2
$\begingroup$

In a comment to the question Finitely many group extensions? Derek Holt mentioned that for the Lamplighter Group $L = C_2\wr\mathbb{Z}$ and the cyclic group $C_2$ with two elements there should be many isomorphism classes of $G$ satisfying the exact sequence

$1\to C_2\to G \to L \to 1$.

What do these groups G look like?
$\endgroup$
3
  • $\begingroup$ I don't have a clue, but that wreath symbol is driving me nuts! Why not \wr \mathbb{Z}? $\endgroup$
    – pjs36
    Feb 19, 2015 at 19:07
  • $\begingroup$ @pjs36: Sorry for driving you nuts. Earlier I didn't have time to look up "\wr" and just copied and pasted from the other question. $\endgroup$
    – j.p.
    Feb 19, 2015 at 21:08
  • $\begingroup$ No offense intended, I just found it funny, that's all! Why mixing high-level math with ASCII is amusing, I'm not sure... $\endgroup$
    – pjs36
    Feb 19, 2015 at 21:24

1 Answer 1

3
$\begingroup$

Let $n>0$, let $N_n$ be the group defined by the presentation $\langle X \mid R \rangle$, where $X = \{ x_i :i \in {\mathbb Z}\} \cup \{t\}$, and $$R = \{t^2\} \cup \{x_i^2:i \in {\mathbb Z} \} \cup\{[x_i,t]:i \in {\mathbb Z} \} \cup \{ [x_i,x_{i+n}]t^:i \in {\mathbb Z}\} \cup \{[x_i,x_{i+k}]: i,k \in {\mathbb Z}, k>0,k \ne n \}.$$ So $N_n$ is a central extension by $C_2=\langle t \rangle$ by the infinite elementary abelian $2$-group spanned by the $x_i$. The map $x_i \mapsto x_{i+1}$, $t \mapsto t$ prerves $R$ so it induces an automorphism $\alpha$ of $N_n$, and the semidirect product $G_n = N_n \rtimes_{\alpha} {\mathbb Z}$ is a central extension of $C_2 = \langle t \rangle$ by the Lamplighter Group L=$C_2 \wr {\mathbb Z}$.

We claim that $G_n \not\cong G_m$ for $n \ne m$. Since any such isomorphism would have to mat $N_n$ to $N_m$, it would be induced by an automorphism $\beta$ of $L$. But any such $\beta$ either induces the identity on the quotient ${\mathbb Z}$ of $L$ and translates the $x_i$ by $\beta(x_i) = x_{i+m}$ for some $m$, or it inverts the quotient ${\mathbb Z}$ and relects the $x_i$ by $\beta(x_i) = x_{m-i}$ for some $m$, and in either case, $\beta$ induces an automorphism of each $G_n$. (If $\beta(x_0)$ is a sum of of more than one $x_i$ then it is not hard to see that $x_0 \not\in {\rm im}(\beta)$, so $\beta$ cannot be an automorphism.)

$\endgroup$
1
  • $\begingroup$ Very nice, thank you. I'm still struggling to understand if the extra-special $2$-groups $N_n$ are isomorphic or not (if you happen to know, a simple yes/no would be appreciated a lot), but the $G_n$ are different. $\endgroup$
    – j.p.
    Feb 20, 2015 at 7:23

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .