# How many non-isomorphic extensions with kernel $S^1$ and quotient cyclic of order $p$?

I want to determine how many non-isomorphic extensions (as group they are non-isomorphic) are possible of the form $$1 \to \mathbb{S}^1 \to G \to (\mathbb{Z}_p)^k \to 1$$, where $$G$$ is a compact lie group and $$\Bbb{Z}_p$$ is the cyclic group of order $$p$$? The only possible homomorphism $$(\Bbb{Z}_p)^k \to Aut(\mathbb{S}^1)=\Bbb{Z}_2$$ is the trivial one if $$p\neq 2$$. If $$p=2$$, then we have $$(\Bbb{Z}_2)^k$$ possible homomorphism $$(\Bbb{Z}_2)^k \to \Bbb{Z}_2$$. Then we have to calculate the group $$H^2((\Bbb{Z}_p)^k; \mathbb{S}^1)$$. I don't know how to proceed further. A detailed answer will be very helpful.

Thank you very much in advance.

• I guess you're interested in extension of topological groups. – YCor Apr 30 at 8:46
• Oh, you mean $\mathbf{Z}_p$ cyclic of order $p$? it traditionally means the $p$-adic group (I started thinking about the question in this way) it's confusing. – YCor Apr 30 at 8:51
• I modified the question. Thanks – mathstudent Apr 30 at 8:53

## 2 Answers

Let's only consider central extensions (so we only miss a few cases when $$p=2$$). I denote by $$C_p=\mathbf{Z}/p\mathbf{Z}$$, to avoid confusion with $$p$$-adics.

So extensions are classified by $$H^2(C_p^k,S^1)$$. The commutator map yields a canonical homomorphism $$\phi$$ from $$H^2(C_p^k,S^1)$$ onto $$\mathrm{Hom}(\Lambda^2C_p^k,S^1)$$, and the latter is isomorphic to $$C_p^{k(k-1)/2}$$ (more canonically, the dual of $$\Lambda^2C_p^k$$). The kernel of this homomorphism consists of those 2-cocycle defining an abelian extension; since $$S^1$$ is an injective $$\mathbf{Z}$$-module the only abelian extension is split so $$\phi$$ is an isomorphism. Hence, the extensions are classified by $$C_p^{k(k-1)/2}$$, more precisely the space of alternating bilinear forms on $$C_p^k$$.

Next, the isomorphism class of the groups thus obtained are classified by the quotient of the latter by the action of $$\mathrm{GL}_k(\mathbf{Z}/p\mathbf{Z})$$. This quotient has exactly $$1+\lfloor k/2\rfloor$$ elements (number of types of alternating forms in dimension $$k$$, regardless of the field, including characteristic 2).

Note: For $$k=1,2,3$$ this makes $$1$$, $$2$$, $$2$$ isomorphism types of Lie groups. That for $$k=2,3$$ it does not match with Ian's answer (3, 8) is because he also considers quotients of order $$p^k$$ that are not $$p$$-elementary.

• Thank you very much. So if $k=4$, then there are 3 non-isomorphic groups. Is there any general way to determine all these groups(maybe the presentation of such groups). In general, for a given k is there any way to determine what are these $1+[k/2]$ groups (maybe how to determine their presentation)? – mathstudent May 4 at 8:14
• For the choice of non-degenerate alternating form, denote by $G_{2\ell}$ the corresponding central extension of $S^1$ by $C_p^{2\ell}$. Then the groups are just $G_{2\ell}\times C_p^m$ for all $\ell,m$ such that $2\ell+m=k$. For instance for $k=4$ these are $G_0\times C_p^4$ (the direct product), $G_2\times C_p^2$, and $G_4$. – YCor May 4 at 9:08
• Thanks. Yes, now I understand it clearly. I need to determine the possible presentation of such groups. So from your answer, I understand that for a given $k$ if we can determine the presentation of $G_k$ (following your notation), then all others can also be deduced. It will very helpful if you please help me to determine the presentation of such groups. – mathstudent May 4 at 14:38

There is a short exact sequence of coefficients $$\mathbb{Z}\rightarrow \mathbb{R}\rightarrow \mathbb{R}/\mathbb{Z}=S^1$$. This gives you a long exact sequence of cohomology groups. For a finite group $$Q$$, $$H^i(Q:\mathbb{R})=0$$ for $$i>0$$, so the long exact sequence collapses to an isomorphism $$H^2(Q;S^1)\cong H^3(Q;\mathbb{Z})$$.

You can use this to classify the isomorphism types of 1-dimensional compact non-connected Lie groups as in your question. For $$p$$ odd, there is only the direct product for $$S^1\rightarrow G\rightarrow C_p$$. For $$p$$ odd, there are two isomorphism types of group $$S^1\rightarrow G \rightarrow (C_p)^2$$, and there are two isomorphism types of group $$S^1\rightarrow G\rightarrow (C_p)^3$$. Of course it gets more complicated as $$k$$ increases, and $$p=2$$ is more complicated than odd $$p$$.

For $$p=2$$ there are already three groups $$S^1\rightarrow G\rightarrow C_2$$: the direct product, the orthogonal group $$O(2)$$, and the subgroup of the unit quaternions generated by the circle $$\cos(\theta)+i\sin(\theta)$$ and $$j$$.

In the first chapter of my PhD thesis (available on ArXiv in an extended version as https://arxiv.org/abs/0711.5020) I classified 1-dimensional compact Lie groups with $$p^k$$ components for each prime $$p$$ and each $$k\leq 3$$. The numbers that I got were 1, 3, 8 for $$p$$ odd and 3, 8, 29 for $$p=2$$. I presume that this was already known and I did not try to publish it.