# generalized quaternion group

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

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. 1.

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

2. 2.

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

3. 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. 4.

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

5. 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

###### 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. 1.

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

2. 2.

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

3. 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 GeneralizedQuaternionGroup 2013-03-22 16:27:41 2013-03-22 16:27:41 Algeboy (12884) Algeboy (12884) 7 Algeboy (12884) Derivation msc 20A99 quaternion groups DihedralGroupProperties generalized quaternion