Open neighborhood in Lie group

Question

The following is from "An Introduction to Lie Groups and Lie Algebras" by Kirillov.

Let $G$ be a connected Lie group and $U$ be a neighborhood of $1$. Then if $H$ is the subgroup generated by $U$, $H$ is open because $h\cdot U$ is a neighborhood of $h$ in $G$.

I am not seeing why the last part is true. If the product in $G$ were an open map it would be clear that $h\cdot U$ is a neighborhood of $h$, but why is it true anyways?

Answer 1

Multiplication by $h$ on the left is a homeomorphism from $G$ to itself. In particular, $U$ open implies $hU$ open.

Answer 2

Let $h\in H$, and $L_h(x)=hx$. Since $L_h$ is a diffeomorphism, it is an open map. We deduce that $L_h(U) =\{hu, u\in U\} \subset H$ is open. Since $h=h. 1\in L_h(U) $, we deduce that $H$ contains an open neighbourhood of $h$ and henceforth is open.