The strategy is to verify the three subgroup axioms for $\operatorname{Im}(\vartheta) = \{\vartheta(g) : g \in G\}$ directly, using the homomorphism property to transfer each condition from $H$ back to $G$. **Step 1: Identity.** By the [Homomorphisms Preserve Identity](/theorems/768) theorem, $\vartheta(e_G) = e_H$, so $e_H \in \operatorname{Im}(\vartheta)$. **Step 2: Closure.** Let $a, b \in \operatorname{Im}(\vartheta)$. Then there exist $c, d \in G$ with $\vartheta(c) = a$ and $\vartheta(d) = b$. Since $G$ is a [group](/page/Group), $cd \in G$. By the homomorphism property: \begin{align*} ab = \vartheta(c)\vartheta(d) = \vartheta(cd) \in \operatorname{Im}(\vartheta). \end{align*} **Step 3: Inverses.** Let $a \in \operatorname{Im}(\vartheta)$, so $a = \vartheta(b)$ for some $b \in G$. Since $G$ is a group, $b^{-1} \in G$ and $\vartheta(b^{-1}) \in \operatorname{Im}(\vartheta)$. We verify this is the inverse of $a$: \begin{align*} \vartheta(b^{-1})\vartheta(b) = \vartheta(b^{-1}b) = \vartheta(e_G) = e_H, \end{align*} and similarly $\vartheta(b)\vartheta(b^{-1}) = e_H$. So $a^{-1} = \vartheta(b^{-1}) \in \operatorname{Im}(\vartheta)$. Associativity is inherited from $H$. Since $\operatorname{Im}(\vartheta) \subseteq H$, it follows that $\operatorname{Im}(\vartheta) \leq H$.