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I found a proof in the book Fourier Analysis On Number Fields that the closure of any subgroup is a subgroup, using the continuity argument along with the nets.
Nevertheless, the following proof seems also plausible, but is it valid?

Statement: If $H$ is a subgroup of a topological group $G$, then so is its closure in $G$.
Denote the closure of $H$ by $H^{closure}$. Let h,h' be two elements in $H^{closure}$, and let $U$ be a neighbourhood of the unity in $G$. Since h and h' lie in $H^{closure}$, there are elements a, a' in $U$ such that
ha$\in H$,
h'a'$\in H$.
Then, as the congugation function is continuous, the left and the right toplogies are the same, and hence there are b, b' $\in U$ such that
hah'a'= bhh'a'= bh"=h"b' $\in H$, where h" is in $H^{closure}H$ $\subseteq H^{closure}$.
As to $H^{closure}H$ $\subseteq H^{closure}$, take two elements e $\in H^{closure}$, f $\in H$, and $v$ in $V$, an arbitrary neighbourhood of unity in $G$. Then there is an element $g$ in $V$ such that
$efv=(eg)f \in HH \in H$.
So $H^{closure}H$ $\subseteq H^{closure}$.

The proof for the inversion is similar; is there any problem in the argument? Thank a lot here.

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P.S. C.Q.F.D. – awllower Nov 23 '11 at 8:11
Are you not supposed to show $H^{closure}H^{closure} \subseteq H^{closure}$ and ${H^{closure}}^{-1} \subseteq H^{closure}$ ? – Matthias Klupsch Nov 23 '11 at 8:15
I showed the first, by means of showing that $H^{closure}H \subseteq H^{closure}$, but then was tired, and the proof is similar, so I will edit again, if needed. – awllower Nov 23 '11 at 8:21
up vote 12 down vote accepted

Sorry... this is not the answer to your question, but notice that the product with inversion $f: G \times G \to G$ defined by $f(x,y) = xy^{-1}$ is continuous.

Therefore, $f^{-1}(\overline{H})$ is closed. Now, notice that $H \times H \subset f^{-1}(\overline{H})$. So, taking closures, $$ \overline{H \times H} \subset f^{-1}(\overline{H}). $$ Now, you just have to show that $\overline{H} \times \overline{H} \subset \overline{H \times H}$, to conclude that $$ f(\overline{H} \times \overline{H}) \subset \overline{H}. $$

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It is extraordinary!! I think even though you did not tell me if the above answer is correct, you did indeed tell me that there is a proof without the concept of nets, which was my initial incentive for this question. Thus I am going to accept the answer. Regards. – awllower Nov 24 '11 at 2:51
@awllower: Glad to help! :-) – André Caldas Nov 24 '11 at 3:02

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