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 $G' \cong C_3 \rtimes C_4$.
