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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

  1. $\{ h\star k : h \in H \, \land \, k \in K \}=\colon HK \le G$
  2. $HK \trianglelefteq G$
  3. $HK =G$
  4. $H \cap K = \{\varepsilon\}$
  5. $G$ is finite

sufficient to prove that $H$ and $K$ are normal?

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    $\begingroup$ I think a subtle difference is being assumed between "to be the direct product of two subgroups" and "to be isomorphic to the direct product of two subgroups". In the example provided below the answer in the linked question there seems to be some confusion between "exterior direct product" and interior one, and the last comment seems to be trying to do some order there. In general, I'd say that if a group is (isomorphic) to the direct product of two of its subgroups then they must be both normal and have trivial interesection, otherwise no (interior) direct product is possible. Hopefully... $\endgroup$ – DonAntonio Mar 4 '16 at 0:08
  • $\begingroup$ Someone else can add som e light to this. Thank you. $\endgroup$ – DonAntonio Mar 4 '16 at 0:08
  • $\begingroup$ It is not very precise but saying that $G$ is abstractly isomorphic to $H\times K$ where $H,K$ are subgroups of $G$ does not give much more information than $|G|=|H||K|$. I don't think (I may be wrong though) that any of these conditions would imply that $H$ and $K$ is normal in $G$. You should try to find counter-example to $3$ and then for $4$. $\endgroup$ – Clément Guérin Mar 4 '16 at 8:23
  • $\begingroup$ @Joanpemo. If $G = \left< a,b \mid a^6= b^2 = id ,\ bab=a^{-1} \right>$, then $ G \cong \left< a^2,b\right> \times \left< b \right> \cong \left< a^2,b\right> \times \left< a^3 \right> $, but $\left< b \right>$ is not normal in $G$. Surely $G$ has two normal subgroups isomorphic to $H$ and $K$ respectively and with trivial intersection, yet one cannot say that $H$ and $K$ are normal (without additional information). $\endgroup$ – Τίμων Mar 4 '16 at 13:33
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    $\begingroup$ I think the confusion is in your first sentence. I believe it should read "Let $G$ be a group and $H$, $K$ two subgroups such that $G\cong H^{\prime}\times K^{\prime}$ with isomorphism $\phi$ where $\phi(H)= H^{\prime}$ and $\phi(K)= K^{\prime}$." Then each of $HK=G$, $H\cap K=1$, $H\lhd G$ and $K\lhd G$ follow immediately. If you slacken the "$\phi(H)= H^{\prime}$ and $\phi(K)= K^{\prime}$" condition and just say "$H\cong H^{\prime}$ and $K\cong K^{\prime}$" then these results do not follow. $\endgroup$ – user1729 Mar 18 '16 at 15:03
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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 $$\begin{align*} (1\;2\;3) &\mapsto \big( (),\ (1\;2\;3) (9\;10\;11) \big) \\ (4\;5\;6\;7\;8) &\mapsto \big((),\ (8\;7\;6\;5\;4)\big) \\ (5\;8)(6\;7) &\mapsto \big((),\ (4\;5)(6\;8)\big) \\ (9\;10\;11) &\mapsto \big((9\;10\;11), \ ()\big) \\ (9\;10) &\mapsto \big((5\;8)(6\;7)(9\;10),\ ()\big) \end{align*}$$

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