Lagrange Theorem

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One of the statements in group theory states that H is a subgroup of a group G which is finite; the order of G will be divided by order of H. Here the order of one group means the number of elements it has. This theorem is named after Joseph-Louis Lagrange and is called the Lagrange Theorem. 

This theorem can be expressed as follows. 

|G| = [G : H] · |H|

Where G is the infinite variant, provided that |H|, |G| and [G : H] are all interpreted as cardinal numbers. 

It is essential to understand the terminology and its three lemmas before learning how to get into its proof. 


What is a Coset?

In group theory, if G is a group and H is its subgroup, H might be used to decompose the underlying set "G" into equal-sized decomposed parts called cosets. 

For example, if g is an element in G and h an element in H then

gH = {gh} which is the left coset of H in the group G in respect to its element.

And, 

Hg = {hg} is the right coset of H under the same logic. 


What are the Three Lemmas?

A lemma is a minor proven logic or argument that helps one to find results of larger and more complicated equations. In the Lagrange theorem, there are three lemmas. Let us understand them under the condition that G is a group and H is its subgroup. 


Lemma no.1 

If the statement above is true, H and any of its cosets will have a one to one correspondence between them. 


Lemma no.2  

If the above statement is true, the left coset relation, g1~ g2 but that is only if g1 × H = g2 × H has an equivalence relation. 


Lemma no.3 

Suppose S is a set and ~ is an equivalence relation on S. If there are two equivalent classes A and B with A ∩ B = ∅, then A = B. 


Proof

Using the previous statement about the relationship between H and g where G is a finite group and H is a subgroup of the order n. 

Now suppose each cost of bH comprises n number of different elements. 

Let H = {h\[_{1}\], h\[_{2}\]..........., h\[_{n}\]}, so b\[_{1}\], bh\[_{2}\]......, bh\[_{n}\] are n distinct members of bH. 

If, bh\[_{i}\] = bh\[_{j}\] ⇒ h\[_{i}\] = h\[_{j}\] is taken to be the cancellation law of G

Since G is finite the number of left cosets will be finite as well, let's say that is k. So, nk is the total number of elements of all cosets. This is also equal to the complete number of elements in G. So one can assume 

r = nk

k = r/n

This shows that the order of H, n is a divisor of m which is the order of group G. It is also clear that the index k is also a divisor of the group's order. 

Hence it is proved that 

|G| = |H|

This Lagrange theorem has been discussed and refined further by several mathematicians and has resulted in several other theorems. 


Lagrange Interpolation Theorem

To understand this theorem, one first needs to realise what is an interpolation. It is a way of finding new data points that are within a range of discrete data points. So, this theorem is a method of constructing a polynomial which goes through a desired set of points as well as takes on certain values at arbitrary points. To put it more precisely, it provides a constructive proof of the following theorem as well. 

If there is a point (2,5), how can one find a polynomial that can represent it?

P (x) = 5

P (2) = 5

If there is a sequence of points, that is (2,5), (3,6), (4,7). How can one find a polynomial that can represent it?

P(x) = \[\frac{(x-3)(x-4)}{(2-3)(3-4)}\] х 5 + \[\frac{(x-2)(x-4)}{(3-2)(3-4)}\] х 6 + \[\frac{(x-2)(x-3)}{(4-2)(4-3)}\] х 7

P(2) = 5

P(3) = 6

P(4) = 7

This can be written in a general form, like

P(x) = \[\frac{(x-x_{2})(x-x_{3})}{(x_{1}-x_{2})(x_{1}-x_{3})}\] х y\[_{1}\] + \[\frac{(x-x_{1})(x-x_{3})}{(x_{2}-x_{1})(x_{2}-x_{3})}\] х y\[_{2}\] + \[\frac{(x-x_{1})(x-x_{2})}{(x_{3}-x_{1})(x_{3}-x_{2})}\] х y\[_{3}\]

P(x) = \[\sum_{1}^{3}\] P\[_{i}\] (x) y\[_{i}\]

Here the theorem states that given n number of real values x\[_{1}\], x\[_{2}\],........,x\[_{n}\] and n number of real values which are not distinct y\[_{1}\], y\[_{2}\],........,y\[_{n}\], there is a unique polynomial P that has real coefficients. This coefficient satisfies the equation 

P(x\[_{i}\]) = y\[_{i}\] for i ∈ {1, 2, …..,n} , such that deg deg(P) <n

To understand how this theorem is proven and how to apply this as well as Lagrange theorem avail Vedantu's live coaching classes. Learn to visualise mathematical problems and solve them in a smart and precise way. Also, you can get sample sheets to practice mathematics at home. 

FAQ (Frequently Asked Questions)

1. What is the Relationship Between Rolle's Theorem and Lagrange's Theorem?

Ans. Lagrange's mean value theorem is one of the most essential results in real analysis, and the part of Lagrange theorem that is connected with Rolle's theorem. One of its crucial uses is to provide proof of the Fundamental Theorem of Calculus. At the same time, one of the particular cases of Lagrange's mean value theorem that satisfies specific conditions is called Rolle's theorem. 

This mean value theorem is also known as LMVT theorem, and it states that 

If on a closed interval [j,k] there is a defined function a which satisfies the following statements.

  1. a is continuous on [j,k].

  2. On the open interval (j,k) a is differentiable.

There exists a value x = l in a way that

f’(l) = [f(k) – f(a)]/(k-a)

Rolle's theorem further adds another statement that is 

  1. If f(j) = f(k), then there should exist at least one value of x. if it is assumed that this value is l and it lies between j and k in a way that f' (l) = 0.  

2. Do I Have to Study Lagrange's Theorem to Understand Rolle's Theorem?

Ans. Rolle's theorem or Rolle's lemma are extended sub clauses of a mean value through which certain conditions are satisfied. The mean value in concern is the Lagrange's mean value theorem; thus, it is essential for a student first to grasp the concept of Lagrange theorem and its mean value theorem.  One cannot understand what Rolle's lemma is stating if their basics about the LMVT theorem is not strong.

3. Is Lagrange Theorem Easy?

Ans. Lagrange theorem and its three lemmas are significantly easy to understand and grasp if practised daily. This theorem is the basis of several other theorems such as the LMVT theorem and Rolle's theorem. This also helps to prove the fundamentals of Calculus and helps mathematicians in solving more critical problems. 


Also, with the right guidance and self-study, no subject in the world is difficult to understand. So it is ideal to learn such critical topics only from experienced tutors.