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Let's define the operation $\odot$ on $\mathbb Z_{6}$ as follows:

$$\begin{aligned}x\odot y=x+y+xy\end{aligned}$$

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  • $(\mathbb Z_6, \odot)$ 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?

  • 1
    $\mathbb{Z}_6$ is an abelian group, so it is customary to write its identity element as $0$ - in this case, it is the residue class of $0\mod 6.$ Now there is no guarantee that the identities for $ \odot $ and modular addition will coincide, but why don't you try it and see what you get.2012-07-09
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    Further, it's impossible to answer the last two questions without verification of the first. The notion of invertability for this monoid is not the same as for multiplication of residues.2012-07-09
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    I think the OP is (confusingly!) using $1_{\mathbb{Z}_6}$ to refer to a hypothetical identity of the structure under consideration, not the identity of $\mathbb{Z}_6$.2012-07-09

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