# Chapter 8 Cayley Theorem and Puzzles

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```Chapter 8
Cayley Theorem and Puzzles
“As for everything else, so for a mathematical theory: beauty can
be perceived but not explained.”(Arthur Cayley)
We have seen that the symmetric group Sn of all the permutations of n
objects has order n!, and that the dihedral group D3 of symmetries of the
equilateral triangle is isomorphic to S3 , while the cyclic group C2 is isomorphic to S2 . We now wonder whether there are more connections between
finite groups and the group Sn . There is in fact a very powerful one, known
as Cayley Theorem:
Theorem 15. Every finite group is isomorphic to a group of permutations
(that is to some subgroup of Sn ).
This might be surprising but recall that given any finite group G =
{g1 , g2 , . . . , gn }, every row of its Cayley table
g1 = e g2
g3
···
gn
gr g1
gr g3
···
gr gn
g1
g2
..
.
gr
..
.
gr g2
gn
is simply a permutation of the elements of G (gr gs ∈ {g1 , g2 , . . . , gn }).
171
172
CHAPTER 8. CAYLEY THEOREM AND PUZZLES
Groups and Permutation Groups
• We saw that D3=S3 and C2=S2.
• Is there any link in general between a given group G and
groups of permutations?
• The answer is given by Cayley Theorem!
Cayley Theorem
Theorem Every finite group is isomorphic to a group of
permutations.
This means a
subgroup of some
symmetric group.
One known link: for a group G, we can consider its
multiplication (Cayley) table. Every row contains a permutation
of the elements of the group.
173
Proof. Let (G, ·) be a group. We shall exhibit a group of permutations (Σ, ◦)
that is isomorphic to G. We have seen that the Cayley table of (G, ·) has
rows that are permutations of {g1 , g2 , . . . , gn }, the elements of G. Therefore
let us define
Σ = {σg : G → G, σg (x) = gx, ∀x ∈ G}
for g ∈ G. In words we consider the permutation maps given by the rows of
the Cayley table. We verify that Σ is a group under map composition.
1. To prove that Σ is closed under composition, we will to prove that
σg2 ◦ σg1 = σg2 g1 , g1 ∈ G, g2 ∈ G.
Indeed, for every x ∈ G,
σg2 (σg1 (x)) = σg2 (g1 x) = g2 (g1 x) = (g2 g1 )x = σg2 g1 (x) ∈ Σ
since g2 g1 ∈ G.
2. Map composition is associative.
3. The identity element is σe (x) = ex, since
σg ◦ σe = σg·e = σg , σe ◦ σg = σe·g = σg .
4. The inverse. Consider g and g −1 , we have gg −1 = g −1 g = e. From
σg2 ◦ σg1 = σg2 g1
we have
σg ◦ σg−1 = σe = σg−1 ◦ σg .
P
Now we claim that (G, ·) and ( , ◦) are isomorphic, where the group
isomorphism is given by
φ : G → Σ, g 7→ σg .
Clearly if σg1 = σg2 then g1 e = g2 e ⇒ g1 = g2 . If g1 = g2 , then σg1 = σg2 .
Hence the map is one-to-one and onto, by construction!
Let us check that φ is a group homomorphism. If g1 , g2 ∈ G,
φ(g1 g2 ) = σg1 g2 = σg1 ◦ σg2 = φ(g1 ) ◦ φ(g2 ),
and hence we are done, φ is an isomorphism between (G, ·) and a permutation
group!
174
CHAPTER 8. CAYLEY THEOREM AND PUZZLES
Proof of Cayley Theorem (I)
• We need to find a group Σ of permutations isomorphic to G.
• Define Σ={ σg : G →G, σg(x)=gx , g in G}
• The set Σ forms a group of permutations:
These are the
permutations given by
the rows of the Cayley
table!
o It is a set of permutations (bijections).
o The identity is σ1 since it maps x to x.
o Associativity is that of map composition.
o Closure: we have that σg1 σg2 = σg1g2.
o Inverse: we have that σg σg(-1) = σ1.
Proof of Cayley Theorem (II)
• Left to prove: G and Σ are isomorphic.
• We define a group isomorphism φ: G →Σ , φ(g)=σg.
o The map φ is a bijection.
o The map φ is a group homomorphism: φ(g1g2)=φ(g1) φ(g2).
[Indeed: φ(g1g2)= σg1g2 = σg1 σg2 =φ(g1) φ(g2).]
175
Now that we saw that all finite groups are subgroups of Sn , we can understand better why we could describe the symmetries of bounded shapes
by the cyclic group Cn or the dihedral group Dn which can be mapped in a
natural way to permutations of the vertex locations in the plane.
Example 30. Consider the group of integers modulo 3, whose Cayley table
is
0
1
2
0
0
1
2
1
1
2
0
2
2
0
1
We have σ0 (x) = x + 0 corresponding to the permutation identity ().
Then σ1 (x) = x + 1 corresponding to the permutation (123), σ2 (x) = x + 2
corresponding to (132).
Since we have a group homomorphism, addition in G = {0̄, 1̄, 2̄} corresponds to composition in Σ = {σ0 , σ1 , σ2 }. For example
1̄ + 1̄ = 2̄ ⇐⇒ (123)(123) = (132).
We next illustrate how the techniques we learnt from group theory can
be used to solve puzzles. We start with the 15 puzzle. The goal is to obtain
a configuration where the 14 and 15 have been switched.
Since this puzzle involves 16 numbers, we can look at it in terms of permutations of 16 elements.
Let us assume that when the game starts, the empty space is in position
16. Every move consists of switching the empty space 16 and some other
piece. To switch 14 and 15, we need to obtain the permutation (14 15) as
a product of transpositions, each involving the empty space 16. Now the
permutation (14 15) has parity -1, while the product of transpositions will
always have parity 1, since 16 must go back to its original position, and thus
no matter which moves are done, the number of vertical moves are even, and
the number of horizontal moves are even as well.
176
CHAPTER 8. CAYLEY THEOREM AND PUZZLES
Example
Take G={0,1,2} the group of integers mod 3.
0
1
2
0
0
1
2
1
1
2
0
(123)
2
2
0
1
(132)
()
• You can check the consistency of the operations! (homomorphism)
• For example: 1+1 =2 ↔ (123)(123)=(132)
This is a subgroup of S3.
A Historical Point of View
Lagrange ~1770
Galois ~1830
Cauchy~1820-1840
Kronecker~1870
Klein ~1880
Lie ~1880
Number Theory
Geometry
Permutations
(Jordan ~1880)
Group Theory
[The symmetric group is complicated! Needs more tools.]
Cayley ~1854
(modern definition
of group)
177
Some Applications
• Symmetries
• Cryptography
• Puzzles
Symmetries
One of the main focuses of this class
• Symmetries of finite planar shapes (cyclic and dihedral groups)
• Symmetries of some infinite planar shapes (Frieze groups, later!)
One could also study symmetries of 3-dimensional shapes!
A tetrahedral AB4 molecule (ex. methane CH4)
with symmetric group A4.
178
CHAPTER 8. CAYLEY THEOREM AND PUZZLES
15 Puzzle
15 14
• 1870, New England
• 1890, price of 1000\$ to who could solve it.
Impossibility of the 15 Puzzle (I)
Every move involves switching the empty space (say 16) and
some other piece.
11
12
11
15
15
(12 16)
11
12
15
(11 16)
15
12
(15 16)
Solving the puzzle means we can write:
(14 15)= (an 16)(an-1 16) …(a2 16)(a1 16)
11
15
12
12
(12 16)
11
179
We next consider a solitaire puzzle. The goal of the game is to finish
with a single stone in the middle of the board. This does not seem very
easy! We might ask whether it would be easier to finish the game by having
a single stone anywhere instead. To answer this question, we consider the
Klein group, and label every position of the board with an element of the
Klein group, such that two adjacent cells multiplied together give as result the
label of the third cell (this is done by horizontally and vertically). The value
of the board is given by multiplying all the group elements corresponding to
board positions where a stone is. The key observation is that the value does
not change when a move is made.
When the game starts, and only one stone is missing in the middle, the
total value of the board is h (with the labeling shown on the slides). Since
a move does not change the total value, we can only be left with a position
containing an h. Since the board is unchanged under horizontal and vertical
reflections, as well as under rotations by 90, 180, and 270 degrees, this further
restricts the possible positions containing a valid h, and in fact, the easiest
version is as hard as the original game!
Other applications of group theory can be found in the area of cryptography. We already saw Caesar cipher, and affine ciphers. We will see some
more: (1) check digits and (2) the RSA cryptosystem.
180
CHAPTER 8. CAYLEY THEOREM AND PUZZLES
Impossibility of the 15 Puzzle (II)
Solving the puzzle means we can write:
(14 15)= (an 16)(an-1 16) …(a2 16)(a1 16)
parity = -1
parity = 1
place, thus both
number of horizontal
and vertical moves
are even!
Solitaire (I)
181
Solitaire (II)
• A move = pick up a marble, jump it horizontally or vertically
(but not diagonally) over a single marble into a vacant hole,
removing the marble that was jumped over.
• A win = finish with a single marble left in the central hole.
• Would it be easier if a win = finish with a single marble anywhere?
Solitaire (III)
• G={1,f,g,h} = Klein group
• Label the board such that
labels of two cells multiplied
together give the label of the
third cell.
Binary
operation of the
Klein group
182
CHAPTER 8. CAYLEY THEOREM AND PUZZLES
Solitaire (IV)
• total value of the board = the group element obtained by
multiplying together the labels of all of the holes that have
marbles in them.
• the total value does not change when we make a move!
Solitaire (V)
Total value
=h
Total value= ?
• (fgh)15= fgh=e
• without h, we have fg=h.
Since a move does not
change the total value, we
can only be left with h!
183
Solitaire (VI)
A Solitaire board is unchanged under reflection in the
horizontal and vertical axes, and rotation through 90°,
180°, 270° and 360°.
if this h is
valid, then f
is valid!
Solitaire (VII)
If we can solve this
position, then we can
solve the middle one!
We just shown: the “easiest
version” is as hard!!
184
CHAPTER 8. CAYLEY THEOREM AND PUZZLES
Cryptography: Modular Arithmetic
Modular arithmetic (integers modulo n) enables
• Caesar’s cipher
eK: x → eK(x)=x+K mod 26, K=3
• Affine ciphers
eK: x → eK(x)=K1x+K2 mod 26, (K1,26)=1, K=(K1,K2)
• RSA cryptosystem
eK: x → eK(x)=xe mod n, K=(n,e)
Cryptography: Discrete Log Problem
• “Regular” logarithm: loga(b) is defined as the solution x of the
equation ax = b.
• Example: : log2(8)=3 since 23=8.
• Discrete logarithm: let G be a finite cyclic group, take g and h
in G, logg(h) in G is defined as a solution x of the equation gx = h.
• Example: log3(13)=x in the group of invertible integers
modulo 17 means that 3x ≡ 13 (mod 17) , and x=4 is a solution.
Need to check this is a cyclic group!
This is useful in cryptography because solving the discrete log
problem is hard!
185
Cryptography: Check Digit (I)
Take a message formed by a string of digits.
A check digit consists of a single digit, computed from the other
digits, appended at the end of the message.
It is a form of redundancy to enable error detection.
We will look at the Check Digit introduced by J. Verhoeff in
1969, based on the dihedral group D5.
Cryptography: Check Digit (II)
Multiplication table of D5 with 0=do-nothing, 1-4=rotations, 59=reflections, *=binary operation in D5.
*
0
1
2
3
4
5
6
7
8
9
0
0
1
2
3
4
5
6
7
8
9
1
1
2
3
4
0
6
7
8
9
5
2
2
3
4
0
1
7
8
9
5
6
3
3
4
0
1
2
8
9
5
6
7
4
4
0
1
2
3
9
5
6
7
8
5
5
9
8
7
6
0
4
3
2
1
6
6
5
9
8
7
1
0
4
3
2
7
7
6
5
9
8
2
1
0
4
3
8
8
7
6
5
9
3
2
1
0
4
9
9
8
7
6
5
4
3
2
1
0
186
CHAPTER 8. CAYLEY THEOREM AND PUZZLES
Cryptography: Check Digit (III)
How does it work?Let σ be a permutation in S10. To any string
a1a2… an-1 of digits, we append the check digit an so that
σ(a1)* σ2(a2)* …*σn-1(an-1)* σn(an)=0.
Binary operation
Composition of
the permutation σ of D5
Single-digit errors are detected: if the digit a is replaced by
b, then σi(a) is replaced by σi(b) (σi(a)≠ σi(b) when a≠b) thus
the check digit is changed and an error is detected.
Cryptography: Check Digit (IV)
Example. Take σ=(1,7,9)(2,5,10,4,6) and the digit 12345 (n-1=5).
[23456]
2
3
4
5
• σ(2)=5, σ (3)=3, σ (4)=5, σ (5)=2, σ (6)=6.
• 5*3*5*2*6* σ6(a6)=0
5* σ6(a6)=0
σ6(a6)= 5 and a6=2.
• We get [234562] that is 123451.
Check digit 8
on a German
banknote.
187
Application of Euler Theorem: RSA
RSA is an encryption scheme discovered by River, Shamir and
Alice and Bob Story
Alice and Bob want to exchange confidential data in the
presence of an eavesdropper Eve.
188
CHAPTER 8. CAYLEY THEOREM AND PUZZLES
Alice and Bob story by xkcd
RSA Protocol (I)
• Select two distinct large primes p and q (“large” means 100
digits ).
This function counts the
integers coprime to n.
• Compute n=pq.
• The Euler totient function of n is
= (p-1)(q-1).
• Pick an odd integer e such that e is coprime to
• Find d such that ed = 1 modulo
.
e exists because it is
coprime to the Euler
totient function!
Publish e and n as public keys, keep d private.
.
189
RSA Protocol (II)
• Alice: public key = (n,e), d is private.
• Bob sends m to Alice via the following encryption: c =me mod n.
• Alice decrypts: m = cd mod n.
Why can Alice decrypt?
Step 1 cd mod n = (me)d mod n.
Step 2 We have ed =1 +k
.
Step 3 Now (me)d mod n = m 1+k
coprime to n.
= m mod n when m is
190
CHAPTER 8. CAYLEY THEOREM AND PUZZLES
Exercises for Chapter 8
Exercise 40.
• Let G be the Klein group. Cayley’s Theorem says that
it is isomorphic to a subgroup of S4 . Identify this subgroup.
• Let G be the cyclic group C4 . Cayley’s Theorem says that it is isomorphic to a subgroup of S4 . Identify this subgroup.
Exercise 41. Show that any rearrangement of pieces in the 15-puzzle starting from the standard configuration (pieces are ordered from 1 to 15, with
the 16th position empty) which brings the empty space back to its original
position must be an even permutation of the other 15 pieces.
Exercise 42. Has this following puzzle a solution? The rule of the game is
the same as the solitaire seen in class, and a win is a single marble in the
middle of the board. If a win is a single marble anywhere in the board, is
that any easier?
```