Hypothesis: Let $H$ be a normal subgroup of $G$. Let $p: G \rightarrow G/H$ form the universal covering over $G/H$. Let $\phi$ be an arbitrary deck transformation of $G$.
Question: Why is it that $\phi(1) \in H$? Do we have also that, necessarily, $\phi(1) = 1$?
If $\phi$ is a deck transformation, then by definition, $p \circ \phi = p.$ Hence, $p(\phi(1)) = 1H$, and hence $\phi(1) \in H.$
On the other hand, the universal cover is always a regular covering space and hence, the space of deck transformations act transitively on the fiber of any point. Hence, since the fiber above $1H$ is all of $H$, for any $h \in H$, there is a deck transformation $\phi$ such that $\phi(1) = h.$