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.

$\blacksquare$

Sources

• 1996: John F. Humphreys: A Course in Group Theory ... (previous) ... (next): Chapter $11$: The Sylow Theorems: Example $11.7$