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, $\paren 1$ and $\paren 2$, 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 $-a$, $\set {\pm a} = \set {a, -a}$ for simplicity. === Lemma 1 === {{:Non-Abelian Order 8 Group with One Order 2 Element is Quaternion Group/Lemma 1}} By Lagrange's Theorem and Cosets are Equivalent, let: :$G / \set {\pm 1} = \set {\set {\pm 1}, \set {\pm a}, \set {\pm b}, \set {\pm c} }$ === Lemma 2 === {{:Non-Abelian Order 8 Group with One Order 2 Element is Quaternion Group/Lemma 2}} Now draw up the incomplete Cayley table for $G / \set {\pm 1}$: :$\begin{array}{c|cccc} & \pm 1 & \pm a & \pm b & \pm c \\ \hline \pm 1 & \pm 1 & \pm a & \pm b & \pm c \\ \pm a & \pm a & \pm 1 & & \\ \pm b & \pm b & & \pm 1 & \\ \pm c & \pm c & & & \pm 1 \\ \end{array}$ 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 the 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: :$\tuple {1, a, b, c} = \tuple {1, c', b', a'} = \tuple {\mathbf 1, \mathbf i, \mathbf j, \mathbf k}$ {{qed}}