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  1. $(Nx)(Ny)=Nxy$
  2. $N(x)(Nx^{-1})=(Nx^{-1})(Nx)=N$
  3. $N(Nx)=(Nx)N=Nx$

Deduce that the right cosets of $N$ in $G$ form a group under the given multiplication.

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Where are you stuck? – lhf Dec 9 '12 at 21:17
OK - so what is a normal subgroup (how defined on your course or in your textbook) - then show how you might try to apply the definition to the three equations you are given. Note that the concept of a normal subgroup is valuable, precisely because all these things are true - so you should work at understanding why this works. – Mark Bennet Dec 9 '12 at 21:17
I understand a normal to subgroup (N say) to be one where gN=Ng for all g in G. – Mathlete Dec 9 '12 at 21:19
For the first, could I say LHS=(Nx)(N(x^-1)xy)=(Nx)((x^-1)Nxy)=Nxy? I don't understand what to do about the second N though, as I'd be left with NNxy... – Mathlete Dec 9 '12 at 21:21
up vote 3 down vote accepted

Something which I like to do is using the monoid of nonempty subsets of $G$. Let $A,B\subset G$, we will define $AB=\{ab|a\in A,b\in B\}$. It is easy to show that multiplication of sets is associative. Note that the cosets $Nx,xN$ are $N\{x\},\{x\}N$ respectively. Also note that for any subgroup of $H$ of $G$ we have $HH=H$ In the next proof I will use the facts :$NN=N$ ,$N\{x\}=\{x\}N$ and the fact that multiplication of subsets is associative.

a) $(N\{x\})(N\{y\})=N(\{x\}N\{y\})=N(N\{x\}\{y\})=NN(\{x\}\{y\})=N\{xy\}$

b),c) can be proved similarly.

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Could you possibly explain how you got the last three steps? – Mathlete Dec 9 '12 at 21:29
you mean step a) ? – Amr Dec 9 '12 at 21:29
OK I noted a typing error – Amr Dec 9 '12 at 21:30

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