Proof explanation on a group of order $595$ having a normal Sylow $17$-subgroup.

Question

Prove that a group of order 595 has a normal Sylow 17-subgroup.

The proof is as follows:

By Sylow, $n_{17} = 1$ or $35$. Assume $n_{17} = 35$. Then the union of the Sylow $17$-subgroups has $561$ elements. By Sylow, $n_5 = 1$. Thus, we may form a cyclic subgroup of order $85$ (from a previous theorem) But then there are $64$ elements of order $85$. This gives too many elements.

My question is: Where does this $64$ come from?

Answer 1

It comes from Euler's totient function, i.e., $$ \phi(85)=64. $$ Indeed, a cyclic group $C_n$ of order $n$ has exactly $\phi(n)$ generators, i.e., elements of order $n$.