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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.

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Just to see if I am understanding the definition of a zebra group right. So if we take two steps in any series of chief factors, we get groups which alternatingly have normal Sylow-2-subgroups but not Sylow-p-subgroups and vice versa? –  Tobias Kildetoft Dec 20 '10 at 10:01
    
Yes, N(2i+2)/N(2i) has a normal Sylow p-subgroup N(2i+1)/N(2i), but no normal non-identity 2-subgroups, and N(2i+1)/N(2i-1) has a normal Sylow 2-subgroup N(2i)/N(2i-1), but no normal non-identity p-subgroups. G has a unique chief series, since minimal normal subgroups commute. An example of a zebra group with 2 stripes is the symmetric group on 4 points. –  Jack Schmidt Dec 20 '10 at 12:26

2 Answers 2

up vote 4 down vote accepted

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.

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Exactly. I also did not check if there were better choices, but the list of possible choices must be pretty short, since the prime factorization of |GL(5,2)| is pretty short. Do you know of anything relating the p-length to the dimension of a faithful module in characteristic p? –  Jack Schmidt Dec 20 '10 at 12:50
    
If it helps, the obvious route gives 4 stripes of type 2⋉3⋉2^2⋉3^3⋉2^6⋉3^9⋉2^18. It seems like the requirements grow very quickly. –  Jack Schmidt Dec 20 '10 at 12:53
    
Thanks, I used the derived length instead of the p-length, and got a theoretical answer that agreed with experiments. –  Jack Schmidt Dec 20 '10 at 15:52

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).

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