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.

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    $\begingroup$ $\Bbb Z_8$ has a normal subgroup of order $4$, yet it isn't of the form $H_4\times L$. $\endgroup$ – Arthur Jan 25 at 22:16
  • $\begingroup$ @Arthur I see, so my hypothesis is false. Thanks for clarifying! $\endgroup$ – lamasabachthani Jan 25 at 22:18
  • $\begingroup$ @Arthur Is there anything meaningful I can say about groups like this, or is it too vague a question unless I know something about the order of the parent group? $\endgroup$ – lamasabachthani Jan 25 at 22:19
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    $\begingroup$ Any group of the form $A \times B$, where $A$ is a group of order $3^m$ and $B$ is a group of order $2^n$ with $n \ge 2$ has this property. Since it is widely believed that (in this standard sense) most finite groups have order $2^n$, this amounts to a lot of examples! $\endgroup$ – Derek Holt Jan 26 at 7:52
  • $\begingroup$ You already have asked this question yesterday with more context from an "old" exam sheet. This would be helpful here, too, because then we have more context (of a possible different interpretation, for example). You should not delete questions, but rather edit them. $\endgroup$ – Dietrich Burde Jan 26 at 11:04

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 $12$:

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

  2. $Q_8\times C_3$ 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 $N=\langle x^2\rangle\rtimes C_2$.

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    $\begingroup$ Since $N_1\cap N_2=1$ by order considerations, we immediately get that $N_1N_2=N_1\times N_2$, and this is abelian since $N_1$ and $N_2$ are. (So we only need to know about groups of order 3 and 4, not 12.) $\endgroup$ – verret Jan 26 at 3:43
  • $\begingroup$ @verret: Quite right. $\endgroup$ – Arturo Magidin Jan 26 at 23:10

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.


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