Group of Order 15 is Cyclic Group/Proof 1
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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: $1.4$