Proposition 1.39, Hatcher - Quotient By A Non-Normal Subgroup

Question

This is from Hatcher's Algebraic Topology. In the red box highlighted above, we're not requiring $H$ to be a normal subgroup of $\pi_1(X,x_0)$, so I don't understand how the quotient can be a group - the group of deck transformations on $\tilde{X}$.

Answer 1

(I am just elaborating on things written in the comments.) If $H$ is a subgroup of a group $G$ then $N(H)$ (in Hatcher's notation) is the normalizer of $H$. That is, $N(H)=\{g\in G: gHg^{-1}=H\}.$ Then $H\subseteq N(H)$ and is in particular a normal subgroup of $N(H)$ by definition. In conclusion, it makes perfect sense to consider the quotient group $N(H)/H$.