What the set $I$ of invertible elements in $\mathbb{Z}_{9}$? $5.2 = 1$
The set of invertible elements in $\mathbb{Z}_{9}$ is $I = \{1,2,5\}$.
Q1: Is it correct? 
I was also asked to show that it is not isomorphic to $S_{3}$
How can I show that it is not isomorphic? Thank you. 
 A: $$\begin{array}{c|c}
\times&0&1&2&3&4&5&6&7&8\\\hline
0&0&0&0&0&0&0&0&0&0\\\hline
1&0&\color{red}1&2&3&4&5&6&7&8\\\hline
2&0&2&4&6&8&\color{red}1&3&5&7\\\hline
3&0&3&6&0&3&6&0&3&6\\\hline
4&0&4&8&3&7&2&6&\color{red}1&5\\\hline
5&0&5&\color{red}1&6&2&7&3&8&4\\\hline
6&0&6&3&0&6&3&0&6&3\\\hline
7&0&7&5&3&\color{red}1&8&6&4&2\\\hline
8&0&8&7&6&5&4&3&2&\color{red}1
\end{array}$$
Therefore, the invertible elements of $\Bbb Z_9$ are $\{1,2,4,5,7,8\}$.
Indeed, $S_3$ has $6$ elements while $\Bbb Z_9$ has $9$ elements, so they are not isomorphic to each other.

In general, for $\Bbb Z_n$, $x$ is invertible iff $\gcd(x,n)=1$.
By Bézout's Lemma, $\gcd(x,n)=1 \iff \exists a,b:ax+bn=1 \iff \exists a:ax=1$.
A: Yes, you are correct. Clearly two groups that are not the same order cannot be isomorphic to each other...
We wish to find all elements that have a multiplicative inverse in $\mathbb{Z}_9$ (as indicated by your question). 9 is not prime, so some inverses may not exist, luckily for you, $\mathbb{Z}_9$ is a relatively small group to check.
$$1, 2\cdot5, 4 \cdot 7, 8 \cdot 8$$
Those are the multiplicative inverse pairings
