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, $\left({1}\right)$ and $\left({2}\right)$, no elements in $G$ have order $8$, i.e. they all have order $1,2$ or $4$.

Let the identity element be $1$ and the one with order $2$ be $-1$.

Also denote $1 \circ a$ as $+a$, $-1 \circ a$ as $-a$, $\left\{ {\pm a}\right\} = \left\{ {a, -a}\right\}$ for simplicity.

Lemma 1

 * $\left\{ {\pm 1}\right\}$ is a normal subgroup of $G$.

By Lagrange's Theorem and Cosets are Equivalent, let $G/\left\{ {\pm 1}\right\} = \left\{ { \left\{ {\pm 1}\right\}, \left\{ {\pm a}\right\}, \left\{ {\pm b}\right\}, \left\{ {\pm c}\right\} }\right\}$.

Lemma 2
Now draw up the imcomplete Cayley Table for $G/\left\{ {\pm 1}\right\}$:
 * $\begin{array}{c|cccc}

& \pm 1 & \pm a & \pm b & \pm c \\ \hline \pm 1 & \pm 1 & \pm a & \pm b & \pm c \\ \pm a & \pm a & \pm 1 & & \\ \pm b & \pm b & & \pm 1 & \\ \pm c & \pm c & & & \pm 1 \\ \end{array}$

By Group has Latin Square Property the Cayley Table can be completed:
 * $\begin{array}{c|cccc}

& \pm 1 & \pm a & \pm b & \pm c \\ \hline \pm 1 & \pm 1 & \pm a & \pm b & \pm c \\ \pm a & \pm a & \pm 1 & \pm c & \pm b \\ \pm b & \pm b & \pm c & \pm 1 & \pm a \\ \pm c & \pm c & \pm b & \pm a & \pm 1 \\ \end{array}$

Now from the above draw up the incomplete Cayley Table for $G$:
 * $\begin{array}{c|cccccccc}

& 1 & -1 & a & -a & b & -b & c & -c \\ \hline 1 & 1 & -1 & a & -a & b & -b & c & -c \\ -1 & -1 & 1 & -a & a & -b & b & -c & c \\ a & a & -a & -1 & 1 & & & & \\ -a & -a & a & 1 & -1 & & & & \\ b & b & -b & & & -1 & 1 & & \\ -b & -b & b & & & 1 & -1 & & \\ c & c & -c & & & & & -1 & 1 \\ -c & -c & c & & & & & 1 & -1 \\ \end{array}$

By Group has Latin Square Property and $\left({G1}\right):$ Associativity, the Cayley Table can be completed in two ways:
 * $\begin{array}{c|cccccccc}

& 1 & -1 & a & -a & b & -b & c & -c \\ \hline 1 & 1 & -1 & a & -a & b & -b & c & -c \\ -1 & -1 & 1 & -a & a & -b & b & -c & c \\ a & a & -a & -1 & 1 & c & -c & -b & b \\ -a & -a & a & 1 & -1 & -c & c & b & -b \\ b & b & -b & -c & c & -1 & 1 & a & -a \\ -b & -b & b & c & -c & 1 & -1 & -a & a \\ c & c & -c & b & -b & -a & a & -1 & 1 \\ -c & -c & c & -b & b & a & -a & 1 & -1 \\ \end{array}$

or:
 * $\begin{array}{c|cccccccc}

& 1 & -1 & a' & -a' & b' & -b' & c' & -c' \\ \hline 1 & 1 & -1 & a' & -a' & b' & -b' & c' & -c' \\ -1 & -1 & 1 & -a' & a' & -b' & b' & -c' & c' \\ a' & a' & -a' & -1 & 1 & -c' & c' & b' & -b' \\ -a' & -a' & a' & 1 & -1 & c' & -c' & -b' & b' \\ b' & b' & -b' & c' & -c' & -1 & 1 & -a' & a' \\ -b' & -b' & b' & -c' & c' & 1 & -1 & a' & -a' \\ c' & c' & -c' & -b' & b' & a' & -a' & -1 & 1 \\ -c' & -c' & c' & b' & -b' & -a' & a' & 1 & -1 \\ \end{array}$

Referring to the Cayley Table of Quaternion Group:
 * $\begin{array}{r|rrrrrrrr}

& \mathbf 1 &  \mathbf i & -\mathbf 1 & -\mathbf i &  \mathbf j &  \mathbf k & -\mathbf j & -\mathbf k \\ \hline \mathbf 1 & \mathbf 1 &  \mathbf i & -\mathbf 1 & -\mathbf i &  \mathbf j &  \mathbf k & -\mathbf j & -\mathbf k \\ \mathbf i & \mathbf i & -\mathbf 1 & -\mathbf i &  \mathbf 1 &  \mathbf k & -\mathbf j & -\mathbf k &  \mathbf j \\ -\mathbf 1 & -\mathbf 1 & -\mathbf i & \mathbf 1 &  \mathbf i & -\mathbf j & -\mathbf k &  \mathbf j &  \mathbf k \\ -\mathbf i & -\mathbf i & \mathbf 1 &  \mathbf i & -\mathbf 1 & -\mathbf k &  \mathbf j &  \mathbf k & -\mathbf j \\ \mathbf j & \mathbf j & -\mathbf k & -\mathbf j &  \mathbf k & -\mathbf 1 &  \mathbf i &  \mathbf 1 & -\mathbf i \\ \mathbf k & \mathbf k &  \mathbf j & -\mathbf k & -\mathbf j & -\mathbf i & -\mathbf 1 &  \mathbf i &  \mathbf 1 \\ -\mathbf j & -\mathbf j & \mathbf k &  \mathbf j & -\mathbf k &  \mathbf 1 & -\mathbf i & -\mathbf 1 &  \mathbf i \\ -\mathbf k & -\mathbf k & -\mathbf j & \mathbf k &  \mathbf j &  \mathbf i &  \mathbf 1 & -\mathbf i & -\mathbf 1 \end{array}$

The results follow by identifying $\left({1, a, b, c}\right) = \left({1, c', b', a'}\right) = \left({\mathbf 1, \mathbf i, \mathbf j, \mathbf k}\right)$.