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For a group $G$, if $K\unlhd H\leq G$ and $N\unlhd G$, then $NK\unlhd NH$? How come?

If $NK/N\unlhd NH/N$, then by correspondence theorem, $NK\unlhd NH$, but why is $NK/N\unlhd NH/N$? I know homomorphisms like the natural projection preserve normal subgroups, so wouldn't $\pi(K)\unlhd\pi(H)$, or $K/N\unlhd H/N$? Why is there the extra $N$ on top?

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2 Answers 2

But $N$ is not necessarily a subgroup of either $K$ or $H$, so that $K/N$ or $H/N$ may not even be defined. $NK$ can be thought of as the smallest subgroup that contains both $N$ and $K$, so that in this case $NK/N$ makes sense (and of course the same is true for $NH/N$).

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$N$ is not a subgroup of $K$, so what sense would you give to $K/N$?

About your first question: since $N$ is invariant under conjugation by any element of $G$ and $K$ is invariant under conjugation by any element of $H$, just write down $(nh)^{-1}n^\prime k(nh)$ with $n, n^\prime\in N$,$h\in H$, $k\in K$ and convince yourself (after some manipulation) that you land in $NK$.

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