The strategy is to assume the existence of two candidates in each case and use the [group](/page/Group) axioms to show the candidates must coincide. For the identity (Step 1), the trick is to evaluate $e_1 * e_2$ using each identity axiom in turn, obtaining $e_1 = e_1 * e_2 = e_2$. For inverses (Step 2), the same "sandwich" idea applies: given two purported inverses $y$ and $z$ of $g$, we insert the identity $e = g * z$ into $y * e$ and use associativity to collapse the expression to $z$. The only tools required are associativity, the identity axiom, and the inverse axiom. **Step 1: Uniqueness of the identity.** [claim:No Two Distinct Identities] Suppose $e_1, e_2 \in G$ both satisfy the identity axiom, so that $a * e_1 = a = e_1 * a$ and $a * e_2 = a = e_2 * a$ for all $a \in G$. Then $e_1 = e_2$. [/claim] [proof] Evaluate $e_1 * e_2$ in two ways. Since $e_2$ is an identity, $e_1 * e_2 = e_1$. Since $e_1$ is an identity, $e_1 * e_2 = e_2$. Therefore $e_1 = e_2$. [/proof] **Step 2: Uniqueness of inverses.** [claim:No Two Distinct Inverses] Let $g \in G$ and suppose $y, z \in G$ both satisfy the inverse axiom for $g$, so that $g * y = e = y * g$ and $g * z = e = z * g$. Then $y = z$. [/claim] [proof] We compute: \begin{align*} y &= y * e \\ &= y * (g * z) \\ &= (y * g) * z \quad \text{(associativity)} \\ &= e * z \\ &= z. \end{align*} [/proof]