Proof of Uniqueness of the Identity Element and of Inverses in a Group

Throughout we use only the three defining conditions of a group.

Part 1. By condition 2 of Group and Abelian Group at least one identity element exists. Let $e$ and $e'$ both be identity elements of $(G,\ast)$. Applying the defining property of $e$ to the element $e'$ gives

Applying the defining property of $e'$ to the element $e$ gives

Hence $e=e'$, so the identity element is unique. We denote it by $e_G$.

Part 2. Let $a\in G$.

Existence. By condition 3 of Group and Abelian Group there is an identity element $e_0$ of $(G,\ast)$ such that every element of $G$ has an inverse with respect to $e_0$. By Part 1 we have $e_0=e_G$, so there exists $b\in G$ with $a\ast b=e_G$ and $b\ast a=e_G$.

Uniqueness. Suppose $b,b'\in G$ both satisfy

Using the defining property of $e_G$, then $a\ast b'=e_G$, then associativity (condition 1 of Group and Abelian Group) applied to the triple $b,a,b'$, then $b\ast a=e_G$, and finally the defining property of $e_G$ again, we obtain

Hence the inverse of $a$ is unique, and we denote it by $a^{-1}$.

Part 3. Condition 2 of Group and Abelian Group asserts the existence of an element $e\in G$, and by Part 1 that element is $e_G$. Hence $e_G\in G$ and $G$ is nonempty.