How to determine all the invertible elements? I have this excercise, I need your help on the third point:
i) Determine two integers $\alpha$ and $\beta$ such that $12\alpha + 7\beta = 1$
Answer: $\alpha = 3$ and $\beta = -5$
ii) Determine all the solutions of
$$7x\equiv 1 (mod. 12)$$
Answer: $[-5]_{12} = \{-5+12k, k\in\mathbb{Z}\}$
iii) determine invertible elements (for product) for $(\mathbb{Z}_{12}, +, \cdot)$, and zero divisors
Answer: Here I need your help! How can I determine all invertible elements and all zero divisors?
iv) determine, if exists, a class $[a]\in\mathbb{Z}_{12}$ such that $[a][6]=[2].$
Answer: No, doesn't exist. $gcd(6,12)\neq 1$
 A: There is the brute force method: try them all!
2 (2x6=12=0), 3 (3x4=12=0), 4 (4x3=12=0), 6 (6x2=12=0), 8 (8x6=48=0), 9 (9x4=36=0), 10 (10x6=60=0)  are zero divisors
1, 5 (5 x 5 = 25 = 1), 7 (7 x 7 = 49 = 1), 11 (11x11=121=1) are units. 
Notice that the units have gcd 1 with 12...
A: On iv), your answer is right, but the justification is not. For example, it is easy to see that there are $[a]$ such that $[a][6]=[6]$, even though $\gcd(6,12)\ne 1$. For $a=1$ will work, so will $a=3$, as will $a=9$.
The reason that we do not have an $a$ such that $[a][6]=[2]$ is that $\gcd(6,12)$ does not divide $2$.  For if $[a][6]=[2]$, then $6a\equiv 2 \pmod {12}$, or equivalently $12$ divides $6a-2$.  If this happens, then $6$ divides $6a-2$, which implies that $6$ divides $2$. That is clearly not the case.
Or else we could say that there is no $a$ such that $[a][6]=[2]$ because we tried everything from $a=0$ to $a=11$, and nothing worked.  What's fine for a small modulus like $12$, but one would not care to do that with $1200$.   
A: Hint $\rm\ n\:$ invertible in $\rm\mathbb Z_{12}\iff \exists\: a\!:\: an\equiv 1\iff \exists\: a,b\!:\: an+12b = 1\iff gcd(n,12)=1$
Conversely, $\rm\: gcd(n,12) = c > 1\:\Rightarrow\: n(12/c) \equiv 12(n/c) \equiv0\:\Rightarrow\: n\:$ is a zero-divisor, if $\rm n\not\equiv 0$.
A: An element $a$ is invertible in this finite structure if and only if there exists an $n \ge 1$ such that
$$ a^n = 1$$
Here is what I entered in the first row for columns A-F:
1 =mod(A1*$A1,12) =mod(B1*$A1,12) =mod(C1*$A1,12) =countif(A1:D1,1) =if(E1>0,"Invertible","")

So you drag it down putting numbers in the A column.
Result:

So the invertible elements are in the multiplicative subgroup $\{1,5,7,11\}$.
