Lagrange Theorem in Group Theory

In group theory, the Lagrange theorem states that if ‘H’ is a subgroup of the group ‘G,’ then the order of ‘H’ divides the order of ‘G.’ It is named after Joseph-Louis Lagrange, who derived it and is one of the central theorems of abstract algebra. 

Mathematically, it is expressed as

|G|/|H| or o(G)/o(H).

Here, the order of the group represents the number of elements.

While proving the Lagrange theorem, we need to understand some terminologies and lemmas.

Cosets

If ‘G’ is a finite group and ‘H’ is a subgroup of ‘G,’ such that g Є G, then

gH = {gh: h Є H} is the left coset of ‘H’ in ‘G’ with respect to the element of ‘G,’ and Hg = {hg: h Є H} is the right coset of ‘H’ in ‘G’ with respect to the element of ‘G.’ 

For example,

The 3-cycle (0, 1, 2) Є S₃ has order 3, so H = ⟨(0, 1, 2)⟩ equals to {e, (0, 1, 2), (0, 2, 1)} 

Thus,

(1,2)H = {(1, 2), (1, 2) (0, 1, 2), (1, 2) (0, 2, 1)} = {(1, 2), (0, 1), (0, 2)}

(0, 1)H = {(0, 1), (0, 1) (0, 1, 2), (0, 1) (0, 2, 1)} = {(0, 1), (0, 2), (1, 2)}

We conclude that (1,2)H = (2,3)H.

Now, let us explore the lemmas that will help us to prove the Lagrange theorem.

Lemmas 

Lemma 1 with Proof

Statement

If ‘G’ is a group with the subgroup ‘H,’ then there is a one-to-one correspondence between ‘H’ and any coset of ‘H.’

Proof 

Let ‘L’ be a left coset of ‘H’ in ‘G.’ Then there is an element g Є G such that L = g ∗ H, where ‘∗’ represents the binary operation in ‘G.’ 

Now, let us define a function f: H → L by f(x) = g ∗ x 

First, we prove that the function ‘f’ is one-to-one. 

If x₁ ≠ x₂ and since ‘G’ has cancellation, g ∗ x₁ ≠ g ∗ x₂, then f(x₁) ≠ f(x₂)

Now, we prove that the function ‘f’ is an onto function. If l Є L and since L = g ∗ H, there is an element h Є H such that l = g ∗ h. It follows that f(h) = l

Since ‘l’ is arbitrary, the function ‘f’ is onto.

Thus, Lemma 1 is proved.

Lemma 2 with Proof

Statement

If ‘G’ is a group with the subgroup ‘H,’ then the left coset relation, g₁ ∼ g₂ if and only if g₁ ∗ H = g₂ ∗ H is an equivalence relation.

Proof

Since ‘∼’ is defined in terms of set equality and equality for sets is an equivalence relation, we prove that ‘∼’ is an equivalence relation. 

First, we prove that ‘∼’ is reflexive. Let g Є G be given. Then, g ∗ H = {g ∗ h: h Є H}, and this set is well defined. Thus, g ∗ H = g ∗ H.

Now, let us prove that ‘∼’ is symmetric. Let g₁, g₂ Є G with g₁ ∼ g₂ 

By the definition of ‘∼,’ g₁ ∗ H = g₂ ∗ H 

That is, {g₁ ∗ h: h Є H} = {g₂ ∗ h: h Є H}

Since the set equality is symmetric, {g₂ ∗ h: h Є H} = {g₁ ∗ h: h Є H}. This implies g₂ ∼ g₁

Since g₁ and g₂ are arbitrary, ∼ is symmetric. 

Now, we prove that ‘∼’ is transitive. 

Let g₁, g₂, g₃ Є G with g₁ ∼ g₂ and g₂ ∼ g₃ 

Then, g₁ ∗ H = {g₁ ∗ h: h Є H} = {g₂ ∗ h: h Є H} = g₂ ∗ H and 

g₂ ∗ H = {g₂ ∗ h: h Є H} = {g₃ ∗ h: h Є H} = g₃ ∗ H. 

Since the set equality is transitive,

g₁ ∗ H = {g₁ ∗ h: h Є H} = {g₃ ∗ h: h Є H} = g₃ ∗ H, or g1 ∗ H = g3 ∗ H. 

That is, g1 ∼ g3

Since g₁, g₂, and g₃ Є G are arbitrary, ‘∼’ is transitive. 

Thus, Lemma 2 is proved. 

Lemma 3 with Proof

Statement

Let ‘S’ be a set and ‘∼’ be an equivalence relation on ‘S.’ If ‘A’ and ‘B’ are two equivalence classes with A ∩ B = ɸ, then A = B.

Proof

Here, we prove that A ⊂ B and B ⊂ A

Since ‘A’ and ‘B’ are arbitrary, it is sufficient to show the former. 

Let a Є A. Since A ∩ B ≠ ɸ, there is an element c Є A ∩ B 

Since ‘A’ is an equivalence class of ‘∼’ and both ‘a’ and ‘c’ are in ‘A,’ which follows that a ∼ c. Since a ∼ c, c Є B and ‘B’ is an equivalence class of ‘∼,’ which follows that a Є B

Thus, Lemma 3 is proved.

Now, using the three lemmas above, we can prove the Lagrange statement.

Lagrange Theorem Proof 

Let ‘H’ be any subgroup of the order ‘n’ of a finite group ‘G’ of order ‘k.’ 

Here, we consider the coset breakdown of ‘G’ related to ‘H’ by assuming each coset of aH comprises ‘n’ different elements.

If H = {h₁,h₂,…,h_n}, then aH = {ah₁, ah₂, …, ah_n}

Here, ah₁,ah₂,…,ah_n are the ‘n’ distinct members of aH.

Thus, ah_i = ah_j ⇒ h_i = h_j, which is the cancellation law of ‘G.’

Since ‘G’ is a finite group, the number of discrete left cosets is also finite.

Let us consider the number of discrete left cosets of ‘H’ in ‘G’ as ‘p.’

The total number of elements of all cosets is np, equal to the total number of elements of ‘G.’ 

Hence, ${k=np}$

⇒ ${p=\dfrac{k}{n}}$, which proves that ‘n’ is a divisor of ‘k.’ 

⇒ the order of ‘H’ is a divisor of the order of the finite group ‘G.’

Here, we conclude that ‘p’ is also a divisor of the order of the group.

Hence, the Lagrange theorem is proved by o(G)/o(H).

However, its converse may or may not be true. A group satisfying the converse to Lagrange’s theorem is called a CLT group, which states that for a finite group ‘G,’ if ‘n’ divides ‘G,’ then there exists a subgroup H ≤ G of order ‘n.’

Lagrange Theorem Corollaries

Now, let us verify some corollaries related to the Lagrange theorem.

Corollary 1

If ‘G’ is a group of finite order ‘k’ and ‘g’ is an element of the group ‘G,’ then the order of ‘g’ divides the order of ‘G.’ In particular, g^k = e.

Proof

Let the order of the element ‘g’ be ‘p,’ the least positive integer.

Thus, g^p = e

Then g, g², g³, …, g^{p – 1}, g^p = e, the group ‘G’ elements are all distinct and form a subgroup.

Since the subgroup has the order ‘p,’ thus the order of ‘g’ divides the group ‘G.’

We get,

k = np, where ‘n’ is a positive integer.

Now,

g^k = g^np = (g^p)ⁿ = e

⇒ g^k = e, proved.

Corollary 2

If the order of the finite group ‘G’ is a prime number, then it has no proper subgroups.

Proof

Let the prime order of the group ‘G’ be ‘k.’ 

Now, using the prime numbers property, we get only two divisors of ‘k,’ which are 1 and ‘k.’ 

The subgroups of ‘G’ will be {e} and ‘G’ itself. 

Thus, there are no proper subgroups. Hence, it is proven.

Corollary 3

A finite group of prime order is a cyclic group.

Proof

Let ‘G’ be the finite group whose prime order is ‘k’ and g ≠ e Є G.

Since the order of ‘g’ is a divisor of ‘k’ and ‘k’ is a prime number, using the prime numbers property, we get only two divisors of ‘k,’ which are 1 and ‘k.’ 

Thus, the order of ‘g’ is either 1 or ‘k.’

Since g ≠ e, the order of ‘g’ is o(g) ≠ 1. The order of o(g) = p, say.

Thus, the cyclic subgroup of ‘G’ generated by ‘g’ is also of order ‘k,’ which proves that ‘G’ is the same as the cyclic subgroup formed by ‘g.’ 

Thus, ‘G’ is cyclic, which is proved.