I am not sure if this is already posted, though, I hope I can get some help, and thank in advance. This question arises from the proof of the following.
Proposition: Let G be a topological group, of which H is a subgroup. Then, H is closed in G, if, and only if, there exists a neighborhood U of 1 in G, such that the intersection of U and H is closed in G.
During the proof, there is an assumption that confuses me, that is, it assumes a neighborhood V of 1, such that V=$V^{\iota}$, and that V*V is a subset of U. But, as far as I am concerned, there is no use of the assumption that $V^{\iota}$=V, and this is exactly my question.
Is it true that we always can find, for every neighborhood U of 1 in G, a neighborhood V of 1 in G such that V=$V^{\iota}$, and that V*V is a subset of U?
I know that the latter assumption results from the continuity of the multiplication, but I am very confused by the former. Thanks and regards here.