Euler Totient Function (Example 2)

... answers to question posed here.

Example 2

For `n=12`, each of the numbers in `\{1,2,...,12\}` belongs to exactly one of the subsets
     `S_1  = \{1, 5, 7, 11\}`
     `S_2  = \{2, 10\}`
     `S_3  = \{3, 9\}`
     `S_4  = \{4, 8\}`
     `S_6  = \{6\}`
     `S_12 = \{12\}`
 Each subset is in fact an equivalence class of the relation ~ (where `a~b \Leftrightarrow`
`\gcd(a,12)=\gcd(b,12)`).  `S_h` means the equivalence class (or subset) of all the numbers whose highest common factor with `12` is  `h`.  Now observe
     `S_1  = \{1, 5, 7, 11\}`,   `\divide  1`, get `\{1, 5, 7, 11\}  \subseteq \{1,...,12\}`  `\#=\phi(12)`
     `S_2  = \{2, 10\}`,   `\divide  2`, get `\{1, 5\}                       \subseteq \{1,...,6\}`   `\#=\phi(6)`
     `S_3  = \{3, 9\}`,   `\divide  3`, get `\{1, 3\}                     \subseteq \{1,...,4\}`    `\#=\phi(4)`
     `S_4  = \{4, 8\}`,   `\divide  4`, get `\{1, 2\}                         \subseteq \{1,...,3\}`   `\#=\phi(3)`
     `S_6  = \{6\}`,               `\divide  6`, get `\{1}                   \subseteq \{1,...,2\}`   `\#=\phi(2)`
     `S_12 = \{12\}`,            `\divide 12`, get `\{1\}                      \subseteq \{1\}`          `\#=\phi(1)`
Hence
     `\#S_1 + \#S_2 + \#S_3 + \#S_4 + \#S_6 + \#S_12`
=   `\phi(12) + \phi(6) + \phi(4) + \phi(3) + \phi(2) + \phi(1)`
=   `4 + 2 + 2 + 2 + 1 + 1`
=   12