Lagrange’s Theorem: Statement, Proof, Converse

Lagrange’s theorem states that the order of any subgroup of a finite group divides the order of the group. In this page, we will state and prove Lagrange’s theorem along with some applications and solved problems.

Statement

The Lagrange’s theorem for finite groups states the following: Let G be a finite group and H be it’s subgroup. Then the order of H divides the order of the group G.

Proof of Lagrange’s Theorem

As H is a subgroup of G, the set G is partitioned into disjoint union of left cosets of H in G. Since G is finite, we can write

G = g₁H ∪ g₂H ∪ … ∪ g_kH

for some positive integers k and g_i∈G. Here k is the number of distinct left cosets of H in G.

As we know that each left coset has as many elements as H, we obtain that

|G| = |g₁H| + |g₂H| + … + |g_kH|

= |H| + |H| + … + |H| (k times)

= k |H|.

This implies |H| divides |G|. In other words, the order of each subgroup of a finite group divides the order of the group. This completes the proof of the Lagrange’s theorem.

Remark: The number k of distinct left cosets is called the index of H in G and is denoted by [G:H].

Converse of Lagrange’s Theorem

The converse of Lagrange’s theorem says that “If d divides the order of a group G, then G has a subgroup of order d.”

The converse is not true. Here is an example: Note 6 divides the order of the alternating group A₄ as |A₄| = 12. But there is no subgroup of order 6 in A₄.

Proof:

In order to prove the above, we need to prove the following result.

Result: Let H be an index two subgroup of G, i.e, [G:H] = 2. Then $x^2 \in H$ for every $x \in G$. $\underline{\color{blue}\textit{Proof:}}$ As [G:H] = 2, H is normal in G. The two left cosets are H and G-H. Since $|\dfrac{G}{H}|=2$, so (G-H)2 = H, the identity element in G/H. If $x \in H$, then $x^2 \in H$ If $x \in G-H$, then $xH = G-H$ ⇒ $(xH)^2 = H$ ⇒ $x^2 H = H$ ⇒ $x^2 \in H$. Combining, we get that $x^2 \in H$ for every $x \in G$.

If possible, suppose there is a subgroup H of A₄ of order 6. Note [A₄ : H] = 2. Then by the above result, the square of each element of A₄ belong to H.

Now, we list the complete 12 elements of A₄ below.

identity, (1 2)(3 4), (1 3)(2 4), (1 4)(2 3), (1 2 3), (1 3 2), (1 2 4), (1 4 2), (1 3 4), (1 4 3), (2 3 4), (2 4 3).

Lets denote these elements as f₀, f₁, …, f₁₁ respectively.

Now, we compute:

$f_0^2=f_1^2= f_2^2= f_3^2$ = identity.

$f_4^2=f_5$, $f_5^2=f_4$, $f_6^2=f_7$, $f_7^2=f_6$, $f_8^2=f_9$, $f_9^2=f_8$, $f_{10}^2=f_{11}$, $f_{11}^2=f_{10}$.

We see that there are more than 6 squares in A₄ and each of them must belong to H by the above result which is a contradiction to |H| = 6.

Thus, A₄ has no subgroup of order 6. This is an example which proves that the converse of Lagrange’s theorem is not true.

Applications

Lagrange’s theorem for finite groups says that if G is a finite group, then |G| = [G:H] × |H|.

Theorem 1: Every group of prime order is cyclic.

Proof:

Let G be a group of order p, a prime number. As |G| = p>1, so there is an element a in G, such that $\circ$(a) >1. Let H = <a>.

So |H| = $\circ$(a)>1

By Lagrange’s theorem, |H| divides |G| = p.

So |H| = $\circ$(a) = 1 or p.

But $\circ$(a)>1. As a result, |H| = p. This implies G = H = <a>, a cyclic group.

Therefore, we have shown that every group of prime order is a cyclic group.

Theorem 2: The order of each element in a finite group G divides the order of G.

Proof:

Let G be a finite group and let a ∈ G. Set H = <a>. By Lagrange’s theorem, $\circ$(a) = |H| divides the order of G. Hence, the proof follows.

Related Articles:

1. Prove Every Subgroup of Cyclic Group is Cyclic
2. Prove Group of Prime Order is Cyclic