Talk:Subgroup is Normal iff Normal Subset

While it is true that if $N$ is a normal subset, $\forall g \in G: g N = N g$, we still need to prove that $\forall g \in G: g N = N = N g$.

That's what the bit I added was that had been missed. --prime mover (talk) 15:19, 3 February 2013 (UTC)


 * Ignore this. It's rubbish. I've reverted it back to what you put. --prime mover (talk) 15:21, 3 February 2013 (UTC)