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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:

$$R^{2^n}=1$$ $$S^2=-1$$



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?

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@m.k. What do you mean by "overlap"? –  Pedro Tamaroff Mar 17 '13 at 17:14
Actually, never mind that. Notice that $R^{2^{n-1}} = S^2 = -1$. What you have shown is that the group has at most $2^{n+2}$ elements. The exact amount of elements is $2^{n+1}$. –  Mikko Korhonen Mar 17 '13 at 17:19
It is very similar to $D_{2^n}$, the significant difference being that $S^2=-1$ instead of $S^2=1$. This has the effect that the elements outside of the cyclic subgroups generated by $R$ all have order 4, whereas in $D_n$ they all have order 2. In fact this construction works for any even value ($\ge 4$) of the order of $\omega$ - it does not have to be a power of 2. –  Derek Holt Mar 17 '13 at 18:04
The group that is defined as $\mathbb{H}_{n-1}$ here is often denoted by $Q_{2^n}$. You can find similarities between $D_{2^n}$ and $Q_{2^n}$ here (theorems 4.1 and 4.2). One property that is not mentioned there is the fact that $\operatorname{Aut}(D_{2^n}) \cong \operatorname{Aut}(Q_{2^n})$ for $n \geq 4$. –  Mikko Korhonen Mar 17 '13 at 18:29
Also $Q_{2^{(n+1)}}/\langle S^2\rangle \cong D_{2^n}$ –  Jack Schmidt Mar 4 at 19:56

2 Answers 2

up vote 3 down vote accepted

This group can be defined "purely in group-theoretic terms":

$$ R^{2^n}=1, \ R^{2^{n-1}} = S^2, \ S^{-1}RS=R^{-1} $$ ($S^{-1}R=R^{-1}S^{-1}$ is unnecessary). This group is described in W.Burnside, Theory of groups of finite order, sect.105, Theorem VI.

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Thanks for the reference. –  Pedro Tamaroff Mar 17 '13 at 18:08

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.

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They are also the only finite non-cyclic $p$-groups with a unique subgroup of order $p$. I wonder how this is related to the fact you mention. –  Mikko Korhonen Mar 17 '13 at 18:47
@m.k. A good question! IIRC in classifying fixed-point-free groups the fact that all their abelian subgroups must be cyclic is used heavily, and might provide the bridge (it is easy to see that for an abelian group to have a fixed-point-free representation it is necessary and sufficient for that the group to be cyclic). –  Jyrki Lahtonen Mar 17 '13 at 19:24
I guess that's it. If $G$ is a finite $p$-group where every abelian subgroup of $G$ is cyclic, then $G$ has a unique subgroup of order $p$. Proof: let $H \leq G$ be central of order $p$. If $K \leq G$ has order $p$ and $H \neq K$, then $HK$ is a subgroup of $G$ isomorphic to $C_p \times C_p$. –  Mikko Korhonen Mar 17 '13 at 21:15
And the converse (if $G$ is a finite $p$-group with a unique subgroup of order $p$, then every abelian subgroup of $G$ is cyclic) can be proven with the classification of finite abelian groups (there might be an easier way to do this). –  Mikko Korhonen Mar 17 '13 at 21:47

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