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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.?

Bill Dubuque
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Eva Lin
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  • Welcome to Mathematics Stack Exchange. It’s the multiplicative group of integers modulo $n$. I’d suggest looking at the orders of the elements – J. W. Tanner Oct 02 '23 at 03:17
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    $\mathbb{Z}_n^\times$ is called the "unit group of $\mathbb{Z}_n$". This notation is very common, actually. – diracdeltafunk Oct 02 '23 at 03:20
  • Cf. this question – J. W. Tanner Oct 02 '23 at 03:27
  • Expanding on prior comment, if it was cyclic, one of the elements except for $[1]$ would have to generate all of them, so just see if any do (you are overcomplicating by trying to formally set up a contradiction). BTW, it's known for which $n$ this is cyclic, but I won't give it away. – AlgTop1854 Oct 02 '23 at 03:34
  • @J.W.Tanner Thanks for providing the terminology and the link! Both are very helpful. I think I get the most of it now, but last bit of the question that has been tricked me up and it is probably quite naive...I was slightly confused by the fact that the operation is not defined here. So what exactly would, say, $[1]^2$ be? like..$[1+1]$, or $[1*1]$, or it doesn't matter/I am thinking about it in the wrong way? – Eva Lin Oct 02 '23 at 03:58
  • The group of order 4 has only two structures:$Z_4$ and $Z_2 \times Z_2$ .Just calculate $[1]^2,[3]^2,[5]^2,[7]^2$ and show that all elements are of order 2. – Lionel666 Oct 02 '23 at 04:05
  • @Lionel666 Thanks -- and a naive question (repost from another comment), what xactly would, say, $[1]^2$ be? like $[1+1]$, or $[1∗1]$, or it doesn't matter/I am thinking about it in the wrong way? – Eva Lin Oct 02 '23 at 04:14
  • Here $[1]^2=[1\times1]$ – J. W. Tanner Oct 02 '23 at 12:31

1 Answers1

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$\mathbb Z_8^*$ 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.

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