$\mathbb{Z}^*_8$
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!
Best Answer
$\mathbb{Z_8}^*$ 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 $n$.