Unit - 4

Cauchy's theorem for finite abelian groups

Q1) State and prove Cauchy's theorem for finite abelian groups.

A1)

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.

Q2) Define homomorphism.

A2)

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 may exist more than one homomorphism of G onto G’.

Q3) Explain normal subgroup.

A3)

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)

Q4) 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.

A4)

Proof: (a) It is a homomorphism, because

(b) It is not a homomorphism, because .

(c) It is a homomorphism, because*

(d) It is a homomorphism, because

Q5) The following are three equivalent conditions for a subgroup H to be a normal subgroup of a group G.

1.

2.

3.

A5)

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.

Q6) Prove that congruence modulo n is an equivalence relation on Z.

A6)

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.

Q7) Define the Kernel of a homomorphism.

A7)

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.

Q8) 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.

A8)

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}

Q9) The following are three equivalent conditions for a subgroup H to be a normal subgroup of a group G.

1.

2.

3.

A9)

(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: 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.

Q10) Explain Cayley’s theorem.

A10)

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.

Q11) Define isomorphism.

A11)

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).

Q12) List the properties of Isomorphisms Acting on Groups

A12)

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.

Q13) What is the first isomorphism theorem?

A13)

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.

Q14) What is the third isomorphism theorem?

A14)

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,