Recall for $n \in \mathbb{N}$, we let:
$\mathbb{Z}_n^{\times} = \{\alpha \in \mathbb{Z}_n : \exists \beta \in \mathbb{Z}_n s.t. \alpha\beta = [1] \}$
Show that $\mathbb{Z}_8^{\times}$ is not a cyclic group.
I know that in $\mathbb{Z}_8^{\times} = \{ [1], [3], [5], [7] \}$ but do not really know what to do from here. I am thinking about disprove by contradiction, i.e. assume $\exists x \in \mathbb{Z} s.t. <a> = \mathbb{Z}_8^{\times}$. That means, $\exists a,b,c,d \in \mathbb{Z} s.t. x^a = [1]_8, x^b = [1]_8, x^c = [1]_8, x^d = [1]_8$. But now I am stuck here. I feel like it is quite obvious and I am missing something here...
Any suggestions appreciated.
P.S. I have been trying to look on internet but I cannot find any other places other than my professor's notes that uses the notation $\mathbb{Z}_n^{\times}$. Is there a name for this etc.?
$\mathbb Z_8^*$ is a multiplicative group ($\cdot$ ) where the operation $\cdot$ is defined as $[a]\cdot [b]= ab \mod 8$.
Now, $\mathbb Z_8^*=\{[1],[3],[5],[7]\}$.
Here, $[1]^2=[1]\cdot [1]=1.1 \mod 8=1$
$[3]^2=[3]\cdot [3]=3.3 \mod 8=1$
$[5]^2=[5]\cdot [5]=5.5 \mod 8=1$
$[7]^2=[7]\cdot [7]=7.7 \mod 8=1$
Definition: A group $G$ having $n$ elements is cyclic if it has an element of order $n$.
Since all elements in $\mathbb Z_8^*$ have order $2$, but $\mathbb Z_8^*$ has order 4, we can say that $\mathbb Z_8^*$ is not cyclic.