Suppose we have the quotient group $\frac{G}{N}$ and $x,y \in G$. If $Nx \in \frac{G}{N}$ and $Ny \in \frac{G}{N}$, prove that $Nx = Ny \Rightarrow xy^{-1} \in N$.
Why $Nx = Ny \Rightarrow xy^{-1} \in N \textrm{ if } N \lhd G?$
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abstract-algebra
group-theory
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0I think you $\LaTeX$ed too hard. ;) – 2017-01-04
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1@SpamIAm sorry im noob:) – 2017-01-04
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0Anyway, what have you tried? Note that $x = 1x \in Nx = Ny$. – 2017-01-04
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0If $z = nx = n'y$, then $nxy^{-1} = n'$. You can continue from here right? – 2017-01-04
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0The claim does not even require that the subgroup $N$ is normal! – 2017-01-04
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0Recall the equivalence relation that defines the set $G/N $. It is not necessary that $N $ be normal. – 2017-01-04
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0@Nameless thanks. solved:) – 2017-01-04
1 Answers
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Given $Nx=Ny$.
Since $N$ is a subgroup of $G$, $1\in N$.
Therefore, $x=1x\in Nx=Ny$.
This means that $x=ny$ for some $n\in N$.
Multiplying both sides by $y^{-1}$ we get $$xy^{-1}=n\in N$$