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 $\overline{b^{-1}}$?