Explain $\langle \emptyset \rangle=\{1\},\langle 1 \rangle=\{1\}. H\leq G \implies \langle H\rangle=H.$ On page $61$ of the book Algebra by Tauno Metsänkylä, Marjatta Näätänen, it states

$\langle \emptyset \rangle =\{1\},\langle 1 \rangle =\{1\}. H\leq G \implies \langle H \rangle =H$

where $H \leq G$ means that H is the subgroup of G.
Now assumme $H=\emptyset$ so $\langle \emptyset \rangle = \emptyset \not = \{1\}$, contradiction. Please explain p.61 of the book that is the line in orange above.

The notation $H \leq G$ means that $H$ is a subgroup of $G$. Your proposed counterexample fails because $\emptyset$ is not a subgroup of $G$ (it doesn't contain the identity element).
