Slight reordering of composition factors in a finite solvable group of order 4 mod 8

Question

Given a solvable finite group G of order 4 mod 8, I would like to collect the 2-torsion in adjacent positions, if possible. The Sylow 2 subgroup has order 4, and so is either cyclic or the Klein 4-Group $V_4$. In the first case, we know that the Sylow 2 subgroup has a normal complement by Burnsides theorem, so the group is a semi-direct product of a group of odd order, and the cyclic group of order 4. In the other case (Sylow 2 subgroup $V_4$) it is not clear whether there is a composition series with the two $Z/2$ quotients adjacent or at the beginning.

Answer 1

EDIT: Added a few details, to make it less sloppy.

Consider a chief series for $G$. Then each factor is elementary abelian. We may assume that we have two "separated" $C_2$ factors, otherwise we are done. Therefore, somewhere in our normal series, we see $C_2.X.C_2$, where $X$ is a group of odd order (potentially grouping a few odd factors of our series).

Since we started with a normal series, the "bottom" $C_2$ is actually normal in the group $C_2.X.C_2$ and therefore central. Moreover, by Schur-Zassenhaus (or Hall's theorem), the $C_2.X$-extension is actually split, so direct and $C_2.X.C_2=(C_2\times X).C_2$. Now, $X$ is characteristic in $C_2\times X$ so normal in $(C_2\times X).C_2$, so we can write $C_2.X.C_2=C_2.C_2.X$, as required.

(I used Atlas notation for extensions so in a sense this is a more of a sketch than anything, but I hope it was clear. Note that I did not use the shape of the Sylow $2$-subgroup, this covers the cyclic case as well. Of course I did use solvability, which is clearly necessary.)

Addendum 1: We did not use the shape of the Sylow $2$-subgroup, but we do use the fact that it has order $4$. Otherwise, we could reach the following situation : $N_4.X.C_2$, where $N_4$ is some normal subgroup of order $4$, and we cannot proceed as before. ($S_4$ is an example where this happens.) We also use solvability, obviously, otherwise the question barely makes sense (or consider $A_5$).

Addendum 2: Under the hypothesis of the question, we could actually show that either the Sylow $2$-subgroup has a normal complement, or $G=X.A_4$ for some normal subgroup $X$.

Addendum 3: With just a little more work, one could do the same thing for a soluble group with order divisible by $p^2$ but not by $p^3$, for a prime $p$. At some point, we would need to use the fact that no subgroup of $Aut(C_p)$ has an automorphism of order $p$.