Multiplicative Subgroups and Primitive Elements

Introduction

This chapter continues our study of group theory by exploring subgroups and generators. The concept of a primitive element will be introduced at the end. We assume you are already familiar with the definition of a group. If you need a refresher, check out this article.

To build intuition, we begin with additive groups, which are straightforward and help clarify core concepts like subgroups and generators.

We then shift to multiplicative groups of integers modulo $n$. The integers themselves, under multiplication, do not form a group—only $1$ and $-1$ have multiplicative inverses in $\mathbb{Z}$, so the group axioms fail. To address this, we consider multiplication modulo $n$, focusing on the integers less than $n$ that are coprime to it. These elements do have multiplicative inverses modulo $n$, and together they form a well-defined group. This construction plays a central role in number theory and is foundational to many cryptographic systems.

Finally, we examine generators—elements that can produce an entire group or subgroup through repeated multiplication. Understanding generators reveals important subgroup structures, especially when $n$ is prime, and highlights their critical role in cryptographic applications.

1. Additive Groups

Additive groups use addition (often modulo some number) as the operation, with the identity element being $0$ and the inverse of an element $a$ being $-a$, making their structure relatively straightforward. Let’s dive into examples to see how they work.

1.1 Example: $(\mathbb{Z}_6, +)$

To warm up, consider $\mathbb{Z}_6 = \{0, 1, 2, 3, 4, 5\}$ under addition. Start with closure: $$3 + 4 = 7, \text{ or } 5 + 5 = 10$$ takes us outside the set. To fix this, we use addition modulo 6: $$3 + 4 = 7 \equiv 1 \pmod{6} \text{ and } 5 + 5 = 10 \equiv 4 \pmod{6}.$$ Now every result stays in $\{0, 1, 2, 3, 4, 5\}$. Here is the addition table:

Every result is in $\{0, 1, 2, 3, 4, 5\}$, so closure holds. Check the other properties:

• Associativity: Grouping does not change the result:
  • $(2 + 3) + 4 = 5 + 4 = 9 \equiv 3 \pmod{6}$,
  • \$2 + (3 + 4) = 2 + (7 \equiv 1) = 3$.
• Identity: $0$ works as identity element, since $3 + 0 = 3$ (see the table’s first row or column).
• Inverses: Each element has a pair summing to $0 \pmod{6}$:

Thus, $(\mathbb{Z}_6, +)$ is a group with order (the number of elements in a set) $|\mathbb{Z}_6| = 6$.

Exercise 1.1: Check whether $(\mathbb{Z}_{9}, +)$ is a group.

Hint: Try building the addition table like we did for $\mathbb{Z}_6$.

Doing this by hand can be a bit tedious, so here’s a Python script that will generate the full addition table for any $\mathbb{Z}_n$:

def print_addition_table(mod):
    header = ["+ mod " + str(mod)] + list(range(mod))
    print(" | ".join(str(h).rjust(4) for h in header))
    print("-" * (6 * (mod + 1)))
    for row in range(mod):
        line = [str(row).rjust(4)]
        for col in range(mod):
            value = row + col
            result = value % mod
            if value >= mod:
                line.append(f"{value} ≡ {result}".rjust(6))
            else:
                line.append(str(result).rjust(6))
        print(" | ".join(line))


# Try it with Z_9
print_addition_table(9)

Follow-up: After analyzing $\mathbb{Z}_9$, try generating the table for $\mathbb{Z}_{10}$. Can you determine whether $(\mathbb{Z}_{10}, +)$ is also a group?

Thus, $(\mathbb{Z}_n, +)$ is a group with order $n$. Finite groups like this are key in modular arithmetic and cryptography. Next, we turn our attention to how certain elements and subsets within a group can reveal deeper structure—through subgroups and generators.

2. Subgroups and Generators

2.1 Understanding Subgroups

When studying groups, we often encounter subsets that retain the group’s structure under the same operation. These special subsets, called subgroups, behave like miniature versions of the parent group. Not every subset qualifies, though—let’s explore this with examples to see what makes a subgroup.

Example 2.1.1: Subgroups in $(\mathbb{Z}, +)$

Consider $(\mathbb{Z}, +)$, the group of all integers under addition, and two familiar subsets:

• The set of even integers: $\{\ldots, -4, -2, 0, 2, 4, \ldots\}$
• The set of odd integers: $\{\ldots, -3, -1, 1, 3, 5, \ldots\}$

Let’s check the even integers:

• Closure: The sum of two even numbers is even (e.g., $-2 +4= 2$).
• Identity: $0$ is even and included.
• Inverses: The inverse of any even number is also even (e.g., the inverse of $2$ is $-2$).
• Associativity: Inherited from $\mathbb{Z}$.

The even integers satisfy all group properties—this is a valid subgroup.

Now check the odd integers:

• Closure: $1 + 3= 4$, which is even—not part of the set. So closure fails.
• Identity: $0$ is not odd, so the identity element is missing.
• Inverses: For $1$, $-1$ is odd—but since closure and identity already fail, it is not a subgroup.

The odd integers under addition do not satisfy the subgroup criteria—they’re just a subset, not a subgroup.

Example 2.1.2: Subgroups in $(\mathbb{Z}_8, +)$

Now take $(\mathbb{Z}_8, +) = \{0, 1, 2, 3, 4, 5, 6, 7\}$ and test two subsets:

• Evens modulo 8: $\{0, 2, 4, 6\}$
• First half: $\{0, 1, 2, 3\}$

For $\{0, 2, 4, 6\}$:

• Closure: $2 +4= 6$, $4 +6= 10 \equiv2\pmod{8}$ (both in set).
• Identity: $0$ is present.
• Inverses: $2 +6= 8 \equiv 0$, $4 +4= 8 \equiv 0$, $0 + 0 = 0$ (all pairs work).
• Associativity: Holds from $\mathbb{Z}_8$.

This is a subgroup!

For $\{0, 1, 2, 3\}$ :

• Closure: $2+3= 5$ (not in set).
• Identity: $0$ is present.
• Inverses: For $1$, no element in ${0, 1, 2, 3}$ gives $0$ (e.g., $1 +3=4\pmod{8}$).

This fails to be a group, so it is only a subset, not a subgroup.

Definition: A subset $H$ of a group $G$ is a subgroup if it is a group under the same operation, meaning it satisfies closure, contains the identity, has inverses for all its elements, and inherits associativity from $G$.

Exercise 2.1.1: Find all subgroups of $(\mathbb{Z}_{5}, +)$.

2.2 Generators in Additive Groups

Now that we have seen subgroups, let’s explore how single elements can generate them—or even the whole group. We’ll revisit $(\mathbb{Z}_6, +) = \{0, 1, 2, 3, 4, 5\}$, the additive group under modulo $6$ addition, and examine what happens when we repeatedly add an element to itself, like “walking” around the numbers modulo $6$.

Example 2.2.1: Generators in $(\mathbb{Z}_6, +)$

Try 1:

• $1$
• $1 + 1 = 2$
• $2 + 1 = 3$
• $3 + 1 = 4$
• $4 + 1 = 5$
• $5 + 1 = 6 \equiv 0 \pmod{6}$

This gives $\{0, 1, 2, 3, 4, 5\} = \mathbb{Z}_6$. Repeatedly adding $1$ cycles through the entire group.

Try 2:

• $2$
• $2 + 2 = 4$
• $4 + 2 = 6 \equiv 0 \pmod{6}$
• $0 + 2 = 2$

This gives $\{0, 2, 4\}$—only half the group.

Try 3:

• $3$
• $3 + 3 = 6 \equiv 0 \pmod{6}$
• $0 + 3 = 3$

This gives $\{0, 3\}$, even smaller!

Try 5:

• $5$
• $5 + 5 = 10 \equiv 4 \pmod{6}$
• $4 + 5 = 9 \equiv 3 \pmod{6}$
• $3 + 5 = 8 \equiv 2 \pmod{6}$
• $2 + 5 = 7 \equiv 1 \pmod{6}$
• $1 + 5 = 6 \equiv 0 \pmod{6}$

This gives $\{0, 1, 2, 3, 4, 5\} = \mathbb{Z}_6$, covering everything, just like $1$.

Exercise 2.2.1: In $(\mathbb{Z}_9, +)$, which elements generate the entire $\mathbb{Z}_9$? Which elements generate proper subgroups?

2.2.2 Subgroups Generated by Elements

Now consider any group $G$ with a binary operation (like addition or multiplication), and let $g$ be an element of $G$. By repeatedly applying the group operation to $g$, we can form the set of elements it generates. This set is denoted $\langle g \rangle$ and is called the cyclic subgroup generated by $g$.

For example in additive groups:

$$\langle g \rangle = \{ g, g + g, g + g + g, \ldots \} = \{ n \cdot g \mid n \in \mathbb{Z} \}$$

For many elements, $\langle g \rangle$ is a proper subgroup. But when $\langle g \rangle = G$, we say that $g$ is a generator of $G$, and that $G$ is a cyclic group. As shown in Example 2.2.1, we computed the subgroups generated by elements in $(\mathbb{Z}_6, +) = \{0, 1, 2, 3, 4, 5\}$ under addition modulo 6. The subgroups generated by each element are as follows:

• $\langle 0 \rangle = \{0\}$ (proper subgroup)
• $\langle 1 \rangle = \{0, 1, 2, 3, 4, 5\} = \mathbb{Z}_6$
• $\langle 2 \rangle = \{0, 2, 4\}$ (proper subgroup)
• $\langle 3 \rangle = \{0, 3\}$ (proper subgroup)
• $\langle 4 \rangle = \{0, 4, 2\}$ (proper subgroup)
• $\langle 5 \rangle = \{0, 1, 2, 3, 4, 5\} = \mathbb{Z}_6$

Elements $1$ and $5$ generate the entire group $\mathbb{Z}_6$, making them generators, while $0$, $2$, $3$, and $4$ generate proper subgroups.

Example 2.2.3: Generators in $(\mathbb{Z}_5, +)$

Now test $\mathbb{Z}_5 = \{0, 1, 2, 3, 4\}$ under addition modulo 5:

• $\langle1\rangle = \{0, 1, 2, 3, 4\}=\mathbb{Z}_5$
• $\langle2\rangle =\{2, 4, 6 \equiv 1 \pmod 5 , 8 \equiv 3 \pmod 5\} =\{1,2,3,4\}=\mathbb{Z}_5$
• $\langle3\rangle= \{3, 6 \equiv 1 \pmod 5, 9 \equiv 4\pmod 5 , 7 \equiv 2\pmod 5 \}=\{1,2,3,4\}=\mathbb{Z}_5$
• $\langle4\rangle =\{4, 8 \equiv 3 \pmod 5, 12 \equiv 2\pmod 5 , 16 \equiv 1\pmod 5 \}=\{1,2,3,4\}=\mathbb{Z}_5$

Every non-zero element is a generator.

This happens because $\gcd(g, 5) = 1$ for all $g \neq 0$, and $5$ is prime.

Conclusion: In $(\mathbb{Z}_n, +)$, an element $g$ generates the entire group if and only if $\gcd(g, n) = 1$. For prime $n$, all non-zero elements are generators. For composite $n$, only some elements qualify.

2.3 Code: Exploring Additive Generators in $(\mathbb{Z}_n, +)$

def additive_closure(a, n):
    """Generates {0, a, 2a, 3a, ...} mod n until it repeats."""
    result = []
    current = 0
    while current not in result:
        result.append(current)
        current = (current + a) % n
    return sorted(result)  # Sorted for readability

# Show generated sets in (Z_6 , +)
print("Z_6:")
for a in range(6):
    print(f"Generated by {a}: {additive_closure(a, 6)}")

# Show generated sets in (Z_5, +)
print("\nZ_5:")
for a in range(5):
    print(f"Generated by {a}: {additive_closure(a, 5)}")

Output:

(Z_6, +):
Generated by 0: [0]
Generated by 1: [0, 1, 2, 3, 4, 5]
Generated by 2: [0, 2, 4]
Generated by 3: [0, 3]
Generated by 4: [0, 2, 4]
Generated by 5: [0, 1, 2, 3, 4, 5]

(Z_5, +):
Generated by 0: [0]
Generated by 1: [0, 1, 2, 3, 4]
Generated by 2: [0, 1, 2, 3, 4]
Generated by 3: [0, 1, 2, 3, 4]
Generated by 4: [0, 1, 2, 3, 4]

Having explored additive groups, we now turn to their multiplicative counterparts.

3. Multiplicative Groups

Multiplicative groups operate under multiplication (often modulo some number), with the identity element being $1$, though finding inverses can be more nuanced, especially in finite settings. To illustrate this, we’ll examine a concrete example using multiplication modulo $7$.

3.1 Example:

Consider $\mathbb{Z}_7 = \{0, 1, 2, 3, 4, 5, 6\}$:

• Closure holds, e.g., $2 \times5= 10 \equiv 3$, is in the set.
• Identity: The identity element is $1$, but this fails for $0$, which has no inverse;
• Inverses: Each nonzero element has an inverse:
  • $1 \times1= 1$
  • $2 \times4= 8 \equiv1\pmod{7}$
  • $3 \times5= 15 \equiv1\pmod{7}$
  • $4 \times2= 8 \equiv1\pmod{7}$
  • $5 \times3= 15 \equiv1\pmod{7}$
  • $6 \times6= 36 \equiv1\pmod{7}$

Since, $0$ has no inverse, $(\mathbb{Z}_7, \times)$ is not a group. But if we remove $0$ and consider only the nonzero elements — that is, $\mathbb{Z}_7 \setminus \{0\} = \{1, 2, 3, 4, 5, 6\}$ — we do get a group under multiplication. This set is often denoted $\mathbb{Z}_7^*$ and is a group of order 6.

You can use this code to generate the multiplication table for any modulus $n$:

def print_multiplication_table(mod):
    header = ["× mod " + str(mod)] + list(range(mod))
    print(" | ".join(str(h).rjust(4) for h in header))
    print("-" * (6 * (mod + 1)))

    for row in range(mod):
        line = [str(row).rjust(4)]
        for col in range(mod):
            value = row * col
            result = value % mod
            if value >= mod:
                line.append(f"{value} ≡ {result}".rjust(6))
            else:
                line.append(str(result).rjust(6))
        print(" | ".join(line))

print_multiplication_table(5)

3.2 Example: $(\mathbb{Z}_6\setminus \{0\}, \times),$ Is Not a Group

Let’s now test the set $\{1, 2, 3, 4, 5\}$ under multiplication mod 6.

• Closure: Holds.
• Identity:$1$ still works.
• Inverses: Inverses fail. Only $1$ and $5$ are invertible:
  • $1 \times1= 1$
  • $5 \times5= 25 \equiv1\mod 6$

Elements $2$, $3$, and $4$ have no counterparts that yield $1$. This failure arises because $6$ is not prime: it factors as $6 =2\times 3$. The factorization introduces zero divisors—nonzero elements that, when multiplied by other nonzero elements, yield zero modulo $n$. For example, $4 \times3= 12 \equiv 0 \pmod{6}$, even though both $3$ and $4$ are nonzero in $\mathbb{Z}_6$. Since $2$, $3$, and $4$ share common factors with $6$ (namely $2$, $3$, and $2$, respectively), they fail to have multiplicative inverses.

3.3 Example: $(\mathbb{Z}_8\setminus\{0\}, \times)$

As another example, let’s now consider multiplication in $\mathbb{Z}_8\setminus\{0\} = \{1, 2, 3, 4, 5, 6, 7\}$: | $\times$ | $1$ | $2$ | $3$ | $4$ | $5$ | $6$ | $7$ | |———-|—–|—–|—–|—–|—–|—–|—–| | $1$ | $1$ | $2$ | $3$ | $4$ | $5$ | $6$ | $7$ | | $2$ | $2$ | $4$ | $6$ | $8 \equiv 0$ | $10 \equiv 2$ | $12 \equiv 4$ | $14 \equiv 6$ | | $3$ | $3$ | $6$ | $9 \equiv 1$ | $12 \equiv 4$ | $15 \equiv 7$ | $18 \equiv 2$ | $21 \equiv 5$ | | $4$ | $4$ | $8 \equiv 0$ | $12 \equiv 4$ | $16 \equiv 0$ | $20 \equiv 4$ | $24 \equiv 0$ | $28 \equiv 4$ | | $5$ | $5$ | $10 \equiv 2$ | $15 \equiv 7$ | $20 \equiv 4$ | $25 \equiv 1$ | $30 \equiv 6$ | $35 \equiv 3$ | | $6$ | $6$ | $12 \equiv 4$ | $18 \equiv 2$ | $24 \equiv 0$ | $30 \equiv 6$ | $36 \equiv 4$ | $42 \equiv 2$ | | $7$ | $7$ | $14 \equiv 6$ | $21 \equiv 5$ | $28 \equiv 4$ | $35 \equiv 3$ | $42 \equiv 2$ | $49 \equiv 1$ |

Only the elements $\{1, 3, 5, 7\}$—those that are coprime with 8—have multiplicative inverses modulo 8:

• $1 \times1= 1 \pmod 8$
• $3 \times3= 9 \equiv 1 \pmod 8$
• $5 \times5= 25 \equiv 1 \pmod 8$
• $7 \times7= 49 \equiv 1 \pmod 8$

The remaining elements, $\{2, 4, 6\}$, all share a common factor with 8 and therefore do not have inverses. Thus, $\mathbb{Z}_8 \setminus \{0\}$ is not a group under multiplication.

3.4. What Is $\mathbb{Z}_n^*$?

The set *$\mathbb{Z}_n^$** is formally defined as:

$$ \mathbb{Z}_n^* = \{ a \in \mathbb{Z}_n \mid \gcd(a, n) =1\} $$

That is, $\mathbb{Z}_n^*$ consists of all elements in $\mathbb{Z}_n$ that have a multiplicative inverse modulo $n$.

• When $n$ is prime, every nonzero element is invertible, so:
  $$\mathbb{Z}_n^* = \mathbb{Z}_n \setminus \{0\}$$

• When $n$ is composite, only elements coprime to $n$ are invertible. For example:

  • $\mathbb{Z}_6^* = \{1, 5\}$
  • $\mathbb{Z}_8^* = \{1, 3, 5, 7\}$

In both cases, $\mathbb{Z}_n^*$ forms a group under multiplication. However, the full set $\mathbb{Z}_n \setminus \{0\}$ does not always form a group—unless $n$ is prime.

Earlier, we used $\mathbb{Z}_n \setminus \{0\}$ to demonstrate why the absence of inverses (and presence of zero divisors) breaks group structure when $n$ is composite. That distinction is more than technical: it plays a central role in understanding modular arithmetic, cryptographic algorithms, and number theory.

In the next section, we’ll explore how the structure of $\mathbb{Z}_n^*$ changes when $n$ is prime.

3.5 Why Prime Moduli Work

These failures motivate prime moduli. In $\{1,2,\ldots, p-1\}$, where $p$ is prime, every element has an inverse. In fact, this is guaranteed by a classic result from number theory.

Consider the pattern for $\{1, 2, \ldots, n-1\}$ modulo $n$:

• If $n$ is not prime, it is not a group due to zero divisors and missing inverses (e.g., $n = 6, 8$).
• If $n$ is prime, it is a group.

3.6 Python Code for Computing Inverse

One of the simplest and most efficient methods is to use Python’s built-in pow(a, -1, p) to compute modular inverses. Behind the scenes, this works because of a famous number theory result called Fermat’s Little Theorem, which guarantees that an inverse exists and tells us: $$ a^{-1} \equiv a^{p – 2} \pmod{p} $$

when the modulus $p$ is a prime number and $a$ is not divisible by $p$.

Here’s an example:

def mod_inverse(a, p):  
    return pow(a, -1, p)

# Test for Z_7^*
def test_inverses(p):
    print(f"Testing inverses modulo {p}:")
    for a in range(1, p):
        inv = mod_inverse(a, p)
        print(f"Inverse of {a} mod {p} is {inv} (check: {a} * {inv} = {a * inv % p})")

if __name__ == "__main__":
    p = 7
    test_inverses(p)

Output:

Testing inverses modulo 7:
Inverse of 1 mod 7 is 1 (check: 1 * 1 = 1)
Inverse of 2 mod 7 is 4 (check: 2 * 4 = 1)
Inverse of 3 mod 7 is 5 (check: 3 * 5 = 1)
Inverse of 4 mod 7 is 2 (check: 4 * 2 = 1)
Inverse of 5 mod 7 is 3 (check: 5 * 3 = 1)
Inverse of 6 mod 7 is 6 (check: 6 * 6 = 1)

3.7 Exercises

Math Exercises:

1. Find the modular inverse (if it exists) for the following, using Python’s pow(a, -1, n) function or a calculator. For prime moduli, you may optionally use Fermat’s Little Theorem:

   • $3^{-1} \mod 11$
   • $7^{-1} \mod 26$
   • $4^{-1} \mod 15$

2. For which values of $a$ in $\mathbb{Z}_{12}^*$ does the modular inverse exist?

Coding Exercises:

1. Write a function list_all_inverses(n) that returns a dictionary of all elements in $\mathbb{Z}_n^*$ with their inverses (if they exist).

2. Write a program that takes user input a and n, and checks whether a modular inverse exists. If it does, print the inverse. Try it with different values and see what you discover.

3. Challenge: Pick a few small primes p, and write a program that checks whether
   pow(a, p - 1, p) == 1 for all a in {1, 2, ..., p - 1}.
   Does this hold for every a? What happens if p isn’t prime?

4. Generators in Multiplicative Groups

We now turn our attention to generators in multiplicative groups. To build intuition, we begin with concrete examples in $\mathbb{Z}_7^* = \{1, 2, 3, 4, 5, 6\}$, exploring how individual elements can generate the full group—or just a part of it—through repeated multiplication

Example 4.1: $(\mathbb{Z}_7^*, \times)$

Try 3:

• $3^1 = 3$
• $3^2 = 9 \equiv 2 \pmod{7}$
• $3^3 = 27 \equiv 6 \pmod{7}$
• $3^4 = 81 \equiv 4 \pmod{7}$
• $3^5 = 243 \equiv 5 \pmod{7}$
• $3^6 = 729 \equiv 1 \pmod{7}$

$\langle 3 \rangle = \{1, 2, 3, 4, 5, 6\} = \mathbb{Z}_7^*$. Element $3$ is a generator.

Try 2:

• $2^1 = 2$
• $2^2 = 4$
• $2^3 = 8 \equiv 1 \pmod{7}$

$\langle 2 \rangle = \{1, 2, 4\}$, a subgroup, not the full group.

Try Other Elements:

• $4$: \$4, 16 \equiv 2, 8 \equiv 1$ $\Rightarrow \{1, 2, 4\}$
• $5$: \$5, 25 \equiv 4, 20 \equiv 6, 30 \equiv 2, 10 \equiv 3, 15 \equiv 1$ $\Rightarrow \mathbb{Z}_7^*$
• $6$: \$6, 36 \equiv 1$ $\Rightarrow \{1, 6\}$

Summary:

4.1 Primitive Elements

As we see, in the additive group $(\mathbb{Z}_p, +)$, where $p$ is prime, every non-zero element is guaranteed to be a generator. That is, repeatedly adding any non-zero element will eventually cycle through all elements of the group. However, in multiplicative groups $(\mathbb{Z}_p^*, \times)$, only some elements are generators—these are called primitive elements. This contrast highlights a fundamental difference: additive groups over primes are always cyclic with all non-zero elements as generators, while multiplicative groups over primes are cyclic but have only a subset of elements that serve as generators.

Definition: An element $g \in \mathbb{Z}_p^*$ is a primitive element if $$\langle g \rangle = \{g^1, g^2, \ldots, g^{p-1}\} \pmod{p} = \mathbb{Z}_p^*.$$

Take, for example, $\mathbb{Z}_7^*$: $3$ and $5$ are primitive elements, while $2$, $4$, and $6$ are not.

A foundational result in number theory guarantees that for any prime $p$, the group $\mathbb{Z}_p^*$ is cyclic, which means it always contains at least one primitive element. This stands in contrast to the additive group $(\mathbb{Z}_p, +)$, where every non-zero element generates the full group.

Example 4.1.1: $\mathbb{Z}_5^* = \{1, 2, 3, 4\}$

• Element 2: \begin{align*} 2^1 &= 2 \pmod{5}, \\ 2^2 &= 4 \pmod{5}, \\ 2^3 &= 8 \equiv 3 \pmod{5}, \\ 2^4 &= 16 \equiv 1 \pmod{5} \end{align*} $\langle 2 \rangle = \{1, 2, 3, 4\} = \mathbb{Z}_5^*$, so $2$ is a primitive element.

• Element 3: \begin{align*} 3^1 &= 3 \pmod{5}, \\ 3^2 &= 9 \equiv 4 \pmod{5}, \\ 3^3 &= 27 \equiv 2 \pmod{5}, \\ 3^4 &= 81 \equiv 1 \pmod{5} \end{align*} $\langle 3 \rangle = \{1, 2, 3, 4\} = \mathbb{Z}_5^*$, so $3$ is a primitive element too.

Example 4.1.2: $\mathbb{Z}_{11}^* = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}$

• Element 2: \begin{align*} 2^1 &= 2 \pmod{11}, \\ 2^2 &= 4 \pmod{11}, \\ 2^3 &= 8 \pmod{11}, \\ 2^4 &= 16 \equiv 5 \pmod{11}, \\ 2^5 &= 10 \pmod{11}, \\ 2^6 &= 20 \equiv 9 \pmod{11}, \\ 2^7 &= 18 \equiv 7 \pmod{11}, \\ 2^8 &= 14 \equiv 3 \pmod{11}, \\ 2^9 &= 6 \pmod{11}, \\ 2^{10} &= 12 \equiv 1 \pmod{11} \end{align*} $\langle 2 \rangle = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\} = \mathbb{Z}_{11}^*$, so $2$ is a primitive element.

• Element 3: \begin{align*} 3^1 &= 3 \pmod{11}, \\ 3^2 &= 9 \pmod{11}, \\ 3^3 &= 27 \equiv 5 \pmod{11}, \\ 3^4 &= 15 \equiv 4 \pmod{11}, \\ 3^5 &= 12 \equiv 1 \pmod{11} \end{align*} $\langle 3 \rangle = \{1, 3, 4, 5, 9\}$, a subgroup, not the full group.

Example 4.1.3: $\mathbb{Z}_{17}^*$

• Element 3: \begin{align*} 3^1 &= 3 \pmod{17}, \\ 3^2 &= 9 \pmod{17}, \\ 3^3 &= 27 \equiv 10 \pmod{17}, \\ 3^4 &= 81 \equiv 13 \pmod{17}, \\ 3^5 &= 243 \equiv 5 \pmod{17}, \\ 3^6 &= 729 \equiv 15 \pmod{17}, \\ 3^7 &= 2187 \equiv 11 \pmod{17}, \\ 3^8 &= 6561 \equiv 16 \pmod{17}, \\ 3^9 &= 19683 \equiv 14 \pmod{17}, \\ 3^{10} &= 59049 \equiv 8 \pmod{17}, \\ 3^{11} &= 177147 \equiv 7 \pmod{17}, \\ 3^{12} &= 531441 \equiv 4 \pmod{17}, \\ 3^{13} &= 1594323 \equiv 12 \pmod{17}, \\ 3^{14} &= 4782969 \equiv 2 \pmod{17}, \\ 3^{15} &= 14348907 \equiv 6 \pmod{17}, \\ 3^{16} &= 43046721 \equiv 1 \pmod{17} \end{align*} $\langle 3 \rangle = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16\} = \mathbb{Z}_{17}^*$, order $16$ (full group). So, $3$ is a primitive element of $\mathbb{Z}_{17}^*$.

Additional Note: Although we mentioned earlier that $\mathbb{Z}_n^$ is always a group, it is not always cyclic when $n$ is not prime. In contrast, when $n = p$ is prime, $\mathbb{Z}_p^$ is always cyclic. This distinction matters in practice—especially in cryptography—where we prefer to work with cyclic groups, so we often choose prime moduli to ensure that $\mathbb{Z}_p^*$ has this cyclic structure.

4.2 Python Code: Finding Primitive Elements Modulo a Prime

To find primitive elements in $\mathbb{Z}_p^$, we compute the subgroup generated by a candidate element and check whether it includes all elements of $\mathbb{Z}_p^$. The following code verifies whether a given element $g$ is primitive by performing this check. You can use it to explore and test more examples of primitive elements:

def is_primitive_element(g, p):
    """Check if g is a primitive element modulo p."""
    required = set(range(1, p))
    generated = set()
    val = 1
    for _ in range(1, p):
        val = (val * g) % p
        generated.add(val)
    return generated == required

def primitive_elements(p):
    """Find all primitive elements of prime p."""
    elements = []
    for g in range(2, p):
        if is_primitive_element(g, p):
            elements.append(g)
    return elements

# Example usage:
prime = 11
print(f"Primitive elements modulo {prime}: {primitive_elements(prime)}")

Sample Output:

Primitive elements modulo 11: [2, 6, 7, 8]

For a more efficient approach, the galois library can directly compute primitive elements in the multiplicative group of a finite field, which for a prime $p$ corresponds to $\mathbb{Z}_p^*$. First, install the library with pip install galois, then use:

import galois
print(galois.GF(7).primitive_elements)  # Output: [3, 5]

This method is optimized for large primes, making it ideal for practical applications, while the manual code above helps understand the concept of primitive elements.

Exercise

Let’s apply what we’ve learned about generators and primitive elements.

1. Use the Python code above to find a generator of $(\mathbb{Z}_{13}^, \times)$.
   (Hint: You’re looking for an element whose powers generate all elements of $\mathbb{Z}_{13}^$.)

2. Use your generator $g$ to list all elements of $\mathbb{Z}_{13}^$ in the form $g^k$. (Hint: You should get 12 distinct powers.)*

3. For each of the following values of $k \in \{1, 2, 3, 4, 6\}$, write out the subgroup generated by $g^k$.
   (Hint: This approach helps you find all subgroups without brute force. For example, consider the case $k = 2$ below)

   Example:
   If $g = 2$ (a generator of $\mathbb{Z}_{13}^*$), then:

   • $g^2 = 2^2 = 4$
   • The subgroup generated by $4$ is:
     $$ \langle4\rangle = \{4, 4^2 =16\equiv3\pmod{13}, 4^3=64\equiv12 \pmod{13}, \dots\} = \{4, 3, 12, 9, 10, 1\} $$

Now try this for $k = 1, 3, 4, 6$.
(Remember: compute $g^k$ first, then list its successive powers modulo 13 until you loop back to 1.)

4. Use your work above to list all distinct subgroups of $(\mathbb{Z}_{13}^, \times)$. (Hint: Some powers of $g$ generate the same subgroup.)*

Conclusion:
This chapter focused on understanding multiplicative groups modulo a prime and their subgroup structures. We saw that these groups are cyclic, meaning they can be generated by a single element — called a primitive element. We also explored how the powers of a primitive element generate cyclic subgroups, and how different powers produce different subgroups.

Because we haven’t yet introduced formal tools like Lagrange’s Theorem, we approached subgroup discovery manually — especially in the final exercise, where we tried a few values of $k$ to see which subgroups they generate. In the next chapter, you’ll learn the underlying principles that allow you to determine subgroup sizes and generators systematically.