3
$\begingroup$

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.

$\endgroup$
5
  • 1
    $\begingroup$ $\Bbb Z_8$ 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
  • 1
    $\begingroup$ 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
7
$\begingroup$

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

$\endgroup$
2
  • 1
    $\begingroup$ 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
3
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.