How to finish the proof of: If $N\unlhd G=H\times K$, then $N$ is abelian or $N$ intersects $H \times \{e\}$ or $\{e\} \times K$ non trivially Question:  Let $N$ is normal to $G=H\times K$, Prove that $N$ is abelian or $N$ intersects one of $H \times \{e\}$ or $\{e\} \times K$ non trivially.  
Solution:  Since $G=H\times K$, then $G$ can be written as an internal direct product of $H$ and $K$, then $H\cap K=\{e\}$ Suppose $h\in H$ and $k\in K$ and $hkh^{-1}k^{-1}\in H\cap K$, then $hk=kh$, where $e$ denotes the identity of both $H$ and $K$.  Hence $Nhkh^{-1}k^{-1}=N$, $Nhk=Nkh$. Also, $Nhk=NhNk=Nkh=NkNh$. So $NhNk=NkNh$.  Hence $N$ is a normal abelian group.   If we interpret $G=H\times K$ as the external direct product, then $H\times K=(H\times \{e\})\cdot(\{e\}\times K)$.  Also, since $N\leq (H\times K)$, then $H\cong (H\times \{e\})$, $K\cong (\{e\}\times K)$, $N\cong (N\times \{e\})$ and $N\cong (\{e\}\times N)$, so
$$\begin{align} N\times N&\cong (N\times \{e\})\cdot (\{e\}\times N)\\
&\cong N\cdot N \in N.
\end{align}$$
Can I conclude at this point that both $N\cap H$ and $N\cap K$ are non empty?
Thank you in advance
 A: Suppose $N$ is normal, and nonabelian. Since it is nonabelian, there exists an element $(h,k)\in N$ such that either $h$ is not central in $H$, or $k$ is not central in $K$ (if both $h\in Z(H)$ and $k\in Z(K)$, then $(h,k)\in Z(G)$, hence it commutes with everything in $G$, and thus in $N$).
If $h\notin Z(H)$, then there exists $h'\in H$ such that $h'h(h')^{-1}\neq h$. Since $N$ is normal in $G$, then
$(h',e)(h,k)(h',e)^{-1} = (h'h(h')^{-1},k)\in N$. Therefore,
$$(h',e)(h,k)(h',e)(h,k)^{-1} = (h'h(h')^{-1}h^{-1},e)\in N.$$
Since $h'$ and $h$ do not commute, this is a nontrivial element of $H\times \{e\}$. Thus, $N$ intersects $H\times\{e\}$ nontrivially.
If $k\notin Z(K)$, then a symmetric argument shows that $N$ intersects $\{e\}\times K$ nontrivially.
Thus, if $N$ is normal, then either $N$ is abelian, or else $N$ intersects either $H\times\{e\}$ or $\{e\}\times K$ nontrivially.
Note that in fact we established more: the argument holds if $N$ contains an element that is not in $Z(H)\times Z(K) = Z(G)$; so if $N\triangleleft H\times K$, either $N$ is central, or else it intersects $H\times\{e\}$ or $\{e\}\times K$ nontrivially.

If you don't know what the center is, you don't need it. If $N$ is nonabelian find two elements $(h_1,k_1)$ and $(h_2,k_2)$ of $N$ that do not commute. We must have either the $h_i$ nontrivial do not commute, or the $k_i$ nontrivial do not commute. In the first case, conjugate $(h_1,k_1)$ by $(h_2,e)$ and proceed as before. In the latter, conjugate by $(e,k_2)$.
