Show H & \bar{H} Topological Groups if H Subgroup of G

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In the discussion, it is established that if H is a subgroup of G and a subspace of G, then both H and its closure \(\overline{H}\) are topological groups. The key requirement is to demonstrate that the multiplication and inverse operations restricted to H remain continuous. Additionally, it is necessary to prove that \(\overline{H}\) is closed under these operations to confirm its status as a topological group.

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tomboi03
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Let H be a subspace of G. Show that if H is also a subgroup of G, then both H and\bar{H} are topological groups.

So, this is what I've got...

if H is a subgroup of G then H \subset G.
Since H is a subspace of G then H is an open subset.

But, i don't even know if that's right.
How do i do this?

Thanks!
 
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Clearly H is a group and it's a topological space. It need not be an open set on the topology on G (if it's a subspace in the sense I think you mean - a subset with the subspace topology).

What you need to show is that the restrictions of the multiplication and inverse maps to H are still continuous.
The \overline{H} is more tricky. You need to show that it is closed under the operations too.

Let me know if this helps?
 

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