Lagrange Theorem in Discrete mathematics

Joseph- Louis Lagrange developed the Lagrange theorem. In the field of abstract algebra, the Lagrange theorem is known as the central theorem. According to this theorem, if there is a finite group G, which contains a subgroup H, in this case, the order of H will divide the order of G. In a group, we can indicate the number of elements with the help of order of that group. In this section, we are going to learn about the statements and proof of this theorem. We will also learn about the three lemmas, which will be helpful to prove this theorem.

Statements of Lagrange theorem:

According to this statement, the order of group G can be divided with the help of order of subgroup H. We can use the following way to represent the Lagrange theorem:

|G| = |H|

Now we will prove the Lagrange theorem. For this, first, we have to know about three lemmas and some important terminologies, which are described as follows:

What is Coset?

When we want to learn about the Lagrange theorem fully, then we have to know about the coset, which is described as follows:

In a group theory, if there is a finite group G, which has its subgroup H, and if this finite group contains an element g, then:

gH = { hg: h is used to show an element of the subgroup H} is the left coset of H in G for the element of G.

Hg = {hg: h is used to show an element of the subgroup H} is the right coset of H in G for the element of G.

Now we will learn the next topic that will be helpful to prove the Lagrange theorem, i.e., Lemmas, which is described as follows:

Lemmas:

There are three types of lemmas, and we will see them one by one like this:

Lemma 1: If there is a group G that contains its subgroup H, in this case, the subgroup H and any coset of H must have one-to-one correspondence.

Lemma 2: If there is a group G, which contains its subgroup H, in this case, the left coset relation will be g1 ~ g2 iff it contains the following equivalence relation:

g1* H = g2 * H.

Lemma 3: Suppose there is a set S and also an equivalence relation "~" on S. Suppose there are two equivalence classes, A and B, in such a way that A ∩ B = ∅, in this case, A = B.

Proof of Lagrange Theorem:

Now we can use the above three lemmas so that we can prove the statement of Lagrange.

Proof of Lagrange Statement:

Suppose there is a finite group G which has an order m, and it also has a subgroup that has an order n. Now we will consider the coset breakdown of group G related to H. Now we will assume that each coset of aH is made of n different elements.

Suppose H = {h1, h2, h3, …, hn}, then ah1, ah2, ah3, …., ahn will be known as the n distinct members of aH.

Now we will assume ah_i=ah_j⇒h_i=h, which is the cancellation law of group G. Since, we know that group G is a finite group. That's why the number of discrete left coset will also be finite, says p. So, np will be used to indicate the total number of elements of all cosets. In graph G, the total number of elements will also be equal to np. Hence, m = np.

p = m /n

Here n is the order of subgroup H, and m is the order of finite group G

The above equation shows that the order of subgroup is a divisor of the order of subgroup H. It will also show that the index p is also a divisor of the order of group G.

Hence proved, |G| =|H|

Lagrange Theorem Corollary

Here we have three types of corollary in the Lagrange theorems, which are described as follows:

Corollary 1:

If there is a group G that has a finite order m, in this case, the order of group G will be divided by the order of any a ∈ G. Here a^m = e.

Proof: Suppose the order of 'a' is indicated by p, which is known as the least positive integer. So it has the following relation:

a^p = e

In this case, we have the following:

a, a², a³, a⁴, …., a^p-1, a^p = e, the group contains all the distinct elements which are used to form a subgroup. Since p is used to show the order of subgroups, thus group G will be divided by p, the order of a.

So, it can write in the following way:

m = np. Here, n is used to indicate a positive integer.

So,

a^m = a^np = (a^p)ⁿ = e

Hence proved

Corollary 2:

If prime order is contained by the order of finite group G, in this case, the group will have no proper subgroups.

Proof: Suppose m is the prime order of G. Now, according to the prime number property, there is a total of two divisors, 1 and m, in the prime order m. Therefore, the subgroup of group G will be itself G and {e}. So for the finite group G, we don't have any proper subgroups. Hence proved.

Corollary 3:

If there is a group that has prime order, then that group will be a cyclic group.

Proof: Let us assume m is the prime order in the group G and a ≠ e ∈ G.

Since the order of 'a' is calculated by the divisor of m. The order of G can either be m or 1. But, the order of a, o(a) ≠ 1, since a ≠ e. Therefore, the order of o(a) = p and the order m will also be contained by the 'a', which generates the cyclic subgroup of G.

With the above explanation, we have proved that the cyclic subgroup, which is generated by 'a', and the group G, are both similar. So G is cyclic.

Important Notes:

There are some important points that we should know about the Lagrange theorem. These points are described as follows:

• According to this law, the order of group G will be dividend of the order of subgroup H.
• If there is a group G that has a finite order m, the order of group G will be divided by order of any a ∈ G and, in particular, a^m = e.
• If prime order is contained by the order of finite group G, the group will not have any proper subgroups.
• If there is a group that has prime order, then that group will be a cyclic group.