The strategy is to construct the induced map $\bar\vartheta : G/\ker(\vartheta) \to \operatorname{Im}(\vartheta)$ defined by $g\ker(\vartheta) \mapsto \vartheta(g)$, then verify it is well-defined, a homomorphism, injective, and surjective. The key insight is that elements with the same image differ by an element of the kernel, which is precisely the equivalence that the quotient identifies. **Step 1: Define the induced map.** Let $K = \ker(\vartheta)$. By [Kernel Is a Normal Subgroup](/theorems/788), $K \unlhd G$, so $G/K$ is a [group](/page/Group) by the [Quotient Group](/theorems/790) theorem. By [Image of a Homomorphism Is a Subgroup](/theorems/769), $\operatorname{Im}(\vartheta) \leq H$. Define: \begin{align*} \bar\vartheta : G/K &\to \operatorname{Im}(\vartheta) \\ gK &\mapsto \vartheta(g). \end{align*} **Step 2: Well-definedness.** [claim:Induced Map Well Defined] If $aK = bK$, then $\vartheta(a) = \vartheta(b)$. [/claim] [proof] By the [Equal Cosets Criterion](/theorems/786), $aK = bK$ implies $a^{-1}b \in K$, i.e., $\vartheta(a^{-1}b) = e_H$. By the homomorphism property, $\vartheta(a)^{-1}\vartheta(b) = e_H$, so $\vartheta(a) = \vartheta(b)$. [/proof] **Step 3: Homomorphism.** For $aK, bK \in G/K$: \begin{align*} \bar\vartheta(aK \cdot bK) = \bar\vartheta(abK) = \vartheta(ab) = \vartheta(a)\vartheta(b) = \bar\vartheta(aK) \cdot \bar\vartheta(bK). \end{align*} **Step 4: Injectivity.** [claim:Trivial Kernel of Induced Map] $\ker(\bar\vartheta) = \{K\}$ (the identity of $G/K$). [/claim] [proof] Suppose $\bar\vartheta(aK) = e_H$, i.e., $\vartheta(a) = e_H$. Then $a \in K$, so $aK = K$. Conversely, $\bar\vartheta(K) = \vartheta(e) = e_H$. So $\ker(\bar\vartheta) = \{K\}$, which means $\bar\vartheta$ is injective. [/proof] **Step 5: Surjectivity.** For any $v \in \operatorname{Im}(\vartheta)$, there exists $a \in G$ with $\vartheta(a) = v$. Then $aK \in G/K$ and $\bar\vartheta(aK) = \vartheta(a) = v$. Therefore $\bar\vartheta$ is an isomorphism, giving $G/\ker(\vartheta) \cong \operatorname{Im}(\vartheta)$.