Lakhasly

This chapter defines algebraic structures, starting with internal composition laws on a set G, illustrated by addition and multiplication on number sets and composition of maps. Associative, commutative, neutral, and inverse elements are defined, leading to the definition of a group (G,⋆) requiring associativity, a neutral element, and inverses for all elements. Commutative groups are also defined. Key theorems establish the uniqueness of the neutral and inverse elements, and a formula for the inverse of a product. Subgroups H of G are defined, characterized by non-emptiness, closure under the operation and inverses. Equivalent conditions for subgroups are provided, with specific examples for additive and multiplicative groups. The chapter then introduces rings (A,+,.), defined as a commutative group under (+) and an associative operation (∙) distributive over (+). Commutative rings and rings with identity are defined. Subrings are defined and characterized. Finally, fields (𝕜,+,.) are defined as rings where all non-zero elements are invertible; commutative fields are also defined. Subfields are characterized as subrings where inverses of non-zero elements exist, with examples showing (Q,+,.) and (R,+,.) as subfields of (C,+,.).

Chapter III: Algebraic structures
Definitions
Let G be a non-empty set.

1. The internal composition law:
   We call an internal composition law (a binary operation) a map, which is defined as: f:𝐺×𝐺→ 𝐺
   (𝑥,𝑦) → f(𝑥,𝑦)
   Wenotethislawby ⋆,T, +, ×, ∙,o,...
   Remark 1
   In general, we note this law by (⋆) and we write x ⋆ y instead of f (𝑥, 𝑦). So, we have
   ∀ x , y ∈ G : x ⋆ y ∈ G.
   Examples

• The laws (+) and ( ∙ ) are internal composition laws on the sets: N, Z, Q, R and C.


• The law (−) is an internal composition law on the sets: Z, Q, R and C but is not on the set N.


• The law (o) is an internal composition law on the set of maps from G to G.
  Remark 2
  In the following, we suppose that (⋆) is an internal composition law on the set G.

2. The associative law:
   (⋆) is an associative law If :
   Examples
   ∀x,y,z∈G: (x⋆y)⋆z=x⋆(y⋆z).

• Thelaws(+)and(∙)areassociativelawsonthesets:N,Z,Q,RandC.


• The law (−) is not an associative law on the sets: N, Z, Q, R and C.


• The law (o) is an associative law on the set of maps from G to G.

3. The commutative law:
   (⋆) is a commutative law if:
   Examples
   ∀ x , y ∈ G, x ⋆ y = y ⋆ x.

• Thelaws(+)and(.)arecommutativelawsonthesets:N,Z,Q,RandC.


• The law (−) is not a commutative law on the sets: N, Z, Q, R and C.


• The law (o) is not a commutative law on the set of maps from G to G.

4. The neutral element (the identity element):
   We say that the set G has a neutral element with respect to the law (⋆) if:
   ∃e∈G: ∀x∈G,x⋆e=x=e⋆x (eistheneutralelementofG).
   Examples

• e = 0 is the neutral element for the law (+) and e = 1 is the neutral element for the law ( ∙ ) on the sets: N,Z,Q,RandC.


• e = Id is the neutral element for the law (o) on the set of maps from G to G.
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5. The inverse element (the symmetric element):
   We suppose that the set G has a neutral element e with respect to the law (⋆). We say that the element x has an
   inverse on G with respect to the law (⋆) if :
   ∀x∈G∃x′∈G: x⋆x′=e=x′⋆x (x′ istheinverseelementofx).
   Examples

• x ′ = − x is the inverse element for the law (+) on the sets: Z, Q, R and C.


• The inverse element on N does not exist (− x is not in N).


• x ′ = x -1 is the inverse element for the law ( ∙ ) on the sets: Q {0}, R {0} and C {0}.


• x ′ = x -1 is not in N and Z.


• If f is a bijective map then f-1 is the inverse element of f for the law (o) on the set of maps from G to G.
  Group and Subgroup Definition1
  Let G be a non-empty set with an internal composition law (⋆). The pair (G, ⋆) is a group if the following conditions are satisfied:

1. (⋆) is an associative law.


2. Ghasaneutralelementewithrespectto(⋆)


3. Any element x of G has a symmetric .
   If moreover the law (⋆) is commutative, then (G, ⋆) is called a commutative (or abelian) group.
   Examples


4. (R, +), (Z, +), (Q, +) and (C, +) are commutative groups with the usual addition as operation. (N, +) is not a group.


5. (Q {0}, ∙), (R {0}, ∙) and (C {0}, ∙) are commutative groups with the usual multiplication as operation. (N {0}, ∙) and (Z {0}, ∙) are not groups.


6. The set of bijective maps with respect to the law (o) is a non-commutative group.
   Theorem1
   Let (G, ⋆) be a group then:
    The neutral element e is unique.
    An element x ∈ G has only one inverse x′ and (x′ )′= x.  ∀x,y∈G, (x⋆y)′=y′⋆x′
   Definition2
   Let (G, ⋆) be a group. A part H ⊂ G is a subgroup of G if (H, ⋆) is also a group with the law induced by that of G.. We can write H≤ G to indicate that H is a subgroup of G.
   Theorem2
   Let (G, ⋆) be a group. H ⊂ G is a subgroup of G if and only if: • H≠∅.
   •∀x,y∈H, x⋆y∈H, •∀x∈H, x′ ∈H.
   Theorem3
   Let (G, ⋆) be a group. H ≤ G if and only if:
   • H≠∅.
   • ∀x,y∈H, x⋆y′∈H. (y′ istheinverseofy).
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Remark1
Let (G,+)beagroup.H≤Gifand only if: • H≠∅.
• ∀x,y∈H,x−y∈H.(−yisthesymmetricofy). Remark2
Let (G, .) be a group. H ≤ G if and only if: • H≠∅.
• ∀x,y∈H,x.y-1∈H.(y-1 istheinverseofy ). Remark3
If (G, ⋆) is a group with neutral element e, then the subsets {e} and G are always groups, they are called the trivial subgroups.
Example1
Let (Z,+) be an abelian group, show that:
𝐻 = {𝑥 = 3k /k ∈ Z } is a subgroup of Z.

1. 0=3.0=3k ⟹0∈H⟹H≠∅.


2. ∀ x , y ∈ H : x = 3k1 , y = 3 k2 ⟹ x − y = 3(k1− k2)= 3k, k ∈ Z ⟹ x − y ∈ H. So, H≤Z.
   Example2
   Let (R {0}, .) be an abelian group, show that :
   𝐻 = {𝑥 ∈ R /x > 0 } is a subgroup of R {0}.


3. 1>0⟹1∈H⟹H≠∅.


4. ∀ x , y ∈ H : x > 0, y > 0 ⟹ x . y -1 > 0 ⟹ x . y -1 ∈ H. So, H ≤ R {0}.
   Ring and Subring Definition1
   Let A be a non-empty set with two internal composition laws (+) and (∙) , it is said that (A, +, . ) is a ring if the following conditions are satisfied:

1. (A,+)isacommutativegroup.


2. Thelaw(∙)isassociative:∀x,y,z∈A, (x.y).z=x.(y.z). 3. (∙)isdisributivetothelaw(+):
    ∀x,y,z∈A, x.(y+z)=(x.y)+(x.z).  ∀x,y,z∈A, (y+z).x=(y.x)+(z.x).
   Remarque :

• If moreover the operation (∙) is commutative i.e: ∀ x , y ∈ G, x . y = y . x , (A, +, . ) is said to be a commutative ring.


• If A has a neutral element compared to the (∙) law, (A, +, . ) is said to be a ring with identity or a ring with one..
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Example
(Z, +, . ), (Q, +, . ), (R, +, . ), (C, +, . ), are commutative rings with identity. Definition2
Let(A,+,.)bearing. B⊂AisasubgringofAif: • B≠∅.,
•∀x,y∈B, x−y∈B,
•∀x,y∈B, x.y∈B.
Note that a subring (B, +, . ) is also a ring with the law induced by that of A.
Example
(n Z, +, . ) is a subring of the ring (Z, +, . ) where n ≠ 0.
Field and Subfield Definition1
Let (𝕜, +, . ) be a ring with identity . x ∈ 𝕜 is invertible in 𝕜 if : ∃ y∈ 𝕜{0} : x . y = 1 = y . x, where 0 and 1 are the neutral elements of 𝕜 with respect to (+) and (.) respectively.
The set of all invertible elements is noted by:
𝕜*={ x∈𝕜/xisinvertiblein𝕜 }.
Example
x ∈ Z is invertible in Z if : ∃ y∈ Z {0} : x . y = 1 ⟹ 𝒚 = (𝟏) ∈ Z⟹x=∓1⟹ Z = {-1,1 }. 𝒙
Q= Q {0}, R*= R {0}, C*= C {0}. Definition2
Let (𝕜, +, . ) be a ring with identity, we say that (𝕜, +, . ) is a field if all elements of 𝕜{0} are invertible i. e : 𝕜*=𝕜{0}.
Remark
In addition, if the law (. ) is commutative, we said that ( 𝕜 , +, .) is a commutative field.
Examples

1. (R,+,.),(Q,+,.)𝑒𝑡(C,+,.)arecommutativefields. 2. (Z, +, . ) is not a field because Z *= {-1,1} ≠ Z{0}.
   Definition3
   Let (𝕜, +, . ) be a field and L is a subset of 𝕜, then (L, +, . ) is a subfield of (𝕜, +, . ) if: • L≠∅.,
   •∀x,y∈L, x−y∈L,
   • ∀ x , y ∈ L (y≠0), x . y -1 ∈ L. (y -1 is the symmetric element of y). Note that a subfield (L, +, . ) is also a field with the law induced by that of 𝕜.
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Remark
If (𝕜, +, . ) is a field and L is a subset of 𝕜, then (L, +, . ) is a subfield of (𝕜, +, . ) if: 1) (L,+,.)isasubringof(𝕜,+,.).
2) ∀x∈L{0},x--1 ∈L.
Examples

1. (R,+,.),(Q,+,.)aresubfieldsof(C,+,.). 2. (Q,+,.)isasubfieldof(R,+,.).
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