# 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.

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Hint: remember that every group is a subgroup of itself. – Newb Mar 10 '14 at 0:46
@Newb how does that help? There may be more than one subgroup, if the entire group is Abelian then all subgroups are normal. I want to show that if I have a subgroup that HAPPENS to be Abelian, that it is (or is not) a normal subgroup. – Alec Teal Mar 10 '14 at 0:48
groupprops.subwiki.org/wiki/Special:Ask/… the examples are endless :) – r9m Mar 10 '14 at 0:59

It turns out to be false. Consider, e.g., the dihedral group of order $8$. The subgroup generated by a flip is abelian (it has order $2$ after all) but not normal.

In fact, it is pretty easy to find counterexamples, if we utilize the following result:
Proposition. If $N$ is an order $2$ subgroup of $G$, then $N\lhd G$ if and only if $N\subseteq Z(G)$.
Proof. The backward direction follows from the fact that conjugation by any element of $G$ fixes any element of $Z(G)$, including those of $N$. For the forward direction, note that a conjugate of a non-identity elements of $G$ is never the identity. Thus, if $1\ne a\in N$ then $gag^{-1}$ can only possibly be $a$. q.e.d.

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Take the group $S_n$ of permutations. Subgroup generated by a transposition $(12)$ is cyclic, but is not normal.

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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 $G$.

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