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I'm looking for an example of a finite abelian group A and a finite group G acting trivially on A such that there are two extensions $E_1$ and $E_2$ with base A and quotient G (i.e., they are both central extensions, and hence both give corresponding elements of $H^2(G,A)$) and:

  1. $E_1$ and $E_2$ are isomorphic as abstract groups.
  2. Under the natural action of $\operatorname{Aut}(G) \times \operatorname{Aut}(A)$ on $H^2(G,A)$ (by pre- and post-composition with 2-cocycles that then descends to action on cohomology classes), the cohomology classes corresponding to $E_1$ and $E_2$ are not in the same orbit.

Basically condition (2) states that $E_1$ and $E_2$ are not only not congruent extensions, they are not even congruent up to a relabeling of the subgroup A and the quotient G. Another way of putting this is that there is no isomorphism between $E_1$ and $E_2$ that sends the A inside $E_1$ to the A inside $E_2$.

The analogous statement with a nontrivial action of G on A is also of interest to me. In this latter case, though, the entire group $\operatorname{Aut}(G) \times \operatorname{Aut}(A)$ does not act.

I think that examples exist (because of my experience with finding examples for similar specifications) but there may well be a proof to the contrary.

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1 Answer 1

up vote 11 down vote accepted

E = SmallGroup(32,28) is the first example. It has two central subgroups A1 and A2 isomorphic to A ≅ 2 with quotient isomorphic to SmallGroup(16,11), but A1 and A2 are not conjugate in Aut(E).

Examples such as this are reasonably common in p-groups.

Edit: You can even have such an example with G abelian: G = 4×2, A = 4×2, E = SmallGroup(64,3) = 8⋉8, Z(E) = 4×4. E has two central copies of A=4×2 that are not conjugate in Aut(G), but the quotients are both abelian and isomorphic to G.

Edit: Vipul notes you can even have E abelian of order p7.

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I checked this example using GAP and GAP seems to say that the two quotients should be SmallGroup(16,11) each (which is $D_8 \times \mathbb{Z}_2$) rather than SmallGroup(16,3). Perhaps you had some other group in mind? Anyway, this one works. –  Vipul Naik Aug 15 '10 at 21:35
Also, do you think an example of this sort would exist for <em>G</em> abelian as well? –  Vipul Naik Aug 15 '10 at 21:42
@Vipul: Thanks for the typo catch (search was coded up nicely, examining the group was done by hand). I wouldn't be too surprised either way for G abelian: on the one hand you just need some room to work in order for different Aut(G) orbits of Proj(H^2(G,A)) to be abstractly isomorphic, but on the other hand Aut(G) for G elementary abelian is really pretty transitive, so maybe there are not enough orbits. But for G nilpotent of class 2, I would expect a ton of examples. 11% of order 64 have examples. –  Jack Schmidt Aug 15 '10 at 22:44
If you want to generate p-groups systematically by their central series, then the "p-group generation algorithm" (implemented nicely in ANU pq for GAP and magma) is the way to go. This silliness with multiple copies of A goes away if you can "name" A, say as the last term of the lower p-central series. –  Jack Schmidt Aug 15 '10 at 22:46
Also, if you drop the "central" / "trivial action" requirement, then the multiple A silliness is pretty common in all groups (the "general" group only rarely has central factors, so having multiple, non-conjugate central factors is also rare). This problem was one of the reason the Holt–Plesken perfect group project tapered off at the point it did: deciding queer isomorphisms was getting too hard. However, if one was happy only finding the groups with small lower central factors (of the Fitting series of the solvable radical), then one could go much further. –  Jack Schmidt Aug 15 '10 at 22:49

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