Group of Order 15 is Cyclic Group/Proof 1

Theorem

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

Then $G$ is cyclic.

Proof

We have that $15 = 3 \times 5$.

Thus:

$15$ is square-free

$5 \equiv 2 \pmod 3$

$3 \equiv 3 \pmod 5$

The conditions are fulfilled for Condition for Nu Function to be 1.

Thus $\map \nu {15} = 1$ and so all groups of order $15$ are cyclic.

$\blacksquare$

Sources

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