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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 $\overline 5 = \overline {-1}$. – marlu Mar 7 '12 at 14:10

A crucial observation to simplify the study of this structure is that the first and second components are treated entirely independently. So if you can show that the operations $\bar a\cdot_1\bar c=\bar a+\bar c+\bar 3$ and $\bar b\cdot_2\bar d=\bar b\bar d=\overline{bd}$ individually define monoid structures on $\mathbb Z/6\mathbb Z$, then you will have shown that the defined structure is the direct product of these two monoids, and therefore a monoid.

Now the second structure is just multiplication modulo $6$, which defines the multiplicative monoid $(\mathbb Z/6\mathbb Z,\times)$; this is certainly a (commutative) monoid with identity element $\bar1$. The first structure is more like addition modulo $6$, but it is shifted. Once you find the neutral element for this operation, which is easily seen to be $\bar3$, you can see that the shift by $\bar3$ transforms the usual addition modulo $6$ into the operation $\cdot_1$: it is a bijection, and if $x+y=z$ in $\mathbb Z/6\mathbb Z$, then $$(x+\bar3)\cdot_1(y+\bar3)=(x+\bar3)+(y+\bar3)+\bar3=z+\bar9=z+\bar3 $$ in $(\mathbb Z/6\mathbb Z,\cdot_1)$. So the latter structure is a monoid (in fact a commutative group), and $x\mapsto x+\bar3$ is an isomorphism $(\mathbb Z/6\mathbb Z,+)\to(\mathbb Z/6\mathbb Z,\cdot_1)$. This avoids the tedious verification of the associative law and identity property.

One concludes that $(\mathbb Z/6\mathbb Z,\cdot)$ is a commutative monoid as well, with identity $(\bar3,\bar1)$; it is isomorphic to the product monoid $(\mathbb Z/6\mathbb Z,+)\times(\mathbb Z/6\mathbb Z,\times)$ by the isomorphism $(x,b)\mapsto(x+\bar3,b)$. Since the first factor of this product monoid is in fact a group, the invertible elements of the product are those which have a second component that is invertible for $\cdot_2$, in other words that is $\bar1$ or $\bar5$.

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Since you are in $\mathbb Z_6, \overline{b}^{-1}$ only exists for $\overline b= \overline 1$ or $\overline b= \overline5$.

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Looks okay to me, so far. You've shown that, if you want

$$ (a,b)(c,d) = e = (3,1)$$ to hold, you need to have $a + c = 0$ and $bd = 1$.

The first equation is easy to satisfy for all $a$; when does the second have a solution?

If you can answer that question, you will have found the invertible elements. So the problem is not so much "how can I find $b^-1$", but "when is $bd = 1$ solvable for $d$, working modulo 6?".

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