I am trying to find the number of groups $G$ such that $G/\mathbb{Z}_3\cong D_{2n}$, where $\mathbb{Z}_3$ denotes the cyclic group of order $3$ and $D_{2n}$ denotes the dihedral group of order $2n$.

I first consider the case when $G\cong\mathbb{Z}_3:D_{2n}$, a semi-direct product.

I tried to use N/C lemma first: let $\langle a\rangle$ be the normal subgroup of $G$ which is isomorphic to $\mathbb{Z}_3$, then $G/C_G(a) = N_G(a)/C_G(a)$ is isomorphic to a subgroup of $\mathrm{Aut}(\mathbb{Z}_3)\cong\mathbb{Z}_2$. Hence $|C_G(a)| = 3n$ or $6n$. Let $H = \langle b\rangle:\langle c\rangle\cong\mathbb{Z}_n:\mathbb{Z}_2\cong D_{2n}$, a semi-direct product, with $c^{-1}bc = b^{-1}$.

If $|C_G(a)| = 6n$ then $G\cong\mathbb{Z}_3\times D_{2n}$.

Otherwise $|C_G(a)| = 3n$, then $|C_H(a)| = n$, which means a subgroup of $H$ of order $n$ is commutative with $\langle a\rangle$.

If $n$ is odd, then the only subgroup of $H\cong D_{2n}$ of order $n$ is isomorphic to $\mathbb{Z}_n$, which is exactly $\langle b\rangle$. Thus $G = (\langle a\rangle\times\langle b\rangle):\langle c\rangle\cong(\mathbb{Z}_3\times\mathbb{Z}_n):\mathbb{Z}_2$. But how can I find $\mathrm{Aut}(\mathbb{Z}_3\times\mathbb{Z}_n)$? If $\gcd(3,n) = 1$ then the result of the direct product is a cyclic group and so $G\cong D_{6n}$. But what if $\gcd(3,n)\ne 1$?

If $n$ is even, then it is more complicated since there are two more subgroups of $H$ of order $n$, see Normal subgroups of dihedral groups, both of which are isomorphic to $D_n$. Thus there is a possibility that $G\cong(\mathbb{Z}_3\times D_{n}):\mathbb{Z}_2$, and what's that?

What if the extension is not a semi-direct product, and what if $\mathbb{Z}_3$ is replaced by a bigger cyclic group? (May be better if avoid calculating the cohomology group.)

I also wonder if there is any mistake in my analysis.

Thank you in advance!

  • $\begingroup$ What do you mean by "the number of groups"? Are you talking about isomorphism classes of groups or equivalence classes of group extensions? $\endgroup$ – Derek Holt Apr 23 at 8:26
  • $\begingroup$ The only situation in which you can get a nonsplit extension is where $n$ is divisible by $3$, and in that case the only nonsplit extension is the dihedral group $D_{6n}$, so that cas eis not too hard. $\endgroup$ – Derek Holt Apr 23 at 8:29
  • $\begingroup$ @DerekHolt I mean the number of groups up to isomorphism. $\endgroup$ – Hongyi Huang Apr 23 at 8:29
  • $\begingroup$ @DerekHolt Thank you for that. I'll try it in this case. $\endgroup$ – Hongyi Huang Apr 23 at 8:34

The extension of $D_{2n}$ by $C_3$, for $3\nmid n$ is always a semidirect product by the result of Zassenhaus: Every short exact sequence of finite groups $N$ by $Q$ $$ 1\rightarrow N \rightarrow G \rightarrow Q \rightarrow 1 $$ of coprime order ${\rm gcd}(|N|,|Q|)=1$ splits. In general, we have to determine $H^2(Q,N)=H^2(D_{2n},C_3)$, for the equivalence classes of such extensions.

Possible references:

Group cohomology of dihedral groups $H^q(D_{2n},M)$.

Group cohomology of dihedral groups


I know this question has been answered to your satisfaction, but in case it is helpful, here is a list of the number of isomorphism classes of groups $G$ satisfying the conditions in the different cases. I will assume that $n>2$, since $D_4$ is abelian and is not usually called a dihedral group.

$3 \not\!| n$, $n$ odd: $2$ (isomorphism classes of) groups $G$;

$3 \not\!| n$, $n$ even: $3$ groups $G$;

$3|n$, $n$ odd: $3$ groups $G$;

$3|n$, $n$ even: $4$ groups $G$.

  • $\begingroup$ Thank you. That's helpful but how can get this? Do I need to compute $\mathrm{Aut}(\mathbb{Z}_3\times \mathbb{Z}_n)$? Or this is just given by computing $H^2(D_{2n},\mathbb{Z}_3)$? $\endgroup$ – Hongyi Huang Apr 23 at 13:46
  • 1
    $\begingroup$ $H^2(D_{2n},Z_3)$ does not tell you much, In the case when $3|n$, the single nonsplit extrension is $D_{6n}$. All other extensions are split, and you more or less completed the analysis of that case yourself. But in the case when $n$ is even, you correctly observed that $D_{2n}$ has three subgroups of index $2$. They give rise to three equivalence classes of extensions, but the two classes coming from the two extra subgroups you get when $n$ is even are isomorphic as groups. That is why it is important to distinguish between equivalence classes of extensions and isomorphism classes of groups. $\endgroup$ – Derek Holt Apr 23 at 14:38
  • $\begingroup$ Got it. Thank you for spending time helping me! $\endgroup$ – Hongyi Huang Apr 23 at 14:54

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