Is a formula for $gnu(2pq^2)$ known, where $q=2p+1\ $?

Question

Let $p$ be an odd prime such that $q:=2p+1$ is also prime.

Denote $g(p):=gnu(2pq^2)$ = number of groups of order $2pq^2$ upto isomorphy.

The following table shows the first few values

   p                q            g(p)

   3                7            23
   5               11            27
  11               23            39
  23               47            63
  29               59            75
  41               83            99

The calculation with GAP quickly becomes time-consuming, so my question :

Is a formula for $g(p)$ known ?

The Sylow-theorem tells us that a group of order $2pq^2$ must have a normal subgroup of order $q^2$. Does this help to get a formula ?

Answer 1

Following your thoughts. Let G be a group of order $2pq^2$ and $S$ be its unique $q$-Sylow. Then $Q:=G/S$ is a quotient group of order $2p$.

With some basic group theory, there are only two choices for $S$ :

$$S_1:=\mathbb{Z}/q^2\text{ and } S_2:=(\mathbb{Z}/q)^2 $$

and only two choices for $Q$ :

$$Q_1:=\mathbb{Z}/2p\text{ and } Q_2:=D_{2p}\text{ (the dihedral group)}$$

Furthermore, we can use the Schur-Zassenhaus theorem to state that $G$ is isomorphic to $S\rtimes_{\phi}Q$ where $\phi:Q\rightarrow Aut(S)$ is a group morphism.

Now, you just have to compute, for each choice of $(i,j)$, the number of different groups (up to isomorphism) given by $S_i\rtimes_{\phi}Q_j$ where $\phi:Q_j\rightarrow Aut(S_i)$ is a group morphism.

For instance for $(i,j)=(1,1)$, I have four different isomorphism classes and $(i,j)=(1,2)$, I have two different isomorphism classes. Indeed, when $i=1$, we have that $Aut(S_1)$ is cyclic of order $2p(2p+1)$.

When $j=1$, $Q_1$ is generated by one element (let us say it is $\gamma$). If $\phi(\gamma)$ is trivial, we have one isomorphism class. If $\phi(\gamma)$ is of order $2$ then there is a unique element of order $2$ in $Aut(S_1)$, whence a unique isomorphism class. If $\phi(\gamma)$ is of order $p$ then we get $p-1$ choices for $\phi(\gamma)$ but they all lead to the same isomorphism class for the semi-direct product. If $\phi(\gamma)$ is of order $2p$ then we get $p-1$ choices for $\phi(\gamma)$ but they all lead to the same isomorphism class for the semi-direct product. Whence we have four choices.

If $j=2$ then we know that the abelianization of $Q_2$ is $\mathbb{Z}/2$, it follows that there are exactly two group morphisms from $Q_2$ to $Aut(S_1)$ (one for which the image of $\phi$ is trivial and one for which the image of $\phi$ is of order $2$).

Maybe you can handle the more difficult case where $i=2$ (the automorphism group of $S_2$ is more complicated...).

Answer 2

Okay, I have found $2p+17$ groups of order $2pq^2$ when $p,q$ are primes and $2p \mid q-1$. There is just one case where I haven't been able to enumerate the number of groups.

Let $G$ be a group of order $2pq^2$ as above. First, the third Sylow theorem tells us that there is a unique $q$-Sylow group of $G$, call it $Q$. Next, the Schur-Zassenhaus theorem tells us that there is an orthogonal group of order $2p$, call it $P$. $G$ is then the semidirect product of $Q$ by $P$. $Q$ can be two groups, $\mathbb{Z}_q \times \mathbb{Z}_q$ or $\mathbb{Z}_{q^2}$, and $P$ can be two groups, $\mathbb{Z}_{2p}$ or $D_p$.

Case I: $Q = \mathbb{Z}_{q^2}$

In this case, $\text{Aut}(Q) = \mathbb{Z}_{q^2-q} = \mathbb{Z}_{2p(2p-1)}$. If $P = \mathbb{Z}_{2p}$, then the image of $P$ in $\text{Aut}(Q)$ induced by conjugation action of $Q$ by $P$ can be one of four groups: $\mathbb{Z}_{1}, \mathbb{Z}_2, \mathbb{Z}_p, \mathbb{Z}_{2p}$. This leads to four possibilities for $G$. If $P = D_p$, then the image of $P$ in $\text{Aut}(Q)$ can be either $\mathbb{Z}_1$ or $\mathbb{Z}_2$, leading to two more groups. So Case I has six groups.

Case II: $Q = \mathbb{Z}_q \times \mathbb{Z}_q$

Subcase IIa: $P = \mathbb{Z}_{2p}$

Then $Aut(Q) = \text{GL}_2(\mathbb{Z}_q)$. If $P = \mathbb{Z}_{2p}$, then again the image of $P$ in $\text{Aut}(Q)$ can be one of four groups: $\mathbb{Z}_{1}, \mathbb{Z}_2, \mathbb{Z}_p, \mathbb{Z}_{2p}$. For $\mathbb{Z}_1$ there is of course just one possible group. For the rest, remember that for any automorphism of $\mathbb{Z}_q \times \mathbb{Z}_q$ we can find two independent eigenvectors. For $\mathbb{Z}_2$ we get two different groups, one where the nontrivial action acts nontrivially on both eigenvectors, and one where it acts nontrivially on one eigenvector but not the other.

If the image is $\mathbb{Z}_p$, we can look at the action on the two eigenvectors. Suppose $P$ acts nontrivially on both eigenvectors. Then we can fix the action on one of the eigenvectors, and look at the action on the other eigenvector. There are $p-1$ possible actions, but we can pair them off by switching which eigenvector to fix. This will pair off all the $p-1$ actions except when the actions on the eigenvectors are the same or when they are inverse to each other. So we will get $\frac{p-3}{2} +1 + 1 = \frac{p+1}{2}$ possible groups. Also, there is the group that acts on one eigenvector but not the other, so we have $\frac{p+3}{2}$ possibilities in this case.

If the image is $\mathbb{Z}_{2p}$, again we can look at the action on the eigenvectors.

Subcase IIaA: $P$ has an action of order $2p$ on at least one of the eigenvectors.

Subcase IIaA1: $P$'s action on the other eigenvector is also of order $2p$.

In this subcase, we can fix the action on one of the eigenvectors, and the action on the other can be one of $p-1$ possibilities. As above, these pair off except for in two cases, so we get $\frac{p+1}{2}$ possibilities again.

Subcase IIaA2: $P$'s action on the other eigenvector is not of order $2p$.

In this subcase, there are $p-1$ actions of order $p$, one of order $2$, and one of order $1$, for $p+1$ possibilities in total.

Subcase IIaB: neither action of $P$ on an eigenvector is of order $2p$.

In this subcase, the only possibility is that $P$ acts on one eigenvector with order $p$ and the other with order $2$. This adds another group.

Subcase IIb: $P = D_p$.

In this subcase, $P$ can have image in $\text{Aut}(Q)$ of $\mathbb{Z}_1, \mathbb{Z}_2,$ or $D_p$. In the first two cases we again have three possible groups. This leaves just $D_p$, which I am not quite sure how to handle. By my count we have already constructed $2p+16$ different groups, so if I have done everything correctly there will be exactly one group where $D_p$ acts faithfully on $\mathbb{Z}_q \times \mathbb{Z}_q$ by conjugation.

EDIT: I have found a group where $D_p$ acts faithfully on $\mathbb{Z}_q \times \mathbb{Z}_q$: An element of order $p$ acts on the second eigenvector in the inverse manner of the first eigenvector, and the other generator for $D_p$ acts by switching the two eigenvectors. So that's $2p+17$ different groups; all that remains is to show that this is the only possible group in the last case up to isomorphism.