TITLE: The generalized quaternion group $\Bbb H_n$
QUESTION [14 upvotes]: Let $\omega$ be a $2^n$-th primitive root of unity. Let $$R=\begin{pmatrix}\omega & 0\\0&\omega^{-1}\end{pmatrix}$$ and $$S=\begin{pmatrix}0&-1\\1&0\end{pmatrix}$$
Define the subgroup $\langle S,R\rangle =\Bbb H_n$ to be the generalized quaternion group, in $\rm{SL}(2,\Bbb C)$. I have to find the order and list all elements of $\Bbb H_n$. 
Observe that:
$$R^{2^n}=1$$
$$S^2=-1$$
$$SR=R^{-1}S$$
$$S^{-1}R=R^{-1}S^{-1}$$
We can put the last two as $SRSR=RSRS=-1$. I think this suffices to find the order and list the elements. Indeed,  given any string, we can get all the $S$ to the right of the string and all the $R$ consequently to the left using the $3^{\rm rd}$ and $4^{\rm th}$ relations, ending up with something of the form 
$$R^jS^i$$
CORRECTED Now, since $\omega$ is a primitive $2^n$-th root of unity, we can let $j$ range over $1\leq j\leq 2^n$ to get all possible powers $R^j$ for $j\in \Bbb Z$. On the other hand since $S^2=-1,S^3=-S,S^4=1$, and $R^{2^{n-1}}=-1,R^{2^n}=1$, we must restrict $i$ to $0,1$. Thus the order is $2\times 2^n=2^{n+1}$.
NOTE The relations above look pretty similar to those in $D_n$; namely for $R$ a rotation of $2\pi /n$ radians and $S$ a reflection, we have $R^n=1$, $S^2=1$, $SRS=R^{-1}$ and also $D_n=\langle R,S\rangle $. Any comment on this?

REPLY [4 votes]: Yes. You got it right. All the elements of this group are of the form
$$
\left(\begin{array}{cc}\omega^j&0\\0&\omega^{-j}\end{array}\right)\qquad\text{or}\qquad \left(\begin{array}{cc}0&-\omega^j\\\omega^{-j}&0\end{array}\right),
$$
with $0\le j<2^n$.
A curious property of these groups is that they are the only non-cyclic $p$-groups with a fixed-point-free representation (i.e. a group of matrices such that the neutral element is the only one that has one as an eigenvalue). Consequently they play a role in Zassenhaus' classification of fixed-point-free groups and finite near-fields.