Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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
up vote 1 down vote accepted

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.