Group of Order 15 is Cyclic Group/Proof 3

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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.



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