# Is it true that $G$ is abelian?

Suppose $H$ is a normal subgroup of $G$. If $H$ and $G/H$ are abelian, is it true that $G$ is abelian?

I don't think the answer is YES. This is because, $G/H$ is always abelian as $H$ is normal, so this information is nothing new. $H$ is abelian implies that $H\subset Z(G)$ where $Z(G)$ is the centre of $G$.

But I could not think of a counterexample.

• Are you saying that quotient groups are always abelian? I just don't know how to parse "$G/H$ is always abelian as $H$ is normal" – pjs36 Aug 16 '15 at 15:46
• It's also not true that abelian subgroups are in the center of the group. Every subgroup generated by a single element is abelian; thus, by your reasoning, $x \in Z(G)$ since $\langle x \rangle \subseteq Z(G)$ for every group $G$ and every $x \in G$. But certainly not every group is abelian. – pjs36 Aug 16 '15 at 15:52
• My statement that $G/H$ is abelian is wrong, but $H$ being abelian and normal should imply that $H$ is a subset of $Z(G)$. – Landon Carter Aug 16 '15 at 15:58
• That's still not true, unfortunately (you can find counterexamples in small symmetric or alternating groups, among others, like dihedral groups). I'm not sure if you're misinterpreting the $Hg = gH$ criterion for normality or what's leading you to suspect it. – pjs36 Aug 16 '15 at 16:14

Let $G=S_{3}$ and $H=A_{3}$. A counterexample!. Added as PS : A quotient group is not necessarly abelian, take $G=S_{3}$ and $H=\lbrace identity \rbrace$, $H$ is normal, but $G/H \cong S_{3}$ and non abelian.
A typical example are the dihedral groups $D_n$; the set of symmetries of a regular polygon of $n$ sides. It has a normal abelian subgroup $R$ of order $n$ and index $2$, the rotations of angle $\smash[b]{\dfrac{2\pi}n}$. In addition, there are $n$ reflections, which are all conjugates modulo $R$.
The dihedral groups are abelian only for $n=1,2$, since they are generated by two element $r$ and $s$, subject to the relations: $$r^n=e,\quad s^2=e,\quad sr=r^{-1}s.$$