Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Pretty straightforward:

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

share|improve this question
3  
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
2  
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

2 Answers 2

up vote 9 down vote accepted

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.

share|improve this answer
    
Very nice counterexample, thank you. –  Sir Winford Oct 14 '11 at 21:03
1  
@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.

share|improve this answer
1  
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.