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For a subgroup $N$ of $G$, if the product of two right cosets of $N$ in $G$ is again right coset of $N$ in $G$,
then $N$ is normal subgroup of $G$.

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i have proved converse of this statement, which i find very easy. but this proof require some trick. –  White Dwarf Sep 1 '12 at 18:44
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What is your question? –  Code-Guru Sep 1 '12 at 18:44

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Say $aNbN = acN$. Then $Nb = cN$. Write $n_1b = cn_2$ for any $ n_1$ and some $n_2$ in $N$. Take $n_1 = e$ to get $b = cn_2$. Take $n_2 = e$ to get $b = n_1^{-1}c$. Put them together for $cn_2 = n_1^{-1}c$. This is equivalent to $cN = Nc$ and makes $N$ normal.

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Thanks.This is very helpful. –  White Dwarf Sep 2 '12 at 4:47
    
it may be my lack of understanding, but please tell me how you conclude, from aNbN = acN to Nb = cN. –  White Dwarf Sep 2 '12 at 5:15
    
$aNbN = acN$ means that $an_1bn_2 = acn_3$ for any $n_1$, $n_2$ and some $n_3$ in $N$. From that we get $n_1b = cn_3n_2^{-1}$ for any $n_1$ and some $n_3n_2^{-1}$. –  Karolis Juodelė Sep 2 '12 at 6:42

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