A group is isomorphic to the direct product of two subgroups. Conditions for the subgroups to be normal. Let $(G, \star, \varepsilon)$ be a group and $H$ and $K$ two of its subgroups and let
$$\begin{align*}
\diamond\, \colon (H \times K)^2 &\to H \times K \\
\big((h_1,k_1) , (h_2,k_2)\big) & \mapsto (h_1 \star h_2, k_1 \star k_2).
\end{align*}$$
One can easily prove that $ \big(H \times K, \diamond, (\varepsilon, \varepsilon)\big)$ is a group. Let's call it the direct product of the groups $H$ and $K$. Now suppose there exists a bijection $f\colon G \to H \times K$ such that 
$$\forall a,b \!\in\! G\quad f(a \star b) = f(a) \diamond f(b)$$
Let's call it an isomorphism of the groups $(G, \star, \varepsilon)$ and $ \big(H \times K, \diamond, (\varepsilon, \varepsilon)\big)$ and write $$(G, \star, \varepsilon) \cong \big(H \times K, \diamond, (\varepsilon, \varepsilon)\big) \quad \text{or simply} \quad G \cong H \times K$$
Are some of these conditions


*

*$\{ h\star k : h \in H \, \land \, k \in K \}=\colon HK \le G$  

*$HK \trianglelefteq G$  

*$HK =G$  

*$H \cap K = \{\varepsilon\}$  

*$G$ is finite


sufficient to prove that $H$ and $K$ are normal?
 A: The answer is no. Let
$$ \begin{align*}
 &G \colon\!= \langle (1\;2\;3\;4\;5),\ (1\;2)(3\;5),\ (6\;7),\ (6\;7\;8)\rangle \cong D_{10} \times S_3 \\[1ex]
&H \colon\!= \langle (1\;2\;3\;4\;5),\ (2\;5)(3\;4)(7\;8) \rangle \cong D_{10}\\[1ex]
&K \colon\!= \langle (7\;8),\ (6\;7\;8)\rangle \cong S_3
\end{align*}
$$
Then 
$$ H,K \le G, \quad G \cong H \times K, \quad G=HK, \quad H \cap K= \lbrace ()\rbrace $$
But $H \not \trianglelefteq G$. I checked it with GAP. However, $K$ is normal. 

Update
Let 
$$ \begin{align*}
 &G \colon\!= \langle (1 \; 2 \; 3), \ (4 \; 5 \; 6 \; 7 \; 8),\  (5 \; 8)(6 \; 7),\ (9 \; 10 \; 11),\ (9 \; 10) \rangle \cong \mathbb{Z}_3 \times D_{10} \times S_3\\[1ex]
&H \colon\!= \langle (5 \; 8)(6 \; 7)(10 \; 11),\ (9 \; 10 \; 11) \rangle \cong S_3 \\[1ex]
&K \colon\!= \langle (1 \; 2 \; 3)(9 \; 10 \; 11),\ (5 \; 8)(6 \; 7),\ (4 \; 5 \; 6 \; 7 \; 8)\rangle  \cong \mathbb{Z}_3 \times D_{10}
\end{align*}$$
Then 
$$ H,K \le G, \quad G \cong H \times K, \quad G=HK, \quad H \cap K= \lbrace ()\rbrace \quad \text{and} \quad H,K \not\trianglelefteq G$$
And an isomorphism from $G$ to $H \times K$ is given by
$$\begin{align*}
(1\;2\;3) &\mapsto \big( (),\ (1\;2\;3) (9\;10\;11) \big) \\
(4\;5\;6\;7\;8) &\mapsto \big((),\ (8\;7\;6\;5\;4)\big) \\
(5\;8)(6\;7) &\mapsto \big((),\ (4\;5)(6\;8)\big) \\
(9\;10\;11) &\mapsto  \big((9\;10\;11), \ ()\big) \\
(9\;10) &\mapsto  \big((5\;8)(6\;7)(9\;10),\ ()\big)
\end{align*}$$
