1
$\begingroup$

Suppose that $n\ge3$ where n is odd and $D_n$ is a dihedral group of order $2n$. I need to prove or disprove that if $H$ is an abelian finite subgroup of $D_n$, then $H$ is cyclic. I cannot understand what does $H$ being a finite subgroup of dihedral group means. Like, if $D_3=[e,r,r^2,s,sr,sr^2]$, then how can I construct an abelian subgroup of this so that it is not cyclic? Can someone elaborate on this please?

Thanks in advance!

$\endgroup$
  • $\begingroup$ Do you know what the dihedral groups are? $\endgroup$ – Cameron Williams Oct 5 '15 at 21:33
  • $\begingroup$ Yes. Is it the group of functions of rotations and reflections? But I dont know how can I use this definition $\endgroup$ – Jennie Durham Oct 5 '15 at 21:34
  • $\begingroup$ Well, you may ignore the word "finite", it does not add anything, since $D_n$ is of course finite, and so are all its subgroups. As for the rest... well, I guess you just have to derive all subgroups of $D_n$ or look them up somewhere. $\endgroup$ – Ivan Neretin Oct 5 '15 at 21:43
-1
$\begingroup$

Let us recall some definitions:

  • One says that $H$ is a subgroup of $D_n$ (or any group) when $H$ is a subset of $D_n$ this is itself a group with respect to the inherited operations. More explicitly it is a subset that contains the identity element and is closed under products and taking inverses. That it is finite just means that the set $H$ is finite; in this case this is automatic as every subset of $D_n$ is finite.

  • A group (or subgroup) is called abelian when the group law satisfies the claw of commutativity.

Thus what you are supposed to decide is: if $H$ is a (finite) subset of $D_n$ that contains the identity and is closed under multiplication and taking inverses (that is, a subgroup) and such that $gh = hg$ for all $h,g \in H$ (so that it is an abelain subgroup), is it then true that $H$ is cyclic, that is it is can be generated by a single element.

It is true that to disprove this it would suffice to exhibit an example of an abelian subgroup of $D_n$ that is not cyclic.

$\endgroup$
  • $\begingroup$ Thanks for the answer. My problem is, I do not know how to construct an abelian subgroup of $D_3$ $\endgroup$ – Jennie Durham Oct 5 '15 at 21:53
  • $\begingroup$ A dihedral group is a group of symmetries, but how to construct an example of it? $\endgroup$ – Jennie Durham Oct 5 '15 at 21:55
  • $\begingroup$ Sorry if it sounds weird, but I am new to this stuff, it takes some time to get it through my head $\endgroup$ – Jennie Durham Oct 5 '15 at 21:55
  • $\begingroup$ Can you please give an example of a subgroup of $D_3$? I really cannot get this $\endgroup$ – Jennie Durham Oct 5 '15 at 22:13
  • $\begingroup$ To create some abelian subgroup, just take an element and consider the group generated by it. This is always cyclic though. $\endgroup$ – quid Oct 5 '15 at 22:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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