Re: 2-Groups containing every group of order 2^n (n fixed)

The search does not take too long if you check a few simple properties
first.

If the group contains the cyclic group Cn of order n, then it has
exponent at least n.

If the group contains both the dihedral group Dn of order n and
elementary abelian group En of order n, then the subgroup generated by
involutions has at least (n/4)*n elements.

You can also look at how some subgroups behave. If H <= G, then Z(H)
= H n Z(G) and so H/Z(H) is a quotient of H / (H n Z(G)). Similarly
one can use the subgroup generated by central involutions, Soc(G), to
get a few more conditions.

At any rate, using a few of these conditions greatly speeds up the
search, and there are many groups of order 256 containing all groups
of order 16 up to isomorphism as subgroups. SmallGroup(256,384) is
one such.

For quotient groups you can speed up the search using the invariant
Log(Size(G/(G^p[G,G])),p), which is called the rank of the p-group G.
In fact the small groups library is organized by rank, so you can
eliminate large portions. You again have some conditions based on the
exponent and based on the lower central series lower and exponent-p
central series.

For quotients you may also have much better luck in constructing
"small" groups containing all the groups of order n. The p-group
construction method finds ``covering groups'' G such that all groups H
with H/H_n = G/G_n, then H=G/N, where G,H are p-groups of p-class n or
so and H_n is the last nontrivial term of the lower exponent-p central
series.

I doubt these examples would be minimal, but they might give some
better upper bounds than a S_{2^n} style argument (or from a certain
point of view, they might simply be the analogy of the symmetric group
idea for quotients).

On Feb 28, 4:11 pm, "Joe Bohanon" <jbohan...@xxxxxxxxxxx> wrote:
Does anyone know if there is a good bound on the smallest k such that
there exists a group of order 2^k containing every group of order 2^n
(where n is fixed) as a subgroup? An iterated wreath product of 'n'
Z2's works since that's the Sylow Subgroup of S_{2^n}. Being lazy I
decided to "GAP it" and found a group of order 32 containing all 5
groups of order 8. Also, none of the groups of order 128 contain
every group of order 16 (computation becomes quite lengthy after that
point).

A few related questions:
Same question but replace 2 by p

Same question but with quotient groups.

Is there a good upper bound on how many of the groups of order 2^n
that a group of order 2^k can have?

Thanks
Joe


.

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