Group of Order 15 is Cyclic Group/Proof 3

Theorem

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

Then $G$ is cyclic.

Proof

Aiming for a contradiction, suppose $G$ is non-abelian.

Let $n_3$ denote the number of elements of $G$ of order $3$.

From Number of Elements of Order p in Group of Order pq is Multiple of q, $n_3$ is a multiple of $5$.

From Number of Order p Elements in Group with m Order p Subgroups, $n_3$ is a multiple of $2$.

Therefore $n_3$ is a multiple of $10$.

Let $n_5$ denote the number of elements of $G$ of order $5$.

From Number of Elements of Order p in Group of Order pq is Multiple of q, $n_5$ is a multiple of $3$.

From Number of Order p Elements in Group with m Order p Subgroups, $n_5$ is a multiple of $4$.

Therefore $n_5$ is a multiple of $12$.

Together with the identity element which has order $1$, that makes $1 + 12 a + 10 b = 15$ for some positive integers $a$ and $b$.

This is impossible.

Hence by Proof by Contradiction it follows that $G$ must be abelian.

Since $15$ is a product of $2$ distinct primes, by Abelian Group of Semiprime Order is Cyclic, $G$ is cyclic.

$\blacksquare$

Sources

• 1978: Thomas A. Whitelaw: An Introduction to Abstract Algebra ... (previous) ... (next): Chapter $8$: Homomorphisms, Normal Subgroups and Quotient Groups: Exercise $20$