Show that in a metrizable group, every neighborhood $U$ of the neutral element $e$ contains a neighborhood $V \ni e $ such that $V\cdot V^{-1} \subset U$, where $V\cdot V^{-1} = \{x\cdot y^{-1}; x\in V, y\in V\}.$ A neighborhood $V \ni e $ is said to be $\textbf{symmetric}$ when $V = V^{-1}$ , where $V^{-1}= \{x^{-1}; x\in V\}$. And show that every neighborhood of $e$ contains a symmetric neighborhood .
I do not know how to get this issue if they can give you some tips to solve it would be very grateful.
Once you know that whenever $V$ is a neighborhood of the identity element $e$, so is the set $V^{-1} = \{x^{-1} : x \in V \}$ a neighborhood of $e$, then it is natural to consider the intersection $U = V \cap V^{-1}$ of both neighborhoods.
Now try to prove that $U$ is a symmetric neighborhood of $e$, as required. That is,
(a) Show $U$ is a neighborhood of $e$.
(b) Show $x \in U$ implies $x^{-1} \in U$.
I'd point out that this is only using the topology of a group (via continuity of the inverse operation), and so it is more general than only concerning metrizable groups.