Lagrange's Theorem

The most important single theorem in group theory. It helps answer: –How large is the symmetry group of a volleyball? A soccer ball? –How many groups of order 2p where p is prime? (4, 6, 10, 14, 22, 26, …) –Is prime? –Is computer security possible? –etc.

Recall: Let H be a subgroup of G, and a,b in G. 3. aH = bH iff a belongs to bH 4. aH and bH are either equal or disjoint 6. |aH| = |bH|

Lagrange's Theorem If G is finite group and H is a subgroup of G, then a)|H| divides |G|. b)The number of distinct left (right) cosets of H in G is |G|/|H|

My proof: Let H ≤ G with |G| = n, and |H| = k. Write the elements of H in row 1: eh2h2 h3h3 …hkhk

a2a2 a 2 h 2 a 2 h 3 …a 2 h k My proof: Choose any a 2 in G not in row 1. Write a 2 H in the second row. eh2h2 h3h3 …hkhk

a3a3 a 3 h 2 a 3 h 3 …a 3 h k eh2h2 h3h3 …hkhk a2a2 a 2 h 2 a 2 h 3 …a 2 h k My proof: Continue in a similar manner… Since G is finite, this process will end :::: arar a r h 2 a r h 3 a r h k

My proof: Rows are disjoint by (4) Each row has k elements by (6) eh2h2 h3h3 …hkhk a2a2 a 2 h 2 a 2 h 3 …a 2 h k a3a3 a 3 h 2 a 3 h 3 …a 3 h k :::: arar a r h 2 a r h 3 a r h k

My proof: Let r be the number of distinct cosets. Clearly |G| = |H|r, and r = |G|/|H|. eh2h2 h3h3 …hkhk a2a2 a 2 h 2 a 2 h 3 …a 2 h k a3a3 a 3 h 2 a 3 h 3 …a 3 h k :::: arar a r h 2 a r h 3 a r h k

What does Lagrange's Theorem say? Let H ≤ G where |G| = 12. Then |H| could only be… 1, 2, 3, 4, 6, 12: The divisors of 12. G =Z 12 is cyclic, so there is exactly one subgroup of each of these orders. G = A 4 is not cyclic, and there is no subgroup of order 6. The converse of Lagrange's theorem is False!

Definition Let H be a subgroup of G. The number of left (right) cosets of H in G is called the index in G of H and is denoted |G:H|.

|G:H| = |G|/|H| Corollary 1: If G is a finite group and H is a subgroup of G, then |G:H| = |G|/|H|. Proof: This is a restatement of Lagrange's theorem using the definition of the index in G of H.

|a| divides |G| Corollary 2. In a finite group G, the order of each element of the group divides the order of the group. Proof: Let a be any element of G. Then |a| = | |. By Lagrange's Theorem, | | divides |G|.

Groups of prime order Corollary 3. A group of prime order is cyclic. Proof: Let |G| be prime. Choose any a≠e in G. Then | | > 1. Since | | divides |G|, | | = |G| It follows that G = So G is cyclic.

a |G| = e Corollary 4. Let G be a finite group, and let a belong to G. Then a |G| = e. Proof: By corollary 2, |a| divides |G|, so |G| = |a|k for some positive integer k. Hence a |G| = a |a|k = e k = e.

Fermat's little theorem For every integer a and every prime p, a p mod p = a mod p. Proof: To simplify notation, Let a mod p = r. Then a p mod p = (a mod p) p mod p = r p mod p. It remains to show that r p mod p = r for 0 ≤ r < p.

Fermat's little theorem (con't) In case r = 0, 0 p mod p = 0. If r > 0, then r in U(p) = {1, 2, …, p-1}. By corollary 4, r |U(p)| = r p-1 = 1 in U(p). In other words, r p-1 mod p = 1. So, r p mod p = r.

Example: Find mod mod 11 = 50 mod 11 = 6 Check it: = 4,882,812,500,000,000,000 = 11443,892,045,454,454,454+6 So mod 11 = 6

Example: not prime. Suppose, towards a contradiction, that p = is prime. Using Python, we get p = It is easy to calculate p, but factoring is hard!

However 10 p mod p = 10 So 10 p+1 mod p should be 100. To calculate 10 p+1, note that

In Python: p = 2**257-1 t = 10 for n in range(257): t = (t*t)%p print t

Since this number is not 100, p is not prime.