Determine invertible elements of a monoid 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 $\overline{b^{-1}}$?
 A: 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 $\bar5$.
A: Since you are in $\mathbb Z_6, \overline{b}^{-1}$ only exists for $\overline b= \overline 1$ or $\overline b= \overline5$.
A: 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?".
