Proof of Subgroup Criterion and Basic Examples

Claim 1, necessity. Suppose $H$ is a subgroup. Then $e_G\in H$ by condition 1 of that definition, so $H$ is nonempty. If $a,b\in H$, then $b^{-1}\in H$ by condition 3, and hence $ab^{-1}\in H$ by condition 2.

Claim 1, sufficiency. Suppose $H$ is nonempty and $ab^{-1}\in H$ for all $a,b\in H$. We verify the three conditions of Subgroup.

Condition 1. Since $H$ is nonempty, choose $c\in H$. Applying the hypothesis to the pair $a=c$, $b=c$ gives $cc^{-1}\in H$, and $cc^{-1}=e_G$ by Uniqueness of the Identity Element and of Inverses in a Group. Hence $e_G\in H$.

Condition 3. Let $a\in H$. Applying the hypothesis to the pair $e_G$ and $a$, both of which lie in $H$ by the previous paragraph, gives $e_Ga^{-1}\in H$. Since $e_Ga^{-1}=a^{-1}$ by condition 2 of Group and Abelian Group, we get $a^{-1}\in H$.

Condition 2. Let $a,b\in H$. By the previous paragraph $b^{-1}\in H$, so the hypothesis applied to the pair $a$ and $b^{-1}$ gives $a(b^{-1})^{-1}\in H$. By claim 2 of Cancellation Laws and Basic Inverse Identities in a Group we have $(b^{-1})^{-1}=b$, hence $ab\in H$.

Thus $H$ is a subgroup.

Claim 2. Let $H$ be a subgroup. By condition 2 of Subgroup, the set $H$ is closed under $\ast$ in the sense of Binary Operation on a Set, so the restriction of $\ast$ to $H$ is a binary operation on $H$; denote it again by $\ast$. We check the three conditions of Group and Abelian Group for this operation.

Associativity. For $a,b,c\in H\subseteq G$ the identity $(a\ast b)\ast c=a\ast(b\ast c)$ holds in $G$ by condition 1 of Group and Abelian Group, and all four products involved lie in $H$ by closure; hence it holds in $H$.

Identity element. By condition 1 of Subgroup we have $e_G\in H$, and $e_G\ast a=a=a\ast e_G$ for every $a\in H$ because this holds for every element of $G$. So $e_G$ is an identity element of $H$ with the restricted operation.

Inverses. Let $a\in H$. By condition 3 of Subgroup we have $a^{-1}\in H$, and $a\ast a^{-1}=e_G=a^{-1}\ast a$ by Uniqueness of the Identity Element and of Inverses in a Group. So every element of $H$ has an inverse in $H$ with respect to the identity element $e_G$.

Hence $H$ with the restricted operation is a group. Applying Uniqueness of the Identity Element and of Inverses in a Group to this group, its identity element is unique and therefore equal to $e_G$, and for $a\in H$ its inverse in this group is unique and therefore equal to $a^{-1}$.

Claim 3. For $H=G$ the three conditions of Subgroup hold because $e_G\in G$, because $\ast$ takes values in $G$, and because $a^{-1}\in G$ for every $a\in G$, all by Group and Abelian Group and Uniqueness of the Identity Element and of Inverses in a Group.

For $H=\{e_G\}$: condition 1 holds trivially; condition 2 holds because $e_G\ast e_G=e_G$ by condition 2 of Group and Abelian Group; and condition 3 holds because $e_G^{-1}=e_G$ by claim 4 of Cancellation Laws and Basic Inverse Identities in a Group.