generalized quaternion group

The groups given by the presentation

$Q_{4n}=\langle a,b:a^{n}=b^{2},a^{2n}=1,b^{-1}ab=a^{-1}\rangle$

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 quaternion group 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 $0\leq i<2n$ and $j=0,1$ .
4.

$Z(Q_{4n})=\langle a^{n}\rangle\cong\mathbb{Z}_{2}$ .
5.

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

Proof.

Given the relation $b^{-1}ab=a^{-1}$ (rather treating it as $ab=ba^{-1}$ ) then as with dihedral groups we can shuffle 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 $0\leq i<2n$ and $0\leq j<4$ . However we have an added relation that $a^{n}=b^{2}$ so we can write $a^{i}b^{2}=a^{i+2}$ and also $a^{i}b^{3}=a^{i+2}b$ so we restrict to $j=0,1$ . This gives us $4n$ elements of 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 $\langle a^{n}\rangle$ then we have the presentation

$\langle a,b:a^{n}=1,b^{2}=1,b^{-1}ab=a^{-1}\rangle$

which we recognize as the presentation of the dihedral group. Thus $Q_{4n}/\langle a^{n}\rangle\cong D_{2n}$ . 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 $Z(Q_{4n})=\langle a^{n}\rangle=\langle b^{2}\rangle$ . ∎

1 Examples

As mentioned, if $n=1$ then $Q_{4}\cong\mathbb{Z}_{4}$ . If $n=2$ then we have the usual quaternion group $Q_{8}$ . Because of the genesis of quaternions, this group is often denoted with $i, j, k$ relations as follows:

$Q_{8}=\langle-1,i,j,k:i^{2}=j^{2}=k^{2}=-1,ij=k=-1ji\rangle.$

These relations are responsible for many useful results such as defining cross products for three-dimensional manipulations, and are also responsible for the most 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 $\mathbb{C}$ but to treat these as matrices over $\mathbb{R}$ .

$-1=\begin{bmatrix}-1&0\\ 0&-1\end{bmatrix},\quad i=\begin{bmatrix}i&0\\ 0&-i\end{bmatrix},\quad j=\begin{bmatrix}0&i\\ i&0\end{bmatrix},k=\begin{bmatrix}0&-1\\ 1&0\end{bmatrix}.$

A worthwhile additional example is $n=3$ . For this produces 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 only if $n=2$ .

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 $\langle ab\rangle$ . So we require $n\leq 2$ . If $n=1$ then $Q_{4n}$ is cyclic and so trivially Hamiltonian. When $n=2$ we have the usual quaternion group of order 8 which is Hamiltonian by direct inspection: the conjugacy classes are $\{1\}$ , $\{a^{2}\}$ , $\{a,a^{3}\}$ , $\{b,a^{2}b\}$ and $\{ab,a^{3}b\}$ , more commonly described 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 $\langle(1,i)\rangle$ is conjugate to the distinct subgroup $\langle(1,-i)\rangle$ and so such groups are not Hamiltonian.

Proposition 3.

1.

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

Every subgroup of $Q_{4n}$ is either cyclic or a generalized quaternion.
3.

The normal subgroups of $Q_{4n}$ are either subgroups of $\langle a\rangle$ or $n=2^{i}$ and it is maximal subgroups (of index 2) of which there are 2 acyclic ones.

Proof.

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

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

Finally, if $H$ does not contain $Z(Q_{4n})$ then $H$ does not contain an element of the form $a^{i}b$ , so $H\leq\langle a\rangle$ and so it is cyclic.

For the normal subgroup structure, from the relation $b^{-1}ab=a^{-1}$ we see $\langle a\rangle$ is normal. Thus all subgroups of $\langle a\rangle$ are normal as $\langle a\rangle$ is a normal cyclic subgroup. Next suppose $H$ is a normal subgroup not contained in $\langle a\rangle$ . Then $H$ contains some $a^{i}b$ , and so $H$ contains $Z(Q_{4n})$ . Thus $H/Z(Q_{4n})$ is a normal subgroup of $D_{2n}$ . We know this forces $H/Z(Q_{4n})$ to be contained in $\langle a\rangle/Z(Q_{4n})$ , 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 $n=2^{i}$ .

Proof.

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

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

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

Title	generalized quaternion group
Canonical name	GeneralizedQuaternionGroup
Date of creation	2013-03-22 16:27:41
Last modified on	2013-03-22 16:27:41
Owner	Algeboy (12884)
Last modified by	Algeboy (12884)
Numerical id	7
Author	Algeboy (12884)
Entry type	Derivation
Classification	msc 20A99
Synonym	quaternion groups
Related topic	DihedralGroupProperties
Defines	generalized quaternion