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Let $N,H,L$ be subgroups of a group $G$ such that $N$ is normal in $G$, and $L$ is normal in $H$. By using natural map arguments, show that $(LN)/N$ is normal in $(HN)/N$.

Could any one tell me how to solve this one?

The natural maps are $\phi_1: G\rightarrow G/N(g\mapsto gN)$, $\phi_2:H\rightarrow H/L(h\mapsto hL)$

Well, what to do later?

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1 Answer 1

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Since the natural map $HN\rightarrow HN/N$ is surjective it maps normal subgroups to normal subgroups, so it is enough to show that $LN$ is normal in $HN$. Let $x=l\cdot n\in LN$ and $y=hm\in HN$, then $$ yxy^{-1}=hmlnm^{-1}h^{-1}=hlh^{-1}\hat{n} $$ for an element $\hat{n}\in N$, since $N$ is normal in $G$. Since $L$ is normal in $H$, $hlh^{-1}\in L$ and so $yxy^{-1}\in LN$.

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ok I'm not sure if this counts as "natural map argument".. – Michalis Sep 21 '12 at 13:41

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