- 191 Views
- Uploaded on
- Presentation posted in: General

Lagrange's Theorem

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

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.

- 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|

- 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|

- Let H ≤ G with |G| = n, and |H| = k.
- Write the elements of H in row 1:

- Choose any a2 in G not in row 1.
- Write a2H in the second row.

- Continue in a similar manner…
- Since G is finite, this process will end

- Rows are disjoint by (4)
- Each row has k elements by (6)

- Let r be the number of distinct cosets.
- Clearly |G| = |H|•r, and r = |G|/|H|.

- 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!

- 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|.

- 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.

- 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|.

- 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.

- 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.

- 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.

- 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.

- 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

- 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!

- However 10p mod p = 10
So 10p+1 mod p should be 100.

- To calculate 10p+1, note that

- In Python:
p = 2**257-1

t = 10

for n in range(257):

t = (t*t)%p

print t

- 23323117726701610548024580880832227821258735681932676554551014701139464992104
- Since this number is not 100, p is not prime.