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.
 A: $\emptyset$ is not a subgroup of $G$, because it is not a group. Because $\langle X\rangle$ is by definition

The smallest subgroup of $H$ that includes $X$

then if $X=\emptyset$, it is equal to $\{1\}$
A: 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).
A: 
$\langle \emptyset \rangle=\{1\},\langle 1 \rangle=\{1\}. H\leq G \implies \langle H\rangle=H.$ 

Examples.
$G=\mathbb R^*$ without zero. (G,*) is multiplicative group.


*

*If you have $\langle \emptyset \rangle =\{\}=\emptyset$, then this $\emptyset$ cannot be a group because no unit element, failing the definition of the group so the implication false with this.

*$H=0$. $\langle 0\rangle =\langle \emptyset\rangle=\{1\}$

*$H=1$. $\langle 1\rangle =\{1\}$.

*$H=6$. $\langle 6\rangle \not =\langle 2,3\rangle$ because $2\not\in\langle 6\rangle$ while $2\in\langle 2,3\rangle$.

*$H=2$. $\langle 2\rangle =\{2^n\quad |\quad n\in\mathbb Z\}$. This is when $G=\mathbb Q^*$ because $0$ has no multiplicative inverse. This does not mean that the exponent cannot be zero i.e. the identity element $1\in\langle 2\rangle.$
Exponents must be integers (not real numbers)


*

*$H=2$. When $G=\mathbb R^*$, $\langle 2\rangle = \{2^n\quad |\quad n\in\mathbb R\}.$ Wrong, the exponents must be integers. 

*$H=2$. When $G=\mathbb R^*$, $\langle 2\rangle = \{2^n \quad | \quad n\in\mathbb Z\}$.

*$H=3$. $\langle 3\rangle =\{3^n\}$, $n\in\mathbb R$. Wrong, exponents cannot be real numbers.

*$H=4$. $\langle 4\rangle =\langle 2\rangle$ because $\langle k\rangle,k\in\mathbb R^*$ consisting of all multiplies having the factors and their inverses. Wrong exponents cannot be real numbers.

*$H=5$. $\langle 5\rangle =\{5^n\}, n\in\mathbb R$. Wrong because exponents must be integers.
$\langle a\rangle$ is not defined here with sets of sets


*

*What is $\langle \{1\}\rangle$?  $\langle \{1\}\rangle =\{\{1\}\}$? $\{\{k\}\}$ is not a number so undefined? Sets of sets not addressed here.

*Does the statement "$\langle \emptyset \rangle=\{1\},\langle 1 \rangle=\{1\}. H\leq G \implies \langle H\rangle=H.$" work with sets having sets like $\{\{\{1\}\}\}$? No.
Further questions


*

*Multiplicative group $(\mathbb R^*, ×)$ is group but $(\mathbb R, ×)$ is not group, why? where the question is also on additive groups: $(\mathbb R, +)$ is group but $(\mathbb R^*, +)$ is not group.

