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Lagrange theoremstates that in group theory, for any finite group G, the order of subgroup H of group G divides the order of G. The order of the group is nothing but the number of elements present in the group. This theorem was established and named after Joseph-Louis Lagrange. It is given by,
\(|G| =|G: H|· |H|\)
where G is the infinite variant, provided that |H|, |G| and G : H, are all understood as cardinal natural numbers.
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Key Takeaways: Lagrange Theorem, Cardinal Numbers, Order, Coset, Group Theory, Prime Order
Lagrange Theorem Statement
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Lagrange theorem states that the order of the subgroup H is the divisor of the order of the group G. This can be written as;
|G| = |H| where G is a finite group and H is a subgroup of G.
This indicates:
- |H| is the divisor of |G|
- The number of unique cosets of H in G is given by | |G|/ |H| |
Read Also: Limits and Continuity
What is Coset?
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Consider g is an element of a finite group G and h is the element of a subgroup H of the G, then;
gH is the left coset of H in G with respect to the element of G
Hg is the right coset of H in G with respect to the element of G.
The three lemmas that prove the Lagrange theorem are as follows:
Lemma 1: If G is a finite group and H is its subgroup, then there is a one-on-one correspondence between H and any coset of H.
Lemma 2: If G is a finite group and H is its subgroup, then the left coset relation, g1 ~ g2 if and only if g1 H = g2 H is an equivalence relation.
Lemma 3: Let S be a set and ~ be an equivalence relation on S. If A and B are two equivalence classes with A ∩ B = \(\phi\), then A = B.
Read Also: Mean Value Theorem
Lagrange Theorem Proof
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Let H be any subgroup with an order 'n' of a finite group G of order m. Consider the coset breakdown of G with respect to H. Let’s assume that each coset of aH comprises n different elements.
Let H = {h1,h2,…,hn}, then ah1,ah2,…,ahn are the n number of unique elements of aH.
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Suppose, ahi=ahj⇒hi=hj be the cancellation law of G. As G is a finite group, so the number of discrete left cosets(p) will also be finite. So, the total number of elements of all cosets is np and the total number of elements of G is equal.
m = np
p = m/n
This proves that n, the order of H and the index p is a divisor of m, the order of the finite group G. Hence, proved, |G| = |H|
Read Also: Set Theory
Lagrange Theorem Corollary
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Corollary 1: If G is a finite order m group, the order of any aG divides the order of G, and in specific am = e.
Read Also: Polynomial Formula
Proof: Assume p is the order of a, which is the smallest positive integer; hence,
e = ap
As a result, it can be stated as follows:
The elements of group G, a, a2, a3,...., ap-1, ap = e, are all distinct and form a subgroup.
p divides the group G since it is the order of the subgroup a.
As a result, it may be rewritten as
Where n is a positive integer, m = np
(ap)n = e = am = anp
Hence proved.
Corollary 2: If the order of a finite group G is prime, it has no valid subgroups.
Consider the prime order of group G, that is m. Now, m only has two divisors: 1 and m. (prime numbers property). As a result, the subgroups of G will be e and G itself. Hence it is true.
Read More: Degree of Polynomial
Corollary 3: A cyclic group is a group of prime order (the order has only two divisors).
Assume G is the group of prime order of m and an eG.
Because a's order is a divisor of m, it is either 1 or m.
However, the order of a, o(a) 1, because an e.
As a result, the order of o(a) = p, as well as the cyclic subgroup of G formed by a, are both of order m.
It establishes that G is the same as the cyclic subgroup generated by a, implying that G is cyclic.
Read More: Types of Sets
Things to Remember
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- The number of distinct H cosets in G is given by | |G|/ |H| |
- If G is a finite group and H is a subgroup of G, then t g1 ~ g2 if and only if g1 x H = g2 x H.
- |G| =|G:H| |H| Where G is the infinite variant, assuming that |H|, |G|, and G:H are all cardinal numbers.
- Assume S is a set and is an equivalence relation on S. If A and B are two equivalence classes with the formula A ∩ B = \(\phi\), then A = B.
Also Read: Real-Valued Function
Sample Questions
Ques. If G is a finite group then let P be a subgroup of G. Suppose that the order n of P is relatively prime to the index |G:S|=m. Using the Lagrange theorem prove that N = {a∈GIan=e}. (3 Marks)
S,t ∈ Z such that
sn + tm = 1......................... (i)
Also, its given that as the order of the group G/N is |G/N|=|G:N|=m, we have
gmN = (gN)m = N
for any g∈G by Lagrange’ theorem, and therefore
gm∈N.............................(ii)
If a∈{a∈G|an=e}. Then we have an = e.
It follows that
a = asn + tm = asn atm = atm = (at)m∈N by (ii).
This proves that {a∈G|an=e}⊂N.
On the other hand, if a∈N, then we have an=e as n is the order of group N.
Hence N⊂{a∈G|an=e}.
Putting together these inclusions will yield that N={a∈G|an=e} as required.
Ques. Let G be a finite group and let J and K be two distinct Sylow p-group, where p is a prime number and divides the order |G| of G. Using the Lagrange theorem prove that the product JK will never be a subgroup of G. (5 Marks)
That is, we have |G|=pαn,
where p is not a divisor of the integer n.
So the orders of the Sylow p-subgroups J, K will be pα.
It is known that the J∩K is a subgroup of J, the order of J∩K is pβ for some integer β≤α by Lagrange’s theorem.
Since J and K are distinct subgroups, certainly β<α.
Then the number of elements of the product HK is
Thus JK is not a subgroup of G because the order |JK|=p2α−β divides |G| as per Lagrange’s theorem and pα is the highest power of p dividing G.
Hence it is proved that the JK will never be a subgroup of G.
Ques. Verify Lagrange’s mean value theorem for the following functions on the indicated intervals and find a point ‘c’ in the indicated interval as stated by Lagrange’s mean value theorem: f(x) = x (x – 1) on 1,2 (5 Marks)
= x² – x
Every polynomial function is continuous everywhere on (−∞, ∞) and differentiable for all arguments. Here, f(x) is a polynomial function. So it is continuous and differentiable in (1, 2). So both the necessary conditions of Lagrange’s mean value theorem are satisfied.
Therefore, there exist a point c ∈ (1, 2) such that:
f’(c)=f(2)-f(1)/2-1
⇒f’(c)=f(2)-f(1)/1
f (x) = x² – x
Differentiating with respect to x
f’(x) = 2x – 1
For f’(c), put the value of x=c in f’(x):
f’(c)= 2c – 1
For f (2), put the value of x = 2 in f(x)
f (2) = (2)² – 2
= 4 – 2
= 2
For f (1), put the value of x = 1 in f(x):
f (1) = (1)² – 1
= 1 – 1
= 0
∴ f’(c) = f(2) – f(1)
⇒ 2c – 1 = 2 – 0
⇒ 2c = 2 + 1
⇒ 2c = 3(1,2)
c=3/2\(\epsilon\)(1,2)
Hence Lagrange’s mean value theorem is proved
Ques.Verify Lagrange’s mean value theorem for the function:f (x) = 2x – x² on 0,1 (5 Marks)
Every polynomial function is continuous everywhere on (−∞, ∞) and differentiable for all arguments. Here, f(x) is a polynomial function. So it is continuous and differentiable in (0,1). So both the necessary conditions of Lagrange’s mean value theorem are satisfied.
Therefore, there exist a point c ∈ (0, 1) such that:
f’(c)=f(1)-f(0)/1-0
⇒ f’(c) = f(1) – f(0)
f (x) = 2x – x²
Differentiating with respect to x:
f’(x) = 2 – 2x
For f’(c), put the value of x = c in f’(x):
f’(c)= 2 – 2c
For f (1), put the value of x = 1 in f(x):
f (1)= 2(1) – (1)²
= 2 – 1
= 1
For f (0), put the value of x = 0 in f(x):
f (0) = 2(0) – (0)²
= 0 – 0
= 0
f’(c) = f(1) – f(0)
⇒ 2 – 2c = 1 – 0
⇒ – 2c = 1 – 2
⇒ – 2c = – 1
c=1/2 in 0,1
Hence Lagrange’s mean value theorem is proved
Ques. Show that the Lagrange’s mean value theorem is not applicable to the function f(x) = 1/x on –1,1 (3 Marks)
f(x) = 1/x on –1,1
Here, x ≠ 0
⇒ f (x) exists for all values of x except 0
⇒ f (x) is discontinuous at x=0
∴ f (x) is not continuous in –1,1
Hence Lagrange’s mean value theorem is not applicable to the function f (x) = 1/x on –1,1
Ques. Find a point on the parabola y = (x – 4)², where the tangent is parallel to the chord joining (4, 0) and (5, 1). (5 Marks)
This interval a,b is obtained by x – coordinates of the points of the chord.
Every polynomial function is continuous everywhere on (−∞, ∞) and differentiable for all arguments. Here, f(x) is a polynomial function. So it is continuous and differentiable in (4, 5). So both the necessary conditions of Lagrange’s mean value theorem are satisfied.
Therefore, there exist a point c ∈ (4, 5) such that:
f’(c)=f(5)-f(4) / 5-4
⇒ f’(c) = f(5) – f(4)
f (x) = (x – 4)²
Differentiating with respect to x:
f’(x) = 2(x – 4)(x – 4)d/dx
⇒ f’(x) = 2 (x – 4) (1)
⇒ f’(x) = 2 (x – 4)
For f’(c), put the value of x=c in f’(x):
f’(c) = 2 (c – 4)
For f (5), put the value of x=5 in f(x):
f (5) = (5 – 4)²
= (1)²
= 1
For f (4), put the value of x=4 in f(x):
f (4) = (4 – 4)²
= (0)²
= 0
f’(c) = f(5) – f(4)
⇒ 2(c – 4) = 1 – 0
⇒ 2c – 8 = 1
⇒ 2c = 1 + 8
c=9/2=4.5\(\epsilon\)(1,2)
We know that the value of c obtained in Lagrange’s Mean value Theorem is nothing but the value of x – coordinate of the point of the contact of the tangent to the curve which is parallel to the chord joining the points (4, 0) and (5, 1).
Now, put this value of x in f(x) to obtain y:
y = (x – 4)²
y=(9/2-4)²
y=(9-8/2)²
y=(½)²
y=¼
Hence the required points are (9/2,1/4)
Ques. Prove that f (x) = x³ – 5x² – 3x on 1,3 are verified by Lagrange’s Mean value theorem. (5 Marks)
Every polynomial function is continuous everywhere on (−∞, ∞) and differentiable for all arguments. Here, f(x) is a polynomial function. So it is continuous and differentiable in (1, 3). So both the necessary conditions of Lagrange’s mean value theorem are satisfied.
Therefore, there exist a point c ∈ (1, 3) such that:
f’(c)=f(3)-f(1) / 3-1
⇒ f’(c) = f(3) – f(1)/2
f (x) = x³ – 5x² – 3x
Differentiating with respect to x:
f’(x) = 3x² – 5(2x) – 3
= 3x² – 10x – 3
For f’(c), put the value of x=c in f’(x):
f’(c)= 3c² – 10c – 3
For f (3), put the value of x = 3 in f(x):
f (3)= (3)³ – 5(3)²– 3(3)
= 27 – 45 – 9
= – 27
For f (1), put the value of x = 1 in f(x):
f (1)= (1)³ – 5 (1)² – 3 (1)
= 1 – 5 – 3
= – 7
⇒f’(c)=f(3)-f(1)/2
⇒3c²-10c-3=(-27)-(7)/2
⇒3c²-10c-3=(-20)/2
⇒3c²-10c-3=-10
⇒3c²-10c-3+10=0
⇒3c²-7c-3c+7=0
⇒ c (3c – 7) – 1(3c – 7) = 0
⇒ (3c – 7) (c – 1) = 0
c = 7/3 , 1
c = 7/3\(\epsilon\)(1,3)
Hence it is proved.
Ques.Verify f (x) = 2x² – 3x + 1 on 1,3 for Lagrange’s Mean Value theorem. (5 Marks)
f (x) = 2x² – 3x + 1 on 1,3
Every polynomial function is continuous everywhere on (−∞, ∞) and differentiable for all arguments. Here, f(x) is a polynomial function. So it is continuous and differentiable in (1, 3). So both the necessary conditions of Lagrange’s mean value theorem are satisfied.
Therefore, there exist a point c ∈ (1, 3) such that:
f’(c)=f(3)-f(1) / 3-1
⇒ f’(c) = f(3) – f(1)/2
f (x) = 2x² – 3x + 1
Differentiating with respect to x
f’(x) = 2(2x) – 3
= 4x – 3
For f’(c), put the value of x = c in f’(x):
f’(c)= 4c – 3
For f (3), put the value of x = 3 in f(x):
f (3) = 2 (3)² – 3 (3) + 1
= 2 (9) – 9 + 1
= 18 – 8 = 10
For f (1), put the value of x = 1 in f(x):
f (1) = 2 (1)² – 3 (1) + 1
= 2 (1) – 3 + 1
= 2 – 2 = 0
f’(c)=(f(3)-f(1))/2
⇒4c-3=10-0/2
⇒4c=10/2+3
⇒4c=8
⇒c=8/4=2\(\epsilon\)(1,3)
Hence Lagrange’s mean value theorem is verified.
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