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Is it true that a dihedral group is nonabelian?

I'm not sure if the result is true. I checked it for some lower order and I think the result may correct.

But I failed to prove/disprove the result.

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    $\begingroup$ What happens if you first do the smallest possible rotation and then reflect in some line? What happens if you first reflect in that same line and then do that rotation? $\endgroup$ – Tobias Kildetoft Sep 16 '13 at 7:46
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    $\begingroup$ There are many equivalent definition about dihedral group. What definition did you have learn? $\endgroup$ – D. N. Sep 16 '13 at 8:27
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    $\begingroup$ The easier the question, the more identical answers. What a nonsense ... $\endgroup$ – Martin Brandenburg Sep 16 '13 at 10:04
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Yes, the dihedral groups $D_n$ are nonabelian for $n\ge 3$. It is generated by a rotation $r$ with $r^n=1$ and a reflection $s$ with $s^2=1$. However, you can easily check that a rotation and a reflection will not commute in general. We have $sr=r^{-1}s$ instead for $D_n$ with this presentation.

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The dihedral groups for $n=1$ and $n=2$ are abelian; for $n\geq 3$, the dihedral groups are nonabelian (this is mentioned on Wikipedia).

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$D_3$, i.e, the dihedral group of a triangle is isomorphic to $S_3$ which is non-abelian. It can be shown that this is true for $n \geq 3$.

Remark: Some people denote the dihedral group by $D_{2n}$ which is based on the fact the order is $2n$, while some people denote it by $D_n$.

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  • $\begingroup$ That's a good point to mention about notation! $\endgroup$ – David Ward Sep 16 '13 at 7:55
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Dihedral group of order $2n$ has the presentation

$$\langle x,y \mid x^n=y^2=e,yxy=x^{-1}\rangle$$

When $n=1$, we have $x=e$ and thus it is a group of order 2.

When $n=2$, we have $yxy=x$ and so $xy=yx$. Thus it is abelian.

But when $n\geq 3$, then $yxy=x^{n-1}$ and so $xy=yx^{n-1} \neq yx$. Thus it is not abelian.

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    $\begingroup$ You need to explain why the inequality holds. $\endgroup$ – Zhen Lin Sep 16 '13 at 8:12
  • $\begingroup$ $x$ and $y$ are two element of the group and $xy\neq yx$, means they do not commute, and so it is not an abelian group. $\endgroup$ – D. N. Sep 16 '13 at 8:19
  • $\begingroup$ @deibor: You cannot conclude from a group presentation that certain relations do not hold. For example, even when a presentation contains something like $x^3=1$, it doesn't mean that the order of $x$ is three. In fact, other relations might imply that $x=1$. The general word problem for groups is undecidable. $\endgroup$ – Martin Brandenburg Sep 16 '13 at 10:03
  • $\begingroup$ @Martin What you say is true, but in this specific case we are fine. This is because we know that these relations hold and we know that $x$ has order $n$. If $yx^{n-1}=yx$ then we would have $x^2=1$ so $n\leq 2$. $\endgroup$ – user1729 Sep 16 '13 at 10:06
  • $\begingroup$ The presentation $\langle x,y∣x^n=y^2=e,yxy=x−1\rangle$, has been taken as an equivalent definition of dihedral group of order $2n$ by many authors of many text book. Also it is an equivalent definition on wikipedia and groupprop. $\endgroup$ – D. N. Sep 16 '13 at 10:34
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My picture for $D_4$ shows $fr = r^{-1}f$ and not $rf$.
See Zev Chonoles's exquisite pictures at https://math.stackexchange.com/a/686175/53934 too.

enter image description here

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