What is definition of the group $Q_{12}$? I am trying to understand the subgroup lattices of several groups and got stuck in here in the lattice of $Q_{12}$.
My query is: what is the definition of $Q_{12}$ ?
Please help me
 A: Reading the wiki article on the quaternion group and seing that it links to dicyclic groups, it seems like the generalized quaternion group with $12$ elements has representation
$$
\langle a, x\mid a^6 = 1, x^2 = a^3, x^{-1}ax = a^{-1}\rangle
$$
equivalently, it is the subgroup of $GL_2(\Bbb C)$, the invertible $2\times 2$ complex matrices, generated by
$$
\begin{pmatrix}0&-1\\1&0\end{pmatrix} \quad \text{and}\quad \begin{pmatrix}\omega&0\\0&\overline{\omega}\end{pmatrix}
$$
where $\omega=e^{i\pi/3}$ is a complex third root of $-1$.
A: Following the descriptions given by Anjan3 and Arthur, I shall give you more things that will make you never to forget the group.
First of all, think of $D_{2n}$ the dihedral group of order 2n which has the following presentation: $$D_{2n}:=\langle x,y|~x^{n}=y^2=1, xy=yx^{-1} \rangle.$$ It is abelian for $n=2$ and non-abelian for $n\geq 3$. You can also write $D_{2n}:=A \sqcup B$, where $A=\{1,x,x^2,\cdots,x^{n-1}\}$ and $B=\{y,xy,x^2y,\cdots,x^{n-1}y\}$. The set $A$ is the well-known cyclic group of order $n$ while all elements of $B$ have order $2$.
Similarly, we write $$Q_{4n}:=\langle x,y \mid x^{2n}=1,x^{n}=y^2,xy=yx^{-1} \rangle, ~ n\geq 2.$$ Just like the Dihedral groups, we can write $Q_{4n}$ as $Q_{4n}:=A \sqcup B$, where $A=\{1,x,x^2,\cdots,x^{2n-1}\}$ and $B=\{y,xy,x^2y,\cdots,x^{2n-1}y\}$. The set $A$ is the well-known cyclic group of order $2n$ while all elements of $B$ have order $4$.
The smallest group of the form $Q_{4n}$ is gotten when $n=2$, and it is the well-known quaternion group of order $8$. Many people call the group $Q_{4n}$ different names...some call it groups of quaternion type. However, when the group is a $2$-group (i.e., the order of the group is a power of $2$), we all call it the generalised quaternion group.
So, we obtain $Q_{12}$ when $n=3$ in the presentation of $Q_{4n}$ above, and so we have: $$Q_{12}:=\langle x,y \mid x^{6}=1,x^{3}=y^2,xy=yx^{-1} \rangle=\{1,x,x^2,x^3,x^4,x^5\} \sqcup \{y,xy,x^2y,x^3y,x^4y,x^5y\},$$ where elements of the second set all have order $4$, and the first set is the well-known cyclic group of order $6$.
Have fun!
A: It's the generalized quaternion group of order 12.
It has presentation
$$Q_{12}=\langle a, b \ | \ a^3=b^2, \ a^6=1, \ b^{-1}a b = a^{-1}\rangle.$$
