When asked to show that a subgroup $H$ of the abelian group $G$ is normal, does it suffice to say:
- first, $H$ is a subgroup, so it contains the identity element of $G$ and inverses $h^\prime$ for every element $h$ in $H$;
- because $G$ (and thus $H$) is abelian, every pair of elements of $H$ commute with each other. Then every element in $H$ commutes with any element of $G$, as the elements of both $H$ and $G$ all commute with any other element of $G$ (or $H$).
Thus $gH = Hg$ and $H$ is a normal subgroup.
I suspect the above is insufficient. In particular, any potential errors above aside, I was wondering if it is "necessary" to show that $ghg' = h$?