Group of Order 15 has Cyclic Subgroups of Order 3 and Order 5

Theorem

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

Then $G$ has

a cyclic subgroup of order $3$

and:

a cyclic subgroup of order $5$.

Proof

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

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

Thus from the First Sylow Theorem:

$G$ has at least one subgroup $H_3$ of order $3$

and:

$G$ has at least one subgroup $H_5$ of order $5$.

From Prime Group is Cyclic, all such subgroups of order $3$ and order $5$ are cyclic.

$\blacksquare$

Sources

• 1996: John F. Humphreys: A Course in Group Theory ... (previous) ... (next): Chapter $11$: The Sylow Theorems: Example $11.5$