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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.

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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.

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