# Lagrange's Theorem - PowerPoint PPT Presentation

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Lagrange's Theorem. 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 2 257 -1 prime?

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Lagrange's Theorem

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## Lagrange's Theorem

### 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 2257-1 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

• |H| divides |G|.

• 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:

### My proof:

• Choose any a2 in G not in row 1.

• Write a2H in the second row.

### My proof:

• Continue in a similar manner…

• Since G is finite, this process will end

### My proof:

• Rows are disjoint by (4)

• Each row has k elements by (6)

### My proof:

• Let r be the number of distinct cosets.

• Clearly |G| = |H|•r, and r = |G|/|H|.

### What doesLagrange'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 =Z12 is cyclic, so there is exactly one subgroup of each of these orders.

• G = A4 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| = |<a>|. By Lagrange's Theorem, |<a>| 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 |<a>| > 1.

Since |<a>| divides |G|, |<a>| = |G|

It follows that G = <a>

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 = ek = e.

### Fermat's little theorem

• For every integer a and every prime p,

ap mod p = a mod p.

Proof: To simplify notation, Let a mod p = r.

Then ap mod p = (a mod p)p mod p = rp mod p.

It remains to show that

rp mod p = r

for 0 ≤ r < p.

### Fermat's little theorem (con't)

• In case r = 0, 0p mod p = 0.

• If r > 0, then r in U(p) = {1, 2, …, p-1}.

By corollary 4, r|U(p)| = rp-1 = 1 in U(p).

In other words, rp-1 mod p = 1.

So, rp mod p = r.

### Example: Find 5011 mod 11

• 5011 mod 11

= 50 mod 11 = 6

• Check it:

5011 = 4,882,812,500,000,000,000

= 11•443,892,045,454,454,454+6

So 5011 mod 11 = 6

### Example: 2257-1 not prime.

• Suppose, towards a contradiction, that

p = 2257-1 is prime.

Using Python, we get

p = 231584178474632390847141970017375815706539969331281128078915168015826259279871

• It is easy to calculate p, but factoring is hard!

### 2257-1

• However 10p mod p = 10

So 10p+1 mod p should be 100.

• To calculate 10p+1, note that

### 2257-1

• In Python:

p = 2**257-1

t = 10

for n in range(257):

t = (t*t)%p

print t

### 2257-1

• 23323117726701610548024580880832227821258735681932676554551014701139464992104

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