Let $A,B$ be groups, $K\triangleleft A\times B$ be a non-abelian normal subgroup. Show that at least one of the intersection $K\cap A$, $K\cap B$ is non-trivial.
My attempts:
We assume by contradiction that $K\cap A$, $K\cap B$ are both trivial. We can show that the natural homomorphism: $$ \phi: A\times B \rightarrow (A/K\cap A)\times (B/K\cap B) $$ has kernel $\ker\phi=K$, therefore by assumption $A\times B$ is isomorphic to $(A\times B)/K$, I believe this implies that $K$ is Abelian, otherwise there will be some non-trivial commutator lies in K, therefore trivial in $(A\times B)/K$.
I feel a bit unsafe about this proof, I am not sure that '$A\times B$ is isomorphic to $(A\times B)/K$' can happen. Are there any corrections or comments? THANKS!
Edit: This proof seems problematic, maybe $\ker\phi\neq K$ ? Are there any other ideas?
Since $K$ is nonabelian, it contains two elements, $(a,b)$ and $(a',b')$, that do not commute. Thus, either $a$ does not commute with $a'$, or $b$ does not commute with $b'$ (or both). Say, without loss of generality, that $a$ does not commute with $a'$.
If $b$ is trivial, we are done (since $a$ cannot be trivial). Otherwise, conjugating by $(a',1)$ we get $(a'a(a')^{-1},b)\in K$. Multiplying by $(a,b)^{-1}$ we obtain an element of $K\cap (A\times{1})$ which is nontrivial.
The assumption that $K$ is nonabelian is key. Without it, even if $K$ is nontrivial, you could have for example a subgroup of the form $(x,f(x))$ where $x$ ranges over a nontrivial subgroup $H\leq Z(A)$, and $f\colon H\to Z(B)$ is a one-to-one homomorphism.