Condition for Nu Function to be 1

Theorem
Let:
 * $n = \ds \prod_{i \mathop = 1}^s p_i^{m_i}$

where $p_1, p_2, \ldots, p_s$ are distinct primes.

Then:
 * $(1): \quad m_1, m_2, \ldots, m_s = 1$, that is, $n$ is square-free
 * $(2): \quad \forall i, j \in \set {1, 2, \ldots, s}: p_i \not \equiv 1 \pmod {p_j}$


 * every group $G$ of order $n$ is cyclic and so $\map \nu n = 1$.
 * every group $G$ of order $n$ is cyclic and so $\map \nu n = 1$.