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. 

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.
