# If $H$ is a normal subgroup of $G$ and if both $H$ and $G/H$ are abelian, is $G$ abelian?

Pretty straightforward:

If $H$ is a normal subgroup of $G$ and if both $H$ and $G/H$ are abelian, is $G$ abelian?

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As Zev shows, the answer is no. "Solvable" is a generalization of "abelian", and one of the facts about solvable groups, crucial in Galois Theory, is that the answer is "yes" if you replace both occurrences of "abelian" with "solvable". –  Gerry Myerson Oct 14 '11 at 21:04
Groups with the property you describe are called "metabelian" (be careful, though, when reading old papers/books; for a while, especially in Eastern Europe, "metabelian" meant a stronger condition that the commutator subgroup be central). –  Arturo Magidin Oct 15 '11 at 2:26

No, consider $G=S_3$ and $H=A_3=\{e,(123),(132)\}$. Then $H\triangleleft G$, and $H\cong \mathbb{Z}/3\mathbb{Z}$ and $G/H\cong\mathbb{Z}/2\mathbb{Z}$ are both abelian, but $G$ is not abelian.

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Very nice counterexample, thank you. –  Sir Winford Oct 14 '11 at 21:03
@SirWinford Please don't forget to accept an answer :) –  user9413 Oct 14 '11 at 21:20
@Chandrasekhar I was going to right away, but the 15 minute grace period forced me to wait, and it slipped my mind. :) –  Sir Winford Oct 14 '11 at 21:45
As a more general class of examples, consider the group $D_n$ of symmetries on a regular $n$-gon for $n\geq 3$ (of course $D_3 \cong S_3$, so we're really only interested in $n>3$). Let $R$ denote the element of $D_n$ corresponding to a counter-clockwise rotation of $2\pi/n$, and let $H$ be the subgroup of $D_n$ generated by $R$. Then $H \triangleleft D_n$ (since $\vert D_n\,:\,H\vert=2$), and clearly both $H$ and $D_n/H\cong \mathbb{Z}_2$ are abelian. However, $D_n$ is not abelian.
Even more general : if $H$ is any abelian group, and $a \in \mathbb{Z}$ is coprime to $|H|$, then $f : g \mapsto g^a$ is a group automorphism of $H$ of finite order (the order of $a$ in $(\mathbb{Z}/|H|\mathbb{Z})^{\times}$). Consider the semi-direct product $G = H \rtimes \langle f \rangle$. Then $H$ is normal in $G$, and $G/H$ is abelian. If you take $H = \mathbb{Z}/n\mathbb{Z}$ and $a = -1$, you find $G \simeq D_n$. –  Joel Cohen Oct 14 '11 at 22:43