What does Cayley table for a group $(\mathbb{Z}_5^*,\cdot_5)$ tell us? 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.
 A: $(\{0,1,2,3,4\},\cdot_{\bmod 5})$ 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 $x\cdot y =0 \pmod 5$
A: The problem is that $\mathbb{Z}_n$ is not a group with respect to multiplication because some elements (such as $0$) are not invertible. The units are by definition the invertible elements. In general, $a \in \mathbb{Z}_n$ is a unit if and only if $\gcd(a,n) = 1$. In particular, here $n = 5$ is prime and so every non-zero element is a unit. We can check explicitly that $1 \cdot 1 = 1$, $2 \cdot 3 = 6 = 1$, and $4 \cdot 4 = 16 = 1$. Thus $1$ and $4$ are their own inverses, while $2$ and $3$ are mutually inverse.
A: One thing we can see from the Cayley table is which elements have inverses, and which do not. In the table, $a$ has inverse $b$ iff the $(a_{th} \text{ row }, b_{th} \text{ column })$ and $(a_{th} \text{ column }, b_{th}  \text{ row })$ contain $1$.
A unit is a non-zero element that has an (multiplicative) inverse. Then the group of units you mentioned is the set of elements $$\mathbb{Z}_5^* = \{\bar{1},\bar{2},\bar{3},\bar{4}\}$$ That have (multiplicative) inverses in $\mathbb{Z}_5$. 
For example (and we can see this directly from the Cayley table) we have $$4^{-1} = 4 \text{ in } \mathbb{Z}_5^* $$
