# How to show that $|D_{2n}| = 2n$ via the presentation?

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?

• 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 $\langle a\rangle\cap b\langle a\rangle=\{1\}$. – Yadati Kiran Mar 9 at 15:46
• @YadatiKiran: I think you could easily expand that Comment into a good answer, if time permits. – hardmath Mar 9 at 16:18
• 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. – Shaun Mar 9 at 16:28
• You could use the fact that this presentation can be seen quite readily as a semidirect product $\Bbb Z_n\ltimes \Bbb Z_2$. – Shaun Mar 9 at 16:31
• (NB: I might have $\rtimes$ and $\ltimes$ mixed up here; if so, I'm sorry. It's rare when I get it right.) – 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.

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

$$a^n=a^{kq+r}\implies a^r=1\Rightarrow\Leftarrow |a|=k$$ 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)

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 $$\langle a\rangle \cap \langle b\rangle=\{1\}$$.

• How did you prove that $\langle a \rangle \cap \langle b \rangle = 1$? – user370967 Mar 10 at 9:44
• @Math_QED : The edit should suffice. – Yadati Kiran Mar 10 at 15:38
• 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. – Derek Holt Mar 10 at 15:56
• 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. – Christoph Mar 10 at 17:52