# Complete and composition factor-equivalent not implies isomorphic

## 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