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I tried to calculate the number of groups of order $17^3\times 2=9826$ with GAP.

Neither the NrSmallGroups-Command nor the ConstructAllGroups-Command work with GAP. The latter one because of the needed space. Finally, I do not know a formula for $gnu(p^3\times q)$ , with $p,q$ primes and $gnu(n)$=number of groups of order $n$.

Is there any way to calculate such values with GAP ?

Does anyone know a formula for $gnu(p^3\times q)$ for arbitary primes $p,q$ ?

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    $\begingroup$ See icm.tu-bs.de/ag_algebra/software/small/number.html $\endgroup$ – Robert Israel Jan 5 '16 at 23:42
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    $\begingroup$ BTW, though not relevant to this question: one of the newest GAP packages SglPPow gives access to groups of order $p^7$ for $p > 11$, and to the groups of order $3^8$. You need to load it with LoadPackage("sglppow"); of course. $\endgroup$ – Alexander Konovalov Jan 6 '16 at 0:14
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    $\begingroup$ When GAP runs out of memory, it's worth to understand why. If you will enter SetInfoLevel(InfoGrpCon,4); before l:=ConstructAllGroups(9826); then you will see that it is computing extensions of 5 groups of order 4913. It succeeds for 4 of them and then is stuck at last one working with the automorphism group of order 111203278848. We need to understand what's going on there. $\endgroup$ – Alexander Konovalov Jan 6 '16 at 0:23
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    $\begingroup$ I completely agree with Alexander that your efforts seem to be scattered and lead to a lot of duplication. For example, you've already asked about $p^3q$ (and were answered) here and here: math.stackexchange.com/questions/1589461/… math.stackexchange.com/questions/1553589/… $\endgroup$ – verret Jan 6 '16 at 2:19
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    $\begingroup$ The second part of one of your most recent question: math.stackexchange.com/questions/1601119/… had already been answered by Derek Holt in the comments to math.stackexchange.com/questions/1556434/… Perhaps a little more focus would be helpful? $\endgroup$ – verret Jan 6 '16 at 2:19
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Let $G$ be a group of order $p^3 q$, and let $S_p$ and $S_q$ be a $p$-Sylow subgroup and a $q$-Sylow subgroup respectively.

First, let us examine the case when $S_p$ and $S_q$ are both not normal in $G$. By the Sylow theorems, we must have $q \equiv 1 \mod p$, and also we must have $q$ divide $p-1$, $p^2-1$ or $ p^3-1$. The first case is impossible. The second requires requires that $q = p+1$, which means that $p = 2$ and $q = 3$. The third case requires that there are $p^3$ $q$-Sylow groups, which means there are $p^3(q-1)$ elements of order $q$, leaving just $p^3$ elements not of order $q$, which must form the unique $p$-Sylow subgroup.

So we must have $p=2$ and $q=3$. $S_4$ is an example of such a group with no normal Sylow subgroups, and this is the only example, as is nicely shown here.

So for the remainder we can assume either $S_p$ or $S_q$ is normal.

First, let us handle the case $p=2$.

We have five groups of order 8 ($\mathbb{Z}_8, \mathbb{Z}_4 \times \mathbb{Z}_2, \mathbb{Z}_2^3, D_4, Q_8$), so there are five groups that are direct products of a group of order 8 with a group of order $q$.

Otherwise, assume just one of the Sylow subgroups is normal.

Case I: $S_p$ is normal.

a) $S_p \cong \mathbb{Z}_8$

$\text{Aut}(\mathbb{Z}_8) \cong \mathbb{Z}_2 \times \mathbb{Z}_2$, which $S_q \cong \mathbb{Z}_q$ cannot map into except trivially.

b) $S_p \cong \mathbb{Z}_4 \times \mathbb{Z}_2$

$\text{Aut}(\mathbb{Z}_4 \times \mathbb{Z}_2) \cong D_4$, and again $S_q$ can only map trivially to it.

c) $S_p \cong \mathbb{Z}_2^3$

$\text{Aut}(\mathbb{Z}_2^3) \cong \text{GL}_3(\mathbb{F}_2)$, which has order $(8-1)(8-2)(8-4) = 168$. No notrivial mapping unless $q=3$ or $q=7$. Since $3$ and $7$ divide into $168$ exactly once, the image of $S_q$ will be a $q$-Sylow subgroup, and all such groups are conjugate, so we see there is only one distinct action up to isomorphism. So one new group in the case $q=3$ or $q=7$.

d) $S_p \cong D_4$

$\text{Aut}(D_4) \cong \text{Hol}(\mathbb{Z}_4) \cong D_4$. Again no nontrivial mappings.

e) $S_p \cong Q_8$

$\text{Aut}(Q_8) \cong S_4$. So no nontrivial mapping unless $q=3$, in which case we get one new group.

case II: $S_q$ is normal

$\text{Aut}(S_q) \cong \mathbb{Z}_{q-1}$

a) $S_p \cong \mathbb{Z}_8$

We get one new group if $q \equiv 1 \mod 2$ (i.e. always), another new group if $q \equiv 1 \mod 4$, and another new group if $q \equiv 1 \mod 8$.

b) $S_p \cong \mathbb{Z}_4 \times \mathbb{Z}_2$

There are two ways to map $S_p$ nontrivially to $\mathbb{Z}_2$; either the kernel is of the form $\mathbb{Z}_4$, or it is of the form $\mathbb{Z}_2 \times \mathbb{Z}_2$. This gives us two new groups in all cases. In the case $q \equiv 1 \mod 4$, we get one more group by having an element of order $4$ in $S_p$ map to an order for automorphism of $S_q$.

c) $S_p \cong \mathbb{Z}_2^3$

Here the only cyclic image of the group (other than $\mathbb{Z}_1$) is $\mathbb{Z}_2$, and there is only one way of mapping onto it up to isomorphism.

d) $S_p \cong D_4$

Here again the only nontrivial cyclic image is $\mathbb{Z}_2$, but this time there are two ways of mapping onto it; the kernel could be the rotation subgroup, or it could be generated by a 180 degreee rotation and a reflection.

e) $S_p \cong Q_8$

Again the only nontrivial cyclic image is $\mathbb{Z}_2$, and there is only one mapping onto it (all order four subgroups of $Q_8$ are conjugate).

Putting it all together:

$gnu(2^3 3) = 15$

$gnu(2^3 7) = 13$

if $q \equiv 1 \mod 8$, $gnu(2^3 q) = 15$

if $q \equiv 5 \mod 8$, $gnu(2^3 q) = 14$

otherwise, $gnu(2^3 q) = 12$

Next, we handle the case $p > 2$.

As before, there are five groups of order $p^3$, so there are five groups that are direct products of a group of order $p^3$ and a group of order $q$.

Otherwise, assume just one of the Sylow subgroups is normal.

Case I: $S_p$ is normal

a) $S_p \cong \mathbb{Z}_{p^3}$

$\text{Aut}(S_p) \cong \mathbb{Z}_{p^2(p-1)}$. If $q \mid p-1$, we get one new group by mapping a generator of $S_q$ to an automorphism of $S_p$ of order $q$; otherwise there are no new groups.

b) $S_p \cong \mathbb{Z}_{p^2} \times \mathbb{Z}_p$

An automorphism of $S_p$ can map an element $g$ of order $p^2$ to any of the $p^3-p^2$ elements of order $p^2$, and can then map an element $h$ of order $p$ not in $<g>$ to any of the $p^2-p$ such elements, so the automorphism group has order $(p^3-p^2)(p^2-p) = p^3(p-1)^2$. So there will be no nontrivial actions of $S_q$ on $S_p$ unless $ q \mid p-1$.

If $q \mid p-1$, then we can do the following: quotient $S_p$ by the subgroup generated by $pg$, to get a quotient group $H$ isomorphic to $\mathbb{Z}_p^2$. An automorphism $f$ of $S_p$ will quotient down to an automorphism of $H$, and we know an automorphism of $H$ will have two linearly independent eigenvectors. At least one of these eigenvectors will map back to elements of order $p^2$ in $S_p$; choosing a random element $i$, we see that $f$ will take $i$ to $ni$ where $n$ is an element of $\mathbb{Z}_{p^2}^*$. $f$ will also take $h$ to $mi + jh$, where $p \mid m$ and $j$ is an element of $\mathbb{Z}_p^*$. But if $m \neq 0$ then $f$ will have order divisible by $p$, which doesn't work. So $m=0$. So $f$ acts on the groups generated by $i$ and $h$. There are $q-1$ actions on $<i>$ of order $q$, and $q-1$ actions on $<h>$ of order $q$; whatever action we choose for $f$, however, some power of $f$ will perform a prechosen action on $<i>$, so there are really just $q-1$ choices up to isomorphism. So that yields $q-1$ new groups. Also, we get a group where $f$ moves $i$ but not $h$, and a group where $f$ moves $h$ but not $i$, for a total of $q+1$ new groups.

c) $S_p \cong \mathbb{Z}_p^3$

$\text{Aut}(S_p) \cong \text_{GL}_3(\mathbb{F}_p)$, which has order $(p^3-1)(p^3-p)(p^3-p^2) = p^3(p-1)^3(p+1)(p^2+p+1)$. So no new groups unless $q\mid p-1$, $q \mid p+1$, or $q \mid p^2+p+1$. (These are disjoint cases unless $p=3$, $q=2$.) I believe in the latter two cases we get just one new group; however, I don't know how to prove that at this time. In the case $q \mid p-1$, I believe that we can assume that an automorphism of order $q$ will have three linearly independent eigenvectors $i,j,k$ (but again I am unsure how to prove this). Let us first address the case that the action on $i,j,k$ all have order $q$. Then there are $(q-1)^3$ possible automorphisms. However, some of these automorphisms will yield isomorphic groups:

$(i,j,k) \mapsto (x^ai,x^bj,x^ck)$ and $(i,j,k) \mapsto (x^{an} i,x^{bn} j, x^{cn} k)$ with $a,b,c,n \in \mathbb{Z}_q^*$ and $x$ an element of order $q$ in $\mathbb{Z}_p^*$, will yield isomorphic groups, since the latter is just a power of the former.

Permuting $a,b,c$ will yield isomorphic groups, since we can just relabel $i,j,k$.

Finding the number of equivalence classes of $(a,b,c)$ given the above equivalences doesn't seem to be an easy problem. I would welcome any help! (Edit: See below.)

In the case that at least one of of the actions on $i,j,k$ is trivial, we can reduce to the analysis for $\text{GL}_2(\mathbb{F}_p)$ that has been done elsewhere, for example in my previous answer. If only one action is trivial, then we get $\frac{q+1}{2}$ new groups; if two are trivial, we get another new group; the case where all three are trivial has already been counted.

d) $S_p = \mathbb{Z}_{p^2} \rtimes \mathbb{Z}_p$

According to this paper, $\text{Aut}(S_p)$ has order $p^3(p-1)$ and there is an element of order $p-1$. So if $q \mid p-1$, there is cyclic $q$-Sylow subgroup. Since all the $q$-Sylow subgroups are conjugate, and each such group has a unique subgroup of order $q$, all the order $q$ subgroups are conjugate. So we get one new group in the case $q \mid p-1$.

e) $S_p = H \rtimes K \cong (\mathbb{Z}_p \times \mathbb{Z}_p) \rtimes \mathbb{Z}_p$

From the same paper, we see that the group of automorphisms $A$ of $S_p$ has order $p^3(p-1)^2(p+1)$. We also see that there is a normal subgroup $B$ of automorphisms that act trivially on $K$, and thus is isomorphic to the group of automorphisms on $H$. This we know to be a group of order $p(p-1)^2(p+1)$, and therefore has index $p^2$. Thus any element of $A$ not in $B$ will have order divisible by $p$. So any automorphism of order $q$ will belong to $B$. As we saw in my previous answer, if $q \mid p+1$ then we get one new group; if $q \mid p-1$, we get $\frac{q+1}{2}$ groups that move both eigenvectors, and one new group that moves just one.

Case II: $S_q$ is normal.

a) $S_p = \mathbb{Z}_{p^3}$

The automorphism group of $S_q$ is isomorphic to $\mathbb{Z}_{q-1}$, so we get one new group if $p \mid q-1$, one more if $p^2 \mid q-1$, and one more if $p^3 \mid q-1$.

b) $S_p = \mathbb{Z}_{p^2} \times \mathbb{Z}_p$

Let $h$ be an element of order $p^2$, and let $k$ be an element of order $p$ not in $<h>$. If $p \mid q-1$, then we get a new group where $h$ acts trivially and $k$ maps to an automorphism of order $p$, and a new group where $k$ acts trivially and $h$ maps ot an automorphism of order $p$. If they both act nontrivially, then for some $i \in \mathbb{Z}_p^*$, $k+ih$ will act trivially, and we are back to a previous case.

If $p^2 \mid q-1$, we get another group where $k$ acts trivially and $h$ maps to an automorphism of order $p^2$. If instead $k$ maps to an automorphism of order $p$, then for some $i \in \mathbb{Z}_p^*$, $h + pik$ will act trivially, and we are back to the previous case.

c) $S_p \cong \mathbb{Z}_p^3$

If $p \mid q-1$, we will have the kernel of the action of $S_p$ on $S_q$ as a subspace of order $p^2$, and they all will be equivalent. So we get one new group in this case.

d) $S_p \cong \mathbb{Z}_{p^2} \rtimes \mathbb{Z}_p$

If $p \mid q-1$, we have two ways to define the kernel of the action of $S_p$ on $S_q$: take the group isomorphic to $\mathbb{Z}_{p^2}$, or take a group isomorphic to $\mathbb{Z}_p \times \mathbb{Z}_p$. So we get two new groups. $p^2 \mid q-1$ doesn't help in this case since there is now way to get a quotient group of $\mathbb{Z}_{p^2}$.

e) $S_p \cong (\mathbb{Z}_p \times \mathbb{Z}_p) \rtimes \mathbb{Z}_p$

Again, if $p \mid q-1$ we have two ways to define the kernel, one where we take the left group in the semidirect product above, and one where we take the product of the right group above and a subgroup of the left group of order $p$ that the right group fixes. So two more groups in this case.

That completes the analysis, except unfortunately in the case where $\mathbb{Z}_p^3$ is acted on by $\mathbb{Z}_q$. Again, I would welcome any help to complete this final case.

I can give formulas for $p$ odd, except for the case when $q \mid p-1$:

If $q \nmid p+1, q\nmid p^2+p+1, p \nmid q-1$: $gnu(p^3 q) = 5$.

If $q \mid p^2+p+1, p \nmid q-1$: $gnu(p^3 q) = 6$.

If $q \mid p+1$: $gnu(p^3 q) = 7$.

If $p \mid q-1, p^2 \nmid q-1, q \nmid p^2+p+1$: $gnu(p^3 q) = 13$.

If $p \mid q-1, p^2 \nmid q-1, q \mid p^2+p+1$: $gnu(p^3 q) = 14$.

If $p^2 \mid q-1, p^3 \nmid q-1$: $gnu(p^3 q) = 15$.

If $p^3 \mid q-1$: $gnu(p^3 q) = 16$.

EDIT: I have thought about the final case of $\mathbb{Z}_q$ acting on $\mathbb{Z}_p^3$, and I believe I have the answer. First, some terminology:

I'm pretty sure (but could use a reference) that any automorphism of $\mathbb{Z}_p^3$ will have three linearly independent eigenvectors. Denote the eigenvectors $i, j,$ and $k$, and let $g$ be an element of $\mathbb{Z}_p^*$ of order $q$. Then an automorphism of $\mathbb{Z}_p^3$ of order dividing $q$ will take $i$ to $i^{g^x}$, $j$ to $j^{g^y}$, and $k$ to $k^{g^z}$ for some $x, y, z$ in $\mathbb{Z}_q$. Denote that automorphism by $(x,y,z)$. This gives $q^3$ possible automorphisms, but some of these will generate the same groups.

Note that a value of $0$ represents a trivial action on an eigenvector. As stated above, there is one possible group with three trivial actions, one possible group with two trivial actions, and $\frac{q+1}{2}$ possible groups with one trivial action.

That leaves the automorphisms $(x,y,z)$ with $x,y,z \neq 0$.

Note that $(x,y,z)$ and $(a,b,c)$ will generate the same group if we can get from one to the other by multiplying by some nonzero $w \in \mathbb{Z}_q$ (since if $g \in S_q$ gives automorphism $(x,y,z)$, then $g^w$ will give automorphism $(wx,wy,wz)$), or by rearranging the variables (since we can switch the identities of $i,j,k$), or a combination of both. So first, there is a unique group generated by automorphisms where all three variables are the same: (x,x,x). Then there are the automorphisms where two variables are the same and the third is different. We can normalize the former two variables to $1$, and then the third can be anything from $2$ to $q-1$, generating another $q-2$ different groups.

Then, we have the case where all three variables are different. We can take care of the multiplying by $w$ by setting the first variable to $1$. This leaves $(q-2)(q-3)$ possibilities, related by six symmetries. However, we can not necessarily just divide by six, because some automorphisms could be self-symmetric. Note first that there can be no self-symmetry when the symmetry is exchanging two variables, since that leaves one variable the same while changing the other two. So that leaves the two 3-cycles. If $(1,y,z)$ is equivalent to $(y,z,1)$, then the latter must be obtained from the first by multiplication by $y$. So $(y,y^2, yz) = (y,z,1)$, so $z = y^2$ and $y^3 = z^3 = 1$. Similarly, if $(1,y,z)$ is equivalent to $(z,1,y)$ we get the same conditions again. So an automorphism of the form $(1,y,y^2)$ where $y^3=1$ will have two equivalent automorphisms of the form $(1,a,b)$ (including itself); every other automorphism will have six. If $q$ is of the form $3r+1$, $\mathbb{Z}_q^*$ will have two cube roots of unity (which we will call $a$ and $a^2$) other than the identity; otherwise, it will have none other than the identity. So if $q$ is of the form $3r+1$, two of the possible automorphisms correspond to one possible group, and the remaining $(q-3)(q-2)-2$ automorphisms divide into groups of six, resulting in $\frac{(q-3)(q-2)-2}{6}+1 = \frac{q^2 - 5q + 10}{6}$ different groups. If $q$ is not of the form $3r+1$, then there will be $\frac{q^2-5q+6}{6}$ different groups.

So the number of nontrivial semidirect products of $\mathbb{Z}_p^3$ by $\mathbb{Z}_q$ is $\frac{q^2-5q+10}{6} + 1 + (q-2) + \frac{q+1}{2} + 1 = \frac{q^2+4q+13}{6}$ if $q \equiv 1 \mod 3$, and $\frac{q^2+4q + 9}{6}$ otherwise.

All told, if $q \mid p-1, q \equiv 1 \mod 3$ then there are $\frac{q^2+4q+13}{6} + 5 + 1 + (q+1) + 1 + \frac{q+1}{2} + 1 = \frac{q^2 + 13q + 70}{6}$ different groups.

If $q \mid p-1, q \not\equiv 1 \mod 3$, then there are $\frac{q^2+13q+66}{6}$ different groups.

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    $\begingroup$ Although one case is missing, this answer definitely deserves an accept! $\endgroup$ – Peter Jan 6 '16 at 18:44
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Is there a way to calculate this?

Yes: When one runs larger classification tasks, in particular on parameters that were out of reach when the software was written first time it is not uncommon that some routines get called for input parameters that seemed to be implausible when the code for these routines was written.

Indeed this is what happens here: The calculation gets stuck when calculating the rational classes of a group of order 8192 which has over 2000 conjugacy classes and is represented in a permutation representation of degree over 4000 (though it has a faithful representation of degree 64). While this is a perfectly good group it is far from the examples that were in the code authors mind when writing conjugacy tests. It is not hard to add some basic heuristics that will work around such a situation (I will change this for a future release).

With this the calculation (also for some other orders in the same ballpark) will finish in a few minutes.

If you want, send me (by private mail) a list of orders (excluding multiples of $p^7$) on which the calculation fails and I'll have a look whether similar issues remain.

Oh, and since this is probably the answer you really care about:

There are 15 groups of order 9826

While there probably is a reasonable description of groups of order $p^3q$, I suspect a formula would end up messy.

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