Zebra groups and counting stripes 
How many stripes can you paint on a $2$-group of fixed size?

A group of order $2^ap^b$ is solvable, by Burnside's theorem, so its chief factors are either abelian $2$-groups or abelian $p$-groups.  Such a group is a zebra group if its chief factors in any chief series alternate between the $2$ and the $p$.  In other words, if you take a chain of normal subgroups $$1 = N_1 ⊲ N_2 ⊲ \cdots ⊲ N_n = G$$ of maximal length, then the quotient groups $N_{i+1}/N_i$ alternate between being abelian $2$-groups (the stripes) and abelian $p$-groups (the background).

If we fix $a$, say $a=8$, then how many stripes can a zebra group have?

Obviously no more than $8$ $2$-stripes, but with a little work one can see that it can have no more than $4$ $2$-stripes.  Unfortunately, I'm having trouble getting even $3$ stripes.
 A: I doubt whether 3 stripes is possible with $a=8$, although I have not tried to prove it. The problem is that, if you construct the group from the top downwards, then each new layer needs to be a faithful module for the group you have already, which means that the dimensions grow rapidly.
The obvious way to start is a 2 at the top, followed by a 3 and then $2^2$, so we now have $S_4$. The smallest faithful module for $S_4$ over any field has dimension 3, so we can put a $3^3$ under the $S_4$ to get a group $G$ of order $2^33^4$. Now the smallest faithful irreducible module for $G$ over the field of order 2 has dimension 6 (I did that calculation in Magma, but I expect you can do it in GAP), so we get a group of order $2^93^4$ with three stripes, which is a subgroup of ${\rm AGL}(6,2)$, and I would be surprised if you can do better.
A: 
2 is the maximum number of stripes possible on a zebra group with a=8.  S stripes require  a ≥ C⋅9S  for large S.

The number of stripes S of a zebra group gives bounds on the derived length D of a zebra group: 2S - 1 ≤ D ≤ 2S + 1.  The action of G/H on a chief factor H/K must be irreducible and faithful, and so G/H embeds in GL(H/K) = GL(n,q) for q in {2,p} the prime dividing the order q^n of H/K.  However, for large n, the maximal derived length of a solvable subgroup of GL(n,q) is about log(n-2)/log(9), or in other words, n ≥ 9^D.  In particular, as S increases, the "a" from 2a pb increases exponentially.
For small n, the maximal derived length of soluble subgroups of GL(n,2) are: 2, 3, 4, 4.  The maximal derived length of irreducible soluble subgroups of GL(n,2) appear to be 2,2,3,2,6,2,5,4,4.  Hence the minimum dimensions for derived length D=2 is n=2, D=3 is n=4, D=4 is n=6, D=5 is n=6, D=6 is n=6, and D≥7 is n≥11.
The top stripe has order at least 2^1.  By the time H/K is the second stripe, G/H has derived length 2, and so H/K has order at least 2^2.  By the time H/K is the third stripe, G/H has derived length 4, and so H/K has order at least 2^6.
  Since 1+2+6 ≥ 8, this shows there is no three-striped zebra group with a Sylow 2-subgroup of order 2^8.
Three stripes is attainable at a=9.  Four stripes is not attainable for a≤14 and is attainable for a=27.  Roughly speaking, S stripes require a ≥ C⋅9S (though my proof needs S really large).
