$ G=H_1\times ... \times H_k $ and $ K \unlhd G.$ Then If $K$ is not contained in $Z(G)$, there exists $i$ such that $K\cap H_i \neq \{e\}$

Question

Let a group $G$ be a product of groups, i.e $ G=H_1\times ... \times H_k $ and $ K \unlhd G.$ Then If $K$ is not contained in $Z(G)$, there exists $i$ such that $K\cap H_i \neq \{e\}$.

It seems that I should use the fact that with $K$ being normal, $|KH|=\frac{|K||H_i|}{|K\cap H_i|}$. I tried it, but couldn't get any working proof out of it.

Answer 1

Choose some $g = (h_1,\ldots,h_k) \in K \setminus Z(G)$. Then some $h_i \not\in Z(H_i)$ - let's assume $i=1$. So there exists $h \in H_1$ that does not centralize $h_1$. Now consider the commutator of $g$ with $(h,1,1,\ldots,1)$.