The number of groups $G$ such that $G/\mathbb{Z}_3\cong D_{2n}$

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.

• What do you mean by "the number of groups"? Are you talking about isomorphism classes of groups or equivalence classes of group extensions? – Derek Holt Apr 23 at 8:26
• 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. – Derek Holt Apr 23 at 8:29
• @DerekHolt I mean the number of groups up to isomorphism. – Hongyi Huang Apr 23 at 8:29
• @DerekHolt Thank you for that. I'll try it in this case. – 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 $$G$$.

• 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 $H^2(D_{2n},\mathbb{Z}_3)$? – Hongyi Huang Apr 23 at 13:46
• $H^2(D_{2n},Z_3)$ 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. – Derek Holt Apr 23 at 14:38
• Got it. Thank you for spending time helping me! – Hongyi Huang Apr 23 at 14:54