The strategy is to construct a homomorphism $\vartheta : H \times K \to G$ and show it is an isomorphism. The crucial step is showing elements of $H$ and $K$ commute, which follows from normality and the trivial intersection condition. **Step 1: Elements of $H$ and $K$ commute.** [claim:Commuting Subgroups] For all $h \in H$ and $k \in K$, $hk = kh$. [/claim] [proof] Consider the commutator $hkh^{-1}k^{-1}$. Since $K \unlhd G$, $hkh^{-1} \in K$, so $hkh^{-1}k^{-1} = (hkh^{-1})k^{-1} \in K$. Since $H \unlhd G$, $kh^{-1}k^{-1} \in H$, so $hkh^{-1}k^{-1} = h(kh^{-1}k^{-1}) \in H$. Therefore $hkh^{-1}k^{-1} \in H \cap K = \{e\}$, giving $hkh^{-1}k^{-1} = e$, i.e., $hk = kh$. [/proof] **Step 2: Construct the isomorphism.** Define: \begin{align*} \vartheta : H \times K &\to G \\ (h, k) &\mapsto hk. \end{align*} - *Homomorphism:* By Step 1, $ka = ak$ for $k \in K$, $a \in H$. So $\vartheta((h,k)(a,b)) = \vartheta((ha, kb)) = hakb = h(ak)b = h(ka)b = (hk)(ab) = \vartheta((h,k))\vartheta((a,b))$. **Step 3: Injectivity.** Suppose $\vartheta((h,k)) = \vartheta((a,b))$, i.e., $hk = ab$. Then $a^{-1}h = bk^{-1}$. The left side lies in $H$ and the right in $K$, so both lie in $H \cap K = \{e\}$. Therefore $a^{-1}h = e$ and $bk^{-1} = e$, giving $h = a$ and $k = b$. **Step 4: Surjectivity.** By hypothesis (iii), every $g \in G$ can be written as $g = hk$ for some $h \in H$, $k \in K$. So $\vartheta((h,k)) = g$.