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I have the following excercise.

In the set $\mathbb{Z}_6\times\mathbb{Z}_6$ consider the follow: $$(\overline{a}, \overline{b})\cdot(\overline{c}, \overline{d}) = (\overline{a}+\overline{c}+\overline{3}, \overline{b}\overline{d})$$ Prove that $(\mathbb{Z}_6\times\mathbb{Z}_6, \cdot)$ is a monoid, and determine the invertible elements.

Yes, it's a monoid.

Associative: $$\big((\overline{a}, \overline{b})\cdot(\overline{c}, \overline{d})\big) \cdot (\overline{e}, \overline{f})=(\overline{a}, \overline{b})\cdot\big((\overline{c}, \overline{d}) \cdot (\overline{e}, \overline{f})\big)$$

through various steps (and considering $\overline{3}+\overline{3}=\overline{6}=\overline{0} $, is that correct???): $$(\overline{a}+\overline{c}+\overline{e}, \overline{b}\overline{d}\overline{f})=(\overline{a}+\overline{c}+\overline{e}, \overline{b}\overline{d}\overline{f})$$

Identity: $$(\overline{a}, \overline{b})\cdot(\overline{e_1}, \overline{e_2}) = (\overline{a}, \overline{b})\quad \forall (\overline{a}, \overline{b}) \in \mathbb{Z}_6\times\mathbb{Z}_6$$ follow: $$(\overline{a}+\overline{e_1}+\overline{3}, \overline{b}\overline{e_2})=(\overline{a}, \overline{b})$$ thus: $$\overline{a}+\overline{e_1}+\overline{3} = \overline{a}\Rightarrow \overline{e_1}=\overline{-3}$$ and $$\overline{b}\overline{e_2}= \overline{b}\Rightarrow \overline{e_2}=\overline{1}$$ The structure is even commutative, than exist right and left identity.

Invertible elements: $$(\overline{a}, \overline{b})\cdot(\overline{i_1}, \overline{i_2}) = (\overline{-3}, \overline{1})\quad \forall (\overline{a}, \overline{b}) \in \mathbb{Z}_6\times\mathbb{Z}_6$$ so: $$(\overline{a}+\overline{i_1}+\overline{3}, \overline{b}\overline{i_2})=(\overline{-3}, \overline{1})$$ follow (and considering: $\overline{3}+\overline{3}=\overline{6}=\overline{0}$): $$\overline{a}+\overline{i_1}+\overline{3} = \overline{-3}\Rightarrow \overline{a}+\overline{i_1}+\overline{3} +\overline{3} = 0 \Rightarrow \overline{i_1}=\overline{-a}$$ and $$\overline{b}\overline{i_2}=\overline{1} \Rightarrow \overline{i_2}=\overline{b^{-1}}$$

Is (until now) all right? Now, how can I find $\overline{b^{-1}}$?

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    The invertible elements in $\mathbb Z /6\mathbb Z$ are $\overline 1$ and $\overline 5$ since these are the residue classes which are coprime to the modulus 6. We have $\overline 5^{-1} = \overline 5$ because $\overline 5 = \overline {-1}$.2012-03-07

3 Answers 3