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 $a's$ at the beginning and the $b's$ 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 $$ \langle a,b:a^n=1, b^2=1, b^{-1} ab=a^{-1}\rangl $$ 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 $Z(Q_{4n})=\langle a^n\rangle=\langle b^2\rangle$ .

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 $\{1\}$ , $\{a^2\}$ , $\{a,a^3\}$ , $\{b,a^2b\}$ 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.

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: $(a^i b)^2=a^i b a^i b=a^i a^{-i} b^2=b^2$ . So $|a^i b|=4$ .

Now let $H$ be a subgroup of $Q_{4n}$ . If $Z(Q_{4n})\leq H$ 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.