# Characterization of a Normal Subgroup

I am studying a proof of a theorem that states that a subgroup $N$ of a group $G$ is normal in $G$ if and only if $(xA)(yA) = (xy)A$ for all $x,y \in G$. The author goes through a fairly involved element-chase to prove necessity. However, it seems clear that if $N$ is normal in $G$ that $$(Nx)(Ny) = (Nx)(yN) = N(xy)N = (xy)NN = (xy)N$$ where only normality and the nature of the "product" $AB = \{ab | a \in A, b \in B\}$ is used.

Does my argument work or am I overlooking a detail?

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Even quicker: $xAyA = xyAA = xyA$. Your chain looks fine as well, though. –  Dylan Moreland Jan 21 '12 at 2:47
@DylanMoreland Yes, that is better. If you'll post your comment as an answer I'll go ahead and accept it and clear the question. Thanks. –  ItsNotObvious Jan 21 '12 at 3:17
This shows the "only if"; how do you show the "if"? –  Arturo Magidin Jan 21 '12 at 4:20

Your argument is correct, but it requires as a lemma that multiplication of subsets of groups is associative, i.e. if $A$, $B$, and $C$ are subsets of a group, then $(AB)C = A(BC)$. This isn't itself hard to prove, but it explains why your textbook eschews this approach in favor of an argument involving elements.