CryptoBook
Search…
Groups

# Introduction

Modern cryptography is based on the assumption that some problems are hard (unfeasable to solve). Since the we do not have infinite computational power and storage we usually work with finite messages, keys and ciphertexts and we say they lay in some finite sets
$\mathcal{M}, \mathcal{K}$
and
$\mathcal{C}$
.
Furthermore, to get a ciphertext we usually perform some operations with the message and the key.
For example in AES128
$\mathcal{K} = \mathcal{M} = \mathcal{C} = \{0, 1\}^{128}$
since the input, output and key spaces are 128 bits. We also have the encryption and decryption operations:
$Enc: \mathcal{K} \times \mathcal{M} \to \mathcal{C} \\ Dec: \mathcal{K} \times \mathcal{C} \to \mathcal{M}$
The study of sets, and different types of operations on them is the target of abstract algebra. In this chapter we will learn the underlying building blocks of cryptosystems and some of the hard problems that the cryptosystems are based on.

# Definition

A set
$G$
paired with a binary operation
$\cdot:G\times G\to G$
is a group if the following requirements hold:
Closure: For all
$a, b \in G: \$
$a\cdot b \in G$
- Applying the operation keeps the element in the set
Associativity: For all
$a, b, c \in G:$
$(a \cdot b) \cdot c=a\cdot (b\cdot c)$
Identity: There exists an element
$e\in G$
such that
$a\cdot e=e\cdot a=a$
for all
$a\in G$
Inverse: For all elements
$a\in G$
, there exists some
$b\in G$
such that
$b\cdot a=a\cdot b=e$
. We usually denote
$b$
as
$a^{-1}$
For
$n\in\mathbb Z$
,
$a^n$
means
$\underbrace{a\cdot a\dots{}\cdot a}_{n\text{ times}}$
when
$n>0$
and
$\left(a^{-n}\right)^{-1}$
when
$n<0$
. For
$n=0$
,
$a^n=e$
.
If
$ab=ba$
, then
$\cdot$
is commutative and the group is called abelian. We often denote the group operation by
$+$
$\cdot$
and we typically use
$na$
$a^n$
.
Remark
The identity element of a group
$G$
is also denoted with
$1_G$
or just
$1$
if only one groups is present
Examples of groups
Integers modulo
$n$
$= (\mathbb{Z} / n \mathbb{Z}, +)$
.
$\mathbb{Z} / n \mathbb{Z} = \{0, 1, ..., n -1\}$
Let's look if the group axioms are satisfied
1.
$\checkmark$
$\forall a, b \in \mathbb{Z}/ n\mathbb{Z} \text{ let } c \equiv a + b \bmod n$
. Because of the modulo reduction
$c < n \Rightarrow c \in \mathbb{Z}/ n\mathbb{Z}$
2.
$\checkmark$
3.
$\checkmark$
$0 + a \equiv a + 0 \equiv a \bmod n \Rightarrow 0$
is the identity element
4.
$\checkmark$
$\forall a \in \mathbb{Z}/ n\mathbb{Z}$
we take
$n - a \bmod n$
to be the inverse of
$a$
. We check that
$a + n - a \equiv n \equiv 0 \bmod n$
$n - a + a \equiv n \equiv 0 \bmod n$
Therefore we can conclude that the integers mod
$n$
with the modular addition form a group.
1
Z5 = Zmod(5) # Technically it's a ring but we'll use the addition here
2
print(Z5.list())
3
# [0, 1, 2, 3, 4]
4
5
6
# + 0 1 2 3 4
7
# +----------
8
# 0| 0 1 2 3 4
9
# 1| 1 2 3 4 0
10
# 2| 2 3 4 0 1
11
# 3| 3 4 0 1 2
12
# 4| 4 0 1 2 3
13
14
a, b = Z5(14), Z5(3)
15
print(a, b)
16
# 4 3
17
print(a + b)
18
# 2
19
print(a + 0)
20
# 4
21
print(a + (5 - a))
22
# 0
23
Copied!
Example of non-groups
$(\mathbb{Q}, \cdot)$
is not a group because we can find the element
$0$
that doesn't have an inverse for the identity
$1$
.
$(\mathbb{Z}, \cdot)$
is not a group because we can find elements that don't have an inverse for the identity
$1$
Exercise
Is
$(\mathbb{Z} / n \mathbb{Z} \smallsetminus \{0\}, \cdot)$
a group? If yes why? If not, are there values for
$n$
that make it a group?
sɹosᴉʌᴉp uoɯɯoɔ puɐ sǝɯᴉɹd ʇnoqɐ ʞuᴉɥ┴ :ʇuᴉH

## Proprieties

1.
The identity of a group is unique
2.
The inverse of every element is unique
3.
$\forall$
$a \in G \ : \left(a^{-1} \right) ^{-1} = g$
. The inverse of the inverse of the element is the element itself
4.
$\forall a, b \in G:$
$(ab)^{-1} = b^{-1}a^{-1}$
Proof:
$(ab)(b^{−1}a^{−1}) =a(bb^{−1})a^{−1}=aa^{−1}= e.$
1
n = 11
2
Zn = Zmod(n)
3
a, b = Zn(5), Zn(7)
4
print(n - (a + b))
5
# 10
6
print((n - a) + (n - b))
7
# 10
Copied!

# Orders

In abstract algebra we have two notions of order: Group order and element order
Group order
The order of a group
$G$
is the number of the elements in that group. Notation:
$|G|$
Element order
The order of an element
$a \in G$
is the smallest integer
$n$
such that
$a^n = 1_G$
. If such a number
$n$
doesn't exist we say the element has order
$\infty$
. Notation:
$|a|$
1
Z12 = Zmod(12) # Residues modulo 12
2
3
# 12
4
a, b= Z12(6), Z12(3)
5
print(a.order(), b.order())
6
# 2 4
7
print(a.order() * a)
8
# 0
9
10
print(ZZ.order()) # The integers under addition is a group of infinite order
11
# +Infinity
Copied!
We said our messages lay in some group
$\mathcal{M}$
. The order of this group
$|\mathcal{M}|$
is the number of possible messages that we can have. For
$\mathcal{M} = \{0,1\}^{128}$
we have
$|\mathcal{M}| = 2^{128}$
possible messages.
Let
$m \in \mathcal{M}$
be some message. The order of
$m$
means how many different messages we can generate by applying the group operation on
$m$

# Subgroups

Definition
Let
$(G, \cdot)$
be a group. We say
$H$
is a subgroup of
$G$
if
$H$
is a subset of
$G$
and
$(H, \cdot)$
forms a group. Notation:
$H \leq G$
Proprieties
1.
The identity of
$G$
is also in
$H:$
$1_H = 1_G$
2.
The inverses of the elements in
$H$
are found in
$H$
How to check
$H \leq G$
? Let's look at a 2 step test
1.
Closed under operation:
$\forall a, b \in H \to ab \in H$
2.
Closed under inverses:
$\forall a \in H \to a^{-1} \in H$

## Generators

Let
$G$
be a group,
$g \in G$
an element and
$|g| = n$
. Consider the following set:
$\{1, g, g^2, ..., g^{n-1}\} \overset{\text{denoted}}{=} \langle g\rangle.$
This set paired the group operation form a subgroup of
$G$
generated by an element
$g$
.
Why do we care about subgroups? We praise the fact that some problems are hard because the numbers we use are huge and exhaustive space searches are too hard in practice.
Suppose we have a big secret values space
$G$
and we use an element
$g$
to generate them.
If an element
$g \in G$
with a small order
$n$
is used then it can generate only
$n$
possible values and if
$n$
is small enough an attacker can do a brute force attack.
Example
For now, trust us that if given a prime
$p$
, a value
$g \in \mathbb{Z} / p \mathbb{Z}$
and we compute
$y = g^x \bmod p$
for a secret
$x$
, finding
$x$
is a hard problem. We will tell you why a bit later.
1
p = 101 # prime
2
Zp = Zmod(p)
3
H_list = Zp.multiplicative_subgroups() # Sage can get the subgroup generators for us
4
print(H_list)
5
# ((2,), (4,), (16,), (32,), (14,), (95,), (10,), (100,), ())
6
7
g1 = H_list[3][0] # Weak generator
8
print(g1, g1.multiplicative_order())
9
# 32 20
10
11
g2 = Zp(3) # Strong generator
12
print(g2, g2.multiplicative_order())
13
# 3 100
14
15
16
## Consider the following functions
17
def brute_force(g, p, secret_value):
18
"""
19
Brute forces a secret value, returns number of attempts
20
"""
21
for i in range(p-1):
22
t = pow(g, i, p)
23
if t == secret_value:
24
break
25
return i
26
27
def mean_attempts(g, p, num_keys):
28
"""
29
Tries num_keys times to brute force and
30
returns the mean of the number of attempts
31
"""
32
total_attempts = 0
33
for _ in range(num_keys):
34
k = random.randint(1, p-1)
35
sv = pow(g, k, p) # sv = secret value
36
total_attempts += brute_force(g, p, sv)
37
return 1. * total_attempts / num_keys
38
39
## Let's try with our generators
40
print(mean_attempts(g1, p, 100)) # Weak generator
41
# 9.850
42
print(mean_attempts(g2, p, 100)) # Strong generator
43
# 49.200
44
Copied!

# Examples

// subgroups, quotient groups
// cyclic groups