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Given a group $H$ and a finite group $F$. Consider the group extension

$1\to F\to G \to H \to 1$.

My question is, are there only finitely many isomorphism classes of $G$ satisfying the above exact sequence?

Jiayin Pan
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  • @qiaochu Thanks for the example. What if H is finitely generated? – Jiayin Pan Feb 18 '15 at 14:49
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    it would be better to ask that as a separate question. Off the top of my head, I'm not sure. – Qiaochu Yuan Feb 18 '15 at 18:22
  • I think there are infinitely many isomorphism types of extensions of $F=C_2$ by the Lamplighter Group $C_2 \wr {\mathbb Z}$. – Derek Holt Feb 18 '15 at 21:29
  • @DerekHolt: As the extensions proposed in your comment have to be central, how do you hope to put a central element of order $2$ "below" the Lamplighter Group? (I have a hard time finding a second possible $G$.) – j.p. Feb 19 '15 at 13:32
  • It's not easy to give details in a comment! – Derek Holt Feb 19 '15 at 17:59
  • @DerekHolt: You're right. As you still seem to think that your construction works, I'll post it as a question. – j.p. Feb 19 '15 at 18:43

2 Answers2

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No. Take $F = \mathbb{Z}_2$ and

$$H = \bigoplus_{k \ge 1} \mathbb{Z}_{2^k}.$$

There are infinitely many non-isomorphic extensions of $H$ by $F$, namely the extensions

$$G_n = \bigoplus_{k \ge 1, k \le n - 1} \mathbb{Z}_{2^k} \oplus \mathbb{Z}_{2^{n+1}} \oplus \bigoplus_{k \ge n+1} \mathbb{Z}_{2^k}.$$

This follows from the result given in this math.SE question that it's possible to distinguish the number of summands isomorphic to $\mathbb{Z}_{2^k}$ in a direct sum of cyclic groups.

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Qiaochu Yuan
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  • You beat me two it by two minutes, but I think the proof of non-isomorphism is not completely trivial. – Derek Holt Feb 18 '15 at 09:07
  • @Derek: oh, you're right, the thing I said just doesn't work... hmm. Did you also start working through a general group cohomology calculation before just exhibiting the examples? – Qiaochu Yuan Feb 18 '15 at 09:12
  • No I just thought of the example directly, but I then spent about 5 minutes before posting it convincing myself that I could prove that the groups are all non-isomorphic. – Derek Holt Feb 18 '15 at 09:13
  • If you take $F=Z_3$ and pick different homomorphisms from $H$ to $Aut(F)$ (only the $k$th summand of $H$ being not in the kernel) then the proof that the groups are not isomorphic follows easily from looking at the central elements of $G$. – j.p. Feb 18 '15 at 09:18
  • @j.p. I don't see that. The method works, but proving non-isomorphism seems more or less equally hard as in the abelian examples. – Derek Holt Feb 18 '15 at 09:30
  • @DerekHolt: You can distinguish the groups by checking if all elements of order $2^n$ are central or not. – j.p. Feb 18 '15 at 14:39
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No. Let $p$ be a prime, let $F = C_p$, and let $H = \times_{k=1}^ \infty C_{p^k}$ be the direct product of cyclic groups of order $p^i$, with one factor for each $k > 0$.

Then there are infinitely many isomorphism classes of abelian groups $G$ satisfying the hypothesis, because $G$ could isomorphic to $$\times_{k=1}^{i-1}C_{p^k} \times C_{p^{i+1}} \times C_{p^{i+1}} \times_{k=i+2}^\infty C_{p^k}$$ for any $i>0$. (I'll leave it to the reader to prove that these possibilities are mutually non-isomorphic.)

Derek Holt
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