Cosets and Lagrange's Theorem

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



Example:

For the group, (Z₄, +) = ({[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]:

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

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^-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: For aÎ G, the equivalence class [a] is nothing but the left coset a * H. Further G is partitioned into distinct
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: 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: Let G be a finite group and let H be a subgroup then o(H)½o(G).

By Step2, G = a₁ * H È a₂ * H È … È a_t * H, where,

a₁ * H, a₂ * H ,…, a_t * H are the distinct coset of G ( and have no common elements)

Therefore, |G| = |a₁*H| + |a₂ * 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:

If (G, *) is a finite group of order n, then for any a Î G, aⁿ = e, where e is the identity of the group G. (Prove !)