# 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 $x^G\subset x^H$?

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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 $(n_1,n_2) \in N_1 \times N_2 \in H$.

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Try to prove the following: if $x_i \in G_i$ for $i=1,2$, then $(x_1,x_2)^{G_1 \times G_2} = x_1^{G_1} \times x_2^{G_2}$. And in addition $(x_1,x_2)^H = x_1^{H_1} \times x_2^{H_2}$, where $H_i$ is the coordinate projection of $H$ on $G_i$, $i=1,2$. Note that $H_i$ is a normal subgroup of $G_i$.

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Yes, I see. But i have a trouble with the problem, $(x_1 ,x_2)^H=x_1^H_1\times x_2^H_2$. In case, $G_1\times G_2$ is finite, this is true but I can't prove it in case infinite. – FongV Sep 9 '12 at 12:26