Showing that a subgroup of an abelian group is normal—is this sufficient? 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 $ghg' = h$?
 A: It's a good attempt, but your argument is unnecessarily complex and seems to me to use some incorrect logic: 


*

*there's no need to use the fact that $H$ is a subgroup, since $gS=Sg$ for any $g\in G$ and any subset $S\subset G$ whatsoever when $G$ is abelian

*you seem to be saying the fact that every element of $H$ commutes with every other element of $H$ implies that every element of $H$ commutes with every element of $G$, when in fact that is not a correct implication (moreover, both are subsumed by the simple statement "$G$ is abelian")
(Of course, these criticisms of the argument not affecting the fact that the result is true.)
Let's simplify:

For any $g\in G$, we have
  $$gH=\{gh:h\in H\}\overset{\text{(because $G$ is abelian)}\strut}=\{hg:h\in H\}=Hg$$
  and therefore $H$ is normal.

A: Variant: In an abelian group, conjugation is trivial (= identity), hence any subgroup is stable by conjugation, i.e. is normal.
A: Your argument is fine, if a bit wordy. The Reader's Digest version would suffice:
$G$ abelian $\implies gh = hg$ for all $h \in H$, and any $g \in G \implies gH = Hg$.
