# 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.

• $\Bbb Z_8$ has a normal subgroup of order $4$, yet it isn't of the form $H_4\times L$. – Arthur Jan 25 at 22:16
• @Arthur I see, so my hypothesis is false. Thanks for clarifying! – lamasabachthani Jan 25 at 22:18
• @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? – lamasabachthani Jan 25 at 22:19
• 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! – Derek Holt Jan 26 at 7:52
• 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. – 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 $$N=\langle x^2\rangle\rtimes C_2$$.

• 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.) – verret Jan 26 at 3:43
• @verret: Quite right. – 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.