Here are two ways to tell people about the quaternion group:
$\{\,1,i,j,k,-1,-i,-j,-k\,\}$ with $ij=k$, $jk=i$, $ki=j$, $ji=-k$, $kj=-i$, $ik=-j$, $i^2=j^2=k^2=-1$.
$\{\,1,a,a^2,a^3,b,ab,a^2b,a^3b\,\}$ with $a^4=1$, $b^2=a^2$, $ba=a^3b$.
You can see they describe the same group by, for example, using the relations in the first description to show that the elements can also be given as
$\{\,1,i,i^2,i^3,j,ij,i^2j,i^3j\,\}$ with $i^4=1$, $j^2=i^2$, and $ji=i^3j$.
Now your challenge is to do the same thing with a way of listing the elements as 2-by-2 matrices.
As for the question of why use this matrix and not that matrix, my advice is to try both ways and see what happens (and feel free to report back here on your findings).
Look at the orders of the elements of $Q$:
- $1$ has order $1$
- $-1$ has order $2$
- the other elements all have order $4$
Up to isomorphism there are only two types of subgroups of order $4$, namely $\mathbb Z_4$ or $\mathbb Z_2 \times \mathbb Z_2$. The latter has three elements of order $2$, which is impossible in $Q$. So any subgroup of order $4$ must be cyclic. By inspection, the only possibilities are $\langle i \rangle$, $\langle j \rangle$, and $\langle k \rangle$.
Edit to sketch a proof of the statement about the possible types of group of order $4$:
Suppose $G$ is a group of order $4$. If $G$ contains an element of order $4$, then $G$ is cyclic, hence isomorphic to $\mathbb Z_4$.
The only other possibility is that $G$ does not contain an element of order $4$, in which case all of its non-identity elements must have order $2$ (Lagrange). Let $a$ and $b$ be two such elements.
Then $A = \langle a \rangle$ and $B = \langle b \rangle$ are subgroups of order $2$, hence index $2$, hence they are both normal.
Thus $AB$ is also a subgroup, and it contains $a$, $b$, and the identity, so its order is at least $3$. By Lagrange, its order must be $4$, hence $G = AB$. Since $A$ and $B$ are both normal subgroups and $A \cap B = 1$, this is a direct product.
Thus we have exposed $G$ as the direct product $A \times B$. As $A$ and $B$ have order $2$, they are both isomorphic to $\mathbb Z_2$, which shows that $G$ is isomorphic to $\mathbb Z_2 \times \mathbb Z_2$ as claimed.
Best Answer
First of all, I think your desire to remember those $2 \times 2$ matrices is misguided. If you need to know it, then look it up. What should be learned well is the multiplication rules among $i$, $j$, and $k$. That is what everyone knows who has studied quaternions.
I think a better question than "how to remember" those $2 \times 2$ matrices is "where do such matrices comes from"? It is nearly the same question but gets at a more mathematically interesting issue: how would such a matrix representation be discovered? It just sounds like a more compelling issue than "how to remember?" What we are trying to do with the representation as matrices is turn quaternions into a subgroup of ${\rm GL}_2(\mathbf C)$, so we want to make quaternions look like $\mathbf C$-linear maps. Now that is tricky because $\mathbf C$, in its natural embedding into the quaternions, does not commute with all quaternions (unlike the real numbers inside the quaternions). But it can be done and there is a method involved that helps in other situations.
First, let's back up and explain why complex numbers can be viewed as $2 \times 2$ real matrices: associate to $z = a + bi$ in $\mathbf C$ (so $a, b$ are real) the mapping $m_z \colon \mathbf C \to \mathbf C$ given by multiplication by $z$: $m_z(w) = zw$. The function $m_z$ is an $\mathbf R$-linear map, and using the $\mathbf R$-basis $\{1,i\}$ of $\mathbf C$ we can represent $m_z$ as a $2 \times 2$ real matrix by expressing $m_z(1)$ and $m_z(i)$ in terms of the basis $\{1,i\}$: $$ m_z(1) = z = a + bi, \ \ m_z(i) = zi = -b + ai \Longrightarrow [m_z] = \begin{pmatrix}a&-b\\b&\ a\end{pmatrix}, $$ where $[m_z]$ is the matrix representation for $m_z$ with respect to the (ordered) $\mathbf R$-basis $\{1,i\}$ of $\mathbf C$.
Now let's try to do the same thing for the division ring of quaternions $\mathbf H = \mathbf R + \mathbf R i + \mathbf R j + \mathbf R k$. We will view $\mathbf H$ as a complex vector space, but we have to be careful to decide whether to make $\mathbf H$ a left $\mathbf C$-vector space or a right $\mathbf C$-vector space.
Method 1. Let's treat $\mathbf H$ as a left $\mathbf C$-vector space: $$ a + bi + cj + dk = (a + bi) + (c+di)j \in \mathbf C + \mathbf C j. $$ Should we associate to $q \in \mathbf H$ the map $\ell_q \colon \mathbf H \to \mathbf H$ given by left multiplication by $q$ ($\ell_q(s) = qs$) or the map $r_q \colon \mathbf H \to \mathbf H$ given by right multiplication by $q$ ($r_q(s) = sq$)? Left multiplication is not good because it is not $\mathbf C$-linear when we view $\mathbf H$ as a left $\mathbf C$-vector space: $\ell_q(zs) = q(zs) \not= z(qs) = z\ell_q(s)$ in general for complex $z$. But right multiplication by $q$ is $\mathbf C$-linear when $\mathbf H$ is a left $\mathbf C$-vector space thanks to associativity of multiplication: $r_q(zs) = (zs)q = z(sq) = zr_q(s)$.
If we associate to $q \in \mathbf H$ the mapping $r_q \colon \mathbf H \to \mathbf H$, then this is additive in $q$ ($r_{q+q'} = r_q + r_{q'}$), but it is not multiplicative: $r_q(r_{q'}(s)) = (sq')q = s(q'q) = r_{q'q}(s)$, so $r_q \circ r_{q'} = r_{q'q}$. The multiplication got swapped. So we will use conjugation on $\mathbf H$ on top of using right multiplication: associate to $q$ the map "right multiplication by $\overline{q}$": let $m_q \colon \mathbf H \to \mathbf H$ by $m_q(s) = s\overline{q}$. Then $m_q$ is $\mathbf C$-linear since $$ m_q(s + s') = (s+s')\overline{q} = s\overline{q} + s'\overline{q} = m_q(s) + m_q(s') $$ and $$ m_q(zs)= (zs)\overline{q} = z(s\overline{q}) = zm_q(s). $$
Moreover, $m_q$ is additive and multiplicative in $q$: for $q, q' \in \mathbf H$ and $s \in \mathbf H$, $$ (m_q + m_{q'})(s) = m_q(s) + m_{q'}(s) = s\overline{q} + s\overline{q'} = s(\overline{q} + \overline{q'}) = s\overline{q+q'} = m_{q+q'}(s) $$ and $$ (m_q \circ m_{q'})(s) = m_q(m_{q'}(s)) = m_q(s\overline{q'}) = (s\overline{q'})\overline{q} = s(\overline{q'} \ \overline{q}) = s\overline{qq'} = m_{qq'}(s). $$ Therefore $m_{q+q'} = m_q + m_{q'}$ and $m_{qq'} = m_q \circ m_{q'}$. That proves $q \mapsto m_q$ is a ring homomorphism $\mathbf H \to {\rm End}_{\mathbf C}(\mathbf H)$. It is injective since $m_q(1) = \overline{q}$, so we can recover $q$ from $m_q$. Therefore $q \mapsto m_q$ is an embedding of $\mathbf H$ as a subring of ${\rm End}_{\mathbf C}(\mathbf H)$.
For $q = a + bi + cj + dk = (a + bi) + (c + di)j$, we can turn $m_q$ into a $2 \times 2$ complex matrix by using a $\mathbf C$-basis of $\mathbf H$ as a (left) $\mathbf C$-vector space. We'll use the basis $\{1,j\}$: $$ m_q(1) = \overline{q} = a - bi - cj - dk = (a-bi) + (-c - di)j $$ and $$ m_q(j) = j\overline{q} = aj + bk + c - di = c - di + (a + bi)j. $$ Therefore the representation of $m_q$ as a matrix in ${\rm M}_2(\mathbf C)$ using the (ordered) basis $\{1,j\}$ is $$ [m_q] = \begin{pmatrix} a-bi&c-di\\ -c-di&a+bi \end{pmatrix}. $$ For example, $$ [m_i] = \begin{pmatrix} -i&0\\ 0&i \end{pmatrix}, \ \ \ [m_j] = \begin{pmatrix} \ 0&1\\ -1&0 \end{pmatrix}, \ \ \ [m_k] = \begin{pmatrix} \ 0 &-i\\ -i& \ 0 \end{pmatrix}. $$ Comparing this $2 \times 2$ complex matrix representation of $\mathbf H$ to the one in your question, it is not the same.
Method 2. Let's view $\mathbf H$ as a right $\mathbf C$-vector space, so $z \cdot q := qz$ for quaternions $q$ and $z \in \mathbf C$. Then the standard notation $q = a + bi + cj + dk$ looks in the basis $\{1,j\}$ like $$ q = (a + bi) + j(c - di). $$ The left multiplication mapping $\ell_q(s) = qs$ is $\mathbf C$-linear in this new sense (multiplication by $\mathbf C$ is on the right!) and $\ell_{q + q'} = \ell_q + \ell_{q'}$ while $\ell_{qq'} = \ell_q\ell_{q'}$, and $q \mapsto \ell_q$ is injective since $\ell_q(1) = q$.
Since $\ell_q(1) = q = (a + bi) + j(c - di)$ and $\ell_q(j) = qj = -(c+di) + (a+bi)j = -(c+di) + j(a-bi)$, the representation of $\ell_q$ as a matrix in ${\rm M}_2(\mathbf C)$ using the (ordered) basis $\{1,j\}$ is $$ [\ell_q] = \begin{pmatrix} a+bi&-c-di\\ c-di&a-bi \end{pmatrix}. $$ For example, $$ [\ell_i] = \begin{pmatrix} i&\ 0\\ 0&-i \end{pmatrix}, \ \ \ [\ell_j] = \begin{pmatrix} 0&-1\\ 1&\ 0 \end{pmatrix}, \ \ \ [\ell_k] = \begin{pmatrix} \ 0 &-i\\ -i& \ 0 \end{pmatrix}. $$ Comparing this $2 \times 2$ complex matrix representation of $\mathbf H$ to the one in your question, it is (again) not the same.
Method 3. Let's go back to viewing $\mathbf H$ as a left $\mathbf C$-vector space, so each quaternion is written as $z + wj$ with $z, w \in \mathbf C$. For $q \in \mathbf H$, let $f_q \colon \mathbf H \to \mathbf H$ by $$ f_q(z + wj) = q(\overline{z} + \overline{w}j). $$ This mapping $f_q$ is $\mathbf C$-linear: it is easily seen to be an additive function on $\mathbf H$, and for $\alpha \in \mathbf C$, $$ f_q(\alpha(z + wj)) = f_q((\alpha z) + (\alpha w)j) = q(\overline{\alpha z} + \overline{\alpha w}j) = q(\overline{\alpha}\,\overline{z} + \overline{\alpha}\,\overline{w}j) = q\overline{\alpha}(\overline{z} + \overline{w}j). $$ For $q \in \mathbf H$ and $\alpha \in \mathbf C$, $\alpha q = q\overline{\alpha}$, so $$ f_q(\alpha(z + wj)) = \alpha q(\overline{z} + \overline{w}j) = \alpha f_q(z+wj). $$ That completes the proof that $f_q$ is $\mathbf C$-linear.
It is left to you to check that $f_{q+q'} = f_q + f_q$ and $f_{qq'} = f_q \circ f_{q'}$ as functions $\mathbf H \to \mathbf H$. Therefore $q \mapsto f_q$ is a ring homomorphism $\mathbf H \to {\rm End}_{\mathbf C}(\mathbf H)$ and it is injective since $f_q(1) = q$. By picking a $\mathbf C$-basis of $\mathbf H$ (as a left $\mathbf C$-vector space), we can turn this homomorphism into an embedding $\mathbf H \to {\rm M}_2(\mathbf C)$. Using the $\mathbf C$-basis $\{1,j\}$, write $q = a+bi + cj + dk = (a+bi) + (c+di)j$ and then $$ [f_q] = \begin{pmatrix} a+bi&-c-di\\ c+di&a+bi \end{pmatrix}. $$ For example, $$ [f_i] = \begin{pmatrix} i&0\\ 0&i \end{pmatrix}, \ \ \ [f_j] = \begin{pmatrix} 0&-1\\ 1&\ 0 \end{pmatrix}, \ \ \ [f_k] = \begin{pmatrix} 0 &-i\\ i& \ 0 \end{pmatrix}. $$ This is the matrix realization of quaternions in your question. But notice how it corresponds to the rather "peculiar" $\mathbf C$-linear mapping $f_q(z+wj) = q(\overline{z} + \overline{w}j)$. That is part of what makes it pointless to try to "remember" a specific $2 \times 2$ matrix representation of $\mathbf H$. There are many "natural" ways to do this and it's hopeless to try to remember what the results are. If you need a matrix representation, look it up.