Complete and composition factor-equivalent not implies isomorphic
It is possible to have two finite groups and , both of which are Complete group (?)s, such that they are Composition factor-equivalent groups (?): they have the same list of composition factors (i.e., every simple group occurs the same number of times in the composition series of both groups), and such that and are not isomorphic.
- Semidirect product with self-normalizing subgroup of automorphism group of coprime order implies every automorphism is inner
The construction is as follows.
- The group has two non-isomorphic -Hall subgroups, one isomorphic to the symmetric group on four letters and the other isomorphic to a dihedral group of order . Both are self-normalizing subgroups inside .
- The inverse images of these under the natural projection mapping are two non-isomorphic subgroups of of order . Note that they are both solvable, since their quotient by a central subgroup (the center of ) are the solvable groups mentioned above. Call these groups and . Moreover, from step (1), both and are self-normalizing subgroups of . Further, they are non-isomorphic. (For instance, has no element whose order is a multiple of , but does).
- Let be the elementary Abelian group of order and let and (note that both these groups act naturally on since they're subgroups of , its automorphism group).
- Since is Abelian, and both and are solvable, both and are solvable groups. Moreover, since and have the same order, and are solvable groups of the same order, hence they are composition factor-equivalent.
- By fact (1), both and are groups for which every automorphism is inner.
- An easy check shows that both and are centerless. The fact that there are no central elements inside follows from the fact that has no global fixed points under the action of either . The fact that there are no central elements outside follows from the fact that since is Abelian, any element outside acts on by conjugation like some non-identity automorphism in .
- and are not isomorphic: The subgroup is a normal Sylow subgroup in each , hence it is the only subgroup of its order in each . Hence, any isomorphism from to must preserve , and yield an isomorphism . But we know that and , which are isomorphic to the respective quotients, are not isomorphic to each other.
- Combining steps (4)-(7), and are composition factor-equivalent complete groups that are not isomorphic.
To make things a bit easier, first define the IsCompleteGroup function (either using the interactive interface or in a file to be included. Once this is done, the following GAP implementation constructs the example:
gap> T := ElementaryAbelianGroup(121); <pc group of size 121 with 2 generators> gap> A := AutomorphismGroup(T); <group of size 13200 with 4 generators> gap> HallList := Filtered(List(ConjugacyClassesSubgroups(A),Representative),K -> Order(K) = 240); [ <group of size 240 with 6 generators>, <group of size 240 with 6 generators> ] gap> H1 := HallList; <group of size 240 with 6 generators> gap> H2 := HallList; <group of size 240 with 6 generators> gap> G1 := SemidirectProduct(H1,T); <pc group with 8 generators> gap> G2 := SemidirectProduct(H2,T); <pc group with 8 generators> gap> Order(G1) = Order(G2); true gap> IsSolvableGroup(G1); true gap> IsSolvableGroup(G2); true gap> IsCompleteGroup(G1); true gap> IsCompleteGroup(G2); true gap> IsomorphismGroups(G1,G2); fail