$H,K$ be subgroups of a group $G$ such that $G$ is isomorphic with $H \times K$ ; then is $H$ normal in $G$ ? Let $H,K$ be subgroups of a group $G$ such that $G$ is isomorphic with $H \times K$ ; then is $H$ normal in $G$ ? I can prove that $H \times \{e\}$ is normal in $H \times K$ but nothing else ; one thing is obvious is that if $H$ is normal in $G$ , then so is $K$ . Please help . Thanks in advance 
 A: Pick a group $K_0$ and a subgroup $H_0$ of $K_0$ which is not normal. Then let
$$G = H_0 \times K_0,\quad K = \{e\} \times K_0,\quad H = \{e\} \times H_0.$$
By construction, $G$ is isomorphic to $H\times K$ since $H \cong H_0$ and $K \cong K_0$, and $H$ is not a normal subgroup of $K$, hence a fortiori not a normal subgroup of $K$.
The mere existence of some isomorphism $G \cong H\times K$ is a very weak condition. To be able to conclude that $H$ (or $K$) is a normal subgroup of $G$, we need further conditions on how $H$ and $K$ are embedded in $G$. It is not sufficient that $H\cap K$ be trivial, for example. To see that, consider a group that contains an isomorphic copy of itself as a non-normal subgroup. An example of such is the group
$$S = \bigl\{ \pi \colon \mathbb{N} \to \mathbb{N} : \pi \text{ is bijective and } \{ n : \pi(n) \neq n\} \text{ is finite}\bigr\},$$
in which the subgroups $S_k := \{ \pi \in S : \pi(k) = k\}$ are all isomorphic to $S$ but not normal. Then we can take $G = S\times S$, and for example $H = S_1\times \{e\}$, $K = \{e\}\times S_2$. Clearly $H\cap K = \{(e,e)\}$ is trivial, $G \cong H \times K$ since $S_k \cong S$, and neither $H$ nor $K$ are normal in $G$.
A: Let $g$ be an element of $G$, then we have to prove that
$$g^{-1} Hg=H$$
But $G$ is isomorphic to the direct product $H\times K$, thus $g$ correspond to an element of such group that is in the form $(h, k)$. Equally, every $h \in H$ is sent by the isomorphism to an element in the form $(h, 1_K)$.
Therefore the statement we need to prove is 
$$(h^{-1} ,k^{-1} )(H, 1_K)(h,k)=(H,1_K)$$
And this is clearly true since the LHS is equal to
$$(h^{-1} Hh, k^{-1} 1_Kk)$$
And since a subgroup is closed under conjugacy
$$(h^{-1} Hh, k^{-1} 1_Kk) = (H, 1_K)$$
