Transcript Cosets and Lagrange's Theorem

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
My proof:
• Choose any a2 in G not in row 1.
• Write a2H in the second row.
e
a2
h2
a2 h2
h3
a2 h3
…
…
hk
a2 h k
My proof:
• Continue in a similar manner…
• Since G is finite, this process will end
e
a2
h2
a2 h2
h3
a2 h3
…
…
hk
a2 h k
a3
:
ar
a3 h2
:
a r h2
a3 h3
:
ar h3
…
a3 h k
:
ar hk
My proof:
• Rows are disjoint by (4)
• Each row has k elements by (6)
e
a2
a3
:
ar
h2
a2 h2
a3 h2
:
a r h2
h3
a2 h3
a3 h3
:
ar h3
…
…
…
hk
a2 h k
a3 h k
:
ar hk
My proof:
• Let r be the number of distinct cosets.
• Clearly |G| = |H|•r, and r = |G|/|H|.
e
a2
a3
:
ar
h2
a2 h2
a3 h2
:
a r h2
h3
a2 h3
a3 h3
:
ar h3
…
…
…
hk
a2 h k
a3 h k
:
ar hk
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 =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=
231584178474632390847141970017375815
706539969331281128078915168015826259
279871
• 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
2a
2a
2 a 2 a
10 10  10
2 2 a
 10
2 a 1
 10
2257-1
• In Python:
p = 2**257-1
t = 10
for n in range(257):
t = (t*t)%p
print t
2257-1
• 23323117726701610548024580880832
22782125873568193267655455101470
1139464992104
• Since this number is not 100, p is not
prime.