**Proof plan.** Write down the maps $\Theta$ and $\Theta^{-1}$ explicitly, check they are inverses, verify the subgroup property, and note normality is preserved. **Step 1: $\Theta(X) \leq G$ and $K \subseteq \Theta(X)$.** [claim: Theta Lands in Subgroups Containing K] For any $X \leq G/K$, the preimage $\Theta(X) = \{g \in G : gK \in X\}$ is a subgroup of $G$ containing $K$. [/claim] [proof] Since $K \in X$ (as $eK$ is the identity of $G/K$ and $X \leq G/K$), every $k \in K$ satisfies $kK = K \in X$, so $K \subseteq \Theta(X)$. For the subgroup property: if $g_1, g_2 \in \Theta(X)$ then $g_1K, g_2K \in X$, so $(g_1K)(g_2K)^{-1} = g_1g_2^{-1}K \in X$, giving $g_1g_2^{-1} \in \Theta(X)$. Since $eK = K \in X$, we have $e \in \Theta(X)$. [/proof] **Step 2: $\Theta^{-1}(L) = L/K \leq G/K$.** [claim: Inverse Map Is Well-Defined] For any subgroup $L \leq G$ with $K \subseteq L$, the [set](/page/Set) $L/K = \{lK : l \in L\}$ is a subgroup of $G/K$. [/claim] [proof] $L/K$ is non-empty ($eK = K \in L/K$). If $l_1K, l_2K \in L/K$ then $(l_1K)(l_2K)^{-1} = l_1l_2^{-1}K$. Since $L \leq G$, $l_1l_2^{-1} \in L$, so $l_1l_2^{-1}K \in L/K$. [/proof] **Step 3: $\Theta$ and $\Theta^{-1}$ are inverses.** [claim: Mutual Inverses] $\Theta(\Theta^{-1}(L)) = L$ and $\Theta^{-1}(\Theta(X)) = X$. [/claim] [proof] $\Theta(L/K) = \{g \in G : gK \in L/K\} = \{g \in G : g \in L\} = L$ (since $K \subseteq L$). $\Theta^{-1}(\Theta(X)) = \{gK : g \in \Theta(X)\} = \{gK : gK \in X\} = X$. [/proof] **Step 4: Normality is preserved.** [claim: Normality Preserved] $X \trianglelefteq G/K$ if and only if $\Theta(X) \trianglelefteq G$. [/claim] [proof] If $X \trianglelefteq G/K$ and $g \in G$, then $g\Theta(X)g^{-1} = \Theta(gKXKg^{-1}) = \Theta(X)$ since $(gK)X(gK)^{-1} = X$. The reverse is similar. [/proof] Since $\Theta$ is a bijection with the stated properties, the proof is complete. $\square$