Just wonder if there is any known results on the transitive metacyclic groups of $AGL(4,3)$?

Sorry, I should elaborate a bit here. I am working on a graph $\Gamma$ which admits a metacyclic group transitive on its vertices. By some other restricted conditions, I have reduced that $Aut\Gamma$ is contained in $AGL(4,3)$. So it turns out that I need to work on the following problems:

  1. Check whether there exist such metacyclic subgroups in $AGL(4,3)$.
  2. If exist, what are they?

One of my attempt was to inspect subgroups of $AGL(4,3)$ and rule out those impossible cases. But it seems tedious and somehow I still have to seek help from machinery search...

Alternatively, I am hoping to do it on a more elegant and efficient way. That is, to find the generators of the metacyclic groups. We know that a metacyclic group can be generated by two elements. Hence if we can find all the possible generators, then we can construct the groups and verify if these groups are transitive or not. There are some helpful information we already know: written $AGL(4,3)=\mathbb{Z}_3^4{:}GL(4,3)$, so a transitive metacyclic group $G=\langle(a_1,b_1),(a_2,b_2)|a_i\in\mathbb{Z}_3^4,b_i\in GL(4,3),~\mbox{and}~o(b_1),o(b_2)~\mbox{divisible by}~9\rangle$. Then the next question is which exact generators we should choose?

Another thought is that if it is possible to do a pure computer search? as the group is not too big, but I am not too sure how to write up the codes...

Any suggestion would be much appreciated.

  • $\begingroup$ Thank you for both of you answering my question. Both answers are quite close to my desire. I would like to tick Peter's answer however Stefan's computation is also nice. $\endgroup$ – Easy Jan 29 '13 at 10:13

There is no such group. Let $G$ be a transitive metacyclic subgroup of $\text{AGL}(4,3)$. Let $C$ be a cyclic normal subgroup of $G$ with $G/C$ cyclic.

As $G$ permutes transitively the orbits of $C$, the kernels of the action of $C$ on its orbits all have the same size, thus they are equal because $C$ is cyclic. But $C$ acts faithfully on the union of the orbits. We infer that $C$ acts regularly on each orbit. In particular, $\lvert C\rvert$ divides $3^4$.

Note that the Sylow $3$-subgroups of $G$ are transitive too, and subgroups of metacyclic groups are metacyclic too. So we may assume that $G$ is a $3$-group.

On the other hand, $9$ is the maximal order of a $3$-element in $\text{AGL}(4,3)$. (One can see that most easily from the embedding $\text{AGL}(4,3)\le\text{GL}(5,3)$.)

From that we see that $C$ has order $9$, and $G$ is a semidirect product of $C$ with another cyclic group $D$ of order $9$.

View $G=C\rtimes D$ as a subgroup of $\text{GL}(5,3)$. From Jordan's normal form theorem, we see that $\text{GL}(5,3)$ contains two conjugacy classes of elements of order $9$. Let $U$ be the group of upper triangular matrices of $\text{GL}(5,3)$. Consider the two cases of $C$ (corresponding to the Jordan block sizes $4+1$ and $5$, respectively). In both cases, one computes that the exponent of $N_U(C)/C$ is $3$, so there is no room for the cyclic group $D$ of order $9$.

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  • $\begingroup$ Peter, I think there is a better way to show that $G=\mathbb{Z}_9.\mathbb{Z}_9$. Since $G$ is transitive on $3^4$, then so is its Sylow $3$-group $P$, and $81\mid|P|$ by orbit stabiliser theorem. On the other hand, the highest order of $3$-elements in $AGL(4,3)$ is 9. So the only possible structure for $P$ is $\mathbb{Z}_9.\mathbb{Z}_9$. Also I am gonna to partially quote ur argument in my paper. Do you need ur name on? $\endgroup$ – Easy Jan 29 '13 at 10:17
  • $\begingroup$ @generao: Do whatever you want, the argument is so simple that I'm not proud of it ... However, as you raised the question, it would probably be scientifically more correct to mention the two answers. $\endgroup$ – Peter Mueller Jan 29 '13 at 14:47
  • $\begingroup$ @Peter: There was something I didn't reckon at the first time. At your last step, you said one "computes" the exponent is 3. Did you do it by machine or do it manually? I verified that using GAP, but when I tried calculating manually I found that is a huge calculation. Or is there any trick I didn't know? $\endgroup$ – Easy Feb 14 '13 at 7:21
  • $\begingroup$ If $c$ is a generator for $C$, and $u$ normalizes $C$, then $u$ induces an automorphism of order $1$ or $3$ on $C$, hence $cu=uc^m$ with $m=1$, $4$, or $7$. Given $c$ in Jordan normal form, this yields a system of linear equations of $u$, and one verifies that $u^3\in C$. I think this can be done by hand, but I believe that I used Sage for the computation. $\endgroup$ – Peter Mueller Feb 14 '13 at 23:34

There are no such groups. With GAP one can check this as follows:

Construct the group $G := {\rm AGL}(4,3)$:

gap> G := SemidirectProduct(GL(4,3),GF(3)^4);
<matrix group of size 1965150720 with 3 generators>

Move to the natural permutation representation of $G$ on $3^4 = 81$ points:

gap> phi := IsomorphismPermGroup(G);;
gap> H := Image(phi);
<permutation group of size 1965150720 with 3 generators>
gap> DegreeAction(H);

Find all conjugacy classes of $H$ of elements whose order is divisible by 9:

gap> ccl9 := Filtered(ConjugacyClasses(H),
>                     cl->Order(Representative(cl)) mod 9 = 0);;
gap> List(ccl9,Size);
[ 4043520, 36391680, 12130560, 24261120, 36391680 ]
gap> reps := List(ccl9,Representative);;
gap> List(reps,Order);
[ 9, 18, 9, 9, 18 ]

Compute normalizers of conjugacy class representatives in $H$:

gap> normalizers := List(reps,g->Normalizer(H,Group(g)));
[ <permutation group with 7 generators>, <permutation group with 6 generators>,
  <permutation group with 7 generators>, <permutation group with 4 generators>,
  <permutation group with 6 generators> ]
gap> List(normalizers,Size); # the normalizers are nicely small
[ 2916, 324, 972, 486, 324 ]

Search for transitive metacyclic subgroups of $H$:

gap> List([1..5],i->Filtered(AsList(normalizers[i]),
>                            g -> Order(g) mod 9 = 0 and
>                                 IsTransitive(Group(g,reps[i]),[1..81])));
[ [  ], [  ], [  ], [  ], [  ] ]

-- There are none!

However if we allow for two orbits instead of one, there are solutions:

gap> List([1..5],i->ForAny(AsList(normalizers[i]),
>                          g -> Order(g) mod 9 = 0 and
>                               Length(Orbits(Group(g,reps[i]),[1..81])) <= 2));
[ true, true, false, false, false ]
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