Unit 4 | unit 4 cauchys theorem for finite abelian groups

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Unit - 4

Cauchy's theorem for finite abelian groups

4.1 Cauchy's theorem for finite abelian groups

Let G be a finite Abelian group and let p be a prime that divides the order of G. Then G has an element of order p.

Proof:

The statement given above is true for the case in which G has order 2. We prove the theorem by using the Second Principle of Mathematical Induction on |G|. That is, we assume that the statement is true for all Abelian groups with fewer elements than G and use this assumption to show that the statement is true for G as well. Certainly, G has elements of prime order, for if |x| = m and m = qn, where q is prime, then || = q. So let x be an element of G of some prime order q, say.

If q = p, we are finished; so assume that q p. Since every subgroup of an Abelian group is normal, we may construct the factor group G/{x}. Then is Abelian and p divides ||, since | | = |G|/q. By induction, then, has an element- call it y{x} - of order p.

4.2 Group homomorphisms, properties of homomorphisms

Homomorphism- Suppose we have two groups G and G’. We shall suppose that the composition in both the groups are denoted multiplicatively.

Then we say that G is homomorphic to G’ or that G’ is a homomorphic image of G if there exists a mapping f of G onto G’ such that

Also, then f is said to be a homomorphic mapping or a homomorphism of G onto G’.

We express this relation of homomorphism by-

Note- if G is homomorphic to G’, then there is may exist more than one homomorphism of G onto G’.

Normal Sub Group: A Subgroup N of a group G is called a Normal Subgroup of G if it is invariant under conjugation that is the conjugation of an element of N by an element of G is always in N. The usual notation for this relation is NG, and the definition may be written in symbol as

Congruence Relations: If a and b are integers and m is a positive integer, them a is congruent to b modulo m if m divides a-b.

The notation a (mod m) says that a is congruent to b modulo m.
We say that a (mod m) is a congruence and that m is its modulus.
Two integers are congruent mod m if and only if they have the same remainder when divided by m.
If a is not congruent to b modulo m, we write ab (mod m)

Example 1: Let be a group homomorphism, and let H=ker(). Let a Then, the set is the left coset aH of H, and is also the right coset Ha of H, Consequently, the two partitions of G into left cosets and into right cosets of H are the same.

Proof: (a) It is a homomorphism, because

(b) It is not a homomorphism, because .

(d) It is a homomorphism, because

Example 2: The following are three equivalent conditions for a subgroup H to be a normal subgroup of a group G.

Proof: (1) Suppose that H is a subgroup of G such that for all g and all h Then

We claim that actually =H.We must show that H for all g

Let h . Replacing g by in the relation , we obtain .

Consequently,

(3) Suppose that gH=Hg for all g . Then , So for all g

And all By the preceding paragraph, this means that =H for all g.

(2)Conversely, if for all g then = gH But also, giving ,so that hg=g and Hg.

Example:3 Prove that congruence modulo n is an equivalence relation on Z.

Solution:

1) Reflexivity: For any awe have because a-a = 0 is divisible by n. Hence relation is reflexive.

2) Symmetry: suppose a b (mod n)

is divisible by n = k, for some k z a-b = nk

Therefore, b-a = -(a-b) = -nk= n(-k)

Thus, the relation is symmetric.

3) Transitivity: Suppose a b (mod n) and b c (mod n), then = k and

By adding these two equations we get, a-c = n(k+1) = k+l

So, a-c is divisible by n as k+1 Z, i.e., a c (mod n)

Thus, the relation is transitive.

Hence this is an equivalence relation on Z.

Homomorphism into: If f is mapping of a group G, into a group G’, such that

Then we say that f is a homomorphism of G into G’.

Kernel of a homomorphism:

If f is a homomorphism of G onto G’, then the set H of all those elements of G which are mapped on the identity e’ of G’, which means the sub-set is a normal subgroup of the same.

Let so that

e’ being the identity of G’.

We then have

Thus

And hence H is a subgroup of G

Again, if x is any member of G, we have

So that

Hence H is normal subgroup of G.

Definition:

If f is a homomorphism of G onto G’, then the sub-set H consisting of all those elements of G which are mapped on the identity of G’ is a normal sub-group of G, called the Kernel of the homomorphism.

Note-

An injective homomorphism is called a monomorphism and a surjective homomorphism is called an epimorphism. An isomorphism is a homomorphism which is both injective and surjective.
An endomorphism of G is a homomorphism on G and an automorphism of G is an isomorphism of G onto itself.

Example: Let G be (Z, +) i.e., the group of integers under addition and let f: G → G defined by

∅(x) = 3x ∀x ∈G. Prove that f is homomorphism, determine its Kernel.

Solution: We have ∅(x) = 3x ∀x ∈G

∀x, y ∈G ⇒x + y ∈G (∴G is a group under addition)

Now

f (x + y) = 3 (x + y)

= 3x + 3y

= f (x) + f (y)

Hence f is homomorphism.

Kernel of homomorphism consists of half of zero i.e., the integers whose double is zero.

Thus K = {0}

Congruence Relations: If a and b are integers and m is a positive integer, them a is congruent to b modulo m if m divides a-b.

The notation a (mod m) says that a is congruent to b modulo m.
We say that a (mod m) is a congruence and that m is its modulus.
Two integers are congruent mod m if and only if they have the same remainder when divided by m.
If a is not congruent to b modulo m, we write ab (mod m)

Proof: (a) It is a homomorphism, because

(b) It is not a homomorphism, because .

(d) It is a homomorphism, because

Example 2: The following are three equivalent conditions for a subgroup H to be a normal subgroup of a group G.

Proof: (1) Suppose that H is a subgroup of G such that for all g and all h Then

We claim that actually =H.We must show that H for all g

Let h . Replacing g by in the relation , we obtain .

Consequently,

(3) Suppose that gH=Hg for all g . Then , So for all g

And all By the preceding paragraph, this means that =H for all g.

(2)Conversely, if for all g then = gH But also, giving ,so that hg=g and Hg.

Example 3: Prove that congruence modulo n is an equivalence relation on Z.

Solution:

1) Reflexivity: For any awe have because a-a = 0 is divisible by n. Hence relation is reflexive.

2) Symmetry: suppose a b (mod n)

is divisible by n = k, for some k z a-b = nk

Therefore, b-a = -(a-b) = -nk= n(-k)

Thus, the relation is symmetric.

3) Transitivity: Suppose a b (mod n) and b c (mod n), then = k and

By adding these two equations we get, a-c = n(k+1) = k+l

So, a-c is divisible by n as k+1 Z, i.e., a c (mod n)

Thus, the relation is transitive.

Hence this is an equivalence relation on Z.

Key takeaways:

G is homomorphic to G’ or that G’ is a homomorphic image of G if there exists a mapping f of G onto G’ such that

2. if G is homomorphic to G’, then there may exist more than one homomorphism of G onto G’.

3. A Subgroup N of a group G is called a Normal Subgroup of G if it is invariant under conjugation that is the conjugation of an element of N by an element of G is always in N.

4. If a and b are integers and m is a positive integer, them a is congruent to b modulo m if m divides a-b.

5. If f is mapping of a group G, into a group G’, such that

Then we say that f is a homomorphism of G into G’.

4.3 Cayley's theorem

Cayley’s theorem- Every finite group G is isomorphic to a permutation group.

Proof:

Let G be a finite group of order n and that its n elements, in some order are

Let b is any element of G so that it is one of the elements in the set (1),

Then

Are all different elements of G and they are the elements of G written in different order,

In fact

Thus

Is a one-one mapping of the set (1) onto itself so that

Which is a permutation of n degree.

Here we will use the notation for this as below

Hence, we see that we have a permutation of degree n,

corresponding to each

Suppose G’ denotes the set of all permutations corresponding to the elements of the set (1).

Now consider the mapping

This mapping of G onto G’ is one-one for

Again

Thus, we see that we have a one-one mapping f of the group G onto a set G’ of permutations such that

Thus G’ is a group isomorphic to the group G.

Key takeaways

Every finite group G is isomorphic to a permutation group.

4.4 Properties of Isomorphisms

Isomorphism

Let (s, *) and (S’, *) be two semi group. A function f: S S’ is called an isomorphism if s is one to one and onto and if

f(a+b) =f(a)*f(b). For all a, b in S

Procedure to find the isomorphism

Step 1: We do in the function f: S  S’ with domain of f=S

Step 2: We shall show that f is one to one.

Step 3: We shall show that f is onto.

Step 4: We shall show that f(a*b) = f(a)*’ f(b).

Q1. Let S be the set of all even integers. Show that the semi groups (Z, +) and (S, +) are isomorphic

Step 1: We define the function f: Z  S where f(a)=2a

Step 2: Suppose f(a1) =f(a2). Then 2a1=2a2. Hence f is one to one

Step 3: Suppose b is an even integer. Then a=b/2 € Z and f(a)= f(b/2) = 2(b/2) =b. Hence fits onto.

Step 4: We have f (a+b) = 2(a+b) = 2a+2b= f(a)+f(b)

Hence (Z, +) and (S, +) are isomorphic.

An isomorphism from a semi group (S, *) to (S, *) itself is called an automorphism (auto=self) on (S, *)

Properties of Isomorphisms Acting on Elements

Suppose that is an isomorphism from a group onto a group then,

1. carries the identity of G to the identity of

2. For every integer n and for every group element a in G,

3. For any elements a and b in G, a and b commute if and only if

(a) and (b) commute.

4. G = {a} if and only if = (a)}.

5. |a| = | (a)| for all a in G (isomorphisms preserve orders).

6. For a fixed integer k and a fixed group element b in G, the equation b has the same number of solutions in G as does the equation = (b) in

7. If G is finite, then G and have exactly the same number of elements of every order.

Proof:

Observe that property 5 follows from properties 1 and 2, property 6 follows from property 2, and property 7 follows from property 5. For convenience, let us denote the identity in G by e and the identity in by . Then, since e = ee, we have

Also, because (e) , we have (e) = (e), as well. Thus, by cancellation,

= f(e). This proves property 1.

For positive integers, property 2 follows from the definition of an isomorphism and mathematical induction. If n is negative, then 2n is positive, and we have from property 1 and the observation about the positive integer case that

Thus, multiplying both sides on the right by , we have

Property 1 takes care of the case n = 0.

To prove property 4, let G = {a} and note that, by closure { because is onto, for any element b in , there is an element in G such that () = b

Thus, b = and so b {[. This proves that {

Now suppose that {. Clearly, . For any element b in G, we have (b) {. So, for some integer k we have

Because is one-to-one, b = . This proves that {a} = G.

Properties of Isomorphisms Acting on Groups

Suppose that is an isomorphism from a group onto a group then

1. is an isomorphism from onto G.

2. G is Abelian if and only if is Abelian.

3. G is cyclic if and only if is cyclic.

4. If K is a subgroup of G, then (K) = { (k) | k } is a subgroup of G.

Key takeaways:

Let (s, *) and (S’, *) be two semi group. A function f: S S’ is called an isomorphism if s is one to one and onto and if

f(a+b) =f(a)*f(b). For all a, b in S

2. is an isomorphism from onto G.

3. G is cyclic if and only if is cyclic.

4. G is Abelian if and only if is Abelian.

4.5 First, second and third isomorphism theorems.

1. First isomorphism theorem:

This theorem is also known as fundamental theorem of homomorphism.

According to this theorem every homomorphic image of G is isomorphic to a quotient group of G.

In other words,

Let be a homomorphism of groups, then

Hence, in particular, is surjective then

Proof:

Consider the mapping,

Given by

Where K = Ker, for any

Hence, is well defined and injective, further

Hence is a homomorphism. Since is obviously surjective, we conclude that is an isomorphism of groups.

2. Second isomorphism theorem:

Suppose H and N be the sub-groups of G, and , Then

Note- this theorem is also known as diamond isomorphism theorem.

Proof:

Since is a subgroup of G and .

Consider the mapping

Given by

In fact, is the restriction of the natural homomorphism to H.

Hence,

Moreover, is clearly surjective, hence by the first theorem given above,

3. Third isomorphism theorem:

Suppose H and K be normal subgroups of G and then

Note- this theorem is also known as “double quotient isomorphism theorem.”

Proof:

Let the mapping

Given by

The mapping is well defined, for

Further, for all

Hence is a homomorphism. Now is obviously surjective, and

Hence, by the first isomorphism theorem,

Key takeaways:

First isomorphism theorem- every homomorphic image of G is isomorphic to a quotient group of G.
Second isomorphism theorem- Suppose H and N be the sub-groups of G, and , Then