For $\mathbb{Q}_8$ to be a group:
1. it must be closed under it's binary operation.
2. It must be associative.
3. It must have an identity element.
4. Each element must have an inverse.
1. The Cayley Table below shows, that for all arbitrary $x \in \mathbb{Q}_8,$ and $y\in \mathbb{Q}_8,$ then $xy\in \mathbb{Q}_8,$. Thus $\mathbb{Q}_8$ is closed under multiplication.
2. Let's let $\mathbb{M}_8$ be the set of matrices:
(1)
\begin{align} M_1=\begin{pmatrix} -1 & 0 & 0 & 0\\ 0 & -1 & 0 & 0\\ 0 & 0 & -1 & 0\\ 0 & 0 & 0 & -1\\ \end{pmatrix} \end{align}
(2)
\begin{align} M_2=\begin{pmatrix} 0 & -1 & 0 & 0\\ 1 & 0 & 0 & 0\\ 0 & 0 & 0 & 1\\ 0 & 0 & 0-1 & 0\\ \end{pmatrix} \end{align}
(3)
\begin{align} M_3=\begin{pmatrix} 0 & 0 & -1 & 0\\ 0 & 0 & 0 & -1\\ 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ \end{pmatrix} \end{align}
(4)
\begin{align} M_4=\begin{pmatrix} 0 & 0 & 0 & -1\\ 0 & 0 & 1 & 0\\ 0 & -1 & 0 & 0\\ 1 & 0 & 0 & 0\\ \end{pmatrix} \end{align}
(5)
\begin{align} M_5=\begin{pmatrix} 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1\\ \end{pmatrix} \end{align}
(6)
\begin{align} M_6=\begin{pmatrix} 0 & 1 & 0 & 0\\ -1 & 0 & 0 & 0\\ 0 & 0 & 0 & -1\\ 0 & 0 & 1 & 0\\ \end{pmatrix} \end{align}
(7)
\begin{align} M_7=\begin{pmatrix} 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1\\ -1 & 0 & 0 & 0\\ 0 & -1 & 0 & 0\\ \end{pmatrix} \end{align}
(8)
\begin{align} M_8=\begin{pmatrix} 0 & 0 & 0 & 1\\ 0 & 0 & -1 & 0\\ 0 & 1 & 0 & 0\\ -1 & 0 & 0 & 0\\ \end{pmatrix} \end{align}
Let us set up an bijection between $\mathbb{Q}_8$ and $\mathbb{M}_8$, and define it thus:
(9)
\begin{align} -1 \leftrightarrow M_1 -i \leftrightarrow M_2 -j \leftrightarrow M_3 -k \leftrightarrow M_4 1 \leftrightarrow M_5 i \leftrightarrow M_6 j \leftrightarrow M_7 k \leftrightarrow M_8 \end{align}
The Two Cayley Tables below show the isomorphism between $\mathbb{Q}_8$ and $\mathbb{M}_8$
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$Q_8$ |
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$M_8$ |
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$-1$ |
$-i$ |
$-j$ |
$-k$ |
$1$ |
$i$ |
$j$ |
$k$ |
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$M_1$ |
$M_2$ |
$M_3$ |
$M_4$ |
$M_5$ |
$M_6$ |
$M_7$ |
$M_8$ |
$-1$ |
$1$ |
$i$ |
$j$ |
$k$ |
$-1$ |
$-i$ |
$-j$ |
$-k$ |
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$M_1$ |
$M_5$ |
$M_6$ |
$M_7$ |
$M_8$ |
$M_1$ |
$M_2$ |
$M_3$ |
$M_4$ |
$-i$ |
$i$ |
$-1$ |
$k$ |
$-j$ |
$-i$ |
$1$ |
$-k$ |
$j$ |
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$M_2$ |
$M_6$ |
$M_1$ |
$M_8$ |
$M_3$ |
$M_2$ |
$M_5$ |
$M_4$ |
$M_7$ |
$-j$ |
$j$ |
$-k$ |
$-1$ |
$i$ |
$-j$ |
$k$ |
$1$ |
$-i$ |
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$M_3$ |
$M_7$ |
$M_4$ |
$M_1$ |
$M_6$ |
$M_3$ |
$M_8$ |
$M_5$ |
$M_2$ |
$-k$ |
$k$ |
$j$ |
$-i$ |
$-1$ |
$-k$ |
$-j$ |
$i$ |
$1$ |
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$M_4$ |
$M_8$ |
$M_7$ |
$M_2$ |
$M_1$ |
$M_4$ |
$M_3$ |
$M_6$ |
$M_5$ |
$1$ |
$-1$ |
$-i$ |
$-j$ |
$-k$ |
$1$ |
$i$ |
$j$ |
$k$ |
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$M_5$ |
$M_1$ |
$M_2$ |
$M_3$ |
$M_4$ |
$M_5$ |
$M_6$ |
$M_7$ |
$M_8$ |
$i$ |
$-i$ |
$1$ |
$-k$ |
$j$ |
$i$ |
$-1$ |
$k$ |
$-j$ |
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$M_6$ |
$M_2$ |
$M_5$ |
$M_4$ |
$M_7$ |
$M_6$ |
$M_1$ |
$M_8$ |
$M_3$ |
$j$ |
$-j$ |
$k$ |
$1$ |
$-i$ |
$j$ |
$-k$ |
$-1$ |
$i$ |
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$M_7$ |
$M_3$ |
$M_8$ |
$M_5$ |
$M_2$ |
$M_7$ |
$M_4$ |
$M_1$ |
$M_6$ |
$k$ |
$-k$ |
$-j$ |
$i$ |
$1$ |
$k$ |
$j$ |
$-i$ |
$-1$ |
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$M_8$ |
$M_4$ |
$M_3$ |
$M_6$ |
$M_5$ |
$M_8$ |
$M_7$ |
$M_2$ |
$M_1$ |
Now, since the two Cayley Tables show an isomorphism between $\mathbb{Q}_8$ and $\mathbb{M}_8$, and since matrix multiplication is associative, we know that $\mathbb{Q}_8$ is associative.
3. Again from either the Cayley Table or, from multiplication in $\mathbb{R}$, the identity element for $\mathbb{Q}_8$ is 1.
4. Either from the Cayley Table, or from the isomorphism with $\mathbb{M}_8$, it can be shown that the following inverse relationships exist.
$1=e$
$-1$ is inverse to itself
$i$ is inverse to $-i$
$j$ is inverse to $-j$
$k$ is inverse to $-k$
So,
$\mathbb{Q}_8$ is closed under multiplication, it is associative, it contains the identity element, and each element in it has an inverse. Thus,
$\mathbb{Q}_8$ is a group!