# An example of a group of order 336, not isomorphic to $PGL(2,7)$.

I need an example of a finite group $G$ by the following properties:

1) Order $G$ is $336$.

2) For every prime $p$, $G$ has not any elements of $7p$.

3) the number of Sylow $7$-subgroups $G$ is $8$.

4) $G$ is not isomorphic to $PGL(2,7)$.

Can anybody help me!

-
I want a question to be asked, and not commands to be given. – amWhy Dec 28 '12 at 15:22
@amWhy Meh. ${}{}$ – Pedro Tamaroff Dec 28 '12 at 15:35
@amWhy: It sounds like a colored menu. ;-) – Babak S. Dec 28 '12 at 15:37
Use the GAP - Groups, Algorithms, Programming - a System for Computational Discrete Algebra. See gap-system.org – MathOverview Dec 28 '12 at 15:46
I am afraid that there are no groups satisfying 1) - 4). Properties 1), 2), 3) imply that the normalizer of a Sylow 7-subgroup is a Frobenius group of order 42, and hence that $G$ acts 3-transitively on its Sylow 7-subgroups. The only such group is ${\rm PGL}(2,7)$. – Derek Holt Dec 28 '12 at 17:42

This question had already been answered in comments by Derek Holt who gave purely theoretical explanation: a comment of him says "Properties 1), 2), 3) imply that the normalizer of a Sylow 7-subgroup is a Frobenius group of order 42, and hence that $G$ acts 3-transitively on its Sylow 7-subgroups. The only such group is $PGL(2,7)$".

As another comment by @Elias suggests to use GAP, I would also show how to check this in GAP, so that the question may be removed from the unanswered queue:

Step 1: List all groups of order 336:

gap> l:=AllSmallGroups(336);;
gap> Length(l);
228


Step 2: Filter those from condition (2):

gap> l1:=Filtered(l, g -> ForAll(ConjugacyClasses(g),
> c -> not IsPrimeInt(Order(Representative(c))/7)));
[ Group([ (1,4,6,8,5,2,7,3), (1,3,8,6,5,4,7) ]) ]


Observe that this condition already lefts us with only one group.

Step 3: Filter those from condition (3):

gap> l2:=Filtered(l1, g -> Size(First(ConjugacyClassesSubgroups(g),
> c -> Order(Representative(c))=7))=8);
[ Group([ (1,4,6,8,5,2,7,3), (1,3,8,6,5,4,7) ]) ]


Step 4: Check the isomorphism type of the group: it is $PGL(2,7)$ since it thas the same IdGroup:

gap> IdGroup(l2[1]);
[ 336, 208 ]
gap> IdGroup(PGL(2,7));
[ 336, 208 ]


One could also see this from here:

gap> StructureDescription(l2[1]);
"PSL(3,2) : C2"
gap> StructureDescription(PGL(2,7));
"PSL(3,2) : C2"

-