formula e k= cos---hisin^-, where k = 0, 1, ..., P - 1. Follow- p p
Importantly, С„ \u003d (e x) \u003d (e x \u003d 1, e x, ef \u003d e 2, ..., e "-1 \u003d? „_ x). Recall that complex numbers e k, k = 1, ..., P - 1, represented by points on the unit circle that divide it into P equal parts.
- 3. A characteristic example of an additive infinite cyclic group is the additive group of integers Z; it is generated by the number 1, i.e. Z = (1). Geometrically, it is depicted as integer 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 is an integer a f±1 (see fig. 1.3). This similarity of images will be discussed in Section 1.6.
- 4. We choose in an arbitrary multiplicative group G some element A. Then all integer powers of this element form a cyclic subgroup (a) = (a p p e Z)G.
- 5. Let us prove that the additive group of rational numbers Q is not itself cyclic, but any two of its elements lie in a cyclic subgroup.
A. Let us prove that the additive group Q is not cyclic. Assume the contrary: let Q = (-). There is an integer b,
not dividing T. Since - eQ = (-) = sn-|neZ>, then there are
b t/ ( t J
there is an integer rc 0 such that - = n 0 -. But then m = n 0 kb,
where t:b- came to a contradiction.
B. Let's prove that two arbitrary rational numbers -
With „ /1
and - belong to the cyclic subgroup (-), where T there is a d t/
lesser common multiple of numbers b And d. Indeed, let t-bu
, and ai 1 /1 With cv 1/1
and m = av, u, v e Z, then - = - = ai-e(-)u - = - = cv-e(-).
b bu 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)| = |a|.
Proof. 1. Let |a| = ">. Let us prove that all natural powers of an element A different. Assume the contrary: let a k = a t and 0 to Then T - To is a natural number and a t ~ k = e. But this contradicts the fact that | | a =°°. Thus, all natural powers of an element A are different, whence it follows that the group (a) is infinite. Therefore, | (a)| = °° = |a |.
2. Let | a | = n. Let us prove that (a) \u003d (e - a 0, a, a 2,..., a "-1). The inclusion (a 0, a, a 2, ..., o" 1-1) c (a) follows from the definition of a cyclic group. Let us prove the reverse inclusion. An arbitrary element of a cyclic group (A) has the form a t, Where those Z. Divide the schnapps by the remainder: m-nq + r, where 0 p. Since a n = e, That a t = a p i + g \u003d a p h? a r = a r e(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. Assume the contrary: let 0 i P, but a" = A). Then he - e and 0 j - i - came to a contradiction with the condition | a | = P. The theorem has been proven.
Let G- group and element a
G. The order of the element a (denoted ׀а׀) is the smallest natural number n
N, What
a n = a . . . . a =1.
If such a number does not exist, then we say that A is an element of infinite order.
Lemma 6.2. If a k= 1 , then k is divisible by the order of the element A.
Definition. Let G- group and A
G. Then the set
H = (a k ׀ k 
}
is a subgroup of the group G, called the cyclic subgroup generated by the element a (denoted by H =< а >).
Lemma 6.3. Cyclic subgroup H, generated by the element A order n, is a finite order group n, and
H \u003d (1 \u003d a 0, a, ..., a n-1).
Lemma 6.4. Let A is an element of infinite order. Then the cyclic subgroup H
= <A> is infinite and any element from H is written in the form a k
, To
Z, and in a unique way.
The group is called cyclical if it coincides with one of its cyclic subgroups.
Example 1. Additive group Z of all integers is the infinite cyclic group generated by the element 1.
Example 2 The set of all roots n th power of 1 is a cyclic order group n.
Theorem 6.2. Any subgroup of a cyclic group is cyclic.
Theorem 6.3. Every infinite cyclic group is isomorphic to the additive group of integers Z. Any finite cyclic order n isomorphic to the group of all roots n th degree out of 1.
normal subgroup. Group factor.
Lemma 6.5. Let H– subgroup of a group G, for which all left cosets are simultaneously right cosets. Then
aH=Ha,
a 
G.
Definition. Subgroup H groups G called normal in G(denoted HG) if all and left cosets are also right cosets, that is,
aH=Ha,
a
G.
Theorem
6.4.
Let H
G,
G/N is the set of all cosets of the group G by subgroup H. If defined on the set G/N multiplication operation as follows
(aH)(bH) = (ab)H,
That G/N becomes a group, which is called the quotient group of the group G by subgroup H.
Group homomorphism
Definition. Let G 1 and G 2 - groups. Then the mapping f:
G 1 
G 2 is called a homomorphism G 1 in G 2 if
F(ab)
= f(a)f(b)
,
a,b
G 1 .
Lemma 6.6. Let f is a group homomorphism G 1 per group G 2. Then:
1) f(1) - group unit G 2 ;
2) f(a -1)
= f(a) -1
,
a
G 1
;
3) f(G 1) - subgroup of the group G 2 ;
Definition. Let f is a group homomorphism G 1 per group G 2. Then the set
kerf
= {a
G 1
׀f(a)
= 1
G 2
}
is called the kernel of the homomorphism f .
Theorem 6.5.
ker
f
G.
Theorem 6.6. Any normal subgroup of the group G is the kernel of some homomorphism.
Rings
Definition. Non-empty set TO called ring, if two binary operations are defined on it, called addition and multiplication, and satisfying the following conditions:
TO is an abelian group with respect to the addition operation;
multiplication is associative;
distributivity laws hold
x(y+z) = xy+xz;
(x+y)z
= xz+yz,
x,y,z
K.
Example 1. Sets Q And R- rings.
The ring is called commutative, If
xy=yx,
x,y
K.
Example 2
(Comparisons). Let m is a fixed natural number, a And b are arbitrary integers. Then the number A comparable to the number b modulo m if the difference a
– b divided by m(written: a
b(mod m)).
The equation relation is an equivalence relation on the set Z, breaking Z into classes called residue classes modulo m and denoted Z m. A bunch of Z m is a commutative ring with identity.
fields
Definition. A field is a non-empty set R, containing not 2 elements, with two binary operations of addition and multiplication such that:
Example 1. A bunch of Q And R endless fields.
Example 2. A bunch of Z r is the end field.
Two elements a And b fields R other than 0 are called zero divisors if ab = 0.
Lemma 6.7. There are no zero divisors in the field.
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 generatrix is the number 1 (and also 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 p. 32), it follows that the order of any element of a group divides the order of the group (note that all elements of a finite group are elements of a finite order).
Therefore, for any element g of a finite order group, the equality
This simple remark is often helpful.
Indeed, if the group O is cyclic and its generator, then the order of the element is . Conversely, if the group O has an element of order , then among the powers of this element there are different ones, and therefore these degrees 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 the cyclic group of order has exactly generators, where is the number of positive numbers less than and coprime with .
Along with the order, any finite group can be assigned 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 converse is generally not true. Nevertheless, the following assertion holds, which characterizes cyclic groups in the class of finite Abelian groups:
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-one elements of a finite Abelian group O are of order , and let be their least common multiple.
Let's expand the number into a product of powers of various 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 divisible exactly by ie, having the form , where b is coprime with . Let this number be the order of the element g. Then the element has an order (see Corollary 1) on p. 29).
Thus, for any in the group O there is at least one order element. Choosing one such element for each, consider their product. According to the statement proved on pages 29-30, the order of this product is equal to the product of the orders of , that is, it is equal to the number . Since the last number is equal to , this proves that the group O contains an element of order n. Therefore, this group is a cyclic group.
Let now 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 as , where d is some positive or negative integer (generally speaking, it is not uniquely defined). 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 the degree of the element . Indeed, by definition, there exists a number d such that (number d can be negative as well). 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).
We also note that for any order divisor of a finite cyclic group Q in the group O there exists one and only one subgroup H of order (namely, a subgroup with generator
This implies that if a finite cyclic group is simple, then its order is a prime number (or one).
Finally, we note that any quotient group, hence any homomorphic image) of a cyclic group Q is a cyclic group.
For the proof, it suffices to note that the generator of the group is the coset containing the generator of the group O.
In particular, any factor 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 R is a normal divisor. On the other hand, according to what was proved above, the subgroup H is a cyclic group. Since quotient groups by trivial subgroups are known to us, we can consider the subgroup Η to be non-trivial. Let a number be a generator of the subgroup H. We can consider this number to be positive (why?) and, therefore, greater than one.
The subgroup H. obviously consists of all integers divisible by . Therefore, two numbers belong to the same coset with respect to 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, the cosets with respect to the subgroup H are nothing but the classes of numbers that are comparable modulo .
In other words, the factor group of the group Z with respect to the subgroup H is a group (by addition) of classes of numbers that are comparable modulo . 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 the group O. We define a mapping of the group 2 into the group O by setting
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 useful:
subgroup H is generated by a subset M then and
If G = hMi and |M|< ∞, то G называется finitely generated.
A subgroup generated by a single 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 with respect to addition;
2) group Z(n) modulo residues n relative to addition;
her the 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
GZ.
2) If G is a finite cyclic group of order n, then
GZ(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 assumed to be infinity. The order of the element a is denoted by |a|. Note that |hai| = |a|.
2.1. Calculate the element orders of the 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 be of 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| is 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 nontrivial 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 generators in a cyclic group of order n is 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 has exactly one subgroup of order m.
2.20. Find all group generators: 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.
CHAPTER 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
is a special case of a homomorphism). Other varieties of homomorphism are often used:
a monomorphism is an injective homomorphism, an epimorphism is a surjective homomorphism, an endomorphism is a homomorphism into itself, an 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 respectively the kernel and the image of the homomorphism f. 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 the homomorphism is a normal subgroup. The converse is also true: every normal subgroup is the kernel of some homomorphism. To show this, we introduce on the set
16 Section 3. Homomorphisms, factor groups
G/N = (aN | a G) of cosets by a normal subgroup N operation: aN · bN = abN. Then G/N turns into a group, which is called a factor group with respect to the subgroup N. The mapping f: G −→ G/N is an epimorphism, and Kerf = N.
Every homomorphism f: G −→ H is the composition of an epimorphism G −→ G/Kerf, an isomorphism G/Kerf −→ Imf, and a monomorphism Imf −→ H.
3.1. Prove that these maps are homomorphisms.
mami 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 a field), f(x) = ax, a F ; d) f: R → R , f(x) = sgnx;
e) f: R → R , f(x) = |x|; f) f: C → R , f(x) = |x|;
g) f: GL(n, F) → F (where F is a field), f(A) = det A;
h) f: GL(2, F) → G, where G is the 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 the group G is the mapping f: G → G given by the formula
a) g 7→g2 b) g 7→g−1 ,
is a homomorphism?
3.3. Let f: G → H be a homomorphism and 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 groups G1 and G2 antiisomorphic if there exists 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 non-trivial 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. (A 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 the group G forms a composition group.
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 inner automorphisms form a subgroup of Inn G< Aut G.
3.14. Find the group of automorphisms a) Z;
b) a non-cyclic group of order 4 (see Problem 2.16); c) S3;
18 Section 3. Homomorphisms, factor 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 nonzero diagonal elements form a normal subgroup in
e) An C Sn ;
f) 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 element by element.
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);
f) 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 (that is, the order of any of its elements is finite) that 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) (that is, G/C(G) = E).
3.32. We call the commutator of elements x and y of the group G the element := x−1 y−1 xy. The commutator group 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. We 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 factor 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 a 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 factor groups Gk /Gk+1 are Abelian.
3.36. Check that a) abelian groups; b) groups S3 and S4;
c) a subgroup of all upper triangular matrices in GL(n, F) (where F is a field)
are resolvable.
3.37. Let G(n) be a subgroup of G generated by the set (gn | g G). Prove that:
a) G(n) C G;
b) G/G(n) has period n (that is, it satisfies the identity xn = 1);
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?
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.
Cosets, Lagrange's theorem
Let H group subgroup G. The left coset of the element a by subgroup H called a set of elements Ah, Where h belongs H. The left coset is denoted aH. Similarly, the right coset of the element is introduced a by subgroup H, which stands for Ha.
Since a subgroup always has a neutral element, each element a contained in an adjacent class aH (Ha).
Property 2.7. Elements a And b belong to the same left coset by subgroup H if and only if
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) cosets of elements a And b by the subgroup H have a common element, then they coincide.
Proof. Let . Then there will be . Arbitrary element from the left adjacent class aH contained in the left coset bH. Indeed, for , and, therefore, . The inclusion is proved similarly. Thus the theorem is proved.
Corollary 2.1. Left cosets either do not intersect or coincide.
Proof obviously.
Corollary 2.2. The left (right) coset is equicardinal to H.
Proof. Establish a correspondence between the elements of the subgroup H and elements of the adjacent class aH according to the formula . The correspondence is one-to-one. Thus the assertion is proved.
Theorem 2.3 (Lagrange). The order of a finite group is divisible by the order of its subgroup.
Proof. Let G- order group n, A H- subgroup G order k.Equality holds. Let's remove the repeating terms from the right side of the equality. As a result, non-intersecting adjacent classes will remain. Since the number of elements in the adjacent class is , then , where m the number of distinct cosets. This establishes the 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 under a group operation.
A group generated by one element is called cyclic.
Corollary 2.3. Any group contains a cyclic subgroup.
Proof. Let a-group element G. The set is a cyclic subgroup.
Order of a 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. Let's consider a sequence. Since the number of terms in the sequence is infinite, and for the powers of an element a there are a finite number of possibilities, then the same terms will occur in the sequence. Let where k<j And k the first repeating term. Then , and hence the term 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 items. Hence, k=n+1. Since , then .
The order of any element is a divisor of the order of the group, hence a | G | =e for any element of the group.
Corollary 2.4. The order of the group is evenly divisible by the order of any element of the group.
Proof obviously.
Theorem 2.4 (on cyclic groups)
I. For any natural n there is a cyclic order group n.
II. Cyclic groups of the same order are isomorphic to each other.
III. The 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 assertion is proved.
Let the cyclic group G order n generated by the element a, and the cyclic group H, of the same order, is generated by the element b. The correspondence is one-to-one and preserves the operation. The second assertion is proven
Cyclic group of infinite order generated by an element a, is made up of elements. The correspondence is one-to-one and preserves the operation. Thus, the third assertion is proved.
Let H is a subgroup of a cyclic group G generated by the element a. Elements H are degree a. Let's choose in H a. Let this be an element. Let us show that this element is generating in the subgroup H. 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 j, Then k-rj there is a remainder after division k on j and therefore less j. Because in H no elements that are nonzero powers a, less than j, That k-rj= 0, and . The fourth assertion is proved.
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 non-zero power in absolute value a. Let this be an element. Let us show that j=n/m. Item belongs H. Therefore, a non-zero number of the form rj-nv not less than in absolute value j, which is only possible if n divided by j without a trace. The subgroup generated by has the order n/j=m, hence, j=n/m. Since the generating element of a subgroup is determined uniquely by its order, the fifth assertion is proved.