What is this notation? Cyclic group $\mathbb{Z}^*_8$

Question

$\mathbb{Z}^*_8$

As I understand it - $\mathbb{Z}_8$ is the group of integers under addition modulo 8.

So am I correct in thinking its elements are: $\{0,1,2,3,4,5,6,7\}$?

I thought the $*$ meant excluding zero, so I was confused to learn that the elements of $\mathbb{Z}^*_8$ are apparently: $\{1,3,5,7\}$ - missing $2$, $4$ and $6$.

The only possible answer I can think of is that because <1>,<3>,<5> and <7> can generate the whole group - $2$, $4$ and $6$ are omitted from $\mathbb{Z}_8$ in the first place, meaning $*$ does just remove the zero as I thought.

Can someone explain this to me as the notes I'm using aren't helping!

Answer 1

$\mathbb{Z_8}^*$ denotes the multiplicative group of $\mathbb{Z_8}$ as Mark Bennet has said.

You can show that $x \in \mathbb{Z_n}$ has a multiplicative inverse if and only if $(x,n)=1$. The proof is based on a special case of Bezout's theorem that states $(x,n)=1$ if and only if $\exists a,b \in \mathbb{Z}: ax+bn = 1$.

If $(x,n)=1$, then $\exists a,b: ax+bn=1$. This implies that $bn = 1-ax$ or $n \mid 1-ax$ which is the same as $ax \equiv 1 \pmod{n}$.

On the other hand, if there exists $a \in \mathbb{Z_n}$ such that $ax \equiv 1 \pmod{n}$ then $\exists b \in \mathbb{Z}: bn = 1 - ax$. Which gives you the converse.

So, the necessary and sufficient condition for an element in $\mathbb{Z_n}$ to be invertible is that it is relatively prime to $n$.

Answer 2

This is a notational device. The $^*$ is being used to show that the group operation is multiplication, and the elements of the group are the elements of $\mathbb Z_8$ which are coprime to $8$. The identity is $1$.

The integers taken modulo $n$ inherit both addition and multiplication from $\mathbb Z$. If you take the elements coprime to $n$ you get a multiplicative group of order $\varphi (n)$ whose elements satisfy $$x^{\varphi(n)}=1$$This is the Euler-Fermat theorem, a generalisation of Fermat's Little Theorem.