# Extension of a finite group by a connected group necessarily splits?

Suppose that $$G$$ is a compact abelian group. I denote by $$G_0$$ the connected component of the identity in $$G$$.

If $$G_0$$ is open in $$G$$ (equivalently $$G/G_0$$ is finite) is it true that $$G\cong G_0\times G/G_0$$?

I also assume that $$G/G_0$$ is equipped with the quotient topology and that everything is Hausdorff.

*If the title confuses you, the statement above is equivalent to whether the short exact sequence $$0\rightarrow G_0\rightarrow G \rightarrow G/G_0\rightarrow 0$$ splits.

• If $G$ is a Lie group then the answer is yes, if it's any help. – Cronus Dec 19 '18 at 21:49
• If $G$ is a compact abelian Lie group then it is a torus times a finite group. So of course it works. I'm reading your answer though – Yanko Dec 20 '18 at 18:21
• But do you know of a simple proof for that (that a compact abelian Lie group is a torus times a finite group)? – Cronus Dec 21 '18 at 16:29
• Actually, you're right, this fact is not hard to prove. – Cronus Dec 22 '18 at 12:35

## 1 Answer

Yes, this is true. The case for Lie groups is pretty easy (and I explained it here). For the general case, one can prove this using the fact compact groups are inverse limits of Lie groups.

Let $$G$$ be a compact abelian group, and denote $$A=G/G^0$$. By a corollary of Peter-Weyl, every identity neighbourhood contains a subgroup which is co-Lie (i.e. the quotient by it is a Lie group). Thus we can form a net of subgroups $$N_\alpha$$, ordered by reverse inclusion (so $$\alpha \geq \beta$$ if and only if $$N_\beta \subseteq N_\alpha$$), such that $$N_\alpha\subseteq G^0$$ for all $$\alpha$$ (sine $$G^0$$ is open) and such that $$\bigcap N_\alpha=\{1\}$$. Denote $$G_\alpha=G/N_\alpha$$ and denote by $$p_\alpha:G\to G_\alpha$$ the natural quotient map.

Since each $$p_\alpha$$ is open, continuous and surjective, the subgroup $$p_\alpha(G^0)$$ is open and connected, and hence is the connected component of $$G_\alpha$$. Since $$N_\alpha\subseteq G^0$$, we have by the third isomorphism theorem (for topological groups) that $$G_\alpha/G_\alpha^0=(G/N_\alpha)/(G^0/N_\alpha)\cong G/G^0 = A$$. Thus, by the case for Lie groups, we know $$G_\alpha \cong G_\alpha^0\times A_\alpha$$ for $$A_\alpha\cong A$$.

Denote $$A'_\alpha = p_\alpha^{-1}(A_\alpha)$$ and $$A'=\bigcap A'_\alpha$$. I will show that $$A'$$ is mapped isomorphically onto $$A$$ via the quotient $$q:G\to G/G^0$$, which shows the sequence does indeed split.

First let us see that it is mapped injectively into $$A$$ via $$q$$. We have, for every $$\alpha$$, $$p_\alpha(A'_\alpha\cap G^0)\subseteq p_\alpha(A'_\alpha)\cap p_\alpha(G^0)=A_\alpha\cap G_\alpha^0=\{1\}.$$ Thus $$A'_\alpha\cap G^0\subseteq N_\alpha$$ for every $$\alpha$$, so $$\bigcap A'_\alpha\cap G^0\subseteq\bigcap N_\alpha=\{1\}$$. This precisely mean that the intersection of $$A'=\bigcap A'_\alpha$$ with $$\ker q=G^0$$ is trivial, i.e. $$q$$ mapps $$A'$$ injectively into $$A$$.

It remains to show $$q(A')=A$$. I think this is pretty clear. For every $$\alpha$$ we know $$q(A'_\alpha)=A$$. To see this, first note that $$p_\alpha(A'_\alpha\cdot G^0)$$ contains both $$p_\alpha(A'_\alpha)=A_\alpha$$ and $$p_\alpha(G^0)=G_\alpha^0$$ and hence is all of $$G_\alpha=G_\alpha^0\times A_\alpha$$, and therefore $$A'_\alpha\cdot G^0\cdot \ker p_\alpha = G$$, but $$\ker p_\alpha = N_\alpha \subseteq G^0$$, so $$A'_\alpha\cdot G^0=G$$. Since $$\ker q= G^0$$ we have $$q(A'_\alpha)=A$$. But by compactness this implies $$q(A')=A$$ (since if $$a\in A$$ then for every $$\alpha$$ there is $$a_\alpha\in A'_\alpha$$ such that $$q(a_\alpha)=a$$; this net has some converging subnet which converges to some $$a'\in A$$, and by continuity $$q(a')=a$$).

This answer turned out a lot longer than I expected... Using the language of inverse limits it would have been a lot shorter, I think, if you know how to prove $$G\cong \varprojlim G_\alpha$$ in a natural way and not just that every neighbourhood contains a co-Lie subgroup.

• Very nice thank you! – Yanko Dec 20 '18 at 18:27