generalized quaternion group

The groups given by the presentation

$$Q_{4n}=⟨a,b:a^n=b^2,a^{2n}=1,b^{-1}ab=a^{-1}⟩$$

are the generalized quaternion groups. Generally one insists that $n>1$as the properties of generalized quaternions become more uniform at this stage.However if $n=1$ then one observes $a=b^2$ so $Q_{4n}\cong {\mathbb{Z}}_4$.Dihedral group properties are strongly related to generalized quaterniongroup properties because of their highly related presentations. We will see this in many of our results.

Proposition 1.

1. $|Q_{4n}|=4n$.

2. $Q_{4n}$ is abelian if and only if $n=1$.

3. Every element $x\in Q_{4n}$ can be written uniquely as $x=a^i{b}^j$where and $j=0,1$.

4. $Z(Q_{4n})=⟨a^n⟩\cong {\mathbb{Z}}_2$.

5. $Q_{4n}/Z(Q_{4n})\cong D_{2n}$.

Proof.

Given the relation $b^{-1}ab=a^{-1}$ (rather treating it as $ab=b{a}^{-1}$) thenas with dihedral groups we canshuffle words in $\{a,b\}$ to group all the $a^{\prime }s$ at the beginning and the$b^{\prime }s$ at the end. So every word takes the form $a^i{b}^j$. As $|a|=2n$and $|b|=4$ we have and . However we have anadded relation that $a^n=b^2$ so we can write $a^i{b}^2=a^{i+2}$ andalso $a^i{b}^3=a^{i+2}b$ so we restrict to $j=0,1$. This gives us $4n$ elementsof this form which makes the order of $Q_{4n}$ at most $4n$.

As $a^n=b^2$ it follows $[a^n,a^i{b}^j]=[a^n,b^j]=[b^2,b^j]=1$. So$a^n$ is central. If we quotient by $⟨a^n⟩$ then we havethe presentation

$$⟨a,b:a^n=1,b^2=1,b^{-1}ab=a^{-1}⟩$$

which we recognize as the presentation of the dihedral group. Thus$Q_{4n}/⟨a^n⟩\cong D_{2n}$. This prove the order of $Q_{4n}$is exactly $4n$. Moreover, given $a^i{b}^j\in Z(Q_{4n})$ we have

$$1=[a^i{b}^j,b]=b^{-j}{a}^{-i}{b}^{-1}{a}^i{b}^{j}b=b^{-j}{a}^{-i}{a}^{-i}{b}^{-1}{b}^{j}b=b^{-j}{a}^{-2i}{b}^j=a^{2i}.$$

So we have $i=n$. So $a^n{b}^j=b^{j+2}$. Then $1=[b^{j+2},a]$ forces $j=0,2$.This means $Z(Q_{4n})=⟨a^n⟩=⟨b^2⟩$.∎

1 Examples

As mentioned, if $n=1$ then $Q_4\cong {\mathbb{Z}}_4$. If $n=2$ then we have theusual quaternion group $Q_8$. Because of the genesis of quaternions, this group is often denoted with $i,j,k$ relations as follows:

$$Q_8=⟨-1,i,j,k:i^2=j^2=k^2=-1,ij=k=-1ji⟩.$$

These relations are responsible for many useful results such as defining cross products for three-dimensional manipulations, and are also responsible for themost common example of a division ring. As a group, $Q_8$ is a curious specimen of a $p$-group in that it has only normal subgroups yet is non-abelian, it has a unique minimal subgroup and cannot be represented faithfully except by a regular representation – thus requiring degree 8. [To see this note that the unique minmal subgroup is necessarily normal, thus if a proper subgroup is the stabilizer of an action, then the minimal normal subgroup is in the kernel so the representation is not faithful.]

A common work around is to use $2\times 2$ matrices over $ℂ$ but to treat these as matrices over $ℝ$.

A worthwhile additional example is $n=3$. For thisproduces a group order 12 which is often overlooked.

2 Subgroup structure

Proposition 2.

$Q_{4n}$ is Hamiltonian – meaning all a non-abelian group whose subgroups are normal – if and onlyif $n\mathrm{=}\mathrm{2}$.

Proof.

As $Q_{4n}/Z(Q_{4n})\cong D_{2n}$, then if $Q_{4n}$ is Hamiltonian then we require$D_{2n}$ to be as well. However when $n>2$ we know $D_{2n}$ has non-normal subgroups,for example $⟨ab⟩$. So we require $n\le 2$. If $n=1$ then $Q_{4n}$is cyclic and so trivially Hamiltonian. When $n=2$ we have the usual quaterniongroup of order 8 which is Hamiltonian by direct inspection: the conjugacy classesare $\{1\}$, $\{a^2\}$, $\{a,a^3\}$, $\{b,a^{2}b\}$ and $\{ab,a^{3}b\}$, more commonlydescribed by $\{1\}$, $\{-1\}$, $\{i,-i\}$, $\{j,-j\}$ and $\{k,-k\}$. In any case,all subgroups are normal.∎

By way of converse it can be shown that the only finite Hamiltonian groups are$A\oplus Q_8$ where $A$ is abelian without an element of order 4.One sees already in ${\mathbb{Z}}_4\oplus Q_8$ that the subgroup $⟨(1,i)⟩$ is conjugate to the distinct subgroup $⟨(1,-i)⟩$ and so such groups are not Hamiltonian.

Proposition 3.

1. $|a^i|=2n/i$ for and $|a^{i}b|=4$ for all $i$.

2. Every subgroup of $Q_{4n}$ is either cyclic or a generalizedquaternion.

3. The normal subgroups of $Q_{4n}$ are either subgroups of$⟨a⟩$ or $n=2^i$ and it is maximal subgroups (of index 2) ofwhich there are 2 acyclic ones.

Proof.

The order of elements of $⟨a⟩$ follows from standard cyclic group theory.Now for $a^{i}b$ we simply compute: ${(a^{i}b)}^2=a^{i}b{a}^{i}b=a^i{a}^{-i}{b}^2=b^2$.So $|a^{i}b|=4$.

Now let $H$ be a subgroup of $Q_{4n}$. If $Z(Q_{4n})\le H$ then $H/Z(Q_{4n})$ isa subgroup of $D_{2n}$. We know the subgroups of $D_{2n}$ are either cyclic ordihedral. If $H/Z(Q_{4n})$ is cyclic then $H$ is cyclic (indeed it is a subgroupof $⟨a⟩$ or $H=⟨a^{i}b⟩$). So assume that $H/Z(Q_{4n})$ isdihedral. Then we have a dihedral presentation$⟨x,y:x^m=1,y^2=1,y^{-1}xy=x^{-1}⟩$ for $H/Z(Q_{4n})$. Now pullbackthis presentation to $H$ and we find $H$ is quaternion.

Finally, if $H$ does not contain $Z(Q_{4n})$ then $H$ does not contain an elementof the form $a^{i}b$, so $H\le ⟨a⟩$ and so it is cyclic.

For the normal subgroup structure, from the relation $b^{-1}ab=a^{-1}$ wesee $⟨a⟩$ is normal. Thus all subgroups of $⟨a⟩$are normal as $⟨a⟩$ is a normal cyclic subgroup. Next suppose$H$ is a normal subgroup not contained in $⟨a⟩$. Then $H$ containssome $a^{i}b$, and so $H$ contains $Z(Q_{4n})$. Thus $H/Z(Q_{4n})$ is a normalsubgroup of $D_{2n}$. We know this forces $H/Z(Q_{4n})$ to be contained in$⟨a⟩/Z(Q_{4n})$, a contradiction on our assumptions on $H$, or$n=2^i$ and $H/Z(Q_{4n})$ is a maximal subgroup (of index 2).∎

Proposition 4.

$Q_{4n}$ has a unique minimal subgroup if and only if $n\mathrm{=}{\mathrm{2}}^i$.

Proof.

If $p|n$ and $p>2$ then $a^{2n/p}$ has order $p$ and so the subgroup$⟨a^{2n/p}⟩$ is of order $p$, so it is minimal. As the centeris also a minimal subgroup of order 2, then we do not have a unique minimal subgroupin these conditions. Thus $n=2^i$.

Now suppose $n=2^i$ then $Q_{4n}$ is a $2$-group so the minimal subgroups must allbe of order 2. So we locate the elements of order 2. We have shown $|a^{i}b|=4$ forany $i$, and furthermore that ${(a^{i}b)}^2=b^2=a^n$. The only other minimal subgroupswill be generated by $a^i$ for some $i$, and as $|a|=2^{i+1}$ there is a unique minimalsubgroup.∎

It can also be shown that any finite group with a unique minimal subgroup is eithercyclic of prime power order, or $Q_{4n}$ for some $n=2^i$.We note that these groups have only regular faithful representations.