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.