If a subgroup $S$ of $G$ is Abelian (the subgroup, not the entire group!) then is $S$ a normal subgroup? I saw a glimpse of similarity in their definitions and I thought I'd try to prove or disprove it.
I have a feeling that it is true but my attempts to form a proof (that given an Abelian subgroup, that group is a normal subgroup) by using $x\in S$ style to show if it is in the LHS then it is in the RHS has failed because I have to be careful it's not in the not abelian part.
I'd love an example that it is true, or that it is false, my failure to prove might be because it isn't true. 
I have tried searching but the search results are for "every subgroup of an abelian group is normal" which is a totally different thing.
 A: Take the group $S_n$ of permutations. Subgroup generated by a transposition $(12)$ is cyclic, but is not normal.
A: It turns out to be false. Consider, e.g., the dihedral group $D_8$ of order $8$, which can be viewed as consisting of reflections and rotations. The subgroup $H$ generated by any reflection $h$ of order $2$ is abelian but not a normal subgroup of $D_8$. Indeed, if $g\in D_8$ has order $4$, then $g^{-1}Hg$ is a different subgroup of order $2$.
More generally, here is a necessary and sufficient condition for an order $2$ subgroup to be a normal subgroup.
Proposition. Let $G$ be any group. If $N$ is a subgroup of $G$ such that $|N|=2$, then $N\lhd G$ if and only if $N\subseteq Z(G)$, where $Z(G)$ is the center of $G$.
Proof. The backward implication follows from the fact that conjugation by any element of $G$ fixes any element of $Z(G)$, including every element of $N$.
We turn, now, to the forward implication. Let $N\lhd G$, and suppose $N=\{e,x\}$, where $e$ is the identity element of $G$, and $x$ is any non-identity element of $G$. We note that $e\in Z(G)$, because $Z(G)$ is a subgroup of $G$. Next, let $g\in G$. We have, 
$$g^{-1}xg\in g^{-1}Ng=N$$
because $N$ is a normal subgroup of $G$. Thus, $g^{-1}xg=x$ or $g^{-1}xg=e$. If $g^{-1}xg=e$, then $x=gg^{-1}=e$, a contradiction. Hence, $g^{-1}xg = x$. It follows that $x$ commutes with every element of $G$, so $x\in Z(G)$. Since $e$ and $x$ both belong to $Z(G)$, we have $N\subseteq Z(G)$. q.e.d.
A: Why would it? we want to prove that for any $G \in G$ and $h \in S$ $ghg^{-1}\in 
 G$ but this doesn't depend in the structure of $S$ but rather in the structure of $G$.
