Complete and composition factor-equivalent not implies isomorphic

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Statement

It is possible to have two finite groups G_1 and G_2, 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 G_1 and G_2 are not isomorphic.

Facts used

  1. Semidirect product with self-normalizing subgroup of automorphism group of coprime order implies every automorphism is inner

Proof

Construction

The construction is as follows.

  1. The group PGL(2,11) has two non-isomorphic \{2 , 3 \}-Hall subgroups, one isomorphic to the symmetric group on four letters and the other isomorphic to a dihedral group of order 24. Both are self-normalizing subgroups inside PGL(2,11).
  2. The inverse images of these under the natural projection mapping GL(2,11) \to PGL(2,11) are two non-isomorphic subgroups of GL(2,11) of order 240. Note that they are both solvable, since their quotient by a central subgroup (the center of GL(2,11)) are the solvable groups mentioned above. Call these groups H_1 and H_2. Moreover, from step (1), both H_1 and H_2 are self-normalizing subgroups of GL(2,11). Further, they are non-isomorphic. (For instance, H_1 has no element whose order is a multiple of 12, but H_2 does).
  3. Let T be the elementary Abelian group of order 121 and let G_1 = T \rtimes H_1 and G_2 = T \rtimes H_2 (note that both these groups act naturally on T since they're subgroups of GL(2,11), its automorphism group).
  4. Since T is Abelian, and both H_1 and H_2 are solvable, both G_1 and G_2 are solvable groups. Moreover, since H_1 and H_2 have the same order, G_1 and G_2 are solvable groups of the same order, hence they are composition factor-equivalent.
  5. By fact (1), both G_1 and G_2 are groups for which every automorphism is inner.
  6. An easy check shows that both G_1 and G_2 are centerless. The fact that there are no central elements inside T follows from the fact that T has no global fixed points under the action of either H_i. The fact that there are no central elements outside T follows from the fact that since T is Abelian, any element outside T acts on T by conjugation like some non-identity automorphism in H_i.
  7. G_1 and G_2 are not isomorphic: The subgroup T is a normal Sylow subgroup in each G_i, hence it is the only subgroup of its order in each G_i. Hence, any isomorphism from G_1 to G_2 must preserve T, and yield an isomorphism G_1/T \cong G_2/T. But we know that H_1 and H_2, which are isomorphic to the respective quotients, are not isomorphic to each other.
  8. Combining steps (4)-(7), G_1 and G_2 are composition factor-equivalent complete groups that are not isomorphic.

GAP implementation

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[1];
<group of size 240 with 6 generators>
gap> H2 := HallList[2];
<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