Order of the element 7 in the multiplicative group of 21

To find the order of the element \(7\) in the multiplicative group of integers modulo \(21\) (denoted as \(\mathbb{Z}_{21}^*\)), we need to determine the smallest positive integer \(k\) such that:

\[
7^k \equiv 1 \pmod{21}
\]

Let's calculate \(7^k \mod 21\) for increasing values of \(k\):

1. \(7^1 = 7 \equiv 7 \pmod{21}\)
2. \(7^2 = 49 \equiv 49 \mod 21 = 7 \pmod{21}\)
3. \(7^3 = 7 \times 7^2 = 343 \equiv 343 \mod 21 = 7 \pmod{21}\)
4. Continuing this way, it's clear that \(7^k \equiv 7 \pmod{21}\) for any \(k\).

Since \(7^k \equiv 7 \pmod{21}\) never equals 1 for any positive integer \(k\), the element 7 does not have an order in this group. This means that 7 is not an element of the multiplicative group \(\mathbb{Z}_{21}^*\).

**Conclusion**: In the multiplicative group \(\mathbb{Z}_{21}^*\), the element \(7\) is not invertible, and thus it has no order.

Would you like further details or have any questions?

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Here are some related questions:
1. What is the definition of the order of an element in a group?
2. How do you determine if an element is invertible in a multiplicative group modulo \(n\)?
3. What is the structure of the group \(\mathbb{Z}_{21}^*\)?
4. Can every integer modulo \(n\) have an order in \(\mathbb{Z}_n^*\)?
5. What are the other elements in \(\mathbb{Z}_{21}^*\) and their orders?

**Tip**: To check if an element is invertible in \(\mathbb{Z}_n^*\), calculate the greatest common divisor (GCD) of the element and \(n\). If the GCD is 1, the element is invertible.

Learn how to determine the order of the element 7 in the multiplicative group modulo 21 using modular arithmetic and group theory concepts. Detailed step-by-step solution provided.

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