**Step 1: Forward ($\vartheta$ injective $\implies \ker(\vartheta) = \{e_G\}$).** Suppose $g \in \ker(\vartheta)$. Then $\vartheta(g) = e_H = \vartheta(e_G)$. By injectivity, $g = e_G$. So $\ker(\vartheta) = \{e_G\}$. **Step 2: Backward ($\ker(\vartheta) = \{e_G\} \implies \vartheta$ injective).** Suppose $\vartheta(g) = \vartheta(h)$. Then: \begin{align*} \vartheta(h^{-1}g) = \vartheta(h)^{-1}\vartheta(g) = e_H, \end{align*} so $h^{-1}g \in \ker(\vartheta) = \{e_G\}$. Therefore $h^{-1}g = e_G$, giving $g = h$.