Let's define the operation $\odot$ on $\mathbb Z_{6}$ as follows:
$\begin{aligned}x\odot y=x+y+xy\end{aligned}$
Show:
- $(\mathbb Z_6, \odot)$ is a monoid;
- if $3 \in \mathbb Z_6$ is an invertible element, if so use an appropriate congruential equation to determine its invertible element;
- define all the invertible elements in $\mathbb Z_6$.
I've started proving the last two items first. The element $3$ is not an invertible element in $\mathbb Z_6$ because $gcd(3,6) \neq 1$.
In order to define all the invertible elements in $\mathbb Z_6$ I thought it would be enough to enumerate all the elements $[a] \in \mathbb Z_6 : gcd(a,6)=1$. So I concluded all invertible elements in $\mathbb Z_6$ are:
$\begin{aligned}\{1+6k:k\in\mathbb Z\} = 1+6 \mathbb Z\end{aligned}$ and $\begin{aligned}\{5+6k:k\in\mathbb Z\} = 5+6 \mathbb Z\end{aligned}$
In order to prove $(\mathbb Z_6, \odot)$ to be a monoid, we require $\odot$ to be associative and $\exists \mathbb 1_{\mathbb Z_{6}} : x \odot \mathbb 1_{\mathbb Z_{6}} = \mathbb 1_{\mathbb Z_{6}} \odot x = x$
Associativity
$\begin{aligned}(a \odot b) \odot c = a \odot (b \odot c)\end{aligned}$ $\begin{aligned}(a+b+ab) \odot c = a \odot (b+c+bc)\end{aligned}$ $\begin{aligned}a+b+ab+c+ac+bc+abc = a+b+c+bc+ab+ac+abc\end{aligned}$
And the point is proved.
Identity element
$\begin{aligned}a \odot \mathbb 1_{\mathbb Z_{6}} = a\end{aligned}$ $\begin{aligned}1_{\mathbb Z_{6}} + a1_{\mathbb Z_{6}} = 0 \end{aligned}$
... and now I am stuck because I don't know how to proceed in order to evaluate $1_{\mathbb Z_{6}}$.
Is it correct stating $1_{\mathbb Z_{6}} + a1_{\mathbb Z_{6}} = 0 \Rightarrow a1_{\mathbb Z_{6}} = 0$? How do I take it from here?