-1
$\begingroup$

This question already has an answer here:

Let $X$ be a group and $Y$ a normal subgroup of $X$. I am interested in the following two statements:

  1. If $Y$ and $X/Y$ are commutative, then so is $X$.

and

  1. If $Y$ and $X/Y$ are cyclic, then so is $X$.

For 1, I am able to conclude $b^{-1}a^{-1}ba \in Y$ for all $a,b \in X$. Similarly $a^{-1}b^{-1}ab \in Y$. I think the quotient of the dihedral group $D_6$ divided by the group of rotations provide a counterexample to 1.

For 2, I am not aware of any potential counterexamples. If 2 is false, then it immediately follows that 1 is false.

The converse to both of these statements are true.

$\endgroup$

marked as duplicate by Dietrich Burde abstract-algebra Dec 10 '17 at 13:41

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ $A_3\simeq\mathbb Z/3$ and $S_3/A_3\simeq\mathbb Z/2$. $\endgroup$ – Michael Burr Dec 9 '17 at 17:48
  • $\begingroup$ See en.wikipedia.org/wiki/Metabelian_group $\endgroup$ – egreg Dec 9 '17 at 17:57
  • $\begingroup$ For the second question see also Wikipedia: Metacyclic groups. $\endgroup$ – Bernard Dec 9 '17 at 19:41
  • 1
    $\begingroup$ The non-cyclic group of order 4 (Klein's group) is also a counterexample to 2. $\endgroup$ – YCor Dec 9 '17 at 20:21
4
$\begingroup$

A good way to approach such questions is to try this: Let $X$ be the smallest non-Abelian (respectively, non-cyclic) group. If it has a proper normal subgroup, then that must be a solution to your question, because both $Y$ and $X/Y$ will be smaller than $X$, and nothing smaller than $X$ is non-Abelian (respectively, non-cyclic).

Your example, of $D_6$, modding out the group of rotations, works as a counterexample to both, even though it is not the smallest non-cyclic group there is.

(By $D_6$, I’m assuming you mean the dihedral group of order $6$, or the symmetries of an equilateral triangle, sometimes called $D_3$. If you mean the symmetries of a regular hexagon, which is a group of order $12$, sometimes called $D_{12}$, then it still works as a counterexample to both claims.)

$\endgroup$
  • $\begingroup$ @G Tony Jacobs I think most people denote the dihedral group of order 6 by $D_3$ emphasizing the underlying triangle and giving meaning to $D_n$ for all natural numbers $n \in \mathbb N$ $\endgroup$ – Stephen Meskin Dec 10 '17 at 7:02
  • $\begingroup$ @StephenMeskin, I’ve seen both notations commonly enough. See, for example: groupprops.subwiki.org/wiki/Dihedral_group:D8 $\endgroup$ – G Tony Jacobs Dec 10 '17 at 13:18
  • $\begingroup$ This is worth mentioning in my answer, though. I’ll make an edit... $\endgroup$ – G Tony Jacobs Dec 10 '17 at 13:30
  • $\begingroup$ @G Tony Jacobs IMO groupprops is out of the mainstream. But that is off-topic. $\endgroup$ – Stephen Meskin Dec 10 '17 at 15:52
  • $\begingroup$ I just grabbed the first link that came up on Google. I’ve seen it in standard textbooks too. $\endgroup$ – G Tony Jacobs Dec 10 '17 at 16:51
2
$\begingroup$

Your suggestion for the first problem (the dihedral group $D_6$) is correct. And it works for the other problem too. The subgroup $C_3$ of $D_6$ whose elements are the rotations form a normal subgroup and $D_6/C_3\simeq\mathbb{Z}_2$, which is cyclic too. Besides, $C_3$ is cyclic. However, $D_6$ is not cyclic.

$\endgroup$

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