# Intersection of connected components with discrete subgroup

I am currently studying Harmonic Analysis and didn't quite understand part of the proofs for the structure theorems of locally-compact abelia groups (LCA).

Let $$A$$ be an LCA group such that $$\hat{A}/\hat{A}_0$$ is compact, where $$A_0$$ denotes the connected component of the unit in $$A$$ and $$\hat{A}$$ is the dualization of $$A$$.

Now suppose that there is an infinite compact subgroup $$E \leq A$$. We can write this as an exact sequence

$$0 \to E \to A\to A/E\to 0$$

and dualizing it to the exact sequenz $$0 \to B \to \hat{A} \to \hat{E}\to 0 .$$ Where $$\hat{E}$$ is an infinite discrete group and as such non-compact. This is supposedly a contradiction to $$\hat{A}/\hat{A}_0$$ being compact, but I don't see why. By that, we would know that no such group can exist in $$A$$.

The map $$\Phi$$ from $$\hat{A} \to \hat{E}$$ is surjective and continous by construction, and we can restrict it to a map $$\varphi: \hat{A}/\hat{A}_0 \to \hat{E}/(\hat{A}_0\cap\hat{E}).$$ If we knew that $$\hat{A}_0\cap\hat{E}$$ is finite (or even better $$=\{e\}$$) then we would be done, having a surjection from a compact to a non compact group.

Is this the right way to go or am I missing something important? Any help would be greatly appreciated.

## 1 Answer

If by contradiction $$A$$ has an infinite compact subgroup $$E$$, then, as you said, $$\hat{A}$$ has the infinite discrete quotient $$\hat{E}$$. Since $$\hat{A}_0$$ has a trivial image in $$\hat{E}$$, this is an infinite discrete quotient of $$\hat{A}/\hat{A}_0$$. But the latter is compact, so its discrete image should be finite. Contradiction.