Discrete Mathematics-Groups and Rings

Semigroups and unique points

The algebraic system is composed of a non-empty set plus one or several operations.
From this section, we will introduce some special algebraic systems. The so-called special means that the operations in these algebraic systems have special properties. We want to introduce the following algebraic systems:
1. Semigroup:
Definition: Let S be a non-empty set, ★ is a binary operation on S, if ★ satisfies closeness and associativity on S, then it is called <S, *> is a semigroup.
2. Unique point:
Let <M,★> be a semigroup, if the operation of ★ has a unitary, then <M,★> is called a unique point, and it is also called a semigroup.





3. Commutable semigroup.
Suppose <S,★> is a semigroup. If ★ is commutative, then <S,★> is a commutative semigroup.
4. Exchangeable unique points
<M,★> are unique points. If ★ is exchangeable, then <M,★> is exchangeable unique points.
Example: <R,+>, <N,X>, <P(E),∩>, <P(E), symmetric difference> are all commutative semigroups, and they are also commutative unique points.
5.
Subsemigroup <S,★> is a semigroup, B⊆S, if ★ is closed on B, then <B,★> is said to be a subsemigroup of <S,★>.
Example: <N,+> is a sub-semigroup of <I,+>
6. Sub-unique point
<M,★> is a unique point, B⊆M, if ★ is closed on B, and unitary e∈B , Then it is said that <B,★> is the unique point of <M,★>.
Example: <I,+> is the unique point of the child of <R,+>.

Suppose <M,★> is the commutative unique point, and A is the set of all idempotent elements in M, then <A,★> is the child unique point of <M,★>.
Obviously A⊆M, if you want to prove that <A,+> is the sub-different point of <M,★>, according to the sub-different point definition, we only need to prove the element e ∈ A and the closure.

To prove closed, you need to prove that any two elements in the set do an operation, and to prove that this operation is still in A,

Definition and properties of groups

1. Definition of group

Groups are the most important algebraic system in abstract algebra.

1. Definition of the group: Let <G,★> be an algebraic system, if the operation of ★ satisfies closedness and associativity on G, there are unitary elements in <G,★> and each element in G is invertible, then Say <G,★> is a group.
(1) Let <G,★> be a group, if the set G is a finite set, then <G,★> is a finite group. Conversely, it is called an infinite group.
(2) A group containing only unitary elements is called a trivial group.
(3) If ★ operation is commutative, then <G,★> is said to be commutative group or Abel group

Are <R,+> and <P(E),⊕> a group?
<R,+> is a unique point, the unitary is 0, for any real number r, its inverse is -r.
<P(E), ⊕> The unit element is ∅, because for any set A∈P(E), A⊕A=in,
so A-1=A, so they are all groups.

<R,X> is the unique point, the unitary element is 1, and the zero element is 0. Since 0 has no inverse element, <R,X> is not a group.
<P(E),∩> is the unique point, and the unitary element is E. For any set A∈P(E) and A is not equal to the complete set E, is there such a set such that A∩?=E?
There is no such set, that is, A has no inverse element. So <P(E),∩> is not a group.

2. The nature of the group

In addition to the four properties of closed, associable, unitary, and reversible for each element, groups have some other properties.

1. There is no zero element in the group

Set <G, ★> is the group, if | G |> = 2, then G is no zero dollars .
Proof: (Disprove method) Assuming that there is a zero element θ in G, then for any x∈G, there is θ★x=x★θ=θ≠e, so there is no inverse element for the zero element θ, which is the same as <G,★> It is a group contradiction. Therefore, there is no zero element in the group <G,★>.
If an algebraic system has both zero and unitary elements, then unitary and zero elements must not be equal .

2. Each element in the group is an erasable element.

Suppose <G,*> is a group, then for any a,b,c∈G, if there is
(1)a★b=a★c, then b=c.
(2) b★a= c★a then b=c.
The proof can be proved by definition or by theorem.
The theorem can be used as the following theorem
Theorem: Let ★ be a binary operation that can be combined on X, if a ∈ X, and a-1 ∈ X, then a is a cancelable element.

3. Except for the unitary element, there are no other idempotent elements in the group.

Let <G,★> be a group, then there are no idempotent elements in G except for the unitary element.
Proof: (by contradiction) Assume that a∈G is an idempotent element, that is, a★a=a, so a★a=a★e, and a=e due to the elimination, so there is nothing in the group except the unitary element Idempotent.

4. The group equation has a unique solution

Suppose <G,★> is a group, then for any a,b∈G,
(1) there is a unique element x∈G, so that a★x=b… (1)
(2) there is a unique element y∈G, so that y ★a=b …(1)
Thinking: The
solution of the equation a★x=b is a-1★b
What is the solution of the equation y★a=b? b★a-1

5. The characteristics of the finite group operation table

Suppose <G,★> is a finite group, then every element in G must appear in every row (column) in the ★ operation table and only appears once.

<G,★> is a group, for any a, b∈G,
(1) (a-1)-1=a
(2) (a★b)-1= b-1★a-1

The order of the group and the order of the elements in the group

1. Group order:
Definition: Let <G,★> be a group, if |G|=n, then <G,★> is called an n-order group.
When the number of elements contained in G is finite, the order of the group <G,★> is the number of elements contained in G.
When the number of elements contained in G is infinite, the group <G,★> is an infinite group.

It can be seen from the operation table that all first-order groups are isomorphic; all second-order groups are isomorphic;
all third-order groups are isomorphic.

2.
Definition of the order of the elements in the group : Let <G,★> be the group, a∈G, the smallest positive integer k that makes ak=e is called the order of a, denoted by |a|=k, call a as k Order yuan.
If there is no such positive integer k, then the order of a is said to be infinite.
For example, addition on integers is infinite.
For example: the group <I,+> is an infinite group, only the order of unit 0 is 1, and the orders of the other elements are all infinite.
Example: The operation table of <X, o> is shown in the figure below: Is <X, o> a group? If it is a group, find the order of each element.

Let <G,★> be a group, a∈G and |a|=k. Let n be an integer, then
(1)an=e if and only if k/n
(2)|a-1|= |a|

Subgroup and its proof

Definition of group

Definition of subgroup

Let <G,★> be a group, S is a non-empty subset of G, if <S,★> satisfies:
(1) For any a, b∈S, there is a★b∈S; (closed)
(2) Unite e∈S; (there are unitary units)
(3) For any a∈S, if there is a1∈S (invertible),
then <S,★> is said to be a subgroup of <G,★>.
Subgroup: It should be a non-empty subset of the original group, and it should also be a group itself

Any group <G,★> has subgroups, <{e},★> and <G,★> are all subgroups of <G,★>, which are called trivial subgroups of <G,★>.
The trivial group refers to the set of <{e},★>, which has only unitary e.
Example: The algebraic system <R,+> is a group, and the algebraic system <I,+> is a subgroup of <R,+>.
Because I⊆R, the addition of any two integers is still an integer; unitary 0∈l; for each x∈l, its inverse -x∈I

Proof of Subgroup

Use the definition of subgroup to prove:
that is to prove that the operation satisfies the closedness on the non-empty subset, has unitary elements, and each element in the subset is invertible.

Subgroup Judgment Theorem 1: (Limited Closed)

Let <G,★> be a group, B is a finite subset of G, if ★ satisfies the closedness on B, then <B,★> is a subgroup of <G,★> .

(1) First prove that the unitary element e ∈

B (2) Prove that every element in B is invertible, and for any b ∈ B, there is b-1 ∈ B.

In summary, <B,★> is a subgroup of <G,★>.

Subgroup judgment theorem 2:

Let <G,★> be a group, S is a non-empty subset of G, if for any a,b∈S, there is a★b-1∈S, then <S,★> is a child of <G,★> group.

(1) proband identity element e∈S

(2) and then may permit any element in the inverse S

(3) turns out <S, ★> is closed, any of a, b∈S, has a ★ b∈S

mechanized Above, <S,★> is a subgroup of <G,★>.
Exercise: Knowing that <H1,★> and <H2,★> are subgroups of group <G,★>, verify that <H,∩H2,★> is <H1,★>, <H2,★> and <G ,★>'s subgroup.

(1) First prove that H1∩H2 is a non-empty subset of H1, H2 and G.
Obviously H1∩H2⊆H1, H1∩H2⊆H2,
H1∩H2⊆G;
because <H1, ★> and <H2, ★> are Subgroup of group <G,★>, so unitary e∈H1 and e∈H2,
so e∈H1∩H2, that is, H1∩H2≠
∅So H∩H2 is a non-empty subset of H1, H2, and G.
(2) Prove again that for any a,b∈H1∩H2, a★b-1∈H1∩H2

Cosets of Subgroups and Lagrange's Theorem

Cosets of subgroups

1. Definition: Let <H,★> be a subgroup of group <G,★>, a∈G, define the set:
aH={a★h|h∈H}
Ha={h★a|h∈H}
Call aH(Ha) the left (right) coset of H in G determined by a .
We only discuss the left coset, and we have similar conclusions about the right coset

Theorem 1: Two cosets are either equal or disjoint

<H,★> is a subgroup of the group <G,★>, any a, b∈G, there are
(1) aH=bH if and only if a∈bH
(2) aH∩bH= if and only if a ∉bH
a) Necessity, given that aH=bH, because e∈H, so a=a★e∈aH, so a∈bH.



It can be seen from the above theorem that any two left cosets of a subgroup are either equal or disjoint.
When a∈bH, aH=
bH ; when a∉bH, aH∩bH=∅.

Theorem 2: a belongs to only one coset

Let <H,★> be a subgroup of group <G,★>, for any a∈G, a must belong to and belong to only one coset.

Theorem 3: No two elements of the coset are the same

Suppose <G,★> is a finite group, <H,★> is a subgroup of group <G,★>, b∈G, bH is the left coset of <H,★>, then any two elements in bH They are all different.

(By contradiction, suppose two elements in
bH are the same) Suppose b★h1∈bH, b★h2∈bH, (where h1,h2∈H,h1≠h2) make b★h1=b★h2, which can be eliminated Sex has h1=h2, contradictory. So any two elements in bH are not the same.

Lagrange's theorem: the order of a group is an integer multiple of the order of a subgroup

Suppose <G,★> is a finite group, |G|=n, <H,★> is any subgroup of <G,★>, |H|=m, then n=km (k∈l) .
What Lagrange's theorem describes is that the order of the group is an integer multiple
of the order of the subgroup. The order of the group refers to the number of elements in the group.

Lagrange's theorem states that the order of subgroups of an n-th order group is a factor of the order of the group.

The following Corollary 1 explains:
The order of the elements in the group must be a factor of the order of the group .

Cyclic group

1. Definition:

Let <G,★> be a group, if there is an element g∈G, for any x∈G, there is an integer i, so that x=gi, then <G,★> is called a cyclic group. And said g is a generator of G's .
All elements can be generated by one of the element exponents.


Thinking: Is -1 a generator?

2. Types of cyclic groups:

According to the order of the generator g, the cyclic group <G,★> can be divided into two categories:

theorem

Let <G,★> be a finite cyclic group with g as the generator. Then |G|=n
if and only if |g|=n

The number of generators in the cyclic group

Let <G,★> be the cyclic group generated by g.
(1) If G is an infinite cyclic group, then G has only two generators g and g-1.
(2) If G is a cyclic group of order n, then G contains φ(n) generators.
For any positive integer r, if r≤n and relatively prime to n,
then gr is the generator of G.
φ(n) is Euler's function, that is, the number of positive integers less than or equal to n and relatively prime to n.
Proof: (1) If G is an infinite cyclic group, then G has only two generators g and g-1.

Proof: (2) If G is a cyclic group of order n, then G contains φ(n) generators.


Generally speaking, it is not easy to find a subgroup of a group, but for a cyclic group, you can directly find all its subgroups. The group
<G,★> is a cyclic group generated by g, |G|=12, which is
less than or equal to 12 and is relatively prime to 12 positive integers:
1,5,7,11, that is, φ(12)= 4. So <G,★> has 4 generators, namely: g, g5, g7, g11

Suppose <G,+>, G={3a| a∈I}, + is an ordinary addition operation, then <G,+> is an infinite cyclic group, with only two generators: 3 and -3.

Subgroup of cyclic group

(1) If <G,★> cyclic group, the subgroup of <G,*> is still cyclic group. .
(2) If <G, ★> group is an infinite loop, the <G, ★> subgroups except <{e}, ★> is other than an infinite cyclic group.
(3) If <G,★> is a cyclic group of order n, then each positive factor d of n, <G,★> contains exactly one subgroup of order d.

Proof: (1) If <G,★> cyclic group, then the subgroup of <G,*> is still a cyclic group. The
main reason is to find the generator.



Let <G,★> be a group of prime order, then it is nothing more than a trivial subgroup, And it must be a cyclic group.