The idea is to exploit the idempotent property: applying $\vartheta$ to $e_G = e_G \cdot_G e_G$ produces an element of $H$ that squares to itself, and the cancellation law in $H$ forces it to be $e_H$. Step 1 uses the homomorphism property to show $\vartheta(e_G)$ is idempotent in $H$, and Step 2 applies the cancellation law (i.e., multiplication by the inverse) to conclude $\vartheta(e_G) = e_H$. **Step 1: Compute $\vartheta(e_G)$.** Let $a = \vartheta(e_G) \in H$. Then: \begin{align*} a = \vartheta(e_G) = \vartheta(e_G \cdot_G e_G) = \vartheta(e_G) \cdot_H \vartheta(e_G) = a \cdot_H a. \end{align*} **Step 2: Apply cancellation.** Since $H$ is a [group](/page/Group), $a$ has an inverse $a^{-1} \in H$. Left-multiplying both sides of $a = a \cdot_H a$ by $a^{-1}$: \begin{align*} a^{-1} \cdot_H a = a^{-1} \cdot_H (a \cdot_H a) = (a^{-1} \cdot_H a) \cdot_H a = e_H \cdot_H a = a, \end{align*} so $e_H = a$. Therefore $\vartheta(e_G) = e_H$.