generalized quaternion group
The groups given by the presentation^{}
$${Q}_{4n}=\u27e8a,b:{a}^{n}={b}^{2},{a}^{2n}=1,{b}^{1}ab={a}^{1}\u27e9$$ 
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.
Proof.
Given the relation^{} ${b}^{1}ab={a}^{1}$ (rather treating it as $ab=b{a}^{1}$) 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 $$ and $$. 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 $\u27e8{a}^{n}\u27e9$ then we have the presentation
$$\u27e8a,b:{a}^{n}=1,{b}^{2}=1,{b}^{1}ab={a}^{1}\u27e9$$ 
which we recognize as the presentation of the dihedral group. Thus ${Q}_{4n}/\u27e8{a}^{n}\u27e9\cong {D}_{2n}$. 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 $Z({Q}_{4n})=\u27e8{a}^{n}\u27e9=\u27e8{b}^{2}\u27e9$. ∎
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:
$${Q}_{8}=\u27e81,i,j,k:{i}^{2}={j}^{2}={k}^{2}=1,ij=k=1ji\u27e9.$$ 
These relations are responsible for many useful results such as defining cross products for threedimensional 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 nonabelian^{}, 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 $\u2102$ but to treat these as matrices over $\mathbb{R}$.
$$1=\left[\begin{array}{cc}\hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right],i=\left[\begin{array}{cc}\hfill i\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill i\hfill \end{array}\right],j=\left[\begin{array}{cc}\hfill 0\hfill & \hfill i\hfill \\ \hfill i\hfill & \hfill 0\hfill \end{array}\right],k=\left[\begin{array}{cc}\hfill 0\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 0\hfill \end{array}\right].$$ 
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 nonabelian group^{} whose subgroups are normal – if and only if $n\mathrm{=}\mathrm{2}$.
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 nonnormal subgroups, for example $\u27e8ab\u27e9$. 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 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 $\u27e8(1,i)\u27e9$ is conjugate to the distinct subgroup $\u27e8(1,i)\u27e9$ 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 generalized quaternion.

3.
The normal subgroups of ${Q}_{4n}$ are either subgroups of $\u27e8a\u27e9$ 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 $\u27e8a\u27e9$ 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 $H/Z({Q}_{4n})$ 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 $\u27e8a\u27e9$ or $H=\u27e8{a}^{i}b\u27e9$). So assume that $H/Z({Q}_{4n})$ is dihedral. Then we have a dihedral presentation $\u27e8x,y:{x}^{m}=1,{y}^{2}=1,{y}^{1}xy={x}^{1}\u27e9$ 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 $H\le \u27e8a\u27e9$ and so it is cyclic.
For the normal subgroup structure^{}, from the relation ${b}^{1}ab={a}^{1}$ we see $\u27e8a\u27e9$ is normal. Thus all subgroups of $\u27e8a\u27e9$ are normal as $\u27e8a\u27e9$ is a normal cyclic subgroup. Next suppose $H$ is a normal subgroup not contained in $\u27e8a\u27e9$. 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 $\u27e8a\u27e9/Z({Q}_{4n})$, 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 $pn$ and $p>2$ then ${a}^{2n/p}$ has order $p$ and so the subgroup $\u27e8{a}^{2n/p}\u27e9$ 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  20130322 16:27:41 
Last modified on  20130322 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 