Non-Abelian Order 8 Group has Order 4 Element

Theorem

Let $G$ be a non-abelian group of order $8$.

Then $G$ has at least one element of order $4$.

Proof

Let $e \in G$ be the identity of $G$.

Let $g \in G$ be an arbitrary element of $G$ such that $g \ne e$.

From Identity is Only Group Element of Order 1, only $e$ has order $1$.

Thus from Order of Element Divides Order of Finite Group:

$\order g \in \set {2, 4, 8}$

Suppose $\order g = 8$.

Then $G$ is cyclic.

So by Cyclic Group is Abelian, $G$ would be abelian.

So $g$ cannot be of order $8$.

Suppose all elements of $G \setminus \set e$ are of order $2$.

Then by definition $G$ is a Boolean group.

By Boolean Group is Abelian, $G$ would be abelian.

So the only option left is for at least one element of $G$ to be of order $4$.

$\blacksquare$

Sources

• 1978: Thomas A. Whitelaw: An Introduction to Abstract Algebra ... (previous) ... (next): Chapter $7$: Cosets and Lagrange's Theorem: Exercise $18$