What is the dicyclic group of order $12$? (What is $\mathbb{Z}_3\rtimes \mathbb{Z}_4$) I have come across the dicyclic group of order $12$. I can see that this is generated by three elements subject to some relations. Is there a way to realize this group without talking about generators and relations?
For example, is the group isomorphic to $\mathbb{Z}_2\oplus D_3$?

EDIT: I see this answer saying that the group is isomorphic to $\mathbb{Z}_3\rtimes \mathbb{Z}_4$. I am not sure that I understand this. From the definition I have seen of the semi-direct product, one needs subgroups $H$ and $K$ of a group $G$ to write $G = H\rtimes K$. Can someone help me understand.
I also see in that same answer realizing the group as a specific subgroup of $S_7$. Is there a more canonical way of doing this?
 A: Every 3D rotation can be represented as a quaternion. In general, the quaternion
$$q = \cos(\theta/2) + \sin(\theta/2)(xi + yj + zk)$$
represents a rotation of angle $\theta$ about axis $v=xi+yj+zk$, with the condition that $|v|=1$.
Every quaternion can be written in the above form, and every rotation can be represented in the above form in two different ways.
Those two ways come from the fact that if $\theta$ gets increased by $2\pi$ then it results in a different quaternion, but the rotation stays the same.
Consider the group $D_3$. It can be represented as a set of $6$ rotations in 3D, which map a flat equilateral triangle back to itself. These rotations can in turn be represented as quaternions, resulting in a group of $12$ quaternions. This is $\mathrm{Dic}_3$.
Those six rotations in $D_3$ could be chosen to be $0^\circ, 120^\circ$ and $240^\circ$ rotations about axis $i$, along with $180^\circ$ rotations around axes $j$, $\cos(2\pi/3)j+ \sin(2\pi/3)k$ and $\cos(4\pi/3)j + \sin(4\pi/3)k$. These $6$ rotations can in turn be represented as $12$ quaternions:
$$\pm1, \\\pm(\cos(\pi/3) + \sin(\pi/3)i),\\ \pm(\cos(2\pi/3) + \sin(2\pi/3)i), \\ \pm j, \\\pm (\cos(2\pi/3)j+ \sin(2\pi/3)k), \\\pm(\cos(4\pi/3)j + \sin(4\pi/3)k)$$
A similar idea applies to all the other dicyclic groups.
I don't know if there is any intuitive non-quaternionic interpretation of $\mathrm{Dic}_3$. The quaternionic interpretation is intuitive in that you can imagine every element of $\textrm{Dic}_3$ as being an element of $D_3$ with either a $+1$ or $-1$ symbol attached to it. Beware though that this is neither $D_3 \oplus C_2$ nor $D_3 \rtimes C_2$.
A: Suppose you are handed a group of order $12 = 3\cdot 2^2$. In order to apply Sylow's theorems, we are interested in the subgroups of orders $4$ and $3$. The number of $3$-Sylow subgroups is equivalent to $1$ mod $3$ and divides $4$, so there are either $1$ or $4$. The number of $2$-Sylow subgroups (for similar reasons) is either $1$ or $3$.
Notice: Because there is only one group of order 3, but two of order 4, the $3$-Sylow subgroup must be $C_3$, while the $2$-Sylow subgroup could be (a priori) either $C_4$ or $C_2 \times C_2$.
Suppose both subgroups are not normal, i.e. there are 4 $3$-Sylow subgroups and $3$ $2$-Sylow subgroups. Each nonidentity element of $C_3$ has order 3, so there are $4\cdot 2 = 8$ elements of order $3$, meaning there are $3$ elements left to be nonidentity elements of our 3 distinct $2$-Sylow subgroups. Dear reader, check that this is not possible, therefore one of my Sylow subgroups is normal.
The dicyclic group arises in the case where $C_3$ is normal, so I'll leave the cases where the $2$-Sylow subgroup is normal to the reader. Write $H$ for the $2$-Sylow subgroup.
So, $C_3$ is normal in $G$, by order of elements concerns, $C_3\cap H = 1$ and $|C_3H| = 12 = |G|$, so $G = C_3 \rtimes H$. This means there is a homomorphism $\varphi\colon H \to \operatorname{Aut}(C_3) \cong C_2$ such that if $h \in H$ and $k \in C_3$, $hk = \varphi(k)h$. The nontrivial element of $\operatorname{Aut}(C_3)$ inverts each element of $C_3$.
To specify the possibilities for the groups arising, I need to investigate the possibilities for both $H$ and $\varphi$. 
Case 1: $H = C_2\times C_2$. Up to isomorphism, there are two possibilities for $\varphi$: either the image of $\varphi$ is trivial or $\varphi$ is surjective. In the former case, our semidirect product is actually a direct product, so we see that $G \cong C_3\times C_2\times C_2$. If it is surjective, then up to isomorphism (dear reader, convince yourself!) we may assume that it is injective on one $C_2$ factor of $H$ and trivial on the other. Thus we get $G \cong (C_3 \rtimes C_2)\times C_2 \cong S_3\times C_2$. (If we combine $C_3$ and the normal $C_2$ into a normal $C_6$, we see that this group is also $D_6$, the symmetries of the hexagon.)
Case 2: $H = C_4$. Since $C_4$ is cyclic, the homomorphism $\varphi$ is determined by the image of its generator. If that image is trivial, we again get a direct product, $G \cong C_3 \times C_4$. If the image is nontrivial, we get a semidirect product $G \cong C_3 \rtimes C_4 = \langle a, t \mid a^3, t^4, t^{-1}at = a^{-1}\rangle$. Write $x = at^2$, $y = t$ and $z = at$. Then $x^3 = y^2 = z^2 = xyz = t^2$, so $G$ is (a priori) a quotient of the group $\langle x, y, z \mid x^3 = y^2 = z^2 = xyz \rangle$. Seeing that this latter group is actually isomorphic to $G$ seems harder. GroupProps has a proof, which I will talk through in the interest of legibility.

Write $G' = \langle x, y, z \mid x^3 = y^2 = z^2 = xyz \rangle$, and $\alpha = xyz = x^3 = y^2 = z^2$. Since $x,y,z$ generate the group, $\alpha$ is central. We see quickly that $z^2 = xyz$ implies $z = xy$, so $z^2 = y^2 = xyxy$ tells us $y = xyx$, or, equivalently, that $x^{-1} = y^{-1}xy$. Thus $\alpha^{-1} = x^{-3} = (y^{-1}xy)^3 = y^{-1}\alpha y = \alpha$, since $\alpha$ is central. So $\alpha$ has order two, $x^2$ has order $3$, and $y$ has order $4$.
Note that $x^2 = yz = yxy$, so $G'$ is generated by $x^2$ and $y$ ($x = y^{-1}x^2y^{-1}$, $z = xy$), and $y^{-1}x^2y = x^{-2}$, which tells us that $\langle x^2\rangle$ is a normal subgroup of order $3$, and that $G' \cong C_3 \rtimes C_4$.
