**Proof plan.** Define the natural map $\varphi : H \to G/K$ by $\varphi(h) = hK$, apply the [First Isomorphism Theorem for Groups](/theorems/842) to identify kernel and image, and verify normality claims along the way. **Step 1: Setup and homomorphism.** Define \begin{align*} \varphi : H &\to G/K \\ h &\mapsto hK. \end{align*} This is a homomorphism: $\varphi(h_1h_2) = h_1h_2K = (h_1K)(h_2K) = \varphi(h_1)\varphi(h_2)$. **Step 2: Identify the kernel.** [claim: Kernel of Phi] $\ker(\varphi) = H \cap K$. [/claim] [proof] $h \in \ker(\varphi)$ iff $hK = eK$ iff $h \in K$. Since we also require $h \in H$, this gives $\ker(\varphi) = H \cap K$. [/proof] **Step 3: $H \cap K \trianglelefteq H$.** [claim: Intersection Is Normal] $H \cap K \trianglelefteq H$. [/claim] [proof] Since $\ker(\varphi)$ is always normal in the domain of a homomorphism, $H \cap K = \ker(\varphi) \trianglelefteq H$. [/proof] **Step 4: Identify the image.** [claim: Image of Phi] $\operatorname{im}(\varphi) = HK/K = \{hkK : h \in H, k \in K\} = \{hK : h \in H\}$. [/claim] [proof] The image $\{hK : h \in H\}$ corresponds to all $K$-cosets represented by elements of $H$. Since $hkK = hK$ for any $k \in K$, this equals $HK/K$. [/proof] **Step 5: $HK$ is a subgroup of $G$.** [claim: HK Is Subgroup] $HK = \{hk : h \in H, k \in K\}$ is a subgroup of $G$. [/claim] [proof] Let $hk, h'k' \in HK$. Then \begin{align*} h'k'(hk)^{-1} = h'k'k^{-1}h^{-1} = (h'h^{-1})(hk'k^{-1}h^{-1}). \end{align*} The first factor is in $H$. The second factor has $k'k^{-1} \in K$, and since $K \trianglelefteq G$, conjugation by $h$ preserves $K$, so $hk'k^{-1}h^{-1} \in K$. Thus the product lies in $HK$, and $e = e \cdot e \in HK$. [/proof] **Step 6: Conclusion.** The [First Isomorphism Theorem for Groups](/theorems/842) applied to $\varphi$ gives \begin{align*} \frac{H}{H \cap K} = \frac{H}{\ker(\varphi)} \cong \operatorname{im}(\varphi) = \frac{HK}{K}. \qquad \square \end{align*}