Criterion for G to be Isomorphic to H × K When H, K ◁ G

# Criterion for G to be Isomorphic to H × K When H, K are Normal Subgroups of G

 Theorem 1: Let $G$ be a group and let $H$ and $K$ be normal subgroups of $G$. If $G = HK$ and $H \cap K = \{ e \}$ then $G \cong H \times K$.
• Proof: Let $G$ be a group and let $H$, $K$ be normal subgroups of $G$ with $G = HK$ and $H \cap K = \{ e \}$. Let $\phi : H \times K \to G$ be defined for all $(h, k) \in H \times K$ by:
(1)
\begin{align} \quad \phi((h, k)) = hk \end{align}
• We first show that $\phi$ is a homomorphism from $H \times K$ to $G$. Observe that for all $(h_1, k_1), (h_2, k_2) \in H \times K$ we have that:
(2)
\begin{align} \quad \phi((h_1,k_1)(h_2,k_2)) = \phi((h_1h_2, k_1k_2)) = h_1h_2k_1k_2 \end{align}
(3)
\begin{align} \quad \phi((h_1, k_1)) \phi((h_2, k_2)) = h_1k_1h_2k_2 \end{align}
• We want to show that $h_1h_2k_1k_2 = h_1k_1h_2k_2$ for all $h_1, h_2 \in H$ and for all $k_1, k_2 \in K$. To do this, it is sufficient to show that $h_2k_1 = k_1h_2$ for all $h_2 \in H$ and for all $k_1 \in K$, or equivalently, show that $hk = kh$ for all $h \in H$ and for all $k \in K$.
• Let $h' \in H$ and let $k' \in K$. Consider the element $hkh^{-1}k^{-1}$.
• Since $H$ is a normal subgroup of $G$ we have that $ghg^{-1} \in H$ for all $g \in G$ and for all $h \in H$. Since $K \subseteq G$, we have in particular that $khk^{-1} \in H$ for all $k \in K$ and for all $h \in H$. Since $khk^{-1} \in H$ and $h^{-1} \in H$ (and since $H$ is a group) we have that $khk^{-1}h^{-1} \in H$. So $hkh^{-1}k^{-1} \in H$ for all $h \in H$ and for all $k \in K$. In particular:
(4)
• Similarly, since $K$ is a normal subgroup of $G$ we have that $gkg^{-1} \in K$ for all $g \in G$ and for all $k \in K$. Since $H \subseteq G$, we have in particular that $hkh^{-1} \in K$ for all $h \in H$ and for all $k \in K$. Since $hkh^{-1} \in K$ and and $k^{-1} \in K$ for each $k \in K$ (and since $K$ is a group) we have that $hkh^{-1}k^{-1} \in K$ for all $h \in H$ and for all $k \in K$. In particular:
(5)
• From $(*)$ and $(**)$ we see that:
(6)
\begin{align} \quad h'k'h'^{-1}k'^{-1} \in H \cap K = \{ e \} \end{align}
• Thus $h'k'h'^{-1}k'^{-1} = e$, i.e., $h'k' = k'h'$ for all $h' \in H$ and for all $k' \in K$, so the claim is proved and indeed, $\phi$ is a homomorphism from $H \times K$ to $G$.
• Now observe that:
(7)
\begin{align} \quad \ker (\phi) &= \{ (h, k) \in H \times K : \phi((h, k)) = e \} \\ &= \{ (h, k) \in H \times K : hk = e \} \end{align}
• Observe that $hk = e$ implies that $h = k^{-1}$. So $k^{-1} = h \in H$. Thus $k \in H$. So $k \in H \cap K = \{ e \}$, i.e., $k =e$. But then $hk = e$ gives us $h = e$. So $\ker(\phi) = \{ (e, e) \}$, i.e., $\ker(\phi)$ is the trivial group. Thus $(H \times K)/\ker(\phi) \cong H \times K$. So by The First Group Isomorphism Theorem we have that:
(8)
\begin{align} \quad H \times K = (H \times K)/\{(e, e)\} = (H \times K)/\ker(\phi) \cong \phi(H \times K) \end{align}
• Lastly, observe that $\phi$ is surjective since $G = HK$. Indeed, if $g \in G$ then there exists $h \in H$ and $k \in K$ such that $g = hk$. So $(h, k) \in H \times K$ is such that $\phi(h, k) = hk = g$. Thus $\phi(H \times K) = G$, and so from above we conclude that:
(9)