Number of Sylow p-Subgroups in Group of Order 15

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

Then:
 * the number of Sylow $3$-subgroups is in the set $\set {1, 4, 7, \ldots}$
 * the number of Sylow $5$-subgroups is in the set $\set {1, 6, 11, \ldots}$

Proof
Let $G$ be a group of order $15$.

From the Fourth Sylow Theorem:
 * the number of Sylow $p$-subgroups is equivalent to $1 \pmod p$

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

Hence the result.