Multiplication group for $\mathbb Z_n$ modulo $n$ By definition:
Let $\mathbb Z^+_n = \{[0],[1],[2],\ldots,[n−1]\}$
$\mathbb Z^+_4 = \{[0],[1],[2],[3]\},$
but
how $\mathbb Z^*_{12}$ is $\{[1],[5],[7],[11]\}$ ?
how $\mathbb Z^*_{7}$ is $\{[1],[2],[3],[4],[5],[6]\}$ ?

$\require{cancel}$
\begin{align}
\frac 1 {12} & = \frac 1 {12} & \text{No cancellation occurs here.} \\[6pt]
\frac 2 {12} & = \frac {\cancel2\times1}{\cancel2\times6} = \frac 1 6 \\[6pt]
\frac 3 {12} & = \frac {\cancel3\times 1}{\cancel3\times 4} = \frac 1 4 \\[6pt]
\frac 4 {12} & = \frac{\cancel4\times1}{\cancel4\times3} = \frac 13 \\[6pt]
\frac 5 {12} & = \frac 5 {12} & \text{No cancellation occurs here.} \\[6pt]
\frac 6 {12} & = \frac{\cancel6\times1}{\cancel6\times2} = \frac 1 2 \\[6pt]
\frac 7 {12} & = \frac 7 {12} & \text{No cancellation occurs here.} \\[6pt]
\frac 8 {12} & = \frac{\cancel4\times2}{\cancel4\times3} = \frac 2 3 \\[6pt]
\frac 9 {12} & = \frac{\cancel3\times3}{\cancel3\times4} = \frac 3 4 \\[6pt]
\frac{10}{12} & = \frac{\cancel2\times5}{\cancel2\times6} = \frac 5 6 \\[6pt]
\frac{11}{12} & = \frac{11}{12} & \text{No cancellation occurs here.} \\ {}
\end{align}
