Euler Totient Function (Example 2)

... answers to question posed here.

Example 2

For ${\displaystyle n=12}$, each of the numbers in ${\displaystyle \{1,2,...,12\}}$ belongs to exactly one of the subsets
     ${\displaystyle S_1 =\{1,5,7,11\}}$
     ${\displaystyle S_2 =\{2,10\}}$
     ${\displaystyle S_3 =\{3,9\}}$
     ${\displaystyle S_4 =\{4,8\}}$
     ${\displaystyle S_6 =\left\{6\right\}}$
     ${\displaystyle S_{12}=\left\{12\right\}}$
 Each subset is in fact an equivalence class of the relation ~ (where ${\displaystyle a~b\iff }$
${\displaystyle gcd(a,12)=gcd(b,12)}$).  ${\displaystyle S_h}$ means the equivalence class (or subset) of all the numbers whose highest common factor with ${\displaystyle 12}$ is  ${\displaystyle h}$.  Now observe
     ${\displaystyle S_1 =\{1,5,7,11\}}$,   ${\displaystyle ÷ 1}$, get ${\displaystyle \{1,5,7,11\} \subseteq \{1,...,12\}}$  ${\displaystyle \#=\phi \left(12\right)}$
     ${\displaystyle S_2 =\{2,10\}}$,   ${\displaystyle ÷ 2}$, get ${\displaystyle \{1,5\} \subseteq \{1,...,6\}}$   ${\displaystyle \#=\phi \left(6\right)}$
     ${\displaystyle S_3 =\{3,9\}}$,   ${\displaystyle ÷ 3}$, get ${\displaystyle \{1,3\} \subseteq \{1,...,4\}}$    ${\displaystyle \#=\phi \left(4\right)}$
     ${\displaystyle S_4 =\{4,8\}}$,   ${\displaystyle ÷ 4}$, get ${\displaystyle \{1,2\} \subseteq \{1,...,3\}}$   ${\displaystyle \#=\phi \left(3\right)}$
     ${\displaystyle S_6 =\left\{6\right\}}$,               ${\displaystyle ÷ 6}$, get ${\displaystyle \left\{1\right\} \subseteq \{1,...,2\}}$   ${\displaystyle \#=\phi \left(2\right)}$
     ${\displaystyle S_{12}=\left\{12\right\}}$,            ${\displaystyle ÷12}$, get ${\displaystyle \left\{1\right\} \subseteq \left\{1\right\}}$          ${\displaystyle \#=\phi \left(1\right)}$
Hence
     ${\displaystyle \#S_1+\#S_2+\#S_3+\#S_4+\#S_6+\#S_{12}}$
=   ${\displaystyle \phi \left(12\right)+\phi \left(6\right)+\phi \left(4\right)+\phi \left(3\right)+\phi \left(2\right)+\phi \left(1\right)}$
=   ${\displaystyle 4+2+2+2+1+1}$
=   12