Is $\mathbb{Z}_7^*$ cyclic?

Question

Determine whether the following sentence is correct or not.

$$ \mathbb{Z}_7^* \text{ is cyclic. }$$

Is $\mathbb{Z}_7^*$ the same as $\mathbb{Z}$ without $0$??

If it is $\mathbb{Z}_7^*=\{1,2,3,4,5,6\}$, how can it be a group when it does not contain $0$, the identity element??

Or does $\mathbb{Z}_7^*$ mean something else??

Answer 1

Note that In given $G$ we have: $1^{-1}=1$; $2^{-1}=4$; $3^{-1}=5$; $6^{-1}=6$. Now use that in a group $a$ and $a^{-1}$ have same order to conclude that $2$, $4$ have order 3 and $3$, $5$ have order $6$ and $6$ is self invertible element in $G$. More clearly

$$\langle 2 \rangle = \{2, 2^2, 2^3, \ldots \} = \{2, 4, 1 \} = \langle 4 \rangle$$

$$\langle 3 \rangle = \{3, 3^2, 3^3, \ldots \} = \{3, 2,6,4,5,1 \} = \langle 5 \rangle$$

$$\langle 6 \rangle= \{6, 6^2, 6^3, \ldots \} = \{6, 1 \}$$

Here note that we are working under multiplication $\text{mod $7$}$.

Answer 2

$(\mathbb{Z}_7, +, \cdot)$ is a ring. $\mathbb{Z}_7^*$ denotes the units of this ring. This becomes a group with respect to the multiplication.

Now, since we have multiplication, zero is not the identity element. Which is the identity of this group?

Answer 3

This is a group under multiplication.