Understanding Generators of a Cyclic Group

A cyclic group is a group that can be generated by a single element. This element is called a generator of the group. If an element 'g' is a generator of a cyclic group G, it means that every element in G can be written as a power of 'g' (i.e., <g> = G).

The question asks for the number of generators of a cyclic group G that has an order of 8. The order of a group is the number of elements in the group. So, G has 8 elements.

Finding the Number of Generators

For a finite cyclic group, there is a specific way to determine how many generators it has. The number of generators of a cyclic group of order $n$ is equal to the number of positive integers less than or equal to $n$ that are relatively prime to $n$. This quantity is given by Euler's totient function, denoted by $\phi(n)$.

In this case, the order of the cyclic group is $n=8$. We need to find the number of generators of this cyclic group, which is $\phi(8)$.

Calculating Euler's Totient Function $\phi(8)$

Euler's totient function, $\phi(n)$, counts the positive integers up to a given integer $n$ that are relatively prime to $n$. Two integers are relatively prime (or coprime) if their greatest common divisor (GCD) is 1.

To calculate $\phi(8)$, we need to find the positive integers less than or equal to 8 that are relatively prime to 8. Let's list the integers from 1 to 8 and check their GCD with 8:

• GCD(1, 8) = 1 (1 is relatively prime to 8)
• GCD(2, 8) = 2 (2 is not relatively prime to 8)
• GCD(3, 8) = 1 (3 is relatively prime to 8)
• GCD(4, 8) = 4 (4 is not relatively prime to 8)
• GCD(5, 8) = 1 (5 is relatively prime to 8)
• GCD(6, 8) = 2 (6 is not relatively prime to 8)
• GCD(7, 8) = 1 (7 is relatively prime to 8)
• GCD(8, 8) = 8 (8 is not relatively prime to 8)

The positive integers less than or equal to 8 that are relatively prime to 8 are 1, 3, 5, and 7. There are exactly 4 such numbers.

Therefore, $\phi(8) = 4$.

Alternatively, using the formula for $\phi(n)$ based on its prime factorization: If the prime factorization of $n$ is $n = p_1^{k_1} p_2^{k_2} \cdots p_r^{k_r}$, then $\phi(n) = n \prod_{i=1}^{r} \left(1 - \frac{1}{p_i}\right)$.

The prime factorization of 8 is $8 = 2^3$. Here, $p_1 = 2$ and $k_1 = 3$.

So, $\phi(8) = 8 \left(1 - \frac{1}{2}\right) = 8 \left(\frac{1}{2}\right) = 4$.

Both methods show that the number of generators of a cyclic group of order 8 is 4.

Summary of the Number of Generators

For a cyclic group of order $n$, the number of generators is $\phi(n)$. For a cyclic group of order 8, the number of generators is $\phi(8)$, which we calculated to be 4. This means there are 4 distinct elements in the cyclic group of order 8 that can generate the entire group.

Understanding the concept of generators of a cyclic group and how to use Euler's totient function is key to solving such problems in abstract algebra and group theory.