Some more work with semidirect products
Groups of order 12
If $G$ has order $12$, then it has a Sylow subgroup $V$ of order $4$ and a Sylow subgroup $P$ of order $3$. Either $V$ or $P$ is normal. To see this, note that $V\cap P$ is the identity. Suppose $P$ is not normal, so $n_3=4$. Then there are 8 elements of order $3$ in $G$, and together with the identity this accounts for $9$ elements of $G$. So there are at most $3$ elements, plus the identity, available for $V$, so $n_2=1$.
Since either $V$ or $P$ is normal, $G=VP$.
If they are both normal, then $G$ is abelian, and the possibilities are:
 $\Zn{2}\times\Zn{2}\times\Zn{3}$ (invariant factors are $6$ and $2$).
 $\Zn{12}$
If $G$ is not abelian, it is a semidirect product. Also, $P$ is cyclic of order $3$ and $V$ is either cyclic or $\Zn{2}\times\Zn{2}$.

$V$ is normal.
a. $V=\Zn{4}$. Then we need to look at maps $P\to \Aut(V)$. But $P$ has order $3$ and $\Aut(V)$ has order $2$ so there are no nontrivial homomorphisms, so in this case $G$ is cyclic.
b. $V=\Zn{2}\times\Zn{2}$. The automorphisms of $V$ are $\GL_{2}(\Zn{2})$. This group has order $6$ and is isomorphic to $S_{3}$; it acts on $V$ by permuting the three nonzero elements. Therefore we can send the generator of $P$ to an automorphism that cyclically permutes those elements. The resulting semidirect product is $A_4$.

$P$ is normal. The automorphism group of $P$ is $\Zn{2}$, with nontrivial automorphism $x\to x^{1}$ where $x$ is a generator of $P$.
a. $V=\Zn{4}$. We need a map from $V\to \Aut(P)$, and there’s only one, which sends the generator $y\in V$ to the automorphism $x\mapsto x^{1}$. This gives us a group with generators $x$ and $y$ such that $x^3=y^4=1$ and $yxy^{1}=x^{1}$.
b. $V=\Zn{2}\times\Zn{2}$. In this situation there are three nontrivial maps $V\to \Aut(P)$ (either $(1,0)$, $(0,1)$ or $(1,1)$ acts like $x\mapsto x^{1}$.) The resulting group has generators $x,y,z$ where $x^3=y^2=z^2=1$, $y$ and $z$ commute with each other, $x$ and $z$ commute with each other, and $yxy^{1}=x^{1}$. (The actual choice of map $V\to\Aut(P)$ does not affect the isomorphism class of the result). This is the group $D_{12}$ of symmetries of the hexagon.
So there are five isomorphism types of groups of order $12$, two abelian, three nonabelian: $D_{12}$, $A_{4}$, and $\Zn{3}\semi\Zn{4}$.