Find elements in $GL(2,\Bbb C)$ such that $A$ has order $8$, $B$ has order $4$ and $B^2=A^4, B^{-1}AB=A^{-1}$ Find elements in $GL(2,\Bbb C)$ such that $A$ has order $8$, $B$ has order $4$ and $B^2=A^4, B^{-1}AB=A^{-1}$.
So we have to basically find a group of order $16$ where $A$ has order $8$, $B$ has order $4$ and $B^2=A^4, B^{-1}AB=A^{-1}$.
Now I found out and it is also given in wiki, the representation of $Q_8$. Is there any way to extend those matrices?
 A: Consider a group of $2\times 2$ matrices containing two elements $A,B$ that satisfy $B^2=A^4$, $A^8=1$ with $A$ of order $8$, and $ABA=B$. Then we may and do assume that $A$ is diagonal, with diagonal entries $u,v$, where $u$ is a primitive $8$.th root of unit, and Galois theoretically we can even take it to be $(1+i)/\sqrt 2$. Let $a,b,c,d$ be the entries in $B$. Then
$$
ABA - B = 
\begin{bmatrix}
a(u^2-1) & b(uv-1)\\
c(uv-1) & d(v^2-1)
\end{bmatrix}
\ ,
$$
so we have $a=0$ first. There are now two cases (that do not lead to the immediate contraction $a=b=c=d=0$):


*

*The case $v^2=1$ implies $uv\ne 1$, so $b=c=0$, so $B$ is diagonal, so $A,B$ commute, this leads to an immediate contradiction, for instance $A^2=A^2BB^{-1}=ABAB^{-1}=BB^{-1}=1$.

*The case $uv=1$. Then we need $d=0$, so $a=d=0$. Because of $u^4=v^4=-1$, the equation $B^2=A^4$ becomes $B^2=-1$. This gives the only constraint $bc=-1$. We get this way the matrices:
$$
A 
=
\begin{bmatrix}
u &\\&v
\end{bmatrix}
\ ,
\qquad
B=
\begin{bmatrix}
 & b\\
-1/b & 
\end{bmatrix}
\ ,
$$
which verify.



Computer check:
sage: var('b');
sage: u, v = (1+i)/sqrt(2), 1/u
sage: u^8, v^8
(1, 1)
sage: A = matrix( 2, 2, [u, 0, 0, v] )
sage: B = matrix( 2, 2, [0, b, -1/b, 0] )
sage: A^8
[1 0]
[0 1]
sage: B^4
[1 0]
[0 1]

sage: A^4
[-1  0]
[ 0 -1]
sage: B^2
[-1  0]
[ 0 -1]

sage: B^2 - A^4
[0 0]
[0 0]
sage: A*B*A - B
[0 0]
[0 0]

A: $A = \left(\begin{array}{cc}\omega&0\\0&\omega^{-1}\end{array}\right)$,
$B = \left(\begin{array}{cc}0&1\\-1&0\end{array}\right)$, where $\omega$ is a primitive 8th root of $1$. 
