Contents

group theory

# Contents

## Idea

The quaternion group of order 8, $Q_8$, is the finite subgroup of SU(2) $Q_8 \subset SU(2) \simeq S^3 \subset \mathbb{H}$ of unit quaternions which consists of the canonical four basis-quaternions and their negatives:

$Q_8 \;=\; \big\{ \pm 1, \, \pm i, \, \pm j, \, \pm k \big\} \,.$ This is isomorphic to the binary dihedral group of the same order $Q_8 \simeq 2 D_4$. As such, the Dynkin diagram that corresponds to $Q_8$ under the ADE-classification of finite subgroups of SU(2) is D4, the triality-invariant one.

graphics grabbed from Wikipedia here

This order-8 quaternion group $Q_8$ is the first in a row of generalized quaternion groups, $Q_{2^n}$, which are also examples of dicyclic groups, which class forms part of an even larger family. We will treat both general dicyclic groups and the specific example of the quaternion group together.

## Definitions

The dicyclic of order $4n$, $n\geq 2$, is the group $Dic_n$ defined by the presentation $\langle x,y | x^{2n}= x^{n} y^{-2}=y^{-1}x y x=1\rangle$.

The quaternion group (of order 8) is then $Dic_n$ for $n=2$.

The generalised quaternion group of order $2^{k+1}$ is $Dic_n$ with $n= 2^{k-1}$.

## Properties

### General

• $Dic_n$ is (isomorphic to) a finite subgroup of $\mathbb{H}^\ast$ as can be seen by taking generators $x=j$ and $y=\cos(\pi/n) + i\sin(\pi/n)$. For $n=2$ this simply yields the subgroup generated by $i$ and $j$.

• $Dic_n$ has another presentation as $\langle R, S, T | R^2=S^2=T^n=R S T\rangle$. $R S T$ as a power of each of the generators is central and $Dic_n/\langle R S T\rangle= D_{2n}$, where $D_{2n}=\langle R, S, T | R^2=S^2=T^n=R S T=1\rangle$ is the dihedral group of order $2n$.

• $Q_8$ is a Hamiltonian group i.e. a non-abelian group such that every subgroup is normal. Moreover, a general structure theorem for Hamiltonian groups by Baer (1933) says that every Hamiltonian group has a direct product group-decomposition containing $Q_8$ as a factor hence, in particular, every Hamiltonian group contains $Q_8$ as a subgroup! (cf. Scott (1987, p.253))

• $Q_8$ is the multiplicative part of the quaternionic near-field $J_9$. (cf. Weibel (2007))

###### Proposition

(inclusion of $Q_8$ into finite subgroups of SU(2))

Among the finite subgroups of SU(2) (hence among all “finite quaternion groups”) the quaternion group of order 8, $Q_8$ is a proper subgroup precisely of the three exceptional cases:

(e.g. Koca-Moc-Koca 16, p. 8, pointing to Coxeter-Moser 65 and Coxeter 73)

### Character table

linear representation theory of binary dihedral group $2 D_4$

$=$ dicyclic group $Dic_2$ $=$ quaternion group $Q_8$

$\,$

group order: ${\vert 2D_4\vert} = 8$

conjugacy classes:124A4B4C
their cardinality:11222

$\,$

splitting field$\mathbb{Q}(\alpha, \beta)$ with $\alpha^2 + \beta^2 = -1$
field generated by characters$\mathbb{Q}$

character table over splitting field $\mathbb{Q}(\alpha,\beta)$/complex numbers $\mathbb{C}$

irrep124A4B4CSchur index
$\rho_1$111111
$\rho_2$11-11-11
$\rho_3$111-1-11
$\rho_4$11-1-111
$\rho_5$2-20002

character table over rational numbers $\mathbb{Q}$/real numbers $\mathbb{R}$

irrep124A4B4C
$\rho_1$11111
$\rho_2$11-11-1
$\rho_3$111-1-1
$\rho_4$11-1-11
$\rho_5 \oplus \rho_5$4-4000

References

### Matrix representation

There are lots of different ways of defining $Q:=Q_8$. One is that it is the subgroup of $Gl(2,\mathbb{C})$ generated by the matrices

$\xi = \left(\array{i&0\\0&-i}\right)$

and

$\eta =\left(\array{0&-1\\1&0}\right).$

In this form it is a nice exercise to derive a presentation of $Q_8$. Clearly $\xi^4=1$ and $\eta$ is not in $\langle \xi\rangle$ as is easiy checked, so the order of this group must be at least 8.

We note that $\eta^2 = \xi^2$ and that $\eta \xi \eta^{-1}= \xi^{-1}$, so a guess for a presentation would be

$\langle x,y : x^4=1, y^2=x^2, y x y^{-1}=x^{-1}\rangle.$

Let us call $G$ the group presented by this presentation, then there is an obvious epimorphism from $G$ to $Q$ sending $x$ to $\xi$ and $y$ to $\eta$. This is an isomorphism as will be clear if we show that the order of $G$ is less than of equal to 8. Now every element of $G$ can be written in the form $x^i y^j$ with $0\leq i\leq 3$ and $0\leq j\leq 1$, since $y x=x^{-1}y$ so powers of $y$ can be shifted to the right in any expression and then if the resulting power of $y$ is greater than 2 we can use $y^2=x^2$ to replace even powers of $y$ by powers of $x$. We must therefore have that the group $G$ must contain at most 8 elements so the above presentation is a presentation of $Q_8$.

### Subgroup lattice

The following shows the subgroup lattices of the first few generalized quaternion groups: ### Group cohomology

The group cohomology of the generalized quaternion groups: see Tomoda & Zvengrowski 2008

• Kenneth S. Brown, Cohomology of Groups , GTM 87 Springer Heidelberg 1982. (pp.98-101)

• H. S. M. Coxeter, The binary polyhedral groups, and other generalizations of the quaternion group , Duke Math. J. 7 no.1 (1940) pp.367–379.

• T. Y. Lam, Hamilton’s Quaternions , pp.429-454 in Handbook of Algebra III , Elsevier Amsterdam 2004. (preprint)

• W. R. Scott, Group Theory , Dover New York 1987. (pp.189-194, 252-254)

• Charles Weibel, Survey of Non-Desarguesian Planes , Notices of the AMS 54 no.10 (2007) pp.1294–1303. (pdf)

• Mehmet Koca, Ramazan Koç, Nazife Ozdes Koca, Two groups $2^3.PSL_2(7)$ and $2^3:PSL_2(7)$ of order 1344 (arXiv:1612.06107)

• H.S.M. Coxeter, W. O. J. Moser, Generators and Relations for Discrete Groups, (Springer Verlag, 1965);

• H.S.M. Coxeter, Regular Complex Polytopes (Cambridge; Cambridge University Press, 1973).

• Yunze Lu, On the $RO(G)$-graded coefficients of $Q_8$ equivariant cohomology, Topology and its Applications, 2021 (doi:10.1016/j.topol.2021.107921)