conjugate closed group Let H is a normal subgroup of G.
For any $x\in H$, $x^H=\{h^{-1}xh:h\in H\}$ . 
A normal subgroup H of a group G is said to be conjugate closed, if $x^G=x^H$  for $x\in H$.
A group G is said to be conjugate closed if every normal subgroup of G is conjugate closed.
Prove that: A direct product of conjugate closed groups is conjugate closed group.
i.e.:  If $G_1$ and $G_2$ is conjugate closed group, $G=G_1\times G_2$ is conjugate closed group!
Of course, we have $x^H\subset x^G$, so what $x^G\subset x^H$?
I try to prove it, but i can't. Please help me!

I think this is quite hard, but perhaps I am missing something easier!
Following Nicky Heckster's suggestion, let $H_i$ be the projection of $H$ on $G_i$. If $(x_1,x_2)$ and $(y_1,y_2)$ are conjugate in $G$ then, since $H_i \unlhd G_i$, they are conjugate in $H_1 \times H_2$, so we might as well assume that $H_i=G_i$.
Let $N_i = H \cap G_i$, so $N_i \unlhd G_i$ and $G_1/N_1 \cong G_2/N_2$. Then $H \unlhd G$ implies that $G_1/N_1$ is abelian.
So $\langle (x_1,x_2) \rangle (N_1 \times N_2) = \langle(y_1,y_2) \rangle (N_1 \times N_2) \unlhd G$, and hence $(x_1,x_2)$ and $(y_1,y_2)$ are conjugate in $\langle (x_1,x_2) \rangle (N_1 \times N_2)$. So they are conjugate by an element of the form $(x_1,x_2)^r(n_1,n_2)$ for some $r$ with $n_i \in N_i$, and hence they are conjugate by $(n_1,n_2) \in N_1 \times N_2 \in H$.
