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I am thinking on this problem:

If $N\lhd H×K$ then either $N$ is abelian or $N$ intersects one of $H$ or $K$ nontrivially.

I assume; $N$ is not abelian so, there is $(n,n')$ and $(m,m')$ in $N$ such that $([n,m],[n',m'])\neq 1$. But I can’t go further. Hints are appreciated. Thanks.

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    I've edited those extra $N$s into $K$s, but I'll let OP deal with the rest of anon's comment.2012-07-17
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    I assume that the $N$ which appears in the direct product is supposed to be $K$. Also, I adopt the usual convention of identifying $H$ with $H \times 1,$ etc.. Hint: If $N \cap H =1,$ then $[N,H] =1$ (that is, $N$ and $H$ centralize each other) and similarly for $K$.2012-07-17
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    @GeoffRobinson: Gerry edited it correctly. It was my fault.2012-07-17

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Remember: in general, $\,N\lhd G\Longrightarrow [G,N]\leq N$ , so in your case: $$N\lhd H\times K\Longleftrightarrow [H\times K:N]\leq N\Longrightarrow \,\,\text{in particular}\,\,[H:N]\,,\,[K:N]\leq N$$

where we identify $\,H\cong H\times 1\,\,,\,K\cong 1\times K\,$

Now suppose $\,N\,$ intersects both $\,H\,,\,K\,$ trivially, so $\,[H,N]\subset H\cap N =1\,$ and etc...can you take it from here?

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    I say, since $[N,H]=1$ so if $(n,n')$ and $(m,m')$ be in $N$ then $nmn^{-1}m^{-1}=1$. We have the last identity because we can regard $m$ as an element in $H$. The rest is similar for $N$ and $K$. Am I right? Thanks.2012-07-17
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    Lo @BabakSorouh, I meant what I wrote. Why should I write the same thing twice?2012-07-17
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    Indeed, @BabakSorouh: so $\,N\,$ commutes with both $\,H\,,\,K\,$, so...?2012-07-17
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    So, $N$ is an abelian subgroup. Thanks Don.2012-07-17
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    Excuse me. Does $[G, N]$ mean the normalizer of $N$ in $G$?2018-06-06
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    @bfhaha No, it is their commutator: $$[G,N]:=\langle\,[g,n]\;|\;g\in G,\,n\in N\,\rangle $$2018-06-06
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This is my solution which avoids using the notation $[-,-]$. It is the same as DonAntonio's solution and Geoff Robinson's hint essentially.

Lemma. If $A\lhd G, B\lhd G, A\cap B=\{e\}$, then $ab=ba$ for all $a\in A, b\in B$.

Suppose that $N\cap H=\{e\}=N\cap K$. Note that $H\lhd G, K\lhd G$. Then by the lemma, $nh=hn$ and $nk=kn$ for any $n\in N, h\in H, k\in K$. It follows that $N\subseteq Z(G)$ because $G=H\times K$.

This question also appears in Hungerford's Algebra (Exercise I.8.7).