Non-Abelian Order 8 Group with One Order 2 Element is Quaternion Group

Theorem
Let $G$ be a group with the following properties:


 * $(1): \quad G$ is non-abelian.


 * $(2): \quad G$ is of order $8$.


 * $(3): \quad G$ has precisely one element of order $2$.

Then $G$ is isomorphic to the quaternion group $Q$.

Proof
From Order of Element Divides Order of Finite Group all the elements in $G$ have order $1,2,4$ or $8$.

From Cyclic Group is Abelian no elements in $G$ have order $8$, i.e. they all have order $1,2$ or $4$.

Let the identity element be $e$ and the one with order $2$ be $x$.

Then, $\left\{e,x\right\}$ is a subgroup of $G$.