## Tuesday, August 12, 2008

### Subgroups

Let be a group and S Î G be such that it satisfies the following conditions :
1. e Î S, where e is the identity of
2. For any a Î S, a-1 Î S.
3. 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:
1. (Q , +) is a subgroup of (R, +).
2. (Dn , à) is a subgroup of (Sn , à).
3. ({[0], [2], [4]} , +) and ({[0], [3]} , +) are subgroups of (Z6 , +).
4. Let a be an element of a group , then G must contain all the integral powers of a, ar 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 if and only if for any pair of elements a, b Î S, a * b-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-1Î S, then we have a*(b-1)-1 = a * b Î S.
Hence is a subgroup of .

Let and D> be two groups. A mapping g : G ® H is called a group homomorphism from to
D>, if for any a, b Î G,
g(a * b) = g(a) D g(b).

Examples:
1. 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.
2. Let G = (Z , +) and let H = (Zn ,+).
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 eG and eH are the identities of the groups and D> respectively.
Let g : G ® H be a group homomorphism.

Now g(eG * eG) = g (eG) = g (eG) D g(eG), that is, g(eG) is idempotant.

But the only idempotant element of a group is its identity, therefore, g(eG) = eH.

Property 2: Group homomorphism preserves inverses.

Let a Î G and a-1 be its inverse.

Then, g(a*a-1) = g (eG) = eH = g (a) D g (a-1)

Similarly, g(a-1 *a ) = g(eG) = eH = g(a-1) D g(a).

Therefore, g(a-1) is the inverse of g(a) in D>.

Property 3: Group homomorphism preserves subgroups.

Let g : ® D> be a group homomorphism.

Let be a subgroup of .

Therefore eH = g(eG) 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, D> is a subgroup of D>.

A group homomorphism g is called 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 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 eH, 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 to D> is a subgroup of .

Proof:
Let g : ® D> be a homomorphism.

Since g(eG) = eH, eG Î ker (g)

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

As g(a * b) = g(a) D g(b) = eH D eH = eH, we have a * b Î ker(g).

Finally, if a Î ker (g), then g(a-1) = [g(a)]-1 = eH-1 = eH.

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 : ® by g(n) = an.

Then, g(m+n) = am+n = am * an = g(m) * g(n).

Therefore, g is a homomorphism.

As is an infinite group , if m ¹ n then am ¹ an .

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 = am .

Thus, g(m) = am = b.

Hence, g is onto.

Exercise:

Prove that any cyclic group of order n is isomorphic to Zn.

Theorem 3.2-4:
Every finite group of order n is isomorphic to a permutation group of degree n.

Proof:

Let be a group of order n. We know that every row and column in the composition table of represent a
permutation of the element of G. Corresponding to an element a Î G we denote pa, the permutation given by the column under
‘a’ in the composition table.

Thus, pa(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, pe Î P and peà pa = pa à pe = pa, for any a Î G.

Also, for any a Î G, à pa = pe.

Further, for any a, b Î G, pa à pb = p a*b ………… (1)

The equation (1) follows from the fact that

(pa à pb)(c) = (c * a) * b (by right composition)
= c * (a *b) (by associativity)
= pa*b(c)

Thus (P , à) is a group.

Consider the mapping f : G ® P given by f(a) = pa for any a Î G.

Then, f(a * b) = pa*b = pa à pb = f(a) à f(b).

Thus f is a homomorphism.

If a ¹ b then pa ¹ pb, that is, f(a) ¹ f(b).

Hence f is one-one.

Further, for every paÎ P, there exists a Î G such that f(a) = pa.

Thus f is onto.

Hence f is an isomorphism.

Hence the theorem.