Groups

Authors: Ariana, Zademn
Reviewed by:
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 M,K\mathcal{M}, \mathcal{K}M,K
 and C\mathcal{C}C
.
Furthermore, to get a ciphertext we usually perform some operations with the message and the key.
For example in AES128 K=M=C={0,1}128\mathcal{K} = \mathcal{M} = \mathcal{C} = \{0, 1\}^{128}K=M=C={0,1}128
since the input, output and key spaces are 128 bits. We also have the encryption and decryption operations:
Enc:K×M→CDec:K×C→MEnc: \mathcal{K} \times \mathcal{M} \to \mathcal{C} \\  Dec: \mathcal{K} \times \mathcal{C} \to \mathcal{M}Enc:K×M→CDec:K×C→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 setGGG
paired with a binary operation ⋅:G×G→G\cdot:G\times G\to G⋅:G×G→G
is a group if the following requirements hold:
Closure: For all a,b∈G: a, b \in G: \ a,b∈G: 
 a⋅b∈Ga\cdot b \in Ga⋅b∈G
 - Applying the operation keeps the element in the set
Associativity: For all a,b,c∈G:a, b, c \in G: a,b,c∈G:
 (a⋅b)⋅c=a⋅(b⋅c)(a \cdot b) \cdot c=a\cdot (b\cdot c)(a⋅b)⋅c=a⋅(b⋅c)
​
Identity: There exists an elemente∈Ge\in Ge∈G
such that a⋅e=e⋅a=aa\cdot e=e\cdot a=aa⋅e=e⋅a=a
for all a∈Ga\in Ga∈G
​
Inverse: For all elements a∈Ga\in Ga∈G
, there exists some b∈Gb\in Gb∈G
such that b⋅a=a⋅b=eb\cdot a=a\cdot b=eb⋅a=a⋅b=e
. We usually denotebbb
as a−1a^{-1}a−1
​
For n∈Zn\in\mathbb Zn∈Z
, ana^nan
means a⋅a…⋅a⏟n times\underbrace{a\cdot a\dots{}\cdot a}_{n\text{ times}}n timesa⋅a…⋅a​​
when n>0n>0n>0
and (a−n)−1\left(a^{-n}\right)^{-1}(a−n)−1
when n<0n<0n<0
. For n=0n=0n=0
, an=ea^n=ean=e
.
If ab=baab=baab=ba
, then ⋅\cdot⋅
is commutative and the group is called abelian. We often denote the group operation by +++
instead of ⋅\cdot⋅
and we typically use nanana
instead of ana^nan
.
Remark
The identity element of a group GGG
 is also denoted with 1G1_G1G​
 or just 111
 if only one groups is present
Examples of groups
Integers modulo nnn
 (remainders) under modular addition  =(Z/nZ,+)= (\mathbb{Z} / n \mathbb{Z},  +)=(Z/nZ,+)
. 
Z/nZ={0,1,...,n−1}\mathbb{Z} / n \mathbb{Z} = \{0, 1, ..., n -1\}Z/nZ={0,1,...,n−1}

Let's look if the group axioms are satisfied
1.
​✓\checkmark✓
 ∀a,b∈Z/nZ let c≡a+b mod n\forall a, b \in \mathbb{Z}/ n\mathbb{Z} \text{ let }  c \equiv a + b \bmod n∀a,b∈Z/nZ let c≡a+bmodn
. 
Because of the modulo reduction c<n⇒c∈Z/nZc < n \Rightarrow c \in \mathbb{Z}/ n\mathbb{Z} c<n⇒c∈Z/nZ
 
2.
​✓\checkmark✓
Modular addition is associative
3.
​✓\checkmark ✓
​0+a≡a+0≡a mod n⇒00 + a \equiv a + 0 \equiv a \bmod n \Rightarrow 00+a≡a+0≡amodn⇒0
 is the identity element
4.
​✓\checkmark✓
​∀a∈Z/nZ\forall a \in \mathbb{Z}/ n\mathbb{Z} ∀a∈Z/nZ
we take n−a mod nn - a \bmod nn−amodn
to be the inverse of aaa
. We check that
 a+n−a≡n≡0 mod na + n - a \equiv n \equiv 0 \bmod na+n−a≡n≡0modn
​
​n−a+a≡n≡0 mod nn - a + a \equiv n \equiv 0 \bmod nn−a+a≡n≡0modn
​
Therefore we can conclude that the integers mod nnn
 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
print(Z5.addition_table(names = 'elements'))
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
​(Q,⋅)(\mathbb{Q}, \cdot)(Q,⋅)
 is not a group because we can find the element 000
 that doesn't have an inverse for the identity 111
.
(Z,⋅)(\mathbb{Z}, \cdot)(Z,⋅)
is not a group because we can find elements that don't have an inverse for the identity 111
​
Exercise
Is (Z/nZ∖{0},⋅)(\mathbb{Z} / n \mathbb{Z} \smallsetminus \{0\}, \cdot)(Z/nZ∖{0},⋅)
a group? If yes why? If not, are there values for nnn
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∈G :(a−1)−1=ga \in G \ :  \left(a^{-1} \right) ^{-1} = ga∈G :(a−1)−1=g
. The inverse of the inverse of the element is the element itself
4.
​∀a,b∈G:\forall a, b \in G: ∀a,b∈G:
 (ab)−1=b−1a−1(ab)^{-1} = b^{-1}a^{-1}(ab)−1=b−1a−1
​
Proof: (ab)(b−1a−1)=a(bb−1)a−1=aa−1=e.(ab)(b^{−1}a^{−1}) =a(bb^{−1})a^{−1}=aa^{−1}= e.(ab)(b−1a−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 GGG
is the number of the elements in that group. Notation: ∣G∣|G|∣G∣
​
Element order
The order of an element a∈Ga \in Ga∈G
 is the smallest integer nnn
 such that an=1Ga^n = 1_Gan=1G​
. If such a number nnn
 doesn't exist we say the element has order ∞\infty∞
. Notation: ∣a∣|a|∣a∣
​
1
Z12 = Zmod(12) # Residues modulo 12
2
print(Z12.order()) # The additive order 
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 M\mathcal{M}M
. The order of this group ∣M∣|\mathcal{M}|∣M∣
 is the number of possible messages that we can have. For M={0,1}128\mathcal{M} = \{0,1\}^{128}M={0,1}128
we have ∣M∣=2128|\mathcal{M}| = 2^{128}∣M∣=2128
 possible messages.
Let m∈Mm \in \mathcal{M}m∈M
be some message. The order of mmm
 means how many different messages we can generate by applying the group operation on mmm
​
Subgroups
Definition
Let (G,⋅)(G, \cdot)(G,⋅)
 be a group. We say HHH
is a subgroup of GGG
 if HHH
 is a subset of GGG
 and (H,⋅)(H, \cdot)(H,⋅)
forms a group.
Notation: H≤GH \leq GH≤G
​
Proprieties
1.
The identity of GGG
 is also in H:H: H:
​1H=1G1_H = 1_G1H​=1G​
​
2.
The inverses of the elements in HHH
are found in HHH
​
How to check H≤GH \leq GH≤G
? Let's look at a 2 step test
1.
Closed under operation: ∀a,b∈H→ab∈H\forall a, b \in H \to ab \in H∀a,b∈H→ab∈H
​
2.
Closed under inverses: ∀a∈H→a−1∈H\forall a \in H \to a^{-1} \in H∀a∈H→a−1∈H
​
Generators
Let GGG
be a group,g∈Gg \in Gg∈G
an element and ∣g∣=n|g| = n∣g∣=n
. Consider the following set:
{1,g,g2,...,gn−1}=denoted⟨g⟩.\{1, g, g^2, ..., g^{n-1}\} \overset{\text{denoted}}{=} \langle g\rangle.{1,g,g2,...,gn−1}=denoted⟨g⟩.
 This set paired the group operation form a subgroup of GGG
generated by an element ggg
. 
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 GGG
and we use an element ggg
to generate them.
If an elementg∈Gg \in Gg∈G
with a small order nnn
 is used then it can generate only nnn
 possible values and if nnn
 is small enough an attacker can do a brute force attack.
Example
For now, trust us that if given a prime ppp
, a value g∈Z/pZg \in \mathbb{Z} / p \mathbb{Z}g∈Z/pZ
 and we compute y=gx mod py = g^x \bmod py=gxmodp
 for a secret xxx
, finding xxx
 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
Sieves
Another take on groups
Last modified 5mo ago
Export as PDF
Copy link
Contents