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!

  • $\begingroup$ 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

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.

  • $\begingroup$ 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

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