Non-Abelian Group has Order Greater than 4/Proof 2
Theorem
Let $\struct {G, \circ}$ be a non-abelian group.
Then the order of $\struct {G, \circ}$ is greater than $4$.
Proof
It follows from Trivial Group is Cyclic Group and Prime Group is Cyclic that groups of order less than $4$ are cyclic.
Therefore, by Cyclic Group is Abelian, all groups of order less than $4$ are abelian.
Let $G$ have order $4$.
From Order of Element Divides Order of Finite Group, every element of $G$ has order that divides $4$.
Therefore any element of $G$ has order $1$, $2$ or $4$.
Let $g \in G$ have order $4$.
Then $G$ is generated by $g$.
Therefore $G$ is cyclic, and therefore abelian by Cyclic Group is Abelian.
Let $G$ be such that it has no element of order $4$.
From Identity is Only Group Element of Order 1, all the elements apart from its identity must have order $2$.
That is, $G$ is a Boolean group.
By Boolean Group is Abelian, $G$ is abelian.
The result follows.
$\blacksquare$
Sources
- 1978: Thomas A. Whitelaw: An Introduction to Abstract Algebra ... (previous) ... (next): $\S 44.3$ Some consequences of Lagrange's Theorem