**Proof plan.** Define the surjective homomorphism $\varphi : G/K \to G/L$ by $\varphi(gK) = gL$, check well-definedness, identify its kernel as $L/K$, and invoke the [First Isomorphism Theorem for Groups](/theorems/842). **Step 1: Well-definedness.** [claim: Well-Defined] The map $\varphi : G/K \to G/L$ defined by $\varphi(gK) = gL$ is well-defined. [/claim] [proof] If $gK = g'K$ then $g^{-1}g' \in K \subseteq L$, so $gL = g'L$, i.e. $\varphi(gK) = \varphi(g'K)$. [/proof] **Step 2: $\varphi$ is a surjective homomorphism.** [claim: Surjective Hom] $\varphi$ is a surjective group homomorphism. [/claim] [proof] It is a homomorphism since $\varphi(gK \cdot g'K) = \varphi(gg'K) = gg'L = (gL)(g'L) = \varphi(gK)\varphi(g'K)$. It is surjective since every coset $gL \in G/L$ equals $\varphi(gK)$. [/proof] **Step 3: Identify the kernel.** [claim: Kernel] $\ker(\varphi) = L/K$. [/claim] [proof] $\varphi(gK) = eL$ iff $gL = L$ iff $g \in L$. So $\ker(\varphi) = \{gK : g \in L\} = L/K$. [/proof] **Step 4: Conclusion.** By the [First Isomorphism Theorem for Groups](/theorems/842) applied to $\varphi$, \begin{align*} \frac{G/K}{L/K} \cong \operatorname{im}(\varphi) = G/L. \qquad \square \end{align*}