The Order of a Nonabelian Group is At Least 6

Proposition 1: Let $(G, \cdot)$ be a nonabelian group. Then $|G| \geq 6$.

Later we will see that the above proposition is rather weak. In fact, it will become easy later to sho that if $(G, \cdot)$ is a nonabelian group then $|G| \geq 6$ by simply writing down all groups of finite order.

• Proof: Let $(G, \cdot)$ be a nonabelian group. Then there exists $a, b \in G$ such that $a \cdot b \neq b \cdot a$.

• Let $e \in G$ be the identity element of $G$. Note that $a \neq e$, otherwise $a \cdot b \neq b \cdot a$ implies that $b \neq b$. Similarly, note that $b \neq e$, otherwise $a \cdot b \neq b \cdot a$ implies $a \neq a$. Lastly note that $a \neq b$, otherwise $a \cdot b \neq b \cdot a$ implies that $a^2 \neq a^2$. So $a$, $b$, and $e$ are distinct elements of $G$. So immediately $|G| \geq 3$.

• Now consider the elements $a \cdot b, b \cdot a \in G$.

• First, suppose that $a \cdot b, b \cdot a \in \{ e, a, b \}$. Note that $a \cdot b \neq a$, otherwise $b = e$. Similarly, $a \cdot b \neq b$, otherwise $a = e$. So $a \cdot b = e$. A similar argument shows that $b \cdot a = e$. But this is a contradiction, since $a \cdot b \neq b \cdot a$.

• So the set $\{ e, a, b, a \cdot b, b \cdot \} \subset G$ consists of $5$ distinct elements. Thus $|G| \geq 5$.

• Lastly, it is easy to check that there is only one group of order $5$, namely the group $\mathbb{Z}/5\mathbb{Z} = \mathbb{Z}_5$, which is abelian. Thus if $G$ is a nonabelian group then $|G| \geq 6$. $\blacksquare$