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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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    @GeoffRobinson: Gerry edited it correctly. It was my fault.2012-07-17

2 Answers 2

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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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    @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).