Let $K = \ker(\vartheta) = \{g \in G : \vartheta(g) = e_H\}$. The proof has two parts: $K$ is a subgroup, then $K$ is normal. **Step 1: $K$ is a subgroup.** - *Identity:* By [Homomorphisms Preserve Identity](/theorems/768), $\vartheta(e_G) = e_H$, so $e_G \in K$. - *Closure:* If $a, b \in K$, then $\vartheta(ab) = \vartheta(a)\vartheta(b) = e_H \cdot e_H = e_H$, so $ab \in K$. - *Inverses:* If $a \in K$, then $\vartheta(a^{-1}) = \vartheta(a)^{-1} = e_H^{-1} = e_H$, so $a^{-1} \in K$. **Step 2: $K$ is normal.** We verify condition (iii) of the [Equivalent Definitions of Normality](/theorems/787). Let $k \in K$ and $g \in G$. Then: \begin{align*} \vartheta(gkg^{-1}) = \vartheta(g)\vartheta(k)\vartheta(g^{-1}) = \vartheta(g) \cdot e_H \cdot \vartheta(g)^{-1} = e_H. \end{align*} So $gkg^{-1} \in K$. Therefore $K \unlhd G$.