Classify all groups with normal subgroups of order 3 and 4 Suppose a group $G$ has normal subgroups of size 3 and 4. I want to classify all such groups.
The subgroups are clearly $\mathbf Z_3$ and $\mathbf Z_4$ or $K_4=\mathbf Z_2\times\mathbf Z_2$. If $H_4$ denotes one of $\mathbf Z_4$ and $K_4$, then any group of the form
$$\mathbf Z_3\times H_4\times L$$
where $L$ is any group, satisfies the required condition.
I suspect that these are the only groups which satisfy this this condition, but I don't know how to prove it. I am familiar with the following result for obtaining equations of direct products like what I'm after:

If $H,K\leqslant G$, $H\cap K = \{1\}$ and the elements of $H$ and $K$ commute, then $HK\simeq H\times K$.

But I don't believe I can apply it to my situation since I don't know if elements commute.
I appreciate any help with this.
 A: I think this question is too broad/vague. It is equivalent to asking for all (finite) groups that have an abelian normal subgroup of order $12$.
Indeed, if $N_1$ is normal of order $3$ and $N_2$ is normal of order $4$, then $N_1N_2$ is normal of order $12$, and has a normal Sylow $3$-subgroup and a normal Sylow $2$-subgroup. But the nonabelian groups of order $12$ are $A_4$ (which does not have a normal Sylow $3$-subgroup), the dihedral group of order $12$, $D_6$ (which does not have a normal Sylow $2$-subgroup), and the dicyclic group $\langle a,s\mid a^6=e, s^2=a^3, sas^{-1}=a^{-1}\rangle$ (which does not have a normal Sylow $2$-subgroup). Thus, $N_1N_2$ is abelian so $G$ has a normal abelian subgroup of order $12$. Conversely, if $G$ has a normal abelian subgroup $N$ of order $12$, then the Sylow subgroups of $N$ are characteristic in $N$, hence normal in $G$, so $G$ has a normal subgroup of order $4$ and one of order $3$.
However, classifying all groups with a normal abelian subgroup of order $12$ (or any given order, for that matter) seems rather broad. The group need not split into $A\times K$ with $A$ the normal abelian group of order $12$; it need not even split into $A\rtimes K$ with $A$ the normal abelian group of order $12$:

*

*$C_{24}$ does not split into $A\times C_2$ with $A$ abelian of order $12$.


*$Q_8\times C_3$ has a normal abelian subgroup of order $12$, namely $\{1,-1,i,-i\}\times C_3$, which is cyclic of order $12$ (being the direct product of two cyclic groups of relatively prime order), but it is not a semidirect product $C_{12}\rtimes C_2$, because $Q_8\times C_3$ has three subgroups of order $12$ (depending which subgroup of order $4$ of $Q_8$ you take), but $C_{12}\rtimes C_2$ only has two: if $N\triangleleft C_{12}\rtimes C_2 = \langle x\rangle\rtimes C_2$ has order $12$, the intersection with $\langle x\rangle$ has index at most $2$, hence is either $\langle x\rangle$, so $N=\langle x\rangle$, or else $N\cap\langle x\rangle = \langle x^2\rangle$ and then $N=\langle x^2\rangle\rtimes C_2$.
A: When $G$ is abelian your condition is equivalent to having $12 \mid |G|$.
Necessity stems from Lagrange's theorem, sufficiency by using the following remark for $3$ and $2$ Sylow subgroups,

If $P$ is a group of order $p^n$, with $p$ a prime, it has a normal subgroup of order $p^m$ for all $m \leq n$.

Moreover, in that case such groups are of the form
$$
\Bbb Z_3 \oplus \Bbb Z_2 \oplus \Bbb Z_2 \oplus H, \quad \Bbb Z_3 \oplus \Bbb Z_{2^n} \oplus H \quad (n \geq 2)
$$
with $H$ an abelian group.
