Lagrange Theorem
2026-02-28 23:46 Diff
https://brightchamps.com/en-us/math/algebra/lagrange-theorem

According to Lagrange’s theorem, the order of a subgroup H divides the order of the group G. This relationship can be expressed as:

|G| = [G :H] .|H|

What is Coset?

A coset is the set of all products obtained by multiplying a fixed element of the group with each element of a subgroup.

In group theory, if g is an element of G, H is a subgroup of G, and G is a finite group, then
The left coset of H in G with respect to the element of G is g ∈ H is defined as:

gH ={gh \ h ∈ H}. And, The right coset of H in G with respect to the element g ∈ G is the set Hg={gh \ h ∈ H}.

Let us now discuss the lemmas that support the Lagrange theorem.

Lemma 1: In a group G, every coset of a group H has the same number of elements as H, and there’s a one to one matching between them.

Lemma 2: The left coset relation, g1 ~ g2, is an equivalence relation if and only if, g1*H=g2*H, provided that G is a group with subgroup H.

Lemma 3: Assume that S is a set and that ~ is an equivalence relation on S. If A and B are two equivalence classes, then  A ∩ B = Ø .

Lagrange Theorem Proof

The Lagrange statement is easily proven, with the help of the three lemmas. That is listed above.
Lagrange statement proof:

Let G be a finite group of order m, and let H be any subgroup of order n. Let us look how G can be divided into cosets of H.

Let’s now examine how each coset of gH is made up of n distinct elements.

Let H = {h1, h2, …, hn}, then ah 1, ah 2,..., ahn are the n different members of aH.

Assume that G satisfies the cancellation law: if ghi=ghj, then hi=hj.  

The number of distinct left cosets, let’s say p, will likewise be finite since G is a finite group. Therefore, np, or the total number of elements of G, is the sum of the elements of all cosets. Thus, m = np
p = m/n

This shows that m, the order of the finite group G, divides n, the order of H.

Additionally, we observe that the index of a subgroup, which is, p divides the order of the group. 

Thus, it was demonstrated that |G| = |H|

Lagrange Theorem Corollary

Corollary 1: The orders of any element a ∈ G divides the order of the group G, and in particular, am=e, if G is a group of finite order m.

Proof: Since a is the least positive integer and p is its order,
ap = e

Then we can state,
Group G’s elements, a,a2,a3, ..., ap-1, ap = e, are all unique and constitute a subgroup.

Since the element a creates a subgroup with p element and the whole group G also has p elements. That means the subgroup is actually the entire group. So, a generates the whole group.

Thus, we can write it as, If n is a positive integer, then write it as m=np.

Thus, the
am=anp=(ap)n = e
Thus, it is satisfying.

Corollary 2: A finite group G has no proper subgroups if its order is prime order.

Proof: Let us assume that group G has prime order m. The prime numbers property states that there are now only two divisors of m: 1 and m. Hence, the subgroups of G itself. As a result, no proper subgroups exist.

Corollary 3: A prime order group is cyclic if it has only two divisors.

Proof: Assume that a ≠ e ∈ G and that G is the prime order group of m.
The order of a is either 1 or m since it is a divisor of m.

However, since a ≠ e, the order of a, o(a) ≠ 1.

Consequently, the cyclic subgroup of G that is produced by a is of order m, a is the order of o(a)=p.
It establishes that G is cyclic, meaning that it is the same as the cyclic subgroup that a forms.