Proving H and its closure are topological groups when H is a subgroup

[tomboi03]

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!

 

[olliemath]

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?