In this section we prove a very important theorem, popularly called "Lagrange’s Theorem", which had influenced to initiate the

study of an important area of group theory called "Finite Groups". Finite groups have great applications in the study of finite

geometrical and combinational structures.

Let (G, *) be a group and let H be a subgroup G.

*A*

**left coset**corresponding to an element aÎ G, denoted by aH is the set*{a * H / h Î H}.*

Similarly, a

Similarly, a

**right coset**corresponding to an element aÎ G, denoted by Ha is the set*{h * a / h Î H}.*

**Example:**

For the group, (Z

_{4}, +) = ({[0], [1], [2], [3]} , +) and for the subgroup H = {[0], [2]}, we have the following left cosets.

[3] + H = {[3], [1]}.

[2] + H = {[2], [0]}= H.

[1] + H = {[1], [3]}.

[0] + H = {[0], [2]} = H.

Now we shall prove our main result, the Lagrange’s theorem.

Theorem [Lagrange’s Theorem]:

Theorem [Lagrange’s Theorem]:

In a finite group order of any subgroup divides the order of the group.

Proof:

Step 1:

Proof:

Step 1:

*Let (G, *) be a group and let H be a subgroup of G.*

*Define, for all a, bÎ G, a b(mod H) if and only if*

b

b

^{-1}* a Î H. Then the*relation, º (modH), "congruence modulo H", is an equivalence relation.*

For all a Î G, we have, a

^{-1}* a = e Î H.

So, a º a (mod H)

Thus, º is reflexive.

Let a º b (mod H).

Then, b

^{-1}* a Î H.

So, (b

^{-1}* a)

^{–1}Î H.

That is, a

^{-1}* (b

^{-1})

^{-1}Î H.

That is, a

^{-1}* b Î H.

So, by definition of "º ", b º a (mod H)

Thus, º is symmetry.

Let a º b (mod H) and b º c(mod H).

Then, b

^{-1}* a Î H and c

^{-1}* b Î H

Therefore, (c

^{-1}* b) * (b

^{-1}* a) Î H.

That is, c

^{-1}* a Î H

So, a º c (mod H).

Implies, " º " is transitive.

Hence, the step 1.

Step 2:

Step 2:

*For aÎ G, the equivalence class [a] is nothing but the left coset a * H.*

*Further G is partitioned into distinct*

cosets.

cosets.

By definition, [a] = { x Î G / x º a (mod H) }

= { x Î G / a

^{-1}* x Î H}

= {x Î G / h = a

^{-1}* x, for h Î H}

= { x Î G / a * h = x, for h Î H}

= { a * h / for h Î H}

= a * H.

Thus, each equivalence class [a] is the left coset a * H of G corresponding to ‘a’.

Since " º " is an equivalence relation G, distinct equivalence classes partitions the whole group.

Hence the Step 2.

Step 3:

Step 3:

*Let G be a group and let H be a subgroup of H. Then, | a * H | = | H |.*

Let h

_{i}and h

_{j}Î H be such that h

_{i}¹ h

_{j}. Suppose a * h

_{i}= a * h

_{j}.

Then, a

^{-1 }* (a * h

_{i}) = a

^{-1}* ( a* h

_{j}).

That is, (a

^{-1 }* a) * h

_{i}= (a

^{-1}* a)* h

_{j }

That is, e * h

_{i}= e * h

_{j }.

That is, h

_{i}= h

_{j}.

A contradiction to the fact that h

_{i}¹ h

_{j}.

Thus, if h

_{i}and h

_{j}are distinct in H, then a * h

_{i}and a* h

_{j}are distinct in a * H.

Hence, | a * H | = | H |.

Step 4:

Step 4:

*Let G be a finite group and let H be a subgroup then o(H)*½

*o(G).*

By Step2, G = a

_{1}* H È a

_{2}* H È … È a

_{t}* H, where,

a

_{1}* H, a

_{2}* H ,…, a

_{t}* H are the distinct coset of G ( and have no common elements)

Therefore, |G| = |a

_{1}*H| + |a

_{2}* H| + … + |a

_{t}* H|

By Step 3, |G| = |H| + |H| + … + |H| (t times)

So, |G| = t |H|

That is, |H| ½ |G| = t (is an integer).

Hence the order of H divides the order of G.

Hence the Lagrange’s Theorem.

*The number of left cosets of H in a group G in called the*

**index**of H in G.From the Lagrange’s Theorem for a subgroup H, we have the index k of G given by

k = | H | ½ | G |

Corollary:

Corollary:

If (G, *) is a finite group of order n, then for any a Î G, a

^{n}= e, where e is the identity of the group G. (Prove !)

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