Quaternions
Definition. The quaternions, denoted by $Q_8$, is the set of matrices
\[\Set{ \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}, \begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix}, \begin{pmatrix} i & 0 \\ 0 & -i \end{pmatrix}, \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}, \begin{pmatrix} 0 & i \\ i & 0 \end{pmatrix}, \begin{pmatrix} -i & 0 \\ 0 & i \end{pmatrix}, \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}, \begin{pmatrix} 0 & -i \\ -i & 0 \end{pmatrix} }\]which forms a group under matrix multiplication.
We have $Q_8 \le \text{SU}_2 \le \text{SL}_2(\mathbb{C}) \le \text{GL}_2(\mathbb{C})$.
Property. $Q_8$ is not abelian.
Proof.
\[\begin{align*} \begin{pmatrix} i & 0 \\ 0 & -i \end{pmatrix} \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix} &= \begin{pmatrix} 0 & i \\ i & 0 \end{pmatrix} \\ \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix} \begin{pmatrix} i & 0 \\ 0 & -i \end{pmatrix} &= \begin{pmatrix} 0 & -i \\ -i & 0 \end{pmatrix} \end{align*}\]
Hence, it is not isomorphic to $C_8$, $C_4 \times C_2$ and $C_2 \times C_2 \times C_2$.
Proposition. $Q_8 \not \cong D_8$.
Proof.
There are three elements $\Set{r^2, s, sr^2}$ of order $2$ in $D_8$ but only one element $\begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix}$ of order $2$ in $Q_8$.
The rationale behind $Q_8$ can be found in the study of groups of order 8.
Generator
Proposition. $Q_8$ is generated by $a$ and $b$ satisfying $a^4 = e$, $b^2 = a^2$ and $bab^{-1} = a^{-1}$.
Proof.
The elements
\[a = \begin{pmatrix} i & 0 \\ 0 & -i \end{pmatrix} \quad \text{and} \quad b = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}\]satisfy the above condition.
Alternatively, we can picture $Q_8$ having $\Set{\pm 1}$ together with 3 axes which are anticommutative, i.e.
\[Q_8 = \Set{1, -1, i, -i, j, -j, k, -k}\]with
\[i^2 = j^2 = k^2 = -1 \quad \text{and} \quad ij = k, jk = i, ki = j \quad \text{but} \quad ji = -k, kj = -i, ik = -j\]Hence,
| Label | Generated By | Element | Order |
| $1$ | $e$ | $\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$ | 1 |
| $-1$ | $a^2 = b^2$ | $\begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix}$ | 2 |
| $i$ | $a$ | $\begin{pmatrix} i & 0 \\ 0 & -i \end{pmatrix}$ | 4 |
| $j$ | $b$ | $\begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}$ | 4 |
| $k$ | $ab$ | $\begin{pmatrix} 0 & i \\ i & 0 \end{pmatrix}$ | 4 |
| $-i$ | $a^3 = ab^2 = b^2a$ | $\begin{pmatrix} -i & 0 \\ 0 & i \end{pmatrix}$ | 4 |
| $-j$ | $b^3 = a^2b = ba^2$ | $\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$ | 4 |
| $-k$ | $ab^3 = a^3b$ | $\begin{pmatrix} 0 & -i \\ -i & 0 \end{pmatrix}$ | 4 |