[Math] What does Cayley table for a group $(\mathbb{Z}_5^*,\cdot_5)$ tell us

Question

Firsly, I would like you to explain me what $\mathbb{Z}_5^*$ means. My teacher told me it is a group of units of $\mathbb{Z}_5$, but I'm not sure what a group of units is.
If $\mathbb{Z}_5^*=\{0,1,2,3,4\}$, what is $\mathbb{Z}_8^*$?

Now, what bothers me is what Cayley table tells us about the group. If $\cdot_5$ is a binary operation defined as:
$$a\cdot_5b=(a\cdot b)\mod5$$
They Cayley table should look like this:
$$\begin{array}{c|c|c|c|c|c}
\cdot_5 & 0 & 1 & 2 & 3 & 4 \\ \hline
0 & 0 & 0 & 0 & 0 & 0 \\ \hline
1 & 0 & 1 & 2 & 3 & 4 \\ \hline
2 & 0 & 2 & 4 & 1 & 3 \\ \hline
3 & 0 & 3 & 1 & 4 & 2 \\ \hline
4 & 0 & 4 & 3 & 2 & 1
\end{array}$$
What does it tell us about the group $(\mathbb{Z}_5^*,\cdot_5)$?
Thank you for your time.

Answer 1

$(\{0,1,2,3,4\},\cdot_{\bmod 5})$ is not a group: $0$ has no inverse.

Anyway, as the table is symmetric, you can see that the operation is commutative.

And also, if you remove the $0$ from the set, you do get a group.

Units are just invertible elements, from your table you can see that for every number $x$ in $\{1,2,3,4\}$, there exists another number $y$ in the set such that $x\cdot y =0 \pmod 5$