*Let* be a group and S Î G be such that it satisfies the following conditions :

- e Î S, where e is the identity of
- For any a Î S, a
^{-1}Î S. - For a, b Î S, a * b Î S.

*Then*~~ is called a ~~.

**subgroup**of

*For any group*, < {

*e*}

*, *> and* are trivial subgroups. All other subgroups are called

**proper subgroups**.

**Examples:**

- (Q , +) is a subgroup of (R, +).
- (D
_{n }, à) is a subgroup of (S_{n}, à). - ({[0], [2], [4]} , +) and ({[0], [3]} , +) are subgroups of (Z
_{6}, +). - Let a be an element of a group
, then G must contain all the integral powers of a, a ^{r }G, for r Z, thus, the cyclic group generated by an element ‘a’ is a subgroup of.

**Theorem 3.2-1:**

A non-empty subset S of G is a subgroup of

^{-1 }Î S.

**Proof:**

Assume that S is a subgroup. For any pair a, b Î S, we have a, b

^{-1}Î S, thus a * b

^{-1}Î S.

To prove the converse, let us assume that for any pair a, b Î S, a * b

^{-1 }Î S.

Since S is nonempty, let a Î S. Then by taking b = a , we have a * a

^{-1}= e Î S.

For every a Î S, consider the pair a, e in S, then we have, e*a

^{-1}Î S, that is, a

^{-1}Î S.

Finally, for any a, b Î S, consider the pair a, b

^{-1}Î S, then we have a*(b

^{-1})

^{-1}= a * b Î S.

Hence

*Let* and D> be two groups. A mapping g : G ® H is called a to

D>, if for any a, b Î g(a * b) = g(a) D g(b).

**group homomorphism**from

*G,*

**Examples:**

- Let G = (Z , +) and let H = ({1, -1} , •), where • is the usual multiplication.

Define f : G ® H by

Then, f(m + n) = f(m) • f(n), for m, n Î Z.

Thus, f is a homomorphism. - Let G = (Z , +) and let H = (Z
_{n},+).

Define f : G ® H by f(a) = [a (mod n)].

Then, f(a+b) = [(a+b)(mod n)]

= [a(mod n)] + [b(mod n)]

= f(a) + f(b).

Thus, f is a homomorphism.

**Property 1:**

*Group homomorphism preserves the identity element.*

Let e

_{G}and e

_{H}are the identities of the groups

Let g : G ® H be a group homomorphism.

Now g(e

_{G}* e

_{G}) = g (e

_{G}) = g (e

_{G}) D g(e

_{G}), that is, g(e

_{G}) is idempotant.

But the only idempotant element of a group is its identity, therefore, g(e

_{G}) = e

_{H}.

**Property 2:**

*Group homomorphism preserves inverses.*

Let a Î G and a

^{-1}be its inverse.

Then, g(a*a

^{-1}) = g (e

_{G}) = e

_{H}= g (a) D g (a

^{-1})

Similarly, g(a

^{-1}*a ) = g(e

_{G}) = e

_{H}= g(a

^{-1}) D g(a).

Therefore, g(a

^{-1}) is the inverse of g(a) in

**Property 3:**

*Group homomorphism preserves subgroups.*

Let g :

Let

Therefore e

_{H}= g(e

_{G}) Î g(S).

For any a Î S, a

^{-1}Î S and g(a) as well as g(a

^{-1}) = [g (a)]

^{-1}are in g (S).

Also for a, b Î S, a * b Î S and hence

g(a), g(b) Î g(S) and g(a*b) Î g(S).

Since g is a homomorphism, we have, g(a) D g(b) = g(a * b) Î g(S).

Thus,

*A group homomorphism g is called*

"one-one" or "onto" or "one-one and onto" respectively.

**monomorphism**or**epimorphism**or**isomorphism**depending on whether g is"one-one" or "onto" or "one-one and onto" respectively.

*A homomorphism from a group (G , *) to (G , *) is called*

is called

**endomorphism**while an isomorphism of (G, *) into (G, *)is called

**automorphism**.*Let g be a group homomorphism from* to D>. The set of element of G which are mapped onto e

identity of H, is called the

_{H}, the

identity of H, is called the

**kernel of the homomorphism**and it is denoted by

**ker(g)**.

**Theorem 3.2-2:**

The kernel of a homomorphism g from a group

**Proof:**

Let g :

Since g(e

_{G}) = e

_{H}, e

_{G}Î ker (g)

Also, if a, b Î ker(g), then, g(a) = g(b) = e

_{H}.

As g(a * b) = g(a) D g(b) = e

_{H }D e

_{H}= e

_{H}, we have a * b Î ker(g).

Finally, if a Î ker (g), then g(a

^{-1}) = [g(a)]

^{-1}= e

_{H}

^{-1}= e

_{H}.

Thus, for a Î ker(g), a

^{-1}Î ker (g).

Hence ker (g) is a subgroup of (G, *).

**Theorem 3.2-3:**

Any infinite cyclic group is isomorphic to Z.

**Proof:**

Let G = (a) be an infinite cyclic group.

Define g :

^{n}.

Then, g(m+n) = a

^{m+n}= a

^{m}* a

^{n}= g(m) * g(n).

Therefore, g is a homomorphism.

As

^{m}¹ a

^{n}.

Hence, if m ¹ n, g(m) ¹ g(n) .

Thus, g is one-one.

For every b Î G, there exists some m Î Z such that b = a

^{m}.

Thus, g(m) = a

^{m}= b.

Hence, g is onto.

**Exercise:**

Prove that any cyclic group of order n is isomorphic to Z

_{n}.

**Theorem 3.2-4:**

Every finite group of order n is isomorphic to a permutation group of degree n.

**Proof:**

Let

permutation of the element of G. Corresponding to an element a Î G we denote p

_{a}, the permutation given by the column under

‘a’ in the composition table.

Thus, p

_{a}(c) = c * a, for any c Î G.

Let the set of permutation be denoted by

**P**.Then

**P**has n elements. We shall show that <

**P**, à> is a group where à denotes the

right composition of the permutation of

**P**.

Since e Î G, p

_{e}Î

**P**and p

_{e}à p

_{a}= p

_{a}à p

_{e}= p

_{a}, for any a Î G.

Also, for any a Î G, à p

_{a}= p

_{e}.

Further, for any a, b Î G, p

_{a}à p

_{b}= p

_{a*b}………… (1)

The equation (1) follows from the fact that

(p

_{a}à p

_{b})(c) = (c * a) * b (by right composition)

= c * (a *b) (by associativity)

= p

_{a*b}(c)

Thus (

**P**, à) is a group.

Consider the mapping f : G ®

**P**given by f(a) = p

_{a}for any a Î G.

Then, f(a * b) = p

_{a*b}= p

_{a}à p

_{b}= f(a) à f(b).

Thus f is a homomorphism.

If a ¹ b then p

_{a}¹ p

_{b}, that is, f(a) ¹ f(b).

Hence f is one-one.

Further, for every p

_{a}Î

**P**, there exists a Î G such that f(a) = p

_{a}.

Thus f is onto.

Hence f is an isomorphism.

Hence the theorem.

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