How to remember elements of Quaternion group $Q_8$ Quaternion group acts a great example. But how to remember this? Any shortcut? i know this is silly question but just wanted to know. Each 2x2 matrix element is represented by $\{ 1,-1,i,-i,j,-j,k,-k\}$ What i am asking is how to remember matrix representation correspondence with 1, i, j, k

 A: 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.
A: The common mnemonic for remembering quaternion multiplication is to write $i,j,k$ clockwise around a circle.
To multiply two consecutive elements in the clockwise order, you get the third element not chosen. If you multiply consecutive elements in counterclockwise order, you get $-1$ times the third.

to remember matrix representation

Well, the matrix representation isn’t unique.  Find one you like, and you only have to remember three matrices for $i,j,k$.  Remembering three things does not really merit a special trick...  You could even get away with just memorizing $i,j$ and multiply them to get $k$.
The standard representation of quaternions as 2 by 2 matrices looks like this
$\begin{bmatrix}a&b\\-\bar{b}&\bar{a}\end{bmatrix}$.  The three matrices can be all that form.
A: Let me reveal you my secret: The quaternion group $Q_8$ is the four dimensional analogue of $2$-dimensional complex unit group $$\{\pm1, \pm i : i^2=-1\}$$ with two additional complex directions $j$ and $k.$ Moreover it is not commutative but anti-commutative. Altogether it is $$\{\pm1, \pm i, \pm j, \pm k : i^2=j^2=k^2=ijk=-1\}.$$ Note that using $ijk=-1$ with other relations you can prove that $$ij=-ji=k,\qquad jk=-kj=i,\qquad ki=-ik=j.$$ Put them in a clockwise cycle make an easier to memorize all these at onece.
A: Here's a trick you can use to remember the matrix correspondence you give.
On the one hand, multiplying $\mathbb{C}^2$ by $i$ has matrix $[\begin{smallmatrix} i&0\\0&i\end{smallmatrix}]$ ("obviously").
On the other hand, multiplying by $i$ is an anticlockwise $90^{\circ}$ rotation in the complex plane $\mathbb{C}\cong\mathbb{R}^2$, and you should memorize the matrix of such a rotation is $[\begin{smallmatrix}0&-1\\1&0\end{smallmatrix}]$ (seriously, very useful); you can extend this to a linear operator on $\mathbb{C}^2$ and call it $j$ instead of $i$.
Then you can get $k$ from $k=ij$.
A: These are the representations you obtain if you consider elements $g \in Q_8$ as linear maps $g : \mathbb{C} \oplus \mathbb{C}j \to \mathbb{C} \oplus \mathbb{C}j$ with $g(a + bj) = ag + bgj$, where we have identified $1 \in \mathbb{C}$ with $e \in Q_8$ and $i \in \mathbb{C}$ with $i \in Q_8$ (and likewise for $\overline{e}$ and $\overline{i}$).
To determine the matrix of $j$ in this representation, it suffices to look at its action on basis elements.
\begin{align}
j(1 + 0j) &= 1j + 0j^2 = 1j + 0(\overline{e}) = 1j + 0 = 0 + 1j\\
j(0 + 1j) &= 0j + 1j^2 = 0j + 1(-1) = -1 + 0j
\end{align}
This means
$$
j \mapsto \begin{pmatrix} 0 & -1 \\ 1 & 0\end{pmatrix}
$$
This all might seem very complicated, but I'm just being pedantic. Another more compressed example might make things clearer.
We have $k(1) = 1k = k = ij$ and $kj = -i$. Thus
$$
k \mapsto \begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix}
$$
This is the same trick that is used to represent complex numbers as matrices in $M_2(\mathbb{R})$. A trick worth remembering if you ever study complex analysis.
