Talk:Subgroup is Superset of Conjugate iff Normal

The proof of Normal Subgroup Test being so much more compact than the one it is to be merged with, I'm unwilling to delete it till it's been looked over carefully to see whether there is material which is taken for granted that needs to be explained carefully. But the result is identical to the one to be merged with. --prime mover (talk) 11:47, 3 February 2013 (UTC)


 * How about we create a Subgroup is Superset of Conjugate iff Normal/Corollary that reads:


 * ... $N$ is normal in $G$ :
 * $\forall g \in G: g \circ N \circ g^{-1} \subseteq N$


 * and merge Normal Subgroup Test into this new corollary?


 * I am indifferent to how the proof of the theorem is written. If User:Usagiop wants a shorter proof, we can create Subgroup is Superset of Conjugate iff Normal/Proof 2 for that. --Anghel (talk) 11:26, 27 September 2022 (UTC)


 * I just wanted to point out that it is already explained as $\paren 1$ in the proof that it suffices to verify one of them. If the repetition is an intention, it is OK. --Usagiop (talk) 16:01, 27 September 2022 (UTC)


 * You are right that the copy-paste arguments of the proof is just expanding the statement $\paren 1$ ...which is shown by, for example, setting $h := g^{-1}$ and substituting. --Anghel (talk) 22:12, 27 September 2022 (UTC)