Condition for Nu Function to be 1

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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

$(2): \quad \forall i, j \in \set {1, 2, \ldots, s}: p_i \not \equiv 1 \pmod {p_j}$

if and only if:

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

Proof

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Examples

Group of Order $15$ is Cyclic Group

Let $G$ be a group whose order is $15$.

Then $G$ is cyclic.

Sources

• 1978: John S. Rose: A Course on Group Theory ... (previous) ... (next): $1$: Introduction to Finite Group Theory: $1.4$