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I have a group that I'm trying to prove is isomorphic to the Dihedral group.

I know that it is finite, that it is generated by two elements $\alpha$ and $\beta$ such that: $\alpha^2=\beta^n=1$ and that $\alpha\beta\alpha=\beta^{-1}$. EDIT: also, $\alpha\neq \alpha^2$ and $\beta\neq \beta^2\neq\ldots\neq\beta^n$.

I also know that is has at least $2n$ unique elements.
EDIT: Is this assumption redundant?
EDIT: It is redundant for $n$>2. For $n=2$, $\alpha\neq\beta$ is enough (i.e. $D_2\cong C_2\times C_2$).

Is this enough in order to imply that this group is the Dihedral group with $2n$ elements?

Will appreciate any help :)

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  • A group generated by such $\alpha$ and $\beta$ is isomorphic to a quotient of the dihedral group of order $2n$. – Angina Seng May 20 '18 at 17:34
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    That is enough information. To prove it's the dihedral group you need a definition of the dihedral group. (Sometimes what you are assuming is the definition). Related: https://math.stackexchange.com/questions/2205865/some-subgroup-of-dihedral-group-is-normal/2766448#2766448 – Ethan Bolker May 20 '18 at 17:35
  • @LordSharktheUnknown, I think that you're implying that the group might have more relations. But wouldn't that imply that the order must be smaller than $2n$? (my wishful thinking :)) – Daugmented May 20 '18 at 19:22
  • @EthanBolker Doesn't the definition require that there aren't other relations? – Daugmented May 20 '18 at 19:24
  • Hint: $\alpha$ corresponds to a flip; $\beta$ corresponds to a rotation. – Shaun May 20 '18 at 20:34
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    @Daugmented: If you can, check out chapter 3 of "Examples of Groups" by Michael Weinstein (ISBN-10: 0936428171). It is only 10 pages long but a real eye-opener for questions such as yours above. – Moritz May 20 '18 at 21:25
  • @Moritz thanks, did you mean chapter 2? – Daugmented May 20 '18 at 21:34
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    @Daugmented: no, chapter 2 (in my 2nd edition) is about matrix groups. Chapter 3 ist about free groups and presentations. See also chapter 5 (examples of finite groups): example 5.2 is about the dihedral groups. The book is really fantastic. – Moritz May 22 '18 at 06:33
  • Thanks, I'll take a look (probably a little more than that). It seems like the second edition swapped the order of chapters 2 and 3. The Dihedral example 4.1 is the first on the finite groups chapter 4. – Daugmented May 22 '18 at 10:18

1 Answers1

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Adding my own proof with sufficient conditions:

Let $n>2$ and let $\alpha$ and $\beta$ be two elements in a group such that:

  1. $o(\alpha)=2$ and $o(\beta)=n$, where $o(g)$ is the minimal power of $g$ such that $g^{o(g)}=1$.
  2. $\alpha\beta=\beta^{-1}\alpha$.

Then $G=\left<\alpha, \beta\right>\cong D_n$, the dihedral group of order $2n$.

Proof.

It is sufficient to show that $G$ has no other relations. We show that $|G|=2n$.
Since $\alpha\beta=\beta^{-1}\alpha$, every element in $G$ can be written in the form $\beta^i\alpha^j$ where $i\in[0,n-1]$ and $j\in[0,1]$.
Assume to the contrary that there exists $1\leq k\leq n-1$ such that $\beta^k\alpha=Id$. This implies that $\alpha=\beta^{n-k}$. Therefore $k=\frac{n}{2}$, since $\alpha^2=1$.
Since $\alpha\beta=\beta^{-1}\alpha$, we have $\beta^{\frac{n}{2}+1}=\beta^{\frac{n}{2}-1}$ which is true only for $n=2$.

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