Confusion about creating Cayley table for ($\mathbb{Z}_{18}^*,\times)$ On a test I ran into a question about a Cayley table.
The question was "Given the group $(\mathbb{Z}_{18}^*,\times)$, construct the Cayley table."
It also said that this group is sometimes referred to as $(U(18),\cdot)$.
What exactly is the set $\mathbb{Z}_{18}^*$? Because the Wikipedia says it's used as a set of non-zero integers, or non-negative integers. But I've also seen it used as a set of all primes up to 18?
And if I were to create a Cayley table using all primes up to 18, would the following table be the correct result?
\begin{array}{|l|l|l|l|l|l|l|}
\hline
×  & 1  & 5  & 7   & 11  & 13  & 17  \\ \hline
1  & 1  & 5  & 7   & 11  & 13  & 17  \\ \hline
5  & 5  & 25 & 35  & 55  & 65  & 85  \\ \hline
7  & 7  & 35 & 49  & 77  & 91  & 119 \\ \hline
11 & 11 & 55 & 77  & 121 & 143 & 187 \\ \hline
13 & 13 & 65 & 91  & 143 & 169 & 221 \\ \hline
17 & 17 & 85 & 119 & 187 & 221 & 289 \\ \hline
\end{array}
I realize this is a very basic question, but finding consistent information through Googling has not been fruitful for me.
 A: Maybe you are using a convention which is just strangely unfamiliar to me, but I do not think that $\mathbb{Z}_{18}^*$ is meant to indicate the set of all primes up to 18.
For what I know, $\mathbb{Z}_n$ is mostly used as a short for $\mathbb{Z}/n\mathbb{Z}$, which means the set of integers modulo $n$.
For example $(\mathbb{Z}_4,+) := (\mathbb{Z}/4\mathbb{Z},+) = \{0,1,2,3\}$ for the integer with addition.
The star is usually used to indicate the substructure of units, that is the group of those elements which admit inverses.
For example the set $(\mathbb{Z}_4,\times) := (\mathbb{Z}/4\mathbb{Z},\times)$ does not have an inverse for each element (which is the reason why it is not a group, besides the fact that the operation is not even closed). Take a look at the element $\bar{3}$. Then you got
$$
1\cdot 3 \equiv 3 \mod 4 \\
2\cdot 3 \equiv 2 \mod 4 \\
3\cdot 3 \equiv 3 \mod 4
$$
but you will not find an $x \in \mathbb{Z}_4$ such that
$$
x\cdot 3 \equiv 1 \mod 4 \; .
$$
The Cayley table is just a way to encapture the structure of a finite group in a nice tabular way. The Cayley table for $(\mathbb{Z}_3,+)$ would for example be 
$$
\begin{array}{|l|l|l|}
\hline
+  & 0  & 1  & 2  \\ \hline
0  & 0  & 1  & 2  \\ \hline
1  & 1  & 2  & 0  \\ \hline
2  & 2  & 0  & 1  \\ \hline
\end{array}
$$
A: You listed the 6 relatively primes to $18$, and not the primes, which is correct.
Your Cayley table is also (somewhat) correct, but note that we are working in $\Bbb Z_{18}$, i.e. modulo $18$. That is, you should rather put the remainders of the calculated products modulo $18$.
For example, the last term is $17^2 =289\equiv 1\pmod{18}$ (because $288$ is even and its digits sum to $18$ which is divisible by $9$).
(By the way, if you keep on adding the digits in the numbers in your table until you reach 1 digit, then if the result is odd, it's just the remainder, if it's even, subtract $9$.)
But that's also a consequence of $17\equiv -1\pmod{18}$ (meaning that their difference is divisible by $18$), and that congruent numbers are interchangeable in modular arithmetic, just like equal numbers are so in normal arithmetic (i.e. one has e.g. $a\equiv b\pmod m\implies ac\equiv bc\pmod m$ and thus if also $c\equiv d$, then $ac\equiv bc\equiv bd$.)
So, here is a simplified version of the same Cayley table you wrote, but using smallest absolute value representatives:
$$\matrix{\times \\
& 1&5&7&-7&-5&-1\\ 
& 5&7&-1&1&-7&-5\\
& 7&-1&-5&5&1&-7\\
& -7&1&5&-5&-1&7\\
& -5&-7&1&-1&7&5\\
& -1&-5&-7&7&5&1}$$
You can also observe that $5$ generates this group: keeping multiplying by $5$ we receive the following cycle containing all group elements:
$$1\,\mapsto\, 5\,\mapsto \, 7\,\mapsto \, -1\,\mapsto \, -5\,\mapsto \, -7\,\mapsto \, 1\,\mapsto \, \dots $$
