The strategy is to first establish an auxiliary fact — that any idempotent element must be the identity — then use it to show right inverses are also left inverses, and finally that the right identity is also a left identity. **Step 1: Idempotent elements are the identity.** [claim:Idempotent Implies Identity] Let $h \in G$ satisfy $h * h = h$. Then $h = e$. [/claim] [proof] Since $h \in G$, there exists $h^{-1} \in G$ with $h * h^{-1} = e$ (right inverse). Then: \begin{align*} h &= h * e = h * (h * h^{-1}) = (h * h) * h^{-1} = h * h^{-1} = e. \end{align*} [/proof] **Step 2: Right inverses are left inverses.** [claim:Right Inverse Is Left Inverse] For any $g \in G$, if $g * g^{-1} = e$, then $g^{-1} * g = e$. [/claim] [proof] Consider the element $g^{-1} * g$. We compute: \begin{align*} (g^{-1} * g) * (g^{-1} * g) &= g^{-1} * (g * g^{-1}) * g = g^{-1} * e * g = g^{-1} * g. \end{align*} So $g^{-1} * g$ is idempotent. By the claim in Step 1, $g^{-1} * g = e$. [/proof] **Step 3: The right identity is a left identity.** From Step 2, $g * g^{-1} = e$ implies $g^{-1} * g = e$. Then: \begin{align*} g * g^{-1} = e \implies g * g^{-1} * g = e * g \implies g * (g^{-1} * g) = e * g \implies g * e = e * g, \end{align*} and since $g * e = g$ (right identity), we obtain $e * g = g$. So $e$ is also a left identity.