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I am given $G$ locally compact group, and I want to show that there exists a clopen subgroup $H$ of $G$ that is $\sigma$-compact.

So here's what I did so far:

for $e \in U$, where $U$ is a nbhd of the identity element we know that $x \in x\bar{U}$, and $x\bar{U}$ is compact in $G$, now take some finite open cover of this set,$\cup_{i=1}^{n} V_i$, again this set is open nbhd of $x$ so again its closure is compact, and its closure equals the union of $\bar{V_i}$, now I am kind of stuck, I wish I could take this set as the clopen subgroup, but I don't think that I know that it's closed under multiplication, right?

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Hint: Take a compact symmetric neighborhood $K$ of the identity and consider the $\sigma$-compact subgroup $H = \bigcup_{n =1}^{\infty} K^n$.

  1. Prove that $H$ is an open subgroup.

  2. Prove that an open sugroup of a topological group is closed — its complement is a union of (open) cosets.


As additional exercises I suggest:

  1. Prove that the connected component of the identity is a clopen subgroup. (Connected components are open and open subgroups are closed)

  2. Prove that a connected locally compact group $G$ is $\sigma$-compact.

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  • $\begingroup$ Ah, thanks. it does look closed under multiplication. $\endgroup$ – MathematicalPhysicist Dec 9 '11 at 11:43
  • $\begingroup$ Yes, and since $K^{-1} = K$ it is closed under inversion. $\endgroup$ – t.b. Dec 9 '11 at 11:44
  • $\begingroup$ I have another question, which I think uses the same trick, and I don't want to open another post just for it. I am given $G$ a topological group, and I want to show that every compact clopen nbhd of $e$ contains a compact clopen subgroup, $H<G$. Now, here I think I just need to take $K$ which is compact and clopen nbhd of $e$ (which I can always take as being symmetric), and then take $H=\cap_{n=1}^{\infty}K^n$, obviously it's contained in $K$, and it's a subgroup, and it's compact as a countable intersection of compact sets, is this right for this question? $\endgroup$ – MathematicalPhysicist Dec 10 '11 at 11:52
  • $\begingroup$ You're right that the idea is very similar. However, your idea doesn't seem to give an open subgroup in general. See e.g. Hewitt-Ross, Theorem (7.5), p.61 for the standard argument. $\endgroup$ – t.b. Dec 10 '11 at 12:06

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