Consider the dihedral group

$$D_{2n}= \langle a,b \mid a^n = 1 = b^2, b^{-1}ab = a^{-1}\rangle$$

How can I show that $|D_{2n}| = 2n$?

I'm trying to show that we can write every element in the form

$$x = a^i b^j$$ where $i= 0, 1, \dots, n-1$; $b = 0,1$.

I managed to show existence, and it is clear that there are $2n$ such elements, so if I can show that every choice of $i,j$ gives a distinct element, I'll be done.

Any ideas?

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    $\begingroup$ Use $bab=a^{-1}$ to show $a^ib^j\neq a^mb^k$ where $i\neq m,\;j\neq k$. You can also show that $\langle a\rangle$ has index $2$ in $D_{2n}$. So $D_{2n}=\langle a\rangle\cup b\langle a\rangle$ since $\langle a\rangle\cap b\langle a\rangle=\{1\}$. $\endgroup$ – Yadati Kiran Mar 9 at 15:46
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    $\begingroup$ @YadatiKiran: I think you could easily expand that Comment into a good answer, if time permits. $\endgroup$ – hardmath Mar 9 at 16:18
  • $\begingroup$ There's no such thing as the presentation of a group. One can keep adding generators and redundant relations to any given presentation to get a new presentation. $\endgroup$ – Shaun Mar 9 at 16:28
  • $\begingroup$ You could use the fact that this presentation can be seen quite readily as a semidirect product $\Bbb Z_n\ltimes \Bbb Z_2$. $\endgroup$ – Shaun Mar 9 at 16:31
  • $\begingroup$ (NB: I might have $\rtimes$ and $\ltimes$ mixed up here; if so, I'm sorry. It's rare when I get it right.) $\endgroup$ – Shaun Mar 9 at 16:34

Assume $D_{2n}= \langle a,b \mid a^n = 1 = b^2, b^{-1}ab = a^{-1}\rangle$ has this presentation. $D_{2n}\neq\emptyset$ since $1\in D_{2n}$. Let $a,b\in D_{2n}$ such that $a\neq b$, $a,b\neq1$ else this presentation is futile.

$\textit{Claim}$: $|a|=n$ and $|b|=2$.

$b^2=1.$ Then $ |b|\:\Bigg|\:2$. Since $b\neq1\implies |b|=2$.

$a^n=1.$ Then $ |a|\:\Bigg|\:n$. Let $|a|=k,\;k<n$.

By Euclid's algorithm, $\exists!\; q,r\in\mathbb{Z}\;:\;n=kq+r$ with $0\leq r<n$.

$$a^n=a^{kq+r}\implies a^r=1\Rightarrow\Leftarrow |a|=k$$ Hence the claim.

$D_{2n}=\langle a,b\rangle$, every element of $D_{2n}$ has the form $a^ib^j,\;i\in\{0,1,\cdots,(n-1)\} \;\text{and}\;j\in\{0,1\}$.

We prove that the set has distinct elements for all $(i,j)$.

Using $bab=a^{-1}$, we can derive

  • $a^kba^{k}=b$,
  • $ba^kb=a^{-k}\quad$ and subsequently
  • $b^ma^kb^m=a^{((-1)^mk)}$.

We shall also use the fact : $\langle a\rangle \cap \langle b\rangle=\{1\}\tag1$

Suppose $$a^ib^j =a^mb^k\tag2$$ where $i\neq m \;\text{where}\;i,m\in\{0,1,\cdots,(n-1)\},\;j\neq k\;\text{where}\;j,k\in\{0,1\}$. Without loss of generality let $m>i$, from $(2)$ we have the following:

$$\begin{align}a^ib^ja^i =a^mb^ka^i &\implies b^j=a^{m-i}b^k\\ &\implies b^kb^j=b^ka^{((-1)^k(m-i))}b^k=a^{((-1)^k(m-i))}\end{align}$$

i.e. $$b^{k+j}=a^{((-1)^k(m-i))}\tag3$$ Thus, $b^{k+j},a^{((-1)^k(m-i))}=1$. $\; b^{k+j}=1\implies \;k+j\:\Bigg|\:2$. So $k+j=1$ or $2$. $k+j\neq2$ as $j\neq k$ and $j,k\in\{0,1\}$. So $j+k=1$.

$(3)\implies b=a^{((-1)^k(m-i))}\Rightarrow\Leftarrow$ as $b\notin \langle a\rangle$.

For the case in $(2)$ where $i=m$ but $j\neq k$, $$a^ib^j =a^mb^k\implies b^j=b^k\iff j=k$$

For the case in $(2)$ where $i\neq m$ but $j= k$, $$a^ib^j =a^mb^k\implies a^{m-i}=1\;\text{with}\; (m-i)<n \Rightarrow\Leftarrow\;\text{ as}\; |a|=n$$

Since we have shown that $a^ib^j$ is distinct for all $(i,j)$, by simple combinatorics we see that $|D_{2n}|=2n$

Proof for $(1)$ : For suppose $\langle a\rangle \cap \langle b\rangle\neq\{1\}$ then $b\in\langle a\rangle$. If $n$ is odd then, $|b|\not\Bigg|\;n$ giving us a contradiction. If $n$ is even, then $\langle a\rangle$ has two elements of order $2$ namely $b$ and $a^{n/2}$. But this will also contradict the fact that "If $G$ is a group of even order then the number of elements of $G$ of order $2$ is odd." Hence the fact $\langle a\rangle \cap \langle b\rangle=\{1\}$.

  • $\begingroup$ How did you prove that $\langle a \rangle \cap \langle b \rangle = 1$? $\endgroup$ – user370967 Mar 10 at 9:44
  • $\begingroup$ @Math_QED : The edit should suffice. $\endgroup$ – Yadati Kiran Mar 10 at 15:38
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    $\begingroup$ I don't find this proof convincing. You seem to be assuming that $a$ has order $n$ and that $b$ has order $2$, which needs to be proved. You haven't even proved that the group defined by the presentation is nontrivial. $\endgroup$ – Derek Holt Mar 10 at 15:56
  • $\begingroup$ You can't assume $a\neq b$ and $a,b\neq 1$ just by saying "otherwise this presentation is futile". This presentation is a quotient of the free group on $a$ and $b$ by the normal subgroup generated by $a^n$, $b^2$ and $b^{-1}aba$, that's all you can assume. $\endgroup$ – Christoph Mar 10 at 17:52

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