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 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:

e h2 h3 … hk

a2 a2 h2 a2 h3 … a2 hk

My proof:• Choose any a2 in G not in row 1.

• Write a2H in the second row.

e h2 h3 … hk

a3 a3 h2 a3 h3 … a3 hk

e h2 h3 … hk

a2 a2 h2 a2 h3 … a2 hk

My proof:• Continue in a similar manner…

• Since G is finite, this process will end

: : : :

ar ar h2 ar h3 ar hk

My proof:• Rows are disjoint by (4)

• Each row has k elements by (6)

e h2 h3 … hk

a2 a2 h2 a2 h3 … a2 hk

a3 a3 h2 a3 h3 … a3 hk

: : : :

ar ar h2 ar h3 ar hk

My proof:• Let r be the number of distinct cosets.

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

e h2 h3 … hk

a2 a2 h2 a2 h3 … a2 hk

a3 a3 h2 a3 h3 … a3 hk

: : : :

ar ar h2 ar h3 ar hk

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

€

102a ⋅102a =102a +2a

=102⋅2a

=102a+1

2257-1• In Python:

p = 2**257-1

t = 10

for n in range(257):

t = (t*t)%p

print t

2257-1• 23323117726701610548024580880832

227821258735681932676554551014701139464992104

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