Showing that every subgroup of an abelian group is normal I'm working on a proof to show that every subgroup of an abelian group is
also a normal subgroup. Let $G$ be an abelian group and $H$
an arbitrary subgroup of $G$. I want to show that $gHg^{-1} = H$, but I could
use some help with some of the beginning steps to this problem. I've only
recently been getting interested in Abstract Algebra.
 A: First note that $gHg^{-1} = \{ghg^{-1} \mid h \in H\}$.
To show that $H$ is a normal subgroup of $G$, you need to show that, for each $g \in G$, every element of the form $ghg^{-1}$ is an element of $H$ (this shows $gHg^{-1} \subseteq H$) and every element of $H$ is of the form $ghg^{-1}$ for some $h \in H$ (this shows $H \subseteq gHg^{-1}$). 
An early exercise in group theory which makes life a bit easier (which I recommend you try for yourself) is the following:

If $H \subseteq G$, then $gHg^{-1} = H$ for every $g \in G$ if and only if $gHg^{-1} \subseteq H$ for every $g\in G$.

That is, we only need to do the first of the two steps I outlined above, namely, show that every element of the form $ghg^{-1}$ is an element of $H$.
Returning to the specific case at hand, as $G$ is abelian, the elements of the group commute. So what does $ghg^{-1}$ simplify to?

Note, in this case, establishing $gHg^{-1} = H$ is no harder than showing $gHg^{-1} \subset H$, but in general it would be more difficult.
